{"text": "BIOLOGY REVISION AND EXAMINATION TIPS Main reasons why Students Perform Poorly in the Biology Subject  Confusion of biology concepts and terminologies.  Inability to recall ideas about the subject content (subject matter). This could be attributed to lack of interest and concentration in class and poor study habits hence the meaning of the content is never grasped.  Use of poor vocabulary-lack of good English command to express ideas e.g. the relaxation of erector pili muscles cause the hair to fall down instead of lie flat on the skin surface  Assuming too much. This leads to giving sketchy, ambiguous, half answers or giving more information than asked which in most cases is irrelevant.  Not following simple instructions and guidelines in answering questions (examination techniques) particularly graphical and structured questions. Students need to know that most structured questions require a deeper understanding of the concepts behind the question/experiment. Furthermore, most of the answers in the various subsections are tied i.e. the student has to get the first part of the question right first to guarantee him/her a correct response in the subsequent subsections of the question. Possible Remedies  During study sessions, ensure that you understand adequately the subject matter for every topic. This calls for a lot of reasoning, comparison, application, relation, evaluation, analysis and observation. These are core science-process skills that make science therefore Biology a unique discipline. Understanding can be enhanced through having interest in what you are reading (have a positive attitude towards the subject having understood its relevance in your future career as well as individual life and community development); planning and organizing yourself by knowing what to do, where, when and for how long (prepare a realistic study schedule/timetable and follow it); concentrate by being attentive and alert in class; be determined; never give up and have a dream or vision that will help you look into your future. Thorough revision or study ensures information is securely fixed in the long-term memory of the brain (principle of consolidation). The best method is ensuring you review notes right after class. Large chunks of information can be summarized through use of pseudo codes, mnemonics or use of concept maps (schematic mapping).  Arrive in class and examination room in time. This will enable you gather all the verbal and written concepts taught and boost confidence and reduce pre-exam stress respectively.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9235753317720532, "ocr_used": false, "chunk_length": 2562, "token_count": 490}}
{"text": "Large chunks of information can be summarized through use of pseudo codes, mnemonics or use of concept maps (schematic mapping).  Arrive in class and examination room in time. This will enable you gather all the verbal and written concepts taught and boost confidence and reduce pre-exam stress respectively. This can also be enhanced by conferring with other students to predict what might be in the test prior to the start of the examination.  Read through each question at least three times before you decide to write down your answer. Use grammatical clues within a statement as hints for the correct answer. Underline the key words in the question to help you know exactly what is being asked. If you go blank on any of the questions, skip it and go to the next but remember to spare some time later to answer the skipped question (s).  Believe in yourself. Do not try to be perfect; instead try to balance your work by giving the best. Never be in a hurry/panic. The time allocated for each paper has been pretested to be adequate for every student’s level of achievement (intelligence) and it only requires proper planning and approach to fit within the allocated time and even have time to review your work. Budget your time according to what is required of you in each question. Answer questions in a strategic order starting with the easy ones to help you build confidence and familiarize yourself with the vocabulary and concepts that you will deal with, then answer the more difficult questions.  Write your answers neatly, accurately and precisely. Start by writing the points you are sure of first. This is because in Biology, only the first required correct responses are awarded. Any extra point beyond the required is just acknowledged but not awarded even if it is also or is the only correct one yet the earlier stated ones are wrong or not all.  Use proper English and avoid confusing concepts. English is the medium of instruction (lingua franca) of the Biology subject matter and therefore a correct and clear command of the language is paramount to passing the subject.  Remember to write exhaustive answers as half answers and ‘hanging’ statements are never awarded in any Biology examination e.g. for each structure or process, mention the function or reasons respectively depending on how the question is framed.  Concentrate on the unique or distinguishing features or characteristics and avoid obvious things e.g.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9194444444444445, "ocr_used": false, "chunk_length": 2448, "token_count": 487}}
{"text": " Remember to write exhaustive answers as half answers and ‘hanging’ statements are never awarded in any Biology examination e.g. for each structure or process, mention the function or reasons respectively depending on how the question is framed.  Concentrate on the unique or distinguishing features or characteristics and avoid obvious things e.g. if asked for the adaptations of the gills in fish to gaseous exchange, don’t expect to be awarded for mentioning that they are moist yet it is obvious that fish are aquatic animals found in water (moist) bodies.  Strive to answer the questions as they are stated but not to memorize the content read e.g. you might be tempted to state all the adaptations of gaseous exchange structures you know including the structures having a dense network of capillaries…yet the question asked is specifically asking for the adaptations of plant structures. Therefore, get to understand the question asked first before attempting to answer.  Avoid canceling your work by being sure before writing down. Canceling can also be a point of suspicion for cheating. Neatness of the answers also encourages the examiner and implies keenness, clarity and being sure, all being the expected attributes of a modern scientist.  Write all scientific names and technical words/phrases according to the rules of the binomial nomenclature and correct spellings respectively. Avoid using abbreviations, short codes and symbols as most are not conventional (not universally used/accepted by the body of scientists) hence you risk being penalized for their use.  Never add information in brackets if you are not sure e.g. red blood cells (leucocytes). This will not be awarded as it shows confusion of concepts which is penalized.  All diagrams drawn should be clear, neat and proportional sketches (not artistic impressions) and must be true representations of the object/specimen. Follow the simple rules: Labeling lines should not cross each other; the labeling line should touch the structure in question but not hang; use lines not arrows to label, arrows only show direction; do not shade the drawings but use dots or crosses within the part if you want to show the difference; lines must be continuous and not broken; no two structures should have the same name; for cross-sections and plan diagrams, remember to use double outer lines (outline) and only show key/few features avoiding unnecessary details.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9206726825266613, "ocr_used": false, "chunk_length": 2438, "token_count": 492}}
{"text": "This will not be awarded as it shows confusion of concepts which is penalized.  All diagrams drawn should be clear, neat and proportional sketches (not artistic impressions) and must be true representations of the object/specimen. Follow the simple rules: Labeling lines should not cross each other; the labeling line should touch the structure in question but not hang; use lines not arrows to label, arrows only show direction; do not shade the drawings but use dots or crosses within the part if you want to show the difference; lines must be continuous and not broken; no two structures should have the same name; for cross-sections and plan diagrams, remember to use double outer lines (outline) and only show key/few features avoiding unnecessary details.  For essay questions, choose the option you are sure of and comfortable with in terms of raising the maximum points possible. Write down key words as they are fresh in your mind; use the first paragraph as an overview of your essay; when time is up for one question or part of the question, stop writing, leave some space and begin the next question or part of the question. The incomplete answers can be completed later during review. However, remember six incomplete answers will receive more credit (marks) than thee complete ones, so don’t waste time. Compactness, completeness (exhaustiveness) and clarity of a well-organized answer is what is important in essay questions. This can be achieved by ensuring that every sentence earns you at least 2 marks hence a 1-1½ pagelength (A-4 paper/foolscap) essay is more appealing than 3 pages of verbal irrelevancies. Remember also that to know a little and present it well is superior to knowing much and presenting it poorly.  Allow yourself time to review your work. Review allows you to ensure that you have answered all the required questions, not skipped any relevant question, not made some simple mistakes and also gives you time to complete any uncompleted questions as well as correcting the wrongly answered ones.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9113832691694873, "ocr_used": false, "chunk_length": 2037, "token_count": 417}}
{"text": "Remember also that to know a little and present it well is superior to knowing much and presenting it poorly.  Allow yourself time to review your work. Review allows you to ensure that you have answered all the required questions, not skipped any relevant question, not made some simple mistakes and also gives you time to complete any uncompleted questions as well as correcting the wrongly answered ones.  Rules for graphical questions may vary from time to time (year to year) but currently the following apply: Use a thin (well-sharpened) pencil to draw thin clear curves for accuracy; all graphs should be curves drawn using continuous lines regardless of the number; the graph should occupy at least ¾ of the grid/graph paper provided, achieved through appropriate scales; state the title and scale at the top of the graph clearly and avoid using abbreviations (remember to include the given units); ensure the Cartesian plane (X and Y axes) has arrows at the ends to show continuity; use dots/small crosses for plotting; do not extrapolate the curve unless asked for a point beyond the plotted ones; label the axes and curves; be neat; if the range of the data is so wide, break the affected axis/axes but don’t force the curve to begin from the 0 point. Commonly Used First Statements in Biology Questions and their Meanings  State/What -This requires the student to present the said concept in a brief and clear form. If it is an adaptation, then it must be given fully i.e. structure, modification and function.  Explain- Give reasons why something/anything happened or will happen.  Describe-Give a detailed or graphic account of the issue or phenomenon showing all features and characteristics e.g. if it is a structure and function question (adaptation), you are expected to correctly name the structure or feature, explain how it is modified and give the function (s) of the structure. Sketchy explanations are penalized or earn least marks. So give as many detailed points as possible.  Define-Give a precise meaning of a word/phrase normally in one sentence.  Illustrate-Use a figure, diagram or any non-text form as an example to explain or make something.  Name/Give/Mention-Give points only, no explanations  Why-Give clear reasons supported by an argument.  Outline-Give the main features in a sequential manner.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9113068337036743, "ocr_used": false, "chunk_length": 2342, "token_count": 501}}
{"text": " Illustrate-Use a figure, diagram or any non-text form as an example to explain or make something.\n\n Name/Give/Mention-Give points only, no explanations  Why-Give clear reasons supported by an argument.\n\n Outline-Give the main features in a sequential manner.\n\n Distinguish between-Define the terms to clearly bring out the differences in their meanings.\n\nYou only earn a full point (s) or mark (s) when both definitions are correctly given.\n\n Account for-Such questions need a two-prong approach: The first part is to give the trend/situation as it is e.g.\n\nthe volume increases (d) rapidly/gradually with increase in temperature.\n\nThe second portion of the answer involves giving the explanation/reasons for the situation/trend e.g.\n\nbecause the molecules/particles gain kinetic energy, moving/vibrating more rapidly hence occupying more space (hence raising the volume).\n\nThese questions are common in the Paper 2; Section A structured questions and the compulsory question in Section B of the same paper.\n\nThe questions assume prior knowledge in either one or a number of the Biology topics.\n\n Differentiate/Give the differences-Use either a table for the differences or use connectors such as the word ‘while’.\n\nMake sure the responses rhyme to earn a mark e.g.\n\nan artery has a narrow lumen while the capillary has a wide lumen not an artery has a narrow lumen while a capillary’s wall is thin.\n\n Compare (and Contrast)-Outline both the similarities and differences between the given terms/processes.", "metadata": {"source": "TIPS-TO-EXCELLING-IN-BIOLOGY..pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9072571010541718, "ocr_used": false, "chunk_length": 1514, "token_count": 336}}
{"text": "FORM ONE BIOLOGY\nBy the end of form one work, the learner should be able to:\nDefine Biology\nList the branches of Biology\nExplain the importance of Biology\nState and explain some of the characteristics of organisms\nState and explain some of the general characteristics of organisms\nExplain the external features of plants and animals\nWrite down the difference between plants and animals\nDefine classification\nUse the magnifying lens to observe the external features of plants/ animals\nRecord observations of the main external features of plant leaf form\nDraw different types of leaf forms\nObserve, record and draw the main external features of plants\nObserve ,record and draw the main external features of animals\nState the necessity and significance of classification\nName the major units of classification\nName the five kingdoms of living things\nList the taxonomic units in plant and animal kingdoms\nClassify maize and human beings\nDefine Binomial nomenclature\nState the principles of Binomial nomenclature In naming organisms\nUse collecting nets, cutting instructions instruments and hand lens\nPreserve collected specimen\nObserve and group collected and preserved specimen according to their similarities\nDefine a cell\nDraw and label the light microscope\nIdentify parts of the light microscope and state their functions\nDescribe how to care for a light microscope\nDescribe how a light microscope is used\nDraw and label plant and animal cells as seen under a light microscope\nCalculate the magnification of objects as seen under a light microscope\nObserve a prepared slide under a light microscope\nPrepare temporary slide of onion epidermis and observe it under a light microscope\nDraw and label plant and animal cells as seen under electron microscope\nDescribe the structure and function of the cell\nCell wall\nCell membrane\nCytoplasm\nDescribe the structure and function of the cell organelles\nEstimate the size of a cell as seen in the field of view of a microscope\nWrite down the differences between plants and animal cells\nWrite down similarities between plant and animal cells\nList down specialized plant and animal cells\nState the modifications and functions of specialized cells\nDefine tissues, organs and organ systems\nGive examples of tissues organs and organ systems\nDefine the term cell physiology\nDescribe the structure and properties of cell membrane\nDefine diffusion\nCarry out experiments to demonstrate\ndiffusion in liquids\ndiffusion in gasses\nExplain the factors affecting diffusion\nExplain the role of diffusion in living things\nDefine osmosis\nDescribe movement of water molecules across semi-permeable membrane\ndefine and describe the terms used in the study of osmosis such as:\nOsmotic pressure\nOsmotic potential\nIsotonic solution\nHypertonic solution\nHypotonic solution\nTurgor pressure\nHemolysis\nWall pressure\nPlasmolysis\nDeplasmolysis\ncarry out an experiment on selective permeability of membrane\nState factors affecting osmosis\nExplain the role of osmosis in organisms\nExplain the factors affecting osmosis\nDescribe what happens when a plant cell is placed in a hypertonic, hypotonic or isotonic solution\nCarry out an experiment to show plasmolysis in epidermal cells of an onion bulb\nDescribe osmosis of animal cells in a hypertonic solution\nList down factors affecting active transport\nDefine active transport\nDefine the role of active transport in living things\nDefine nutrition\nWrite down the importance of nutrition\nList down the modes of feeding in organisms\nDraw and label the external structure of a leaf\nDraw and label the internal structure of the leaf\nName the parts of a leaf\nState the functions of the parts of a leaf\nDefine photosynthesis\nDraw and label the chloroplast\nDescribe the process of photosynthesis\nList down the importance of photosynthesis\nExplain some of the factors influencing photosynthesis\nExplain the factors affecting photosynthesis\nExplain how the leaf is adapted to the process of photosynthesis\nTest the presence of starch in a green leaf\nInvestigate whether chlorophyll is necessary for photosynthesis\nInvestigate whether light is necessary for photosynthesis\ncarry out an experiment to investigate whether\nCarbon (IV) oxide is necessary for photosynthesis\nOxygen is produced during photosynthesis\nDefine Chemicals of life\nList down types of carbohydrates\nWrite down properties and functions of monosaccharaides\nDefine disaccharides\nList properties and functions of disaccharides\nDefine hydrolysis and condensation\nDefine polysaccharides and lipids\nWrite down the properties of polysaccharides and lipids\ncarry out tests on\nStarch\nReducing sugars\nNon-reducing sugar\nLipids\nProteins\nVitamin c\nWrite down the properties and functions of proteins\nDistinguish between carbohydrates, proteins and lipids\nDefine enzymes\nWrite down the properties and functions of enzymes\nKnow the naming of the enzymes and their substrates\nExplain the importance of enzymes\ncarry out an experiment on\nEffect of temperature on enzymes\nEffects of enzyme concentration on the rate of a reaction\nEffect of PH on enzyme activities\nDefine hetetrophism\nList down the different modes of heterotrophism and describe them\nDefine dentition\nDraw and label different types of teeth\nDescribe the structure of a tooth\nIdentify different types of teeth\nDescribe the adaptations of the teeth to their functions\nDefine dental formulae\nDescribe and write down the dental formulae of herbivore carnivore and omnivore\nWrite down the definition of herbivores, carnivores and omnivores\nExplain the adaptations of dental formulae in various groups of animals, to their mode of feeding\nDraw and label the internal structure of different types of teeth\nWrite down the functions of the different parts of the internal structure of teeth\nName and discuss common dental diseases\nWrite down the adaptations of herbivores to their mode of feeding\nWrite down the adaptations of carnivores to their modes of feeding\nIdentify various organs associated with the digestive system of a rabbit\nDraw and label parts of the human digestive system\nDescribe the regions of the alimentary canal of human digestive system\nExplain the functions of the human digestive system\nDescribe the various regions of the human alimentary canal and their functions\nDescribe how the ileum is adapted to its function\nAnalyze the food content in the alimentary canal of a herbivore\nCarry out an experiment on the breakdown of starch by diastase enzymes\nDescribe how the ileum is farther adapted to its functions\nExplain the end products of the digestion of various food\nExplain the function of the colon\nExplain the process of assimilation of food substances\nWrite down the summary of chemical digestion in alimentary canal\nWrite down the importance of vitamins in human nutrition\nWrite down the sources of vitamins\nState deficiency diseases of various vitamins\nWrite down the importance of mineral salts in human nutrition\nState the source of mineral salts\nState the deficiency diseases of mineral salts\nWrite down the role of roughage in nutrition\nWrite down the role of water in nutrition\nDiscuss factors which determine energy requirements in human beings\nParticipate in group discussions and present findings on factors that determine energy requirements in human beings\nIntroduction To Biology\nBiology derived from Greek words-BIOS meaning LIFE and LOGOS meaning STUDY or KNOWLEDGE.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9372080391091798, "ocr_used": false, "chunk_length": 7364, "token_count": 1500}}
{"text": "FORM ONE BIOLOGY\nBy the end of form one work, the learner should be able to:\nDefine Biology\nList the branches of Biology\nExplain the importance of Biology\nState and explain some of the characteristics of organisms\nState and explain some of the general characteristics of organisms\nExplain the external features of plants and animals\nWrite down the difference between plants and animals\nDefine classification\nUse the magnifying lens to observe the external features of plants/ animals\nRecord observations of the main external features of plant leaf form\nDraw different types of leaf forms\nObserve, record and draw the main external features of plants\nObserve ,record and draw the main external features of animals\nState the necessity and significance of classification\nName the major units of classification\nName the five kingdoms of living things\nList the taxonomic units in plant and animal kingdoms\nClassify maize and human beings\nDefine Binomial nomenclature\nState the principles of Binomial nomenclature In naming organisms\nUse collecting nets, cutting instructions instruments and hand lens\nPreserve collected specimen\nObserve and group collected and preserved specimen according to their similarities\nDefine a cell\nDraw and label the light microscope\nIdentify parts of the light microscope and state their functions\nDescribe how to care for a light microscope\nDescribe how a light microscope is used\nDraw and label plant and animal cells as seen under a light microscope\nCalculate the magnification of objects as seen under a light microscope\nObserve a prepared slide under a light microscope\nPrepare temporary slide of onion epidermis and observe it under a light microscope\nDraw and label plant and animal cells as seen under electron microscope\nDescribe the structure and function of the cell\nCell wall\nCell membrane\nCytoplasm\nDescribe the structure and function of the cell organelles\nEstimate the size of a cell as seen in the field of view of a microscope\nWrite down the differences between plants and animal cells\nWrite down similarities between plant and animal cells\nList down specialized plant and animal cells\nState the modifications and functions of specialized cells\nDefine tissues, organs and organ systems\nGive examples of tissues organs and organ systems\nDefine the term cell physiology\nDescribe the structure and properties of cell membrane\nDefine diffusion\nCarry out experiments to demonstrate\ndiffusion in liquids\ndiffusion in gasses\nExplain the factors affecting diffusion\nExplain the role of diffusion in living things\nDefine osmosis\nDescribe movement of water molecules across semi-permeable membrane\ndefine and describe the terms used in the study of osmosis such as:\nOsmotic pressure\nOsmotic potential\nIsotonic solution\nHypertonic solution\nHypotonic solution\nTurgor pressure\nHemolysis\nWall pressure\nPlasmolysis\nDeplasmolysis\ncarry out an experiment on selective permeability of membrane\nState factors affecting osmosis\nExplain the role of osmosis in organisms\nExplain the factors affecting osmosis\nDescribe what happens when a plant cell is placed in a hypertonic, hypotonic or isotonic solution\nCarry out an experiment to show plasmolysis in epidermal cells of an onion bulb\nDescribe osmosis of animal cells in a hypertonic solution\nList down factors affecting active transport\nDefine active transport\nDefine the role of active transport in living things\nDefine nutrition\nWrite down the importance of nutrition\nList down the modes of feeding in organisms\nDraw and label the external structure of a leaf\nDraw and label the internal structure of the leaf\nName the parts of a leaf\nState the functions of the parts of a leaf\nDefine photosynthesis\nDraw and label the chloroplast\nDescribe the process of photosynthesis\nList down the importance of photosynthesis\nExplain some of the factors influencing photosynthesis\nExplain the factors affecting photosynthesis\nExplain how the leaf is adapted to the process of photosynthesis\nTest the presence of starch in a green leaf\nInvestigate whether chlorophyll is necessary for photosynthesis\nInvestigate whether light is necessary for photosynthesis\ncarry out an experiment to investigate whether\nCarbon (IV) oxide is necessary for photosynthesis\nOxygen is produced during photosynthesis\nDefine Chemicals of life\nList down types of carbohydrates\nWrite down properties and functions of monosaccharaides\nDefine disaccharides\nList properties and functions of disaccharides\nDefine hydrolysis and condensation\nDefine polysaccharides and lipids\nWrite down the properties of polysaccharides and lipids\ncarry out tests on\nStarch\nReducing sugars\nNon-reducing sugar\nLipids\nProteins\nVitamin c\nWrite down the properties and functions of proteins\nDistinguish between carbohydrates, proteins and lipids\nDefine enzymes\nWrite down the properties and functions of enzymes\nKnow the naming of the enzymes and their substrates\nExplain the importance of enzymes\ncarry out an experiment on\nEffect of temperature on enzymes\nEffects of enzyme concentration on the rate of a reaction\nEffect of PH on enzyme activities\nDefine hetetrophism\nList down the different modes of heterotrophism and describe them\nDefine dentition\nDraw and label different types of teeth\nDescribe the structure of a tooth\nIdentify different types of teeth\nDescribe the adaptations of the teeth to their functions\nDefine dental formulae\nDescribe and write down the dental formulae of herbivore carnivore and omnivore\nWrite down the definition of herbivores, carnivores and omnivores\nExplain the adaptations of dental formulae in various groups of animals, to their mode of feeding\nDraw and label the internal structure of different types of teeth\nWrite down the functions of the different parts of the internal structure of teeth\nName and discuss common dental diseases\nWrite down the adaptations of herbivores to their mode of feeding\nWrite down the adaptations of carnivores to their modes of feeding\nIdentify various organs associated with the digestive system of a rabbit\nDraw and label parts of the human digestive system\nDescribe the regions of the alimentary canal of human digestive system\nExplain the functions of the human digestive system\nDescribe the various regions of the human alimentary canal and their functions\nDescribe how the ileum is adapted to its function\nAnalyze the food content in the alimentary canal of a herbivore\nCarry out an experiment on the breakdown of starch by diastase enzymes\nDescribe how the ileum is farther adapted to its functions\nExplain the end products of the digestion of various food\nExplain the function of the colon\nExplain the process of assimilation of food substances\nWrite down the summary of chemical digestion in alimentary canal\nWrite down the importance of vitamins in human nutrition\nWrite down the sources of vitamins\nState deficiency diseases of various vitamins\nWrite down the importance of mineral salts in human nutrition\nState the source of mineral salts\nState the deficiency diseases of mineral salts\nWrite down the role of roughage in nutrition\nWrite down the role of water in nutrition\nDiscuss factors which determine energy requirements in human beings\nParticipate in group discussions and present findings on factors that determine energy requirements in human beings\nIntroduction To Biology\nBiology derived from Greek words-BIOS meaning LIFE and LOGOS meaning STUDY or KNOWLEDGE. Biology means \"life knowledge\".", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9370200108166576, "ocr_used": false, "chunk_length": 7396, "token_count": 1506}}
{"text": "FORM ONE BIOLOGY\nBy the end of form one work, the learner should be able to:\nDefine Biology\nList the branches of Biology\nExplain the importance of Biology\nState and explain some of the characteristics of organisms\nState and explain some of the general characteristics of organisms\nExplain the external features of plants and animals\nWrite down the difference between plants and animals\nDefine classification\nUse the magnifying lens to observe the external features of plants/ animals\nRecord observations of the main external features of plant leaf form\nDraw different types of leaf forms\nObserve, record and draw the main external features of plants\nObserve ,record and draw the main external features of animals\nState the necessity and significance of classification\nName the major units of classification\nName the five kingdoms of living things\nList the taxonomic units in plant and animal kingdoms\nClassify maize and human beings\nDefine Binomial nomenclature\nState the principles of Binomial nomenclature In naming organisms\nUse collecting nets, cutting instructions instruments and hand lens\nPreserve collected specimen\nObserve and group collected and preserved specimen according to their similarities\nDefine a cell\nDraw and label the light microscope\nIdentify parts of the light microscope and state their functions\nDescribe how to care for a light microscope\nDescribe how a light microscope is used\nDraw and label plant and animal cells as seen under a light microscope\nCalculate the magnification of objects as seen under a light microscope\nObserve a prepared slide under a light microscope\nPrepare temporary slide of onion epidermis and observe it under a light microscope\nDraw and label plant and animal cells as seen under electron microscope\nDescribe the structure and function of the cell\nCell wall\nCell membrane\nCytoplasm\nDescribe the structure and function of the cell organelles\nEstimate the size of a cell as seen in the field of view of a microscope\nWrite down the differences between plants and animal cells\nWrite down similarities between plant and animal cells\nList down specialized plant and animal cells\nState the modifications and functions of specialized cells\nDefine tissues, organs and organ systems\nGive examples of tissues organs and organ systems\nDefine the term cell physiology\nDescribe the structure and properties of cell membrane\nDefine diffusion\nCarry out experiments to demonstrate\ndiffusion in liquids\ndiffusion in gasses\nExplain the factors affecting diffusion\nExplain the role of diffusion in living things\nDefine osmosis\nDescribe movement of water molecules across semi-permeable membrane\ndefine and describe the terms used in the study of osmosis such as:\nOsmotic pressure\nOsmotic potential\nIsotonic solution\nHypertonic solution\nHypotonic solution\nTurgor pressure\nHemolysis\nWall pressure\nPlasmolysis\nDeplasmolysis\ncarry out an experiment on selective permeability of membrane\nState factors affecting osmosis\nExplain the role of osmosis in organisms\nExplain the factors affecting osmosis\nDescribe what happens when a plant cell is placed in a hypertonic, hypotonic or isotonic solution\nCarry out an experiment to show plasmolysis in epidermal cells of an onion bulb\nDescribe osmosis of animal cells in a hypertonic solution\nList down factors affecting active transport\nDefine active transport\nDefine the role of active transport in living things\nDefine nutrition\nWrite down the importance of nutrition\nList down the modes of feeding in organisms\nDraw and label the external structure of a leaf\nDraw and label the internal structure of the leaf\nName the parts of a leaf\nState the functions of the parts of a leaf\nDefine photosynthesis\nDraw and label the chloroplast\nDescribe the process of photosynthesis\nList down the importance of photosynthesis\nExplain some of the factors influencing photosynthesis\nExplain the factors affecting photosynthesis\nExplain how the leaf is adapted to the process of photosynthesis\nTest the presence of starch in a green leaf\nInvestigate whether chlorophyll is necessary for photosynthesis\nInvestigate whether light is necessary for photosynthesis\ncarry out an experiment to investigate whether\nCarbon (IV) oxide is necessary for photosynthesis\nOxygen is produced during photosynthesis\nDefine Chemicals of life\nList down types of carbohydrates\nWrite down properties and functions of monosaccharaides\nDefine disaccharides\nList properties and functions of disaccharides\nDefine hydrolysis and condensation\nDefine polysaccharides and lipids\nWrite down the properties of polysaccharides and lipids\ncarry out tests on\nStarch\nReducing sugars\nNon-reducing sugar\nLipids\nProteins\nVitamin c\nWrite down the properties and functions of proteins\nDistinguish between carbohydrates, proteins and lipids\nDefine enzymes\nWrite down the properties and functions of enzymes\nKnow the naming of the enzymes and their substrates\nExplain the importance of enzymes\ncarry out an experiment on\nEffect of temperature on enzymes\nEffects of enzyme concentration on the rate of a reaction\nEffect of PH on enzyme activities\nDefine hetetrophism\nList down the different modes of heterotrophism and describe them\nDefine dentition\nDraw and label different types of teeth\nDescribe the structure of a tooth\nIdentify different types of teeth\nDescribe the adaptations of the teeth to their functions\nDefine dental formulae\nDescribe and write down the dental formulae of herbivore carnivore and omnivore\nWrite down the definition of herbivores, carnivores and omnivores\nExplain the adaptations of dental formulae in various groups of animals, to their mode of feeding\nDraw and label the internal structure of different types of teeth\nWrite down the functions of the different parts of the internal structure of teeth\nName and discuss common dental diseases\nWrite down the adaptations of herbivores to their mode of feeding\nWrite down the adaptations of carnivores to their modes of feeding\nIdentify various organs associated with the digestive system of a rabbit\nDraw and label parts of the human digestive system\nDescribe the regions of the alimentary canal of human digestive system\nExplain the functions of the human digestive system\nDescribe the various regions of the human alimentary canal and their functions\nDescribe how the ileum is adapted to its function\nAnalyze the food content in the alimentary canal of a herbivore\nCarry out an experiment on the breakdown of starch by diastase enzymes\nDescribe how the ileum is farther adapted to its functions\nExplain the end products of the digestion of various food\nExplain the function of the colon\nExplain the process of assimilation of food substances\nWrite down the summary of chemical digestion in alimentary canal\nWrite down the importance of vitamins in human nutrition\nWrite down the sources of vitamins\nState deficiency diseases of various vitamins\nWrite down the importance of mineral salts in human nutrition\nState the source of mineral salts\nState the deficiency diseases of mineral salts\nWrite down the role of roughage in nutrition\nWrite down the role of water in nutrition\nDiscuss factors which determine energy requirements in human beings\nParticipate in group discussions and present findings on factors that determine energy requirements in human beings\nIntroduction To Biology\nBiology derived from Greek words-BIOS meaning LIFE and LOGOS meaning STUDY or KNOWLEDGE. Biology means \"life knowledge\". It is the study of living things/organisms.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9368548387096776, "ocr_used": false, "chunk_length": 7440, "token_count": 1516}}
{"text": "FORM ONE BIOLOGY\nBy the end of form one work, the learner should be able to:\nDefine Biology\nList the branches of Biology\nExplain the importance of Biology\nState and explain some of the characteristics of organisms\nState and explain some of the general characteristics of organisms\nExplain the external features of plants and animals\nWrite down the difference between plants and animals\nDefine classification\nUse the magnifying lens to observe the external features of plants/ animals\nRecord observations of the main external features of plant leaf form\nDraw different types of leaf forms\nObserve, record and draw the main external features of plants\nObserve ,record and draw the main external features of animals\nState the necessity and significance of classification\nName the major units of classification\nName the five kingdoms of living things\nList the taxonomic units in plant and animal kingdoms\nClassify maize and human beings\nDefine Binomial nomenclature\nState the principles of Binomial nomenclature In naming organisms\nUse collecting nets, cutting instructions instruments and hand lens\nPreserve collected specimen\nObserve and group collected and preserved specimen according to their similarities\nDefine a cell\nDraw and label the light microscope\nIdentify parts of the light microscope and state their functions\nDescribe how to care for a light microscope\nDescribe how a light microscope is used\nDraw and label plant and animal cells as seen under a light microscope\nCalculate the magnification of objects as seen under a light microscope\nObserve a prepared slide under a light microscope\nPrepare temporary slide of onion epidermis and observe it under a light microscope\nDraw and label plant and animal cells as seen under electron microscope\nDescribe the structure and function of the cell\nCell wall\nCell membrane\nCytoplasm\nDescribe the structure and function of the cell organelles\nEstimate the size of a cell as seen in the field of view of a microscope\nWrite down the differences between plants and animal cells\nWrite down similarities between plant and animal cells\nList down specialized plant and animal cells\nState the modifications and functions of specialized cells\nDefine tissues, organs and organ systems\nGive examples of tissues organs and organ systems\nDefine the term cell physiology\nDescribe the structure and properties of cell membrane\nDefine diffusion\nCarry out experiments to demonstrate\ndiffusion in liquids\ndiffusion in gasses\nExplain the factors affecting diffusion\nExplain the role of diffusion in living things\nDefine osmosis\nDescribe movement of water molecules across semi-permeable membrane\ndefine and describe the terms used in the study of osmosis such as:\nOsmotic pressure\nOsmotic potential\nIsotonic solution\nHypertonic solution\nHypotonic solution\nTurgor pressure\nHemolysis\nWall pressure\nPlasmolysis\nDeplasmolysis\ncarry out an experiment on selective permeability of membrane\nState factors affecting osmosis\nExplain the role of osmosis in organisms\nExplain the factors affecting osmosis\nDescribe what happens when a plant cell is placed in a hypertonic, hypotonic or isotonic solution\nCarry out an experiment to show plasmolysis in epidermal cells of an onion bulb\nDescribe osmosis of animal cells in a hypertonic solution\nList down factors affecting active transport\nDefine active transport\nDefine the role of active transport in living things\nDefine nutrition\nWrite down the importance of nutrition\nList down the modes of feeding in organisms\nDraw and label the external structure of a leaf\nDraw and label the internal structure of the leaf\nName the parts of a leaf\nState the functions of the parts of a leaf\nDefine photosynthesis\nDraw and label the chloroplast\nDescribe the process of photosynthesis\nList down the importance of photosynthesis\nExplain some of the factors influencing photosynthesis\nExplain the factors affecting photosynthesis\nExplain how the leaf is adapted to the process of photosynthesis\nTest the presence of starch in a green leaf\nInvestigate whether chlorophyll is necessary for photosynthesis\nInvestigate whether light is necessary for photosynthesis\ncarry out an experiment to investigate whether\nCarbon (IV) oxide is necessary for photosynthesis\nOxygen is produced during photosynthesis\nDefine Chemicals of life\nList down types of carbohydrates\nWrite down properties and functions of monosaccharaides\nDefine disaccharides\nList properties and functions of disaccharides\nDefine hydrolysis and condensation\nDefine polysaccharides and lipids\nWrite down the properties of polysaccharides and lipids\ncarry out tests on\nStarch\nReducing sugars\nNon-reducing sugar\nLipids\nProteins\nVitamin c\nWrite down the properties and functions of proteins\nDistinguish between carbohydrates, proteins and lipids\nDefine enzymes\nWrite down the properties and functions of enzymes\nKnow the naming of the enzymes and their substrates\nExplain the importance of enzymes\ncarry out an experiment on\nEffect of temperature on enzymes\nEffects of enzyme concentration on the rate of a reaction\nEffect of PH on enzyme activities\nDefine hetetrophism\nList down the different modes of heterotrophism and describe them\nDefine dentition\nDraw and label different types of teeth\nDescribe the structure of a tooth\nIdentify different types of teeth\nDescribe the adaptations of the teeth to their functions\nDefine dental formulae\nDescribe and write down the dental formulae of herbivore carnivore and omnivore\nWrite down the definition of herbivores, carnivores and omnivores\nExplain the adaptations of dental formulae in various groups of animals, to their mode of feeding\nDraw and label the internal structure of different types of teeth\nWrite down the functions of the different parts of the internal structure of teeth\nName and discuss common dental diseases\nWrite down the adaptations of herbivores to their mode of feeding\nWrite down the adaptations of carnivores to their modes of feeding\nIdentify various organs associated with the digestive system of a rabbit\nDraw and label parts of the human digestive system\nDescribe the regions of the alimentary canal of human digestive system\nExplain the functions of the human digestive system\nDescribe the various regions of the human alimentary canal and their functions\nDescribe how the ileum is adapted to its function\nAnalyze the food content in the alimentary canal of a herbivore\nCarry out an experiment on the breakdown of starch by diastase enzymes\nDescribe how the ileum is farther adapted to its functions\nExplain the end products of the digestion of various food\nExplain the function of the colon\nExplain the process of assimilation of food substances\nWrite down the summary of chemical digestion in alimentary canal\nWrite down the importance of vitamins in human nutrition\nWrite down the sources of vitamins\nState deficiency diseases of various vitamins\nWrite down the importance of mineral salts in human nutrition\nState the source of mineral salts\nState the deficiency diseases of mineral salts\nWrite down the role of roughage in nutrition\nWrite down the role of water in nutrition\nDiscuss factors which determine energy requirements in human beings\nParticipate in group discussions and present findings on factors that determine energy requirements in human beings\nIntroduction To Biology\nBiology derived from Greek words-BIOS meaning LIFE and LOGOS meaning STUDY or KNOWLEDGE. Biology means \"life knowledge\". It is the study of living things/organisms. Branches of Biology\nBotany - study of plants.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9366550895004008, "ocr_used": false, "chunk_length": 7486, "token_count": 1528}}
{"text": "Biology means \"life knowledge\". It is the study of living things/organisms. Branches of Biology\nBotany - study of plants. Zoology - study of animals. Microbiology - study' of microscopic organisms. Morphology - study of external structure of organisms. Anatomy - study of internal structure of organisms. Physiology - study of the functioning or working of the cells or body. Biochemistry - study of the chemistry of materials in living organisms. Cytology - study of cells. Genetics - study of inheritance. Ecology- study of the relationship between organisms and their environment. Taxonomy - sorting out of organisms into groups. Histology - study of fine structure of tissues. Virology - study of viruses. Bacteriology - study of bacteria. Entomology - study of insects. Ichthyology - study of fish. Importance of Biology\nOne learns about the functioning of the human body. One understands the developmental changes that take place in the body. It contributes immensely to improved life. It enables one to enter careers such as:\nMedicine,\nNutrition,\nPublic Health,\nDentistry,\nAgriculture\nEnvironmental Studies. Teaching\nCharacteristics of Living Things\nLife defined through observations of activities carried out by living things;\nNutrition –\nNutrition is the processes by which food/nutrients are acquired/made and utilized by living organisms. Green plants and certain bacteria make their own food. All other organisms feed on complex organic materials. Respiration –\nThis is the breakdown of food to provide energy. The energy released is used for various activities in the organism. Gaseous Exchange –Process throw which respiratory gases(CO2&O2) are taken in and out through a respiratory surface. Excretion –\nExcretion is the removal of metabolic wastes from the body. Substances like urea, carbon dioxide (Carbon (IV) oxide). These substances are poisonous if allowed to accumulate in the body. Growth and Development –\nGrowth means irreversible change in size. All organisms increase in size that is, they grow. Development is irreversible change in complexity. As they do so, they also become differentiated in form. Reproduction-Reproduction is the formation of new individuals of a species to ensure continued existence of a species and growth of its population. Irritability –\nThe ability of organisms to detect and respond to changes in the environment. This is of great survival value to the organism.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9158181252351468, "ocr_used": false, "chunk_length": 2419, "token_count": 489}}
{"text": "Reproduction-Reproduction is the formation of new individuals of a species to ensure continued existence of a species and growth of its population. Irritability –\nThe ability of organisms to detect and respond to changes in the environment. This is of great survival value to the organism. Movement –\nIs the progressive change in position from one place to another. Some organisms are sessile (i.e. fixed to the substratum). The majority of plants move only certain parts. Collection and Observation of Organisms\nBiology as a practical subject is learnt through humane handling of organisms. Materials needed for collection of organisms:-\nKnives to cut portions of plant stem/root or uproot. Polythene bags to put the collected plant or specimens. Insect collecting jars. Insect killing jars. Hand gloves. Sweep nets\nPooters\nTraps\nObservation of Organisms\nObserve the plant/animal in its natural habitat before collecting. Identify the exact place -on surface, under rock, on tree trunk, on branches. What does it feed on? How does it interact with other animals and the environment? How many of that kind of plant or animal are in a particular place? Plant specimens placed on the bench and sorted out into;- seeds/stems/roots/leaves/fruits. Animal specimens may be left inside polythene bags if transparent. Others (killed ones) are put in petri dishes. Use hand lens to observe the external features of small animals. Presenting the Results of Observations\nOrganisms are observed and important features noted down: colour, texture hard or soft; if hairy or not. Size is measured or estimated. Biological Drawings - It is necessary to draw some of the organisms. In making a biological drawing, magnification (enlargement) is noted. Indicate the magnification of your drawing. i.e how many times the drawing is larger/smaller than the actual specimen MG=length of drawing/length specimen\nHow to Draw\nSeveral drawings of one organism may be necessary to represent all features observed, e.g. Anterior view of grasshopper shows all mouth parts properly, but not all limbs. Lateral (side) view shows all the legs. Collection, Observation and Recording of Organisms\nCollection\nPlants and animals collected from the environment, near school or within school compound using nets, bottles and gloves. Animals collected include:-arthropods, earthworms and small vertebrates like lizards/chameleons/ rodents.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9198667221990838, "ocr_used": false, "chunk_length": 2401, "token_count": 506}}
{"text": "Lateral (side) view shows all the legs. Collection, Observation and Recording of Organisms\nCollection\nPlants and animals collected from the environment, near school or within school compound using nets, bottles and gloves. Animals collected include:-arthropods, earthworms and small vertebrates like lizards/chameleons/ rodents. Place in polythene bags and take to the laboratory. Stinging/poisonous insects killed using ether. Other animals are observed live and returned to their natural habitat. Plant specimen collected include:- leaves, flowers and whole plants. Observations are made to show the following:-\nPlants have roots, stems, leaves and flowers. Animals have legs, hair, hard outer covering, feathers, eyes, mouth, limbs and other appendages,\nThe differences between animals and plants collected. Comparison Between Plants And Animals\nClassification I\nIntroduction\nClassification is putting organisms into groups. Classification is based on the study of external characteristics of organisms. It involves detailed observation of structure and functions of organisms. Organisms with similar characteristics are put in one group. Differences in structure are used to distinguish one group from another. The magnifying lens is an instrument that assists in the observation of fine structure e.g. hairs by enlarging them. Using a Magnifying Lens\nA specimen is placed on the bench or held by hand,\nThen the magnifying lens is moved towards the eye until the object is dearly focused and an enlarged image is seen. The magnification can be worked out as follows:\nlength of the drawing\nMagnification = length of the specimen\nNote: magnification has no units. Nececity/need for Classification\nTo be able to identify organisms into their taxonomic groups. To enable easier and systematic study of organisms. To show evolutionary relationships in organisms. Major Units of Classification (Taxonomic Groups)\nTaxonomy is the study of the characteristics of organisms for the purpose of classifying them. The groups are Taxa (singular Taxon). The taxonomic groups include:\nSpecies: This is the smallest unit of classification. Organisms of the same species resemble each other. The number of chromosomes in their cells is the same. Members of a species interbreed to produce fertile offspring. Genus (plural genera): A genus is made up of a number of species that share several characteristics. Members of a genus cannot interbreed and if they do, the offspring are infertile. Family: A family is made up of a number of genera that share several characteristics.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.923527116660164, "ocr_used": false, "chunk_length": 2563, "token_count": 507}}
{"text": "Genus (plural genera): A genus is made up of a number of species that share several characteristics. Members of a genus cannot interbreed and if they do, the offspring are infertile. Family: A family is made up of a number of genera that share several characteristics. Order: A number of families with common characteristics make an order. Class: Orders that share a number of characteristics make up a class. Phylum/Division: A number of classes with similar characteristics make up a phylum (plural phyla) in animals. In plants this is called a division. Kingdom: This is made up of several phyla (in animals) or divisions (in plants). It is the largest taxonomic unit in classification. Kingdoms\nLiving organisms are classified into five kingdoms;\nMonera,\nProtoctista,\nFungi,\nPlantae\nAnimalia. Kingdom Fungi\nSome are unicellular while others are multicellular. They have no chlorophyll. Most are saprophytic e.g. yeasts, moulds and mushrooms. A few are parasitic e.g. Puccinia graminae. Kingdom Monera (Prokaryota)\nThese are very small unicellular organisms. They lack a nuclear membrane\ndo not have any bound membrane organelles. Hence the name Prokaryota. They are mainly bacteria, e.g. Vibrio cholerae. Kingdom Protoctista\nThey are unicellular organisms. Their nucleus and organelles are surrounded by membranes (eukaryotic). They include algae, slime moulds - fungi-like and protozoa\nKingdom Plantae\nThey are all multicellular. They contain chlorophyll and are all autotrophic. They include; Bryophyta (mossplant), Pteridophyta (ferns) and Spermatophyta (seed bearing plants). Kingdom Animalia\nThese are all multicellular and heterotrophic. Examples are annelida (earthworms), mollusca (snails),athropoda, chordata . Example of Arthropods are ticks, butterflies. Members of Chordata are fish, frogs and humans. External Features of Organisms\nIn plants we should look for:-\nSpore capsule and rhizoids in moss plants. Sori and fronds in ferns. Stem, leaves, roots, flowers, fruits and seeds in plants.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9148554336989033, "ocr_used": false, "chunk_length": 2006, "token_count": 489}}
{"text": "External Features of Organisms\nIn plants we should look for:-\nSpore capsule and rhizoids in moss plants. Sori and fronds in ferns. Stem, leaves, roots, flowers, fruits and seeds in plants. In animals, some important features to look for are:\nSegmentation, presence of limbs and, number of body parts, presence and number of antennae. These are found in phylum arthropoda:\nVisceral clefts, notochord, nerve tube, fur or hair, scales, fins, mammary glands, feathers and wings. These are found in chordata. Binomial Nomenclature\nOrganisms are known by their local names. Scientists use scientific names to be able to communicate easily among themselves. This method of naming uses two names, and is called Binomial nomenclature. The first name is the name of the genus: (generic name) which starts with a capital letter. The second name is the name of the species (specific name) which starts with a small letter. The two names are underlined or written in italics. Man belongs to the genus Homo, and the species, sapiens. The scientific name of man is therefore Homo sapiens. Maize belongs to the genus Zea, and the species mays. The scientific name of maize is Zea mays. Practical Activities\nUse of Collecting Nets, Cutting Instruments and Hand Lens. Forceps are used to collect crawling and slow moving animals. Sweep nets are used to catch flying insects. Cutting instrument like scapel is used to cut specimen e.g. making sections. Hand lens is used to magnify small plants and animals. Drawing of the magnified organism are made and the linear magnification of each calculated. Collection and Detailed Observation of Small Plants and Animals\ne.g. moss, ferns, bean. Look for the following:\nMoss plants: Rhizoids and spore capsules. Fern plants: Rhizomes with adventitious roots; large leaves (fronds) with Sori (clusters of sporangia). Seed plants: Tree/shrub (woody) or non-woody (herbs) e.g. bean. Root system - fibrous, adventitious and tap root. Stem - position and length of interrnodes. Type of leaves - simple or compound; arranged as alternate, opposite or whorled.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9110789980732178, "ocr_used": false, "chunk_length": 2076, "token_count": 485}}
{"text": "Root system - fibrous, adventitious and tap root. Stem - position and length of interrnodes. Type of leaves - simple or compound; arranged as alternate, opposite or whorled. Flower - colour, number of parts, size and relative position of each:\nFruits - freshy or dry; edible or not edible. Seeds - monocotyledonous or dicotyledonous. Small animals e.g. earthworms, tick, grasshopper, butterfly, beetles. Observe these animals to see:\nNumber of legs. Presence or absence of wings. Number of antennae. Body covering. Body parts. THE CELL\nIntroduction\nThe cell is the basic unit of an organism. All living organisms are made up of cells. Some organisms are made up of one cell and others are said to be multicellular. Other organisms are made of many cells and are said to be multicellular. Cells are too little to see with the naked eye. They can only be seen with the aid of a microscope. The microscope\nThe microscope is used to magnify objects. Magnification\nThe magnifying power is usually inscribed on the lens. To find out how many times a specimen is magnified, the magnifying power of the objective lens is multiplied by that of the eye piece lens. If the eye piece magnification lens is x10 and the objective lens is x4, the total magnification is x40. Magnification has no units. It should always have the multiplication sign.e.g.x40\nMicroscope parts and their functions\nTo View the Object\nTurn the low power objective lens until it clicks into position. Looking through the eye piece, ensure that enough light is passing through by adjusting the mirror. This is indicated by a bright circular area known as the field of view. Place the slide containing the specimen on stage and clip it into position. Make sure that the specimen is in the centre of the field of view. Using the coarse adjustment knob, bring the low power objective lens to the lowest point. Turn the knob gently until the specimen comes into focus. If finer details are required, use the fine adjustment knob. When using high power objective always move the fine adjustment knob upwards. Care of a Microscope\nGreat care should be taken when handling it. Keep it away from the edge of the bench when using it. Always hold it with both hands when moving it in the laboratory. Clean the lenses with special lens cleaning paper.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9099557424012631, "ocr_used": false, "chunk_length": 2301, "token_count": 499}}
{"text": "Keep it away from the edge of the bench when using it. Always hold it with both hands when moving it in the laboratory. Clean the lenses with special lens cleaning paper. Make sure that the low power objective clicks in position in line with eye piece lens before and after use. Store the microscope in a dust-proof place free of moisture. Cell Structure as Seen Through the Light Microscope\nThe cell as seen above has the following:\nCell membrane (Plasma membrane):\nThis is a thin membrane enclosing cell contents. It controls the movement of substances into and out of the cell. Cytoplasm:\nThis is a jelly-like substance in which chemical processes are carried out. Scattered all over the cytoplasm are small structures called organelles. Like an animal cell, the plant cell has a cell membrane, cytoplasm and a nucleus. vacuole. Plant cells have permanent, central vacuole. It contains cell sap where sugars and salts are stored. Cell wall:\nThis is the outermost boundary of a plant cell. It is made of cellulose. Between the cells is a middle lamella made of calcium pectate. Chloroplasts;\nWith special staining techniques it is possible to observe chloroplasts. These are structures which contain chlorophyll, the green pigment responsible for trapping light for photosynthesis. The Electron Microscope (EM)\nCapable of magnifying up to 500,000 times. The specimen is mounted in vacuum chamber through which an electron beam is directed. The image is projected on to a photographic plate. The major disadvantage of the electron microscope is that it cannot be used to observe living objects. However, it provides a higher magnification and resolution (ability to see close points as separate) than the light microscope so that specimen can be observed in more detail. Cell Structure as Seen Through Electron Microscope\nThe Plasma Membrane\nUnder the electron microscope, the plasma membrane is seen as a double layer. This consists of a lipid layer sandwiched between two protein layers. This arrangement is known as the unit membrane and the shows two lipid layers with proteins within. Substances are transported across the membrane by active transport and diffusion. The Endoplasmic Reticulum (ER)\nThis is a network of tubular structures extending throughout the cytoplasm of the cell. It serves as a network of pathways through which materials are transported from one part of the cell to the other. An ER encrusted with ribosomes it is referred to as rough endoplasmic reticulum.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.919863091029016, "ocr_used": false, "chunk_length": 2487, "token_count": 510}}
{"text": "The Endoplasmic Reticulum (ER)\nThis is a network of tubular structures extending throughout the cytoplasm of the cell. It serves as a network of pathways through which materials are transported from one part of the cell to the other. An ER encrusted with ribosomes it is referred to as rough endoplasmic reticulum. An ER that lacks ribosomes is referred to as smooth endoplasmic reticulum. The rough endoplasmic reticulum transports proteins while the smooth endoplasmic reticulum transports lipids. The Ribosomes\nThese are small spherical structures attached to the ER. They consist of protein and ribonucleic acid (RNA). They act as sites for the synthesis of proteins. Goigi Bodies\nGolgi bodies are thin, plate-like sacs arranged in stacks and distributed randomly in the cytoplasm. Their function is packaging and transportation of glycol-proteins. They also produce lysosomes. Mitochondria\nEach mitochondrion is a rod-shaped organelle. Made up of a smooth outer membrane and a folded inner membrane. The foldings of the inner membrane are called cristae. They increase the surface area for respiration. The inner compartments called the matrix. Mitochondria are the sites of cellular respiration, where energy is produced. Lysosomes\nThese are vesicles containing hydrolytic enzymes. They are involved in the breakdown of micro-organisms, foreign macromolecules and damaged or worn-out cells and organelles .. The Nucleus\nThe nucle s is surrounded by a nuclear membrane which is a unit membrane. The nuclear membrane has pores through which materials can move to the surrounding cytoplasm. The nucleus contains proteins and nucleic acid deoxyribonucleic acid (DNA) and RNA. The chromosomes are found in the nucleus. They are the carriers of the genetic information of the cell. The nucleolus is also located in the nucleus but it is only visible during the non-dividing phase of the cell. The Chloroplasts\nThese are found only in photosynthetic cells. Each chloroplast consists of an outer unit. membrane enclosing a series of interconnected membranes called lamellae. At various points along their length the lamellae form stacks of disc like structures called grana. The lamellae are embedded in a granular material called the stroma. The chloroplasts are sites of photosynthesis.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.926771653543307, "ocr_used": false, "chunk_length": 2286, "token_count": 498}}
{"text": "At various points along their length the lamellae form stacks of disc like structures called grana. The lamellae are embedded in a granular material called the stroma. The chloroplasts are sites of photosynthesis. The light reaction takes place in the lamellae while the dark reactions take place in the stroma. Comparison between animal cell and plant cell\nCell Specialisation\nCells are specialised to perform different functions in both plants and animals. Example;\nPalisade cells have many chloroplasts for photosynthesis. Root hair cells are long and thin to absorb water from the soil. Red blood cells have haemoglobin which transports oxygen. Sperm cells have a tail to swim to the egg. Multicellular organisms cells that perform the same function are grouped together to form a tissue. Each tissue is therefore made up of cells that are specialised to carry out a particular function. Animal Tissues- Examples of animal tissues\nPlant Tissues\nExample of plant tissues\nOrgans\nAn organ is made up of different tissues\ne.g. the heart, lungs, kidneys and the brain in animals and roots, stems and leaves in plants. Organ systems\nOrgans which work together form an organ system. Digestive, excretory, nervous and circulatory in animals and transport and support system in plants. organism\nDifferent organ systems form an organism. Practical Activities\nObservation and Identification of parts of a light microscope and their functions\nA light microscope is provided. Various parts are identified and observed. Drawing and labelling of the microscope is done. Functions of the parts of the mircroscope are stated. Calculations of total magnification done using the formula. Eye piece lens maginification x objective lens rnaginification. Preparation and Observation of Temporary Slides of Plant Cells\nA piece of epidermis is made from the fleshy leaf of an onion bulb. It is placed on a microscope slide and a drop of water added. A drop of iodine is added and a cover slip placed on top. Observations are made, under low and medium power objective. The cell wall and nucleus stain darker than other parts. A labelled drawing is made. The following are noted: Nucleus, cell wall, cytoplasm and cell membrane. Observation of permanent slides of animal cells\nPermanent slides of animal cells are obtained e.g, of cheek cells, nerve cells and muscle cells. The slide is mounted on the microscope and observations made under low power and medium power objectives.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.924979658258747, "ocr_used": false, "chunk_length": 2458, "token_count": 506}}
{"text": "The following are noted: Nucleus, cell wall, cytoplasm and cell membrane. Observation of permanent slides of animal cells\nPermanent slides of animal cells are obtained e.g, of cheek cells, nerve cells and muscle cells. The slide is mounted on the microscope and observations made under low power and medium power objectives. Labelled drawings of the cells are made. A comparison between plant and animal cell is made. Observation and Estimation of Cell Size and Calculation of Magnification of Plant Cells. Using the low power objective, a transparent ruler is placed on the stage of the microscope. An estimation of the diameter of the field of view is made in millimeters. This is converted into micrometres (1mm=1000u)\nA prepared slide of onion epidermal cells is mounted. The cells across the centre of the field of view are counted from left and right and top to bottom. The diameter of field of view is divided by the number of cells lying lengthwise to give an estimate of the length and width of each cell. Cell Physiology\nMeaning of cell physiology\nThe term physiology refers to the functions that occur in living organisms. Cell physiology refers to the process through which substances move across the cell membrane. Several physiological processes take place inside the cell.e.g. respiration. Oxygen and glucose required enter the cell while carbon (IV) oxide and water produced leave the cell through the cell membrane. Structure and properties of cell membrane\nThe cell membrane is the protective barrier that shelter cellular contents. Movement of all substances into and out of the cells takes place across the cell membrane. It is made up of protein and lipid molecules. Lipid molecules have phosphate group attached to it on one end. They are then referred to phospholipids. The phospholipids are arranged to form a double layer. The ends with phosphate group face outwards. the proteins are scattered throughout the lipid double layer. Some of these proteins act as carrier molecules that channel some material in and outside the cells. The cell membrane allows certain molecules to pass through freely while others move through with difficulty and still others do not pass through at all. This is selective permeability and the cell membrane is described as semi-permeable. Properties of cell membrane\nPermeability\nThe cell membrane is semi-permeable. it allows small molecules that are soluble in lipid to pass through with more ease than water soluble molecules. this is due to the presence of the phospholipids double layer.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9219180835695622, "ocr_used": false, "chunk_length": 2547, "token_count": 502}}
{"text": "Properties of cell membrane\nPermeability\nThe cell membrane is semi-permeable. it allows small molecules that are soluble in lipid to pass through with more ease than water soluble molecules. this is due to the presence of the phospholipids double layer. Polarlity\nThe cell membrane has electrical charges across its surface.it has positive charged ions on the outside and negatively charged ions on the inside.this property contributes to electrical impulses sent along nerve cells. Sensitivity to changes in temperature and pH\nVery high temperatures destroy the semi-permeability nature of the cell membrane because the proteins are denatured by extreme pH values have the same effect on the membrane permeability. Physiological processes\nSome of the physiological processes include diffusion, osmosis and active transport. Diffusion\nDiffusion is the movement of molecules or ions from a region of high concentration to a region of low concentration aided by a concentration gradient.. diffusion continues to occur as long as there is a difference in concentration between two regions (concentration gradient). Stops when an equilibrium is reached i.e., when the concentration of molecules is the same in both regions. Diffusion is a process that occurs inside living organisms as well as the external environment.. Does not require energy. Factors Affecting Diffusion\n.~ -\nConcentration Gradient\nAn increase in the concentration of molecules at one region results in a steeper concentration gradient which in turn increases the rate of diffusion. Temperature\nHigh temperature increases kinetic energy of molecules. They move faster hence resulting in an increase in rate of diffusion, and vice versa. Size of Molecules or Ions\nThe smaller the size of molecules or ions, the faster their movement hence higher rate of diffusion. Density\nThe denser the molecules or ions diffusing, the slower the rate of diffusion, and vice versa. Medium\nThe medium through which diffusion occurs also affects diffusion of molecules or ions. For example, diffusion of molecules through gas and liquid media is faster than through a solid medium. Distance\nThis refers to the thickness or thinness of surface across which diffusion occurs. Rate of diffusion is faster when the distance is small i.e., thin surface. Surface Area to Volume Ratio\nThe larger the surface area to volume ratio, the faster the rate of diffusion. For example, in small organisms such as Amoeba the surface area to volume ratio, is greater hence faster diffusion than in larger organisms.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9267583497053046, "ocr_used": false, "chunk_length": 2545, "token_count": 484}}
{"text": "Rate of diffusion is faster when the distance is small i.e., thin surface. Surface Area to Volume Ratio\nThe larger the surface area to volume ratio, the faster the rate of diffusion. For example, in small organisms such as Amoeba the surface area to volume ratio, is greater hence faster diffusion than in larger organisms. Role of Diffusion in Living Organisms\nSome processes that depend on diffusion include the following:\nGaseous exchange: Movement of gases through respiratory surfaces is by diffusion. Absorption of materials into cells Cells obtain raw materials and nutrients from the surrounding tissue fluid and blood through diffusion, e.g., glucose needed for respiration diffuses from blood and tissue fluid into cells. Excretion: Removal of metabolic waste products like carbon (IV) oxide, and ammonia out of cells is by diffusion. Absorption of the end-products of digestion from the intestines is by diffusion. Osmosis\nOsmosis is the movement of water molecules from a region of high water concentration to a region of low water concentration through a semi-permeable membrane. Osmosis is a special type of diffusion that involves the movement of water molecules only and not solute molecules. Osmosis takes place in cells across the cell membrane as well as across non-living membranes\ne.g. cellophane or visking tubing which are also semi-permeable,\nIt is purely a physical process. Factors Affecting Osmosis\nSize of solute molecules-\nOsmosis' occurs only when solute molecules are too large to pass through a semi-permeable membrane. Concentration Gradient . Osmosis occurs when two solutions of unequal solute concentration are separated by a semi-permeable membrane. Temperature ,. High temperatures increase movement of water molecules hence influence osmosis. However, too high temperatures denature proteins in cell membrane and osmosis stops. Pressure\nIncrease in pressure affects movement of water molecules. As pressure increases inside a plant cell, osmosis decreases. Roles of Osmosis in Living Organisms\nThe following processes depend on osmosis in living organisms:\nMovement of water into cells from the surrounding tissue fluid and also from cell to cell. Absorption of water from the soil and into the roots of plants. Support in plants especially herbaceous ones, is provided by turgor pressure, which results from intake of water by osmosis. Absorption of water from the alimentary canal in mammals. Re-absorption of water in the kidney tubules.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9252117789431222, "ocr_used": false, "chunk_length": 2479, "token_count": 507}}
{"text": "Support in plants especially herbaceous ones, is provided by turgor pressure, which results from intake of water by osmosis. Absorption of water from the alimentary canal in mammals. Re-absorption of water in the kidney tubules. Opening and closing stomata. Water Relations in Plant and Animal Cells\nThe medium (solution) surrounding cells or organisms is described by the terms hypotonic, hypertonic and isotonic. A solution whose solute concentration is more than that of the cell sap is said to be hypertonic. A cell placed in such a solution loses water to the surroundings by osmosis. A solution whose solute concentration is less than that of the cell sap is said to be hypotonic. A cell placed in such a solution gains water from the surroundings by osmosis. A solution which has the same solute concentration as the cell sap is said to be isotonic. When a cell is placed in such a solution there will be no net movement of water either into or out of the cell. Osmotic Pressure\nThe term osmotic pressure describes the tendency of the solution with a high solute concentration to draw water into itself when it is separated from distilled water or dilute solution by a semi-permeable membrane. Osmotic pressure is measured by an osmometer. When plant cells are placed in distilled water or in a hypotonic solution, the osmotic pressure in the cells is higher than the osmotic pressure of the medium. This causes the water to enter the cells by osmosis. The water collects in the vacuole which increases in size. As a result the cytoplasm is pushed outwards and it in turn presses the cell membrane next to the cell wall. This builds up water pressure (hydrostatic pressure) inside the cell. When the cell is stretched to the maximum, the cell wall prevents further entry of water into the cell. Then the cell is said to be fully turgid. The hydrostatic pressure developed is known as turgor pressure. Plasmolysis\nWhen a plant cell is placed in a hypertonic medium, it loses water by osmosis. The osmotic pressure of the cell is lower than that of the medium. The vacuole decreases in size and the cytoplasm shrinks as a result of which the cell membrane loses contact with the cell wall. The cell becomes flaccid. The whole process is described as plasmolysis.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9214821349801501, "ocr_used": false, "chunk_length": 2267, "token_count": 504}}
{"text": "The vacuole decreases in size and the cytoplasm shrinks as a result of which the cell membrane loses contact with the cell wall. The cell becomes flaccid. The whole process is described as plasmolysis. Incipient plasmolysis is when a cell membrane just begins to lose contact with the cell wall. Plasmolysis can be reversed by placing the cell in distilled water or hypotonic solution. However, full plasmolysis may not be reversed if cell stays in that state for long. Wilting\nThe term wilting describes the drooping of leaves and stems of herbaceous plants after considerable amounts of water have been lost through transpiration. It is observed in hot dry afternoons or in dry weather. This is when the amount of water lost through transpiration exceeds the amount absorbed through the roots. Individual cells lose turgor and become plasmolysed and the leaves and stems droop. The condition is corrected at night when absorption of water by the roots continue while transpiration is absent. Eventually, wilting plants may die if the soil water is not increased through rainfall or watering. Water Relations in Plants and Animals\nHaemolysis\nHaemolysis is the bursting of cell membrane of red blood cells releasing their haemoglobin. It occurs when red blood cells are placed in distilled water or hypotonic solution. This is because the cell membrane does not resist further entry of water by osmosis after maximum water intake. Crenation\nTakes place when red blood cells are placed in hypertonic solution. They lose water by osmosis, shrink and their shape gets distorted. Animal cells have mechanisms that regulate their salt water balance (osmoregulation) to prevent above processes that lead to death of cells. An Amoeba placed in distilled water, i.e. hypotonic solution, removes excess water using a contractile vacuole. The rate of formation of contractile vacuoles increases. Active Transport\nActive transport is the movement of solutes such as .glucose, amino acids and mineral ions;\nFrom an area of their low concentration to an area of high concentration. It is movement against a concentration gradient and therefore energy is required. As such it only takes place in living organisms. The energy needed comes from respiration. Certain proteins in the cell surface membrane responsible for this movement are referred to as carrier proteins or channel proteins. The shape of each type of carrier protein is specific to the type of substances conveyed through it.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9271111111111112, "ocr_used": false, "chunk_length": 2475, "token_count": 511}}
{"text": "The energy needed comes from respiration. Certain proteins in the cell surface membrane responsible for this movement are referred to as carrier proteins or channel proteins. The shape of each type of carrier protein is specific to the type of substances conveyed through it. It has been shown that the substance fits into a particular slot on the protein molecule,\nAs the protein changes from one form of shape to another the substance is moved across and energy is expended. Factors Affecting Active Transport\nAvailability of oxygen\nEnergy needed for active transport is provided through respiration. An increase in the amount of oxygen results in a higher rate of respiration. If a cell is deprived of oxygen active transport stops . Temperature\nOptimum temperature is required for respiration, hence for active transport. Very high temperatures denature respiratory enzymes. Very low temperatures inactivate enzymes too and active transport stops. Availability of carbohydrates\nCarbohydrates are the main substrates for respiration. Increase in amount of carbohydrate results in more energy production during respiration and hence more active transport. Lack of carbohydrates causes active transport to stop. Metabolic poisons\nMetabolic poisons e.g. cyanide inhibit respiration and stops active transport due to lack of energy. Role of Active Transport in Living Organisms\nProcesses requiring active transport:\nAbsorption of mineral salts from the soil into plant roots. Absorption of end products of digestion e.g. glucose and amino acids from the digestive tract into blood stream. Excretion of metabolic products e.g.urea from the cells. Re-absorption of useful substances and mineral salts back into blood capillaries from the kidney tubules. Sodium-pump mechanism in nerve cells. Re-absorption of useful materials from tissue fluid into the blood stream. Practical Activities\n1.Experiment to Demonstrate Diffusion\nVarious coloured substances such as: dyes, plant extracts and chemicals like potassium pennanganate are used. Potassium manganate (VII) crystals are introduced to the bottom of a beaker filled with water using a glass tubing or drinking straw which is then removed. Observations are made and the disappearance of the crystals and subsequent uniform colouring of water noted. 2.Experiment to Demonstrate Osmosis Using a Visking Thbing\nA strip of visking tubing 8-10 cm is cut and tied at one end using strong thread. About 2 ml of 25% sucrose solution is put inside and the other end tied with thread. The tubing is washed under running water and then blotted to dry.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9212899969023554, "ocr_used": false, "chunk_length": 2589, "token_count": 509}}
{"text": "2.Experiment to Demonstrate Osmosis Using a Visking Thbing\nA strip of visking tubing 8-10 cm is cut and tied at one end using strong thread. About 2 ml of 25% sucrose solution is put inside and the other end tied with thread. The tubing is washed under running water and then blotted to dry. It is immersed in a beaker containing distilled water and left for at least one hour or overnight. It will then be observed that the visking tubing has greatly increased in size and has become firm. A control experiment can be set up using distilled water inside the visking tubing in place of sucrose solution. 3.Experiment to Show Osmosis using Living Tissue\nIrish potato tubers are peeled and scooped out to make hollow space at the centre. Sucrose solution is placed inside the hollow, and the potato tuber placed in a beaker or petri-dish with distilled water. A conttrol is set using a boiled potato. Another one using distilled water inside hollow in place of sugar solution. The experiment is left for 3 hours to 24 hours. 4.Experiment to Demonstrate Turgor and Plasmolysis in Onion Epidermal Cells\nTwo strips of onion epidermis are obtained. One is placed on a slide with distilled water while the other is placed on a slide with 25% sucrose solution and a coverslip placed on top of each. The mounted epidermis is observed under low power microscope and then left for 30 minutes. After 30 minutes, observations are made again. The cells in distilled water have greatly enlarged. Cells in 25% sucrose have shrunk. Nutrition in Plants and Animals\nStructure of the Leaf\nExternal Structure\nThe external structure of the leaf consists of a leaf stalk or petiole and a broad leaf blade or lamina. The lamina has a main vein midrib from which smaller veins originate. The outline of the leaf is the margin and the tip forms the apex. Internal Structure of the Leaf\nEpidermis\nThis is the outer layer of cells, normally one cell thick. It is found in both the upper and lower leaf surfaces. The cells are arranged end to end. The epidermis offers protection and maintains the shape of the leaf. It is covered by a layer of cuticle which reduces evaporation.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9026665043224158, "ocr_used": false, "chunk_length": 2150, "token_count": 489}}
{"text": "The cells are arranged end to end. The epidermis offers protection and maintains the shape of the leaf. It is covered by a layer of cuticle which reduces evaporation. Leaf Mesophyll\nConsists of the palisade layer, next to upper epidermis, and the spongy layer next to the lower epidermis. Palisade Mesophyll Layer\nThe cells are elongated and arranged close to each other leaving narrow air spaces. These contain numerous chloroplasts and are the main photosynthetic cells. In most plants, the chloroplast are distributed fairly uniformly throughout the cytoplasm. In certain plants growing in shaded habitats in dim light, most chloroplasts migrate to the upper region of the palisade cells in order to maximise absorption of the limited light available. Spongy Mesophyll Layer\nThe cells are spherical in shape. They are loosely arranged, with large intercellular spaces between them. The spaces are airfilled and are linked to the stomatal pores. The spongy mesophyll cells have fewer chloroplasts than the palisade mesophyll cells. Vascular Bundles\nThese are made up of the xylem and the phloem tissues. The xylem transports water and mineral salts to the leaves. The phloem transports food manufactured in the leaf to the other parts of the plant and from storage organs to other parts. Adaptations of Leaf for Photosynthesis\nPresence of veins with vascular bundles. Xylem vessels transport water for photosynthesis. Phloem transports manufactured food from leaves to other parts of the plant. Leaf lamina is thin to allow for penetration of light over short distance to reach photosynthetic cells. Broad lamina provides a large surface area for absorption of light and carbon (IV) oxide. Transparent cuticle and epidermal layer allow light to penetrate to mesophyll cells. Palisade cells are close to the upper epidermis for maximum light absorption. Presence of numerous chloroplasts in palisade mesophyll traps maximum light. Chloroplast contain chlorophyll that traps light energy. Spongy mesophyll layer has large intercellular air spaces allowing for gaseous exchange. Presence of stomata for efficient gaseous exchange (entry of carbon (IV) oxide into leaf and exit of oxygen). Mosaic arrangement of leaves to ensure no overlapping of leaves hence every leaf is exposed to light.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.928178243774574, "ocr_used": false, "chunk_length": 2289, "token_count": 498}}
{"text": "Spongy mesophyll layer has large intercellular air spaces allowing for gaseous exchange. Presence of stomata for efficient gaseous exchange (entry of carbon (IV) oxide into leaf and exit of oxygen). Mosaic arrangement of leaves to ensure no overlapping of leaves hence every leaf is exposed to light. Structure and Function of Chloroplasts\nChloroplasts are large organelles (5 um in diameter) found in the cytoplasm of green plant cells. They are visible under the light microscope. They contain chlorophyll, a green pigment and other carotenoids which are yellow, orange and red in colour. Certain plants have red or purple leaves due to abundance of these other pigments. Chlorophyll absorbs light energy and transforms it into chemical energy. The other pigments absorb light but only to pass it onto chlorophyll. The wall of chloroplast consists of an outer and an inner membrane. The two make up the chloroplast envelop. Inner membrane encloses a system of membranes called lamellae. At intervals, the membranes form stacks of fluid filed sacs known as grana (singular granum). Chloroplast and other pigments are attached to the grana. In between the lamellae is a gel-like stroma, that contains starch grains and lipid droplets. Enzymes for the dark stage reaction (light independent stage) are embedded in the stroma. Enzymes for the light dependent stage occur in the grana. Functions\n• .Absorption of light by chlorophyll and other pigments. Light stage of photosynthesis occurs on the grana. (transformation of light energy to chemical energy.)\nCarbon fixation to form carbohydrate takes place in the stroma which has enzymes for dark stage of photosynthesis. Process of Photosynthesis\nPhotosynthesis involves a series of chemical reactions, all of which take place inside chloroplasts. A general equation for photosynthesis is:\nCarbon (IV)Oxide+Water light energy---Glucose+Oxygen\nchlorophyll\n6CO2+6H2O light C6H12O+6O2\nchlorophyll\nThe reaction occurs in two main phases or stages. The initial state requires light and it is called the light dependent stage or simply light stage. It takes place on the lamellae surfaces. Its products are used in the dark stage. The dark stage does not require light although it occurs in the light and is called light independent stage.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9158291946635505, "ocr_used": false, "chunk_length": 2282, "token_count": 505}}
{"text": "It takes place on the lamellae surfaces. Its products are used in the dark stage. The dark stage does not require light although it occurs in the light and is called light independent stage. Light-Stage\nTwo reactions take place that produce raw materials for the dark stage:\nLight energy splits the water molecules into hydrogen and oxygen. This process is called photolysis. The hydrogen is taken up by a hydrogen acceptor called Nicotinamide adenine dinucleotide phosphate (NADP) while oxygen is released as a by-product. 2H2O(l) light energy4H+O2\nphotolysis\nLight energy strikes the chlorophyll molecules and sets in motion a series of reactions resulting in the production of a high energy molecule called adenosine triphophate (ATP). Dark Stage\nThis stage involves the fixation of carbon i.e. the reduction of carbon (IV) oxide by addition of hydrogen to form carbohydrate. It uses the products formed during the light stage. ATP\nCarbon + Hydrogen --- Carbohydrates\n(IV) oxide\nThe synthesis of carbohydrates does not take place in a simple straight line reaction as shown in the equation above. It involves a series of steps that constitute what is known as the Calvin cycle. Carbon (IV) oxide is taken up by a compound described as a carbon (IV) oxide acceptor. This is a 5-carbon compound known as ribulose biphosphate and a six carbon compound is formed which is unstable and splits into two three-carbon compounds. Hydrogen from the light reaction is added to the three carbon compound using energy (ATP) from the light reaction. The result is a three carbon (triose) sugar, (phosphoglycerate or PGA). This is the first product of photosynthesis. Glucose, other sugars as well as starch are made from condensation of the triose sugar molecules. The first product is a 3-carbon sugar which condenses to form glucose (6-C sugar). From glucose, sucrose and eventually starch is made. Sucrose is the form in which carbohydrate is transported from the leaves to other parts of the plant. Starch is the storage product. Other substances like oils and proteins are made from sugars. This involves incorporation of other elements e.g. nitrogen, phosphorus and sulphur. Factors Influencing Photosynthesis\nCertain factors must be provided for before photosynthesis can take place.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9112315814802757, "ocr_used": false, "chunk_length": 2279, "token_count": 497}}
{"text": "This involves incorporation of other elements e.g. nitrogen, phosphorus and sulphur. Factors Influencing Photosynthesis\nCertain factors must be provided for before photosynthesis can take place. The rate or amount of photosynthesis is also influenced by the quantity or quality of these same factors. Carbon(IV) Oxide Concentration\nCarbon (IV) oxide is one of the raw materials for photosynthesis. No starch is formed when leaves are enclosed in an atmosphere without carbon (IV) oxide. The concentration of carbon (IV) oxide in the atmosphere remains fairly constant at about 0.03% by volume. However, it is possible to vary the carbon (IV) oxide concentration under experimental conditions. Increasing the carbon (IV) oxide concentration up to 0.1 % increases the rate of photosynthesis. Further increase reduces the rate. Light Intensity\nLight supplies the energy for photosynthesis. Plants kept in the dark do not form starch. Generally, increase in light intensity up to a certain optimum, increases the rate of photosynthesis. The optimum depends on the habitat of the plant. Plants that grow in shady places have a lower optimum than those that grow in sunny places. Water\nWater is necessary as a raw material for photosynthesis. The amount of water available greatly affects the rate of photosynthesis. The more water available, the more the photosynthetic rate, hence amount of food made. Effect of water on photosynthesis can only be inferred from the yield of crops. It is the main determinant of yield (limiting factor in the tropics). Temperature\nThe reactions involved in photosynthesis are catalysed by a series of enzymes. A suitable temperature is therefore necessary. The optimum temperature for photosynthesis in most plants is around 30\"C. This depends on the natural habitat of the plant. Some plants in temperate regions have 20°C as their optimum while others in the tropics have 45°C as their optimum temperature. The rate of photosynthesis decreases with a decrease in temperature below the optimum. In most plants, photosynthesis stops when temperatures approach O°C although some arctic plant species can photosynthesise at -2°C or even -3°C. Likewise, increase in temperature above the optimum decreases the rate and finally the reactions stop at temperatures above 40°c due to enzyme denaturation. However, certain algae that live in hot springs e.g.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9083193865407629, "ocr_used": false, "chunk_length": 2379, "token_count": 482}}
{"text": "In most plants, photosynthesis stops when temperatures approach O°C although some arctic plant species can photosynthesise at -2°C or even -3°C. Likewise, increase in temperature above the optimum decreases the rate and finally the reactions stop at temperatures above 40°c due to enzyme denaturation. However, certain algae that live in hot springs e.g. Oscilatoria can photosynthesise at 75°C\nChlorophyll\nChlorophyll traps or harnesses the energy from light. Leaves without chlorophyll do not form starch. Chemical Compounds Which Constitute Living Organisms\nAll matter is made up of chemical elements, each of which exists in the form of smaller units called atoms. Some of the elements occur in large amounts in living things. These include carbon, oxygen, hydrogen, nitrogen, sulphur and phosphorus. Elements combine together to form compounds. Some of these compounds are organic. Organic compounds contain atoms of carbon combined with hydrogen and they are usually complex. Other compounds are inorganic. Most inorganic compounds do not contain carbon and hydrogen and they are usually less complex. Cells contain hundreds of different classes of organic compounds. However, there are four classes of organic compounds found in all cells. These are: carbohydrates, lipids, proteins and nucleic acids. Carbohydrates\nCarbohydrates are compounds of carbon, hydrogen and oxygen. Hydrogen and oxygen occur in the ratio of 2: 1 as in water. Carbohydrates are classified into three main groups: monosaccharides, disaccharides and polysaccharides. Monosaccharides\nThese are simple sugars. The carbon atoms in these sugars form a chain to which hydrogen and oxygen atoms are attached. Monosaccharides are classified according to the number of carbon atoms they possess. The most common monosaccharides are:\nGlucose - found free in fruits and vegetables. Fructose - found free in fruits and in bee honey. Galactose - found combined in milk sugar. The general formula for these monosaccharides is (CH2O)n where n is 6. They have the same number of carbon, hydrogen and oxygen molecules i.e. C6H12O6. Properties of Monosaccharides\nThey are soluble in water. They are crystallisable. They are sweet. The are all reducing sugars.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9095197269191078, "ocr_used": false, "chunk_length": 2223, "token_count": 487}}
{"text": "They are crystallisable. They are sweet. The are all reducing sugars. This is because they reduce blue copper (II) sulphate solution when heated to copper oxide which is red in colour and insoluble. Functions of Monosaccharides\nThey are oxidised in the cells to produce energy during respiration. Formation of important biological molecules e.g. deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Some monosaccharides are important metabolic intermediates e.g. in photosynthesis and in respiration. Monosaccharides are the units from which other more complex sugars are formed through condensation. Disaccharides\nThese contain two monosaccharide units. The chemical process through which a large molecule (e.g. a disaccharide) is formed from smaller molecules is called condensation and it involves loss of water. Common examples of disaccharides include sucrose, maltose and lactose. Disaccharides are broken into their monosaccharide units by heating with dilute hydrochloric acid. This is known as hydrolysis and involves addition of water molecules. The same process takes place inside cells through enzymes. Sucrose+water_--hydrolysis-----------------glucose+fructose\nProperties of Disaccharides\nSweet tasting. Soluble in water. Crystallisable. Maltose and lactose are reducing sugars while sucrose is non-reducing sugar. Sucrose is the form in which carbohydrate is transported in plants:\nThis is because it is soluble andjchernically stable. Sucrose is a storage carbohydrate in some plants e.g. sugar-cane and sugar-beet. Disaccharides are hydrolysed to produce monosaccharide units which are readily metabolised by cell to provide energy. Polysaccharides\nIf many monosaccharides are joined together through condensation, a polysaccharide is formed. Polysaccharides may consist of hundreds or even thousands of monosaccharide units. Examples of polysaccharides:\nStarch - storage material in plants. Glycogen is a storage carbohydrate in animals like starch, but has longer chains. Inulin - a storage carbohydrate in some plants e.g. Dahlia. Cellulose - structural carbohydrate in plants. Chitin - forms exoskeleton in arthropods.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9214623142986567, "ocr_used": false, "chunk_length": 2142, "token_count": 492}}
{"text": "Dahlia. Cellulose - structural carbohydrate in plants. Chitin - forms exoskeleton in arthropods. Importance and Functions of Polysaccharides\nThey are storage carbohydrates - starch in plants glycogen in animals. They are hydrolysed to their contituent monosaccharide units and used for respiration. . They form structural material e.g. cellulose makes cell walls. Cellulose has wide commercial uses e.g. Fibre in cloth industry. Cellulose is used to make paper. Carbohydrates combine with other molecules to form important structural compounds in living organisms. Examples are:\nPectins: Combine with calcium ions to form calcium pectate. Chitin: Combine with (NH) group. Makes the exoskeleton of arthropods, and walls of fungi. Lipids\nThese are fats and oils. Fats are solid at room temperature while oils are liquid. They are made up of carbon, oxygen and hydrogen atoms. The structural units of lipids are fatty acids and glycerol. Fatty acids are made up of hydrocarbon chain molecules with a carboxyl group (-COOH) at one end. In the synthesis of a lipid, three fatty acid molecules combine with one glycerol molecule to form a triglyceride. Three molecules of water are lost in the process. This is a condensation reaction and water is given off. Lipids are hydrolysed e.g. during digestion to fatty acids and glycerol, water is added. condensation\n-\nGlycerol + 3 Fatty hydrolysis Lipid + Water acids\nProperties\nFats are insoluble in water but dissolve in organic solvents e.g. in alcohols. They are chemically inactive, hence used as food storage compounds. Functions of Lipids\nStructural materials - as structural material they make up the cell membrane. Source of energy - they are energy rich molecules. One molecule of lipid provides more energy than a carbohydrate molecule. Storage compound - They are stored as food reserves in plants. In animals e.g. mammals, all excess food taken is converted to fats which are stored in adipose tissue, and around internal organs such as the heart and kidneys. Insulation - They provide insulation in animals living in cold climates. A lot of fat is stored under the skin e.g. blubber in seals. Protection - Complex lipids e.g.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9153627448618172, "ocr_used": false, "chunk_length": 2178, "token_count": 492}}
{"text": "A lot of fat is stored under the skin e.g. blubber in seals. Protection - Complex lipids e.g. wax on leaf surfaces protects the plant against water-loss and overheating. Fats stored around some internal organs acts as shock absorbers, thus protecting the organs. Source of Metabolic Water -:-lipids when oxidised produce metabolic water which supplements water requirements in the body. Desert animals e.g. the camel accumulate large quantities of fat in the hump which when oxidised releases metabolic water. Proteins\nProteins are the most abundant organic compounds in cells and constitute 50% of total dry weight. Proteins are compounds which are made up of carbon, hydrogen, nitrogen, oxygen and sometimes sulphur and phosphorus. The structural units of proteins are amino acids. The nature of a protein is determined by the types of amino acids it is made of. There are about 20 common amino acids that make up proteins. Essential and Non-Essential Amino Acids\nEssential amino acids are those which cannot be synthesised in the body of an organism and must therefore be provided in the diet. There are ten amino acids which are essential for humans. These are valine, leucine, phenylalanine, lysine, tryptophan, isoleucine, methionine, threonine, histidine and arginine. Non-essential amino acids are those which the body can synthesise and therefore need not be available in the diet. There are ten of them. These are glycine, alanine, glutamic acid, aspartic acid, serine, tyrosine, proline, glutamine, arginine and cysteine. Proteins are essential in the diet because they are not stored in the body. Excess amino acids are deaminated. Formation of Proteins\nProteins are made up of many amino acid units joined together through peptide bonds. When two amino acids are joined together a dipeptide is formed. The chemical process involved is called condensation and a molecule of water is eliminated . When many amino acids are joined together a polypeptide chain is formed. The nature of a particular protein depends on the types, number and sequence of amino acids from which it is made. Functions of Proteins\nAs structural materials proteins-\nAre the basic building structures of protoplasms. Proteins in conjunction with lipid form the cell membrane.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9178724290932269, "ocr_used": false, "chunk_length": 2260, "token_count": 494}}
{"text": "The nature of a particular protein depends on the types, number and sequence of amino acids from which it is made. Functions of Proteins\nAs structural materials proteins-\nAre the basic building structures of protoplasms. Proteins in conjunction with lipid form the cell membrane. Examples of structural proteins include:\nKeratin (in hair, nails, hoofs, feathers and wool)\nSilk in spider's web. Elastin forms ligaments that join bones to each other. Protective proteins. Antibodies that protect the body against foreign antigens. Fribrogen and thrombin are involved in clot formation, preventing entry of micro-organisms when blood vessel is cut. As functional chemical compounds. Examples are hormones and enzymes that act as regulators in the body. Respiratory pigments. Examples are haemoglobin that transports oxygen in the blood and myoglobin that stores up oxygen in muscles. Contractile proteins - make up muscles, i.e. myosin and actin. Proteins combine with other chemical groups to form important substances e.g. mucin in saliva. Source of energy. Proteins are a source of energy in extreme conditions when carbohydrates and fats are not available e.g. in starvation. Enzymes\nEnzymes are biological catalysts that increase the rate of chemical reaction in the body. They are all produced inside cells. Some are intracellular and they catalyse reactions within the cells . Others are extracellular and are secreted out of the cells where they work. e.g. digestive enzymes. Properties of Enzymes\nEnzymes are protein in nature. Enzymes are specific to the type of reaction they catalyse. This is referred to as substrate specificity. Enzymes work in very small amounts. They remain unchanged after the reaction. They catalyse reversible reactions. They work very fast (high turnover numbers) e.g. the enzyme catalase works on 600 thousand molecules of hydrogen peroxide in one second. Naming of enzymes\nEnzymes are named by adding the suffix -ase to:\nName of substrate that they work on e.g. carbohydrates - carbohydrases e.g.sucrase. Starch (amylose) - amylase\nProtein - proteinase (protease)\nLipids -lipases\nType of chemical reaction catalised e.g.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9179154321158586, "ocr_used": false, "chunk_length": 2156, "token_count": 472}}
{"text": "Naming of enzymes\nEnzymes are named by adding the suffix -ase to:\nName of substrate that they work on e.g. carbohydrates - carbohydrases e.g.sucrase. Starch (amylose) - amylase\nProtein - proteinase (protease)\nLipids -lipases\nType of chemical reaction catalised e.g. Oxidation - oxidase\nReduction - reductase\nHydrolysis - hydrolase\nFactors Affecting Enzyme Action\nTemperature\nEnzymes are sensitive to temperature changes. Generally, the rate of an enzymecontrolled reaction doubles with every 10OC increase in temperature. However, temperatures above 40°C do not favour enzyme reaction. This is because enzymes are denatured by high temperatures. pH\nEvery enzyme has a particular pH range over which it works best. Some enzymes work best in acidic media while others function better in alkaline media. Many enzymes function well under neutral conditions. Enzyme Concentration\nUnder conditions where the substrate is in excess, the rate of an enzyme-controlled reaction increases as the enzyme concentration is increased. Substrate Concentration\nIf the concentration of the substrate is increased while that of the enzyme remains constant, the rate of the reaction will increase for sometime and then become constant. Any further increase in substrate concentration will not result in corresponding increase in the rate of the reaction. Enzyme Inhibitors\nThese are substances that either compete with substrates for enzyme active sites or combine with enzymes and hence they inhibit the enzyme reaction. e.g. certain drugs, cyanide and nerve gas. Co-factors\nMost enzymes require the presence of other compounds known as co-factors which are non-proteins. There are three groups of co-factors. Inorganic ions - e.g. iron, magnesium, copper and zinc. Complex organic molecules known as prosthetic groups are attached to the enzyme\ne.g. flavin adenine dinucleotide (FAD) derived from vitamin B2 (riboflavin). Co-enzymes e.g. coenzyme A is involved in respiration. All co-enzymes are derived from vitamins. Nutrition in Animals=Heterotrophism\nMeaning and Types of Heterotrophism\nThis is a mode of nutrition whereby organisms feed on complex organic matter from other plants or animals. All animals are heterotrophs.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9181019303662764, "ocr_used": false, "chunk_length": 2209, "token_count": 490}}
{"text": "All co-enzymes are derived from vitamins. Nutrition in Animals=Heterotrophism\nMeaning and Types of Heterotrophism\nThis is a mode of nutrition whereby organisms feed on complex organic matter from other plants or animals. All animals are heterotrophs. Their mode of feeding is also said to be holozoic to distinguish it from other special types of heterotrophic nutrition namely:\nsaprophytism\nparasitism. Saprophytism/saprotrophysim- occurs in most fungi and some forms of bacteria. Saprophytes feed on dead organic matter and cause its decomposition or decay. Parasitism is a mode of feeding whereby one organism called the parasite feeds on or lives in another organism called the host and harms it. Modes of Feeding in Animals\nAnimals have developed various structures to capture and ingest food. The type of structures present depend on the method of feeding and the type of food. Carnivorous animals feed on whole animals or portions of their flesh. Herbiverous animals feed on plant material. Omnivorous animals feed on both plants and animal materials. Feeding in Mammals\nThe jaws and teeth of mammals are modified according to the type of food eaten. Mammals have different kinds of teeth. Each type of teeth has a particular role to play in the feeding process. Feeding in Mammals\nThe jaws and teeth of mammals are modified according to the type of food eaten. Mammals have different kinds of teeth. Each type of teeth has a particular role to play in the feeding process. This condition is described as heterodont. The teeth of reptiles and amphibians are all similar in shape and carry out the same function. They are said to be homodont. Types of Mammalian Teeth\nMammals have four kinds of teeth. The incisors are found at the front of the jaw. They are sharp-edged and are used for biting. The canines are located at the sides of the jaw. They are pointed and are used for tearing and piercing. The premolars are next to the canines and the molars are at the back of the jaw. Both premolars and molars are used for crushing and grinding. Teeth are replaced only once in a lifetime. The first set is the milk or deciduous teeth. These are replaced by the second set or the permanent teeth.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9226363636363636, "ocr_used": false, "chunk_length": 2200, "token_count": 487}}
{"text": "Teeth are replaced only once in a lifetime. The first set is the milk or deciduous teeth. These are replaced by the second set or the permanent teeth. Dentition refers to the type of teeth, the number and their arrangement in the jaw. A dental formula shows the type and number of teeth in each half of the jaw. The number of teeth in half of the upper jaw is represented above a line and those on the lower jaw below the line. The first letter of each type of teeth is used in the formula i.e. i = incisors, c = canines, pm = premolars and m = molars. The total number is obtained by multiplying by two (for the two halves of each jaw). Adaptation of Teeth to Feeding\nIn general, incisors are for cutting, canines for tearing while premolars and molars are for grinding. However, specific modifications are observed in different mammals as an adaptation to the type of food they eat. Teeth of Herbivores\nIncisors are long and flat with a sharp chisellike edge for cutting. The enamel coating is thicker in front than at the back so that as the tooth wears out, a sharp edge is maintained. Canines are reduced or absent. If absent, the space left is called the diastema. The diastema allows the tongue to hold food and push it to the grinding teeth at the back of the mouth. Premolars and molars:\nThese are transversely ridged. The ridges on the upper teeth fit into grooves on the lower ones. This gives a sideways grinding surface. The teeth of herbivores have open roots i.e., wide opening into the pulp cavity. This ensures a continued adequate supply of food and oxygen to the tooth. In some herbivores, such as rabbits and elephants, the incisors continue to grow throughout life. Teeth of Carnivores\nIncisors are reduced in size and pointed. They are well suited for grasping food and holding prey. Canines are long, pointed and curved. They are used for piercing and tearing flesh as well as for attack and defence. Premolars and molars: In general, they are long and longitudinally ridged to increase surface area for crushing . Carnassial Teeth: These are the last premolars on the upper jaw and the first molars on the lower one. They are enlarged for cutting flesh. They act as a pair of shears. They also crush bones.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9131838565022422, "ocr_used": false, "chunk_length": 2230, "token_count": 503}}
{"text": "They are enlarged for cutting flesh. They act as a pair of shears. They also crush bones. The teeth of carnivores have closed roots i.e., only a very small opening of the pulp cavity to allow food and oxygen to keep teeth alive. Once broken, no re-growth can take place. Teeth of Omnivores\nIncisors have a wide surface for cutting. Canines are bluntly pointed for tearing. Premolars and molars have cusps for crushing and grinding. The premolars have two blunt cusps while the molars have three to four. Internal Structure of tooth\nThe tooth consists of two main parts:\nCrown: The portion above the gum; it is covered by the enamel. Root: The portion below the gum; it is covered by the cement. The tooth has two roots. Neck: Is the region at the same level with the gum. It forms the junction between the crown and the root. It is covered by enamel. Incisors and canines have one root only. Premolars have one or two roots while molars have two to three roots each. Internally, the bulk of the tooth is made up of dentine which consists of living cells and extends to the root. It is composed of calcium salts, collagen and water. It is harder than bone but wears out with use. This is why it is covered by enamel which is the hardest substance in a mammal's body. Pulp Cavity: Contains blood vessels which provide nutrients to the dentine and remove waste products. It also contains nerve endings which detect heat, cold and pain. Cement: Fixes the tooth firmly to the jaw bone. Common Dental Diseases\nDental Carries\nDental carries are the holes or cavities that are formed as acid corrodes enamel and eventually the dentine. Causes\nThis is caused by bacteria acting on the food left between teeth and on the cusp. Acids are formed that eventually corrode the enamel. The pulp cavity is eventually reached. A lot of pain is experienced then. The bacteria then infect the pulp cavity and the whole tooth decays. Treatment\nTreatment depends on the extent of the dental caries:\nExtraction of Tooth. Filling - this involves replacing the dentine with amalgam, a mixture of hard elements e.g. silver and tin. Root Canal Treatment - This involves surgery and reconstruction. It saves severely damaged teeth. The nerves in the root canal are surgically severed. The tooth is cleaned and filled up with amalgam.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9148806941431671, "ocr_used": false, "chunk_length": 2305, "token_count": 507}}
{"text": "It saves severely damaged teeth. The nerves in the root canal are surgically severed. The tooth is cleaned and filled up with amalgam. Periodontal Diseases\nThese are diseases of the gum. The gum becomes inflamed, and starts bleeding. Progression of the disease leads to infection of the fibres in the periodontal membranes and the tooth becomes loose. This condition is known as pyorrhoea. The diseases are caused by poor cleaning of the teeth. The accumulation of food particles leading to formation of plaque, lack of adequate vitamin A and C in the diet. Treatment\nNutrition - by taking adequate balanced diet rich in vitamins A and C. Antibiotics are used to kill bacteria. Anti-inflamatory drugs are given. Antiseptic is prescribed to use in cleaning the mouth daily to prevent further proliferation of bacteria. The plaque is removed-drilled away - a procedure known as scaling. Care of Teeth\nIn order to maintain healthy teeth the following points should be observed:\nA proper diet that includes calcium and vitamins, particularly vitamin D is essential. The diet should also contain very small quantities of fluorine to strengthen the enamel. Large quantities of fluorine are harmful. The enamel becomes brown, a condition known as dental flourosis. Chewing of hard fibrous foods like carrots and sugar cane to strengthen and cleanse the teeth. Proper use of teeth e.g. not using teeth to open bottles and cut thread. Regular and thorough brushing of teeth after meals. Dental floss can be used to clean between the teeth. Not eating sweets and sugary foods between meals. Regular visits to the dentist for checkup. Washing the mouth with strong salt solution or with any other mouth wash with antiseptic properties. Digestive System and Digestion in Humans\nOrgans that are involved with feeding in humans constitute the digestive system. Digestive System and Associated Glands\nHuman digestive system starts at the mouth and ends at the anus. This is the alimentary canal. Digestion takes place inside the lumen of the alimentary canal. The epithelial wall that faces the lumen has mucus glands (goblet cells). These secrete mucus that lubricate food and prevent the wall from being digested by digestive enzymes. Present at specific regions are glands that secrete digestive enzymes. The liver and pancreas are organs that are closely associated with the alimentary canal. Their secretions get into the lumen and assist in digestions. Digestive system consists of:\nMouth.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9241129032258065, "ocr_used": false, "chunk_length": 2480, "token_count": 506}}
{"text": "The liver and pancreas are organs that are closely associated with the alimentary canal. Their secretions get into the lumen and assist in digestions. Digestive system consists of:\nMouth. Oesophagus. Stomach. Small intestines - consist of duodenum, the first part next to the stomach, ileum - the last part that ends up in a vestigial caecum and appendix which are nonfunctional. Large intestines consist of: colon and rectum that ends in the anus. Ingestion, Digestion and Absorption\nFeeding in humans involves the following processes:\nIngestion: This is the introduction of the food into the mouth. Digestion: This is the mechanical and chemical breakdown of the food into simpler, soluble and absorbable units. Absorption: Taking into blood the digested products. Assimilation: Use of food in body cells. Mechanical breakdown of the food takes place with the help of the teeth. Chemical digestion involves enzymes. Digestion in the Mouth\nIn the mouth, both mechanical and chemical digestion takes place. Food is mixed with saliva and is broken into smaller particles by the action of teeth. Saliva contains the enzyme amylase. It also contains water and mucus which lubricate and soften food in order to make swallowing easy. Saliva is slightly alkaline and thus provides a suitable pH for amylase to act on cooked starch, changing it to maltose. The food is then swallowed in the form of semisolid balls known as boluses. Each bolus moves down the oesophagus by a process known as peristalsis. Circular and longitudinal muscles along the wall of the alimentary canal contract and relax pushing the food along. Digestion in the Stomach\nIn the stomach, the food is mixed with gastric juice secreted by gastric glands in the stomach wall. Gastric juice contains pepsin, rennin and hydrochloric acid. The acid provides a low pH of 1.5-2.0 suitable for the action of pepsin. Pepsin breaks down protein into peptides. Rennin coagulates the milk protein casein. The stomach wall has strong circular and longitudinal muscles whose contraction mixes the food with digestive juices in the stomach. Digestion in the Duodenum\nIn the duodenum the food is mixed with bile and pancreatic juice. Bile contains bile salts and bile pigments.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9173337324751785, "ocr_used": false, "chunk_length": 2227, "token_count": 499}}
{"text": "The stomach wall has strong circular and longitudinal muscles whose contraction mixes the food with digestive juices in the stomach. Digestion in the Duodenum\nIn the duodenum the food is mixed with bile and pancreatic juice. Bile contains bile salts and bile pigments. The salts emulsify fats, thus providing a large surface area for action of lipase. Pancreatic juice contains three enzymes:\nTrypsin which breaks down proteins into peptides and amino acids,\nAmylase which breaks down starch into maltose, and\nLipase which breaks down lipids into fatty acids and glycerol. These enzymes act best in an alkaline medium which is provided for by the bile. Digestion in ileum\nEpithelial cells in ileum secrete intestinal juice, also known as succus entericus. This contains enzymes which complete the digestion of protein into amino acids, carbohydrates into monosaccharides and lipids into fatty acids and glycerol. Absorption\nThis is the diffusion of the products of digestion into the blood of the animal. It takes place mainly in the small intestines though alcohol and some glucose are absorbed in the stomach. The ileum is adapted for absorption in the following ways:\nIt is highly coiled. The coiling ensures that food moves along slowly to allow time for its digestion and absorption. It is long to provide a large surface area for absorption. The epithelium has many finger-like projections called villi (singular villus). They greatly increase the surface area for absorption. Villi have microvilli that further increase the surface area for absorption. The wall of villi has thin epithelial lining to facilitate fast diffusion of products of digestion. Has numerous blood vessels for transport of the end products of digestion. Has lacteal vessels; for absorption of fatty acids and glycerol and transport of lipids. Absorption of Glucose and Amino Acids\nGlucose and other monosaccharides as well as amino acids are absorbed through the villi epithelium and directly into the blood capillaries. First they are carried to the liver through the hepatic portal vein, then taken to all organs via circulatory system. Absorption of Fatty Acids and Glycerol\nFatty acids and glycerol diffuse through the epithelial cells of villi and into the lacteal.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9278542869835629, "ocr_used": false, "chunk_length": 2251, "token_count": 480}}
{"text": "Absorption of Glucose and Amino Acids\nGlucose and other monosaccharides as well as amino acids are absorbed through the villi epithelium and directly into the blood capillaries. First they are carried to the liver through the hepatic portal vein, then taken to all organs via circulatory system. Absorption of Fatty Acids and Glycerol\nFatty acids and glycerol diffuse through the epithelial cells of villi and into the lacteal. When inside the villi epithelial cells, the fatty acids combine with glycerol to make tiny fat droplets which give the lacteal a milky appearance. The lacteals join the main lymph vessel that empties its contents into the bloodstream in the thoracic region. Once inside the blood, the lipid droplets are hydrolysed to fatty acids and glycerol. Absorption of Vitamins and Mineral Salts\nVitamins and mineral salts are absorbed into the blood capillaries in' the villi. Water is mainly absorbed in the colon. As a result the undigested food is in a semi-solid form (faeces) when it reaches the rectum. Egestion: This is removal of undigested or indigestible material from the body. Faeces are temporarily stored in the rectum then voided through the anus. Opening of the anus is controlled by sphincter muscles\nAssimilation: This is the incorporation of the food into the cells where it is used for various chemical processes. Carbohydrates\nused to provide energy for the body. Excess glucose is converted to glycogen and stored in the liver and muscles. Some of the excess carbohydrates are also converted into fat in the liver and stored in the adipose tissue' (fat storage tissue), in the mesenteries and in the connective tissue under the skin, around the heart and other internal organs. Proteins\nAmino acids are used to build new cells and repair worn out ones. They are also used for the synthesis of protein compounds. Excess amino acids are de-aminated in the liver. Urea is formed from the nitrogen part. The remaining carbohydrate portion is used for energy or it is converted to glycogen or fat and stored. Lipids\nFats are primarily stored in the fat storage tissues. When carbohydrates intake is low in the body, fats are oxidised to provide energy. They are also used as structural materials e.g.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.923400447427293, "ocr_used": false, "chunk_length": 2235, "token_count": 505}}
{"text": "Lipids\nFats are primarily stored in the fat storage tissues. When carbohydrates intake is low in the body, fats are oxidised to provide energy. They are also used as structural materials e.g. phospholipids in cell membrane. They act as cushion, protecting delicate organs like the heart. Stored fats under the skin act as heat insulators. Summary of digestion in humans\nImportance of Vitamins, Mineral Salts, Roughage and Water in Human Nutrition\nVitamins\nThese are organic compounds that are essential for proper growth, development and functioning of the body. Vitamins are required in very small quantities. They are not stored and must be included in the diet. Vitamins Band C are soluble in water, the rest are soluble in fat. Various vitamins are used in different ways. Mineral Salts\nMineral ions are needed in the human body. Some are needed in small amounts while others are needed in very small amounts (trace). All are vital to human health. Nevertheless, their absence results in noticeable mulfunction of the body processes. Water\nWater is a constituent of blood and intercellular fluid. It is also a constituent of cytoplasm. Water makes up to 60-70% of total fresh weight in humans. No life can exist without water. Functions of Water\nActs as a medium in which chemical reactions in the body takes place. Acts as a solvent and it is used to transport materials within the body. Acts as a coolant due to its high latent heat of vaporisation. Hence, evaporation of sweat lowers body temperature. Takes part in chemical reactions i.e. hydrolysis. Vitamins, sources, uses and the deficiency disease resulting from their absence in diet\nRoughage\nRoughage is dietary fibre and it consists mainly of cellulose. It adds bulk to the food and provides grip for the gut muscles to enhance peristalsis. Roughage does not provide any nutritional value because humans and all animals not produce cellulase enzyme to digest cellulose. In herbivores symbiotic bacteria in the gut produce cellulase that digests cellulose. Factors Determining Energy Requirements in Humans\nAge: Infants, for instance, need a greater proportion of protein than adults. Sex: males generally require more carbohydrates than females. The requirements of specific nutrients for females depends on the stage of development in the life cycle.", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9196283246977548, "ocr_used": false, "chunk_length": 2316, "token_count": 482}}
{"text": "Factors Determining Energy Requirements in Humans\nAge: Infants, for instance, need a greater proportion of protein than adults.\n\nSex: males generally require more carbohydrates than females.\n\nThe requirements of specific nutrients for females depends on the stage of development in the life cycle.\n\nAdolescent girls require more iron in their diet; expectant and nursing mothers require a lot of proteins and mineral salts.\n\nState of Health: A sick individual requires more of certain nutrients e.g.\n\nproteins, than a healthy one.\n\nOccupation: An office worker needs less nutrients than a manual worker.\n\nBalanced Diet\nA diet is balanced when it contains all the body's nutrient requirements and in the right amounts or proportions.\n\nA balanced diet should contain the following:\nCarbohydrates\nProteins\nLipids\nVitamins\nMineral Salts\nWater\nDietary fibre or roughage\nMalnutrition\nThis is faulty or bad feeding where the intake of either less or more than the required amount of food or total lack of some food components.\n\nDeficiency Diseases\nDeficiency diseases result from prolonged absence of certain components in the diet.\n\nExamples are:\nMarasmus:\nLack of enough food reuslts in thin arms and legs,\nsevere loss of fluid,\ngeneral body wasting\nsunken eyes.\n\nKwashiorkor –\nLack of protein in the diet of children.\n\nThe symptoms of kwashiorkor include wasting of the body, red thin hair, swollen abdomen and scaly skin.\n\nOther deficiency diseases are due to lack of accessory food factors (vitamins and mineral salts.).\n\nSuch diseases include rickets, goitre and anaemia.\n\nTreatment of these deficiency diseases is by supplying the patient with the component missing in the diet.\n\nTHE END\nPractical Activities\nExperiments to show that Carbon (IV) Oxide is necessary for Photosynthesis\nExperiment to Show Effect of Light on Photosynthesis\nExperiment to Show the Effect of Chlorophyll on Photosynthesis\nExperiment To Observe Stomata Distribution in Different Leaves\nTest for Reducing Sugar\nTest for non-reducing sugar\nTest for Lipids;\n(a) Grease Spot Test\n(b) Emulsion Test\nTest for Proteins -Biuret Test\nExperiment To Investigate Presence of Enzyme in Living Tissue\nDissection of a Rabbit to show the Digestive System", "metadata": {"source": "FORM-1-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9229796839729121, "ocr_used": false, "chunk_length": 2215, "token_count": 479}}
{"text": "FORM FOUR BIOLOGY\nDefine the term genetics\nDifferentiate between heredity and variation\nDistinguish between continuous and discontinuous variations\nDescribe continuous and discontinuous variations\nObserve variations in plants and animals\nDescribe the structure, nature and properties of chromosomes\nDescribe the structure, nature and properties of DNA molecule\nDifferentiate between DNA and RNA\nDistinguish between F1 and F2 generation\nDetermine Mendel’s first law of inheritance\nDefine other terms used in inheritance such as phenotype, genotype, dominant gene, recessive gene, haploid and diploid\nDemonstrate monohybrid inheritance in plants and animals\nPredict outcomes of various genetic crosses\nConstruct and make use of pannet squares\nWork out genotypic and phenotypic ratios\nPredict outcomes of various crosses\nDetermine the unknown genotypes in a cross using a test cross\nDescribe albinism as an example of monohybrid inheritance in human beings\nExplain the inheritance of ABO blood groups in human beings\nExplain the inheritance of rhesus factor as an example of monohybrid inheritance in human beings\nPredict the inheritance of blood groups human beings\nDescribe incomplete dominance\nDescribe inheritance of colour in flowers of mirabilis jalapa\nDescribe Inheritance of sickle cell anemia in human beings\nExplain how sex is determined in human beings\nDescribe sex linkages in human beings\nDefine linkage and sex-linkage\nDescribe linkage in human beings e.g.colour blindness and hemophilia\nDescribe colour blindness as an example of sex-linked trait in human beings\nInterpret pedigree of inheritance\nDescribe the Inheritance of hemophilia as an example of sex-linked traits in human beings\nDefine mutation\nDifferentiate between mutations and mutagens\nList down causes of mutations\nState the types of mutations\nList down the various chromosal mutations\nDescribe chromosal mutations\nExplain the Effects of chromosal mutations\nDescribe gene mutations and their effects on organisms\nDescribe areas in which the knowledge of genetics has been applied\nExplain the practical applications of genetics\nDefine evolution\nExplain the current concepts of the origin of life\nExplain the current concepts on origin of life\nDescribe the study of fossils as evidence of organic evolution theory\nDescribe comparative anatomy as evidence of organic evolution\nDescribe occurrence of vestigial structures and geographical distribution of organisms as evidence of organic evolution\nDescribe comparative embryology, cell biology and biochemistry as evidence of organic evolution\nDescribe evolution of hominids\nDescribe Lamarck’s theory\nDescribe and discuss the struggle for existence and survival for the fittest\nDescribe and discuss new concepts of Darwin’s theory\nDescribe natural selection in action\nDescribe natural selection in nature\nDescribe the isolation mechanism in speciation\nDescribe Artificial selection in plants and animals and how it leads to speciation\nExplain the importance of sexual reproduction in evolution\nDefine stimulus\nDefine irritability\nDefine response\nDefine tactic and tropic responses\nList down tactic responses in plants\nList down tropic responses in plants\nDifferentiate between tactic and tropic responses\nDefine geotropism\nDescribe geotropism in roots and shoots of plants\nDifferentiate between Phototropism and geotropism\nCarry out experiments demonstrating both Phototropism and geotropism in a plant seedling\nCarry out experiments to demonstrate tactic responses to light and water\nCarry out experiments to show chemotactic response using fruit juice\nDefine Hydrotropism and thigmotropism\nState the importance of Tactic and tropic responses\nExplain the production of Plant hormones and their effects on plants\nCarry out experiment to investigate hydrotropism\nCarry out experiment to investigate etiolation\nDemonstrate the knee jerk in a reflex action\nDefined Conditioned reflex actions\nDescribe Conditioned reflex action using parlous dog\nCompare simple and conditioned reflex actions\nExplain the role of endocrine system in a human being\nExplain the effect over secretion and under secretion of thyroxin and adrenaline\nIsolate and list the similarities and differences between the endocrine and the nervous system\nState the effects of drug abuse on human health\nDraw and label the mammalian eye\nState the functions of the mammalian eye\nDescribe how the structure of the mammalian eye is adapted to its functions\nDissect and display parts of the mammalian eye\nDescribe how an image is formed and interpreted in the mammalian eye\nDescribe Accommodation in the mammalian eye\nName and explain the Common eye defects\nDescribe Common eye defects and their corrections\nInvestigate the blind spot In the eye\nInvestigate which eye is used more during vision\nName and describe Common eye diseases\nDraw and label the mammalian ear\nDescribe the mammalian ear and how it is adapted to its functions\nDescribe the mechanism of hearing\nDiscuss thick ear drum, damaged cochlea, raptured eardrum, fussed ossicles, otitis media, ostosceleross and tinnitus\nDefine support and movement\nDescribe the necessity of movement in plants and animals\nReview the tissue distribution in monocotyledonous an dicotyledonous plants\nDescribe support in woody and non-woody stems\nDescribe the role of tendrils and tender stems in support\nObserve prepared sections of woody and herbaceous stems\nObserve a wilting plant\nList the types of skeletons\nDescribe the role of exoskeleton in insects\nDescribe the role and components of endoskeleton\nDescribe the role of skeleton in vertebrates\nDraw the structure of a finned fish (tilapia)\nCalculate the tail power\nExplain how locomotion occurs in fish\nName and draw the different fins and state their functions\nDraw the human skeleton and identify the component parts\nIdentify and draw the skull\nIdentify bones of Axial skeleton in the vertebral column\nIdentify the cervical vertebrae\nIdentify the structures of the thoracic vertebrae\nRelate the structure of the thoracic vertebrae to their functions\nIdentify the structures of lumbar, sacral and candal vertebrae\nShow how ribs articulate with thoracic vertebrae\nDraw and label Ribs and sternum\nRelate the structure to their functions\nIdentify components of Appendicular skeleton\nDraw the scapula bone and relate it to its functions\nIdentify the bones of the fore limbs\nDraw the structure of the humerus, radius and ulna\nDraw and label bones of the hand\nDraw the pelvic girdle\nName the bones of The pelvic girdle\nRelate the structure to their functions\nIdentify, draw and label the femur, tibia and tibula bones\nRelate their structure to their functions\nDraw and label the bones of the foot\nRelate the structure of bones of the foot to their functions\nDefine a joint\nList the three types of joints\nDescribe the types of joints\nList examples of movable joints, hinge joints and bell and socket joints\nDefine Immovable joints\nName Immovable joints\nDefine muscles\nExplain the differences between the three types of muscles\nIdentifying biceps and triceps in the arm movement\nGenetics\nIntroduction\nGenetics is the study of inheritance.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9369471209354073, "ocr_used": false, "chunk_length": 7110, "token_count": 1475}}
{"text": "FORM FOUR BIOLOGY\nDefine the term genetics\nDifferentiate between heredity and variation\nDistinguish between continuous and discontinuous variations\nDescribe continuous and discontinuous variations\nObserve variations in plants and animals\nDescribe the structure, nature and properties of chromosomes\nDescribe the structure, nature and properties of DNA molecule\nDifferentiate between DNA and RNA\nDistinguish between F1 and F2 generation\nDetermine Mendel’s first law of inheritance\nDefine other terms used in inheritance such as phenotype, genotype, dominant gene, recessive gene, haploid and diploid\nDemonstrate monohybrid inheritance in plants and animals\nPredict outcomes of various genetic crosses\nConstruct and make use of pannet squares\nWork out genotypic and phenotypic ratios\nPredict outcomes of various crosses\nDetermine the unknown genotypes in a cross using a test cross\nDescribe albinism as an example of monohybrid inheritance in human beings\nExplain the inheritance of ABO blood groups in human beings\nExplain the inheritance of rhesus factor as an example of monohybrid inheritance in human beings\nPredict the inheritance of blood groups human beings\nDescribe incomplete dominance\nDescribe inheritance of colour in flowers of mirabilis jalapa\nDescribe Inheritance of sickle cell anemia in human beings\nExplain how sex is determined in human beings\nDescribe sex linkages in human beings\nDefine linkage and sex-linkage\nDescribe linkage in human beings e.g.colour blindness and hemophilia\nDescribe colour blindness as an example of sex-linked trait in human beings\nInterpret pedigree of inheritance\nDescribe the Inheritance of hemophilia as an example of sex-linked traits in human beings\nDefine mutation\nDifferentiate between mutations and mutagens\nList down causes of mutations\nState the types of mutations\nList down the various chromosal mutations\nDescribe chromosal mutations\nExplain the Effects of chromosal mutations\nDescribe gene mutations and their effects on organisms\nDescribe areas in which the knowledge of genetics has been applied\nExplain the practical applications of genetics\nDefine evolution\nExplain the current concepts of the origin of life\nExplain the current concepts on origin of life\nDescribe the study of fossils as evidence of organic evolution theory\nDescribe comparative anatomy as evidence of organic evolution\nDescribe occurrence of vestigial structures and geographical distribution of organisms as evidence of organic evolution\nDescribe comparative embryology, cell biology and biochemistry as evidence of organic evolution\nDescribe evolution of hominids\nDescribe Lamarck’s theory\nDescribe and discuss the struggle for existence and survival for the fittest\nDescribe and discuss new concepts of Darwin’s theory\nDescribe natural selection in action\nDescribe natural selection in nature\nDescribe the isolation mechanism in speciation\nDescribe Artificial selection in plants and animals and how it leads to speciation\nExplain the importance of sexual reproduction in evolution\nDefine stimulus\nDefine irritability\nDefine response\nDefine tactic and tropic responses\nList down tactic responses in plants\nList down tropic responses in plants\nDifferentiate between tactic and tropic responses\nDefine geotropism\nDescribe geotropism in roots and shoots of plants\nDifferentiate between Phototropism and geotropism\nCarry out experiments demonstrating both Phototropism and geotropism in a plant seedling\nCarry out experiments to demonstrate tactic responses to light and water\nCarry out experiments to show chemotactic response using fruit juice\nDefine Hydrotropism and thigmotropism\nState the importance of Tactic and tropic responses\nExplain the production of Plant hormones and their effects on plants\nCarry out experiment to investigate hydrotropism\nCarry out experiment to investigate etiolation\nDemonstrate the knee jerk in a reflex action\nDefined Conditioned reflex actions\nDescribe Conditioned reflex action using parlous dog\nCompare simple and conditioned reflex actions\nExplain the role of endocrine system in a human being\nExplain the effect over secretion and under secretion of thyroxin and adrenaline\nIsolate and list the similarities and differences between the endocrine and the nervous system\nState the effects of drug abuse on human health\nDraw and label the mammalian eye\nState the functions of the mammalian eye\nDescribe how the structure of the mammalian eye is adapted to its functions\nDissect and display parts of the mammalian eye\nDescribe how an image is formed and interpreted in the mammalian eye\nDescribe Accommodation in the mammalian eye\nName and explain the Common eye defects\nDescribe Common eye defects and their corrections\nInvestigate the blind spot In the eye\nInvestigate which eye is used more during vision\nName and describe Common eye diseases\nDraw and label the mammalian ear\nDescribe the mammalian ear and how it is adapted to its functions\nDescribe the mechanism of hearing\nDiscuss thick ear drum, damaged cochlea, raptured eardrum, fussed ossicles, otitis media, ostosceleross and tinnitus\nDefine support and movement\nDescribe the necessity of movement in plants and animals\nReview the tissue distribution in monocotyledonous an dicotyledonous plants\nDescribe support in woody and non-woody stems\nDescribe the role of tendrils and tender stems in support\nObserve prepared sections of woody and herbaceous stems\nObserve a wilting plant\nList the types of skeletons\nDescribe the role of exoskeleton in insects\nDescribe the role and components of endoskeleton\nDescribe the role of skeleton in vertebrates\nDraw the structure of a finned fish (tilapia)\nCalculate the tail power\nExplain how locomotion occurs in fish\nName and draw the different fins and state their functions\nDraw the human skeleton and identify the component parts\nIdentify and draw the skull\nIdentify bones of Axial skeleton in the vertebral column\nIdentify the cervical vertebrae\nIdentify the structures of the thoracic vertebrae\nRelate the structure of the thoracic vertebrae to their functions\nIdentify the structures of lumbar, sacral and candal vertebrae\nShow how ribs articulate with thoracic vertebrae\nDraw and label Ribs and sternum\nRelate the structure to their functions\nIdentify components of Appendicular skeleton\nDraw the scapula bone and relate it to its functions\nIdentify the bones of the fore limbs\nDraw the structure of the humerus, radius and ulna\nDraw and label bones of the hand\nDraw the pelvic girdle\nName the bones of The pelvic girdle\nRelate the structure to their functions\nIdentify, draw and label the femur, tibia and tibula bones\nRelate their structure to their functions\nDraw and label the bones of the foot\nRelate the structure of bones of the foot to their functions\nDefine a joint\nList the three types of joints\nDescribe the types of joints\nList examples of movable joints, hinge joints and bell and socket joints\nDefine Immovable joints\nName Immovable joints\nDefine muscles\nExplain the differences between the three types of muscles\nIdentifying biceps and triceps in the arm movement\nGenetics\nIntroduction\nGenetics is the study of inheritance. The fact that the offspring of any species resemble the parents indicates that the characters in the parents are passed on to the offspring.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.936768349092549, "ocr_used": false, "chunk_length": 7251, "token_count": 1500}}
{"text": "FORM FOUR BIOLOGY\nDefine the term genetics\nDifferentiate between heredity and variation\nDistinguish between continuous and discontinuous variations\nDescribe continuous and discontinuous variations\nObserve variations in plants and animals\nDescribe the structure, nature and properties of chromosomes\nDescribe the structure, nature and properties of DNA molecule\nDifferentiate between DNA and RNA\nDistinguish between F1 and F2 generation\nDetermine Mendel’s first law of inheritance\nDefine other terms used in inheritance such as phenotype, genotype, dominant gene, recessive gene, haploid and diploid\nDemonstrate monohybrid inheritance in plants and animals\nPredict outcomes of various genetic crosses\nConstruct and make use of pannet squares\nWork out genotypic and phenotypic ratios\nPredict outcomes of various crosses\nDetermine the unknown genotypes in a cross using a test cross\nDescribe albinism as an example of monohybrid inheritance in human beings\nExplain the inheritance of ABO blood groups in human beings\nExplain the inheritance of rhesus factor as an example of monohybrid inheritance in human beings\nPredict the inheritance of blood groups human beings\nDescribe incomplete dominance\nDescribe inheritance of colour in flowers of mirabilis jalapa\nDescribe Inheritance of sickle cell anemia in human beings\nExplain how sex is determined in human beings\nDescribe sex linkages in human beings\nDefine linkage and sex-linkage\nDescribe linkage in human beings e.g.colour blindness and hemophilia\nDescribe colour blindness as an example of sex-linked trait in human beings\nInterpret pedigree of inheritance\nDescribe the Inheritance of hemophilia as an example of sex-linked traits in human beings\nDefine mutation\nDifferentiate between mutations and mutagens\nList down causes of mutations\nState the types of mutations\nList down the various chromosal mutations\nDescribe chromosal mutations\nExplain the Effects of chromosal mutations\nDescribe gene mutations and their effects on organisms\nDescribe areas in which the knowledge of genetics has been applied\nExplain the practical applications of genetics\nDefine evolution\nExplain the current concepts of the origin of life\nExplain the current concepts on origin of life\nDescribe the study of fossils as evidence of organic evolution theory\nDescribe comparative anatomy as evidence of organic evolution\nDescribe occurrence of vestigial structures and geographical distribution of organisms as evidence of organic evolution\nDescribe comparative embryology, cell biology and biochemistry as evidence of organic evolution\nDescribe evolution of hominids\nDescribe Lamarck’s theory\nDescribe and discuss the struggle for existence and survival for the fittest\nDescribe and discuss new concepts of Darwin’s theory\nDescribe natural selection in action\nDescribe natural selection in nature\nDescribe the isolation mechanism in speciation\nDescribe Artificial selection in plants and animals and how it leads to speciation\nExplain the importance of sexual reproduction in evolution\nDefine stimulus\nDefine irritability\nDefine response\nDefine tactic and tropic responses\nList down tactic responses in plants\nList down tropic responses in plants\nDifferentiate between tactic and tropic responses\nDefine geotropism\nDescribe geotropism in roots and shoots of plants\nDifferentiate between Phototropism and geotropism\nCarry out experiments demonstrating both Phototropism and geotropism in a plant seedling\nCarry out experiments to demonstrate tactic responses to light and water\nCarry out experiments to show chemotactic response using fruit juice\nDefine Hydrotropism and thigmotropism\nState the importance of Tactic and tropic responses\nExplain the production of Plant hormones and their effects on plants\nCarry out experiment to investigate hydrotropism\nCarry out experiment to investigate etiolation\nDemonstrate the knee jerk in a reflex action\nDefined Conditioned reflex actions\nDescribe Conditioned reflex action using parlous dog\nCompare simple and conditioned reflex actions\nExplain the role of endocrine system in a human being\nExplain the effect over secretion and under secretion of thyroxin and adrenaline\nIsolate and list the similarities and differences between the endocrine and the nervous system\nState the effects of drug abuse on human health\nDraw and label the mammalian eye\nState the functions of the mammalian eye\nDescribe how the structure of the mammalian eye is adapted to its functions\nDissect and display parts of the mammalian eye\nDescribe how an image is formed and interpreted in the mammalian eye\nDescribe Accommodation in the mammalian eye\nName and explain the Common eye defects\nDescribe Common eye defects and their corrections\nInvestigate the blind spot In the eye\nInvestigate which eye is used more during vision\nName and describe Common eye diseases\nDraw and label the mammalian ear\nDescribe the mammalian ear and how it is adapted to its functions\nDescribe the mechanism of hearing\nDiscuss thick ear drum, damaged cochlea, raptured eardrum, fussed ossicles, otitis media, ostosceleross and tinnitus\nDefine support and movement\nDescribe the necessity of movement in plants and animals\nReview the tissue distribution in monocotyledonous an dicotyledonous plants\nDescribe support in woody and non-woody stems\nDescribe the role of tendrils and tender stems in support\nObserve prepared sections of woody and herbaceous stems\nObserve a wilting plant\nList the types of skeletons\nDescribe the role of exoskeleton in insects\nDescribe the role and components of endoskeleton\nDescribe the role of skeleton in vertebrates\nDraw the structure of a finned fish (tilapia)\nCalculate the tail power\nExplain how locomotion occurs in fish\nName and draw the different fins and state their functions\nDraw the human skeleton and identify the component parts\nIdentify and draw the skull\nIdentify bones of Axial skeleton in the vertebral column\nIdentify the cervical vertebrae\nIdentify the structures of the thoracic vertebrae\nRelate the structure of the thoracic vertebrae to their functions\nIdentify the structures of lumbar, sacral and candal vertebrae\nShow how ribs articulate with thoracic vertebrae\nDraw and label Ribs and sternum\nRelate the structure to their functions\nIdentify components of Appendicular skeleton\nDraw the scapula bone and relate it to its functions\nIdentify the bones of the fore limbs\nDraw the structure of the humerus, radius and ulna\nDraw and label bones of the hand\nDraw the pelvic girdle\nName the bones of The pelvic girdle\nRelate the structure to their functions\nIdentify, draw and label the femur, tibia and tibula bones\nRelate their structure to their functions\nDraw and label the bones of the foot\nRelate the structure of bones of the foot to their functions\nDefine a joint\nList the three types of joints\nDescribe the types of joints\nList examples of movable joints, hinge joints and bell and socket joints\nDefine Immovable joints\nName Immovable joints\nDefine muscles\nExplain the differences between the three types of muscles\nIdentifying biceps and triceps in the arm movement\nGenetics\nIntroduction\nGenetics is the study of inheritance. The fact that the offspring of any species resemble the parents indicates that the characters in the parents are passed on to the offspring. Factors that determine characters (genes) are passed on from parent to offspring through gametes or sex cells.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9365537022269965, "ocr_used": false, "chunk_length": 7362, "token_count": 1521}}
{"text": "FORM FOUR BIOLOGY\nDefine the term genetics\nDifferentiate between heredity and variation\nDistinguish between continuous and discontinuous variations\nDescribe continuous and discontinuous variations\nObserve variations in plants and animals\nDescribe the structure, nature and properties of chromosomes\nDescribe the structure, nature and properties of DNA molecule\nDifferentiate between DNA and RNA\nDistinguish between F1 and F2 generation\nDetermine Mendel’s first law of inheritance\nDefine other terms used in inheritance such as phenotype, genotype, dominant gene, recessive gene, haploid and diploid\nDemonstrate monohybrid inheritance in plants and animals\nPredict outcomes of various genetic crosses\nConstruct and make use of pannet squares\nWork out genotypic and phenotypic ratios\nPredict outcomes of various crosses\nDetermine the unknown genotypes in a cross using a test cross\nDescribe albinism as an example of monohybrid inheritance in human beings\nExplain the inheritance of ABO blood groups in human beings\nExplain the inheritance of rhesus factor as an example of monohybrid inheritance in human beings\nPredict the inheritance of blood groups human beings\nDescribe incomplete dominance\nDescribe inheritance of colour in flowers of mirabilis jalapa\nDescribe Inheritance of sickle cell anemia in human beings\nExplain how sex is determined in human beings\nDescribe sex linkages in human beings\nDefine linkage and sex-linkage\nDescribe linkage in human beings e.g.colour blindness and hemophilia\nDescribe colour blindness as an example of sex-linked trait in human beings\nInterpret pedigree of inheritance\nDescribe the Inheritance of hemophilia as an example of sex-linked traits in human beings\nDefine mutation\nDifferentiate between mutations and mutagens\nList down causes of mutations\nState the types of mutations\nList down the various chromosal mutations\nDescribe chromosal mutations\nExplain the Effects of chromosal mutations\nDescribe gene mutations and their effects on organisms\nDescribe areas in which the knowledge of genetics has been applied\nExplain the practical applications of genetics\nDefine evolution\nExplain the current concepts of the origin of life\nExplain the current concepts on origin of life\nDescribe the study of fossils as evidence of organic evolution theory\nDescribe comparative anatomy as evidence of organic evolution\nDescribe occurrence of vestigial structures and geographical distribution of organisms as evidence of organic evolution\nDescribe comparative embryology, cell biology and biochemistry as evidence of organic evolution\nDescribe evolution of hominids\nDescribe Lamarck’s theory\nDescribe and discuss the struggle for existence and survival for the fittest\nDescribe and discuss new concepts of Darwin’s theory\nDescribe natural selection in action\nDescribe natural selection in nature\nDescribe the isolation mechanism in speciation\nDescribe Artificial selection in plants and animals and how it leads to speciation\nExplain the importance of sexual reproduction in evolution\nDefine stimulus\nDefine irritability\nDefine response\nDefine tactic and tropic responses\nList down tactic responses in plants\nList down tropic responses in plants\nDifferentiate between tactic and tropic responses\nDefine geotropism\nDescribe geotropism in roots and shoots of plants\nDifferentiate between Phototropism and geotropism\nCarry out experiments demonstrating both Phototropism and geotropism in a plant seedling\nCarry out experiments to demonstrate tactic responses to light and water\nCarry out experiments to show chemotactic response using fruit juice\nDefine Hydrotropism and thigmotropism\nState the importance of Tactic and tropic responses\nExplain the production of Plant hormones and their effects on plants\nCarry out experiment to investigate hydrotropism\nCarry out experiment to investigate etiolation\nDemonstrate the knee jerk in a reflex action\nDefined Conditioned reflex actions\nDescribe Conditioned reflex action using parlous dog\nCompare simple and conditioned reflex actions\nExplain the role of endocrine system in a human being\nExplain the effect over secretion and under secretion of thyroxin and adrenaline\nIsolate and list the similarities and differences between the endocrine and the nervous system\nState the effects of drug abuse on human health\nDraw and label the mammalian eye\nState the functions of the mammalian eye\nDescribe how the structure of the mammalian eye is adapted to its functions\nDissect and display parts of the mammalian eye\nDescribe how an image is formed and interpreted in the mammalian eye\nDescribe Accommodation in the mammalian eye\nName and explain the Common eye defects\nDescribe Common eye defects and their corrections\nInvestigate the blind spot In the eye\nInvestigate which eye is used more during vision\nName and describe Common eye diseases\nDraw and label the mammalian ear\nDescribe the mammalian ear and how it is adapted to its functions\nDescribe the mechanism of hearing\nDiscuss thick ear drum, damaged cochlea, raptured eardrum, fussed ossicles, otitis media, ostosceleross and tinnitus\nDefine support and movement\nDescribe the necessity of movement in plants and animals\nReview the tissue distribution in monocotyledonous an dicotyledonous plants\nDescribe support in woody and non-woody stems\nDescribe the role of tendrils and tender stems in support\nObserve prepared sections of woody and herbaceous stems\nObserve a wilting plant\nList the types of skeletons\nDescribe the role of exoskeleton in insects\nDescribe the role and components of endoskeleton\nDescribe the role of skeleton in vertebrates\nDraw the structure of a finned fish (tilapia)\nCalculate the tail power\nExplain how locomotion occurs in fish\nName and draw the different fins and state their functions\nDraw the human skeleton and identify the component parts\nIdentify and draw the skull\nIdentify bones of Axial skeleton in the vertebral column\nIdentify the cervical vertebrae\nIdentify the structures of the thoracic vertebrae\nRelate the structure of the thoracic vertebrae to their functions\nIdentify the structures of lumbar, sacral and candal vertebrae\nShow how ribs articulate with thoracic vertebrae\nDraw and label Ribs and sternum\nRelate the structure to their functions\nIdentify components of Appendicular skeleton\nDraw the scapula bone and relate it to its functions\nIdentify the bones of the fore limbs\nDraw the structure of the humerus, radius and ulna\nDraw and label bones of the hand\nDraw the pelvic girdle\nName the bones of The pelvic girdle\nRelate the structure to their functions\nIdentify, draw and label the femur, tibia and tibula bones\nRelate their structure to their functions\nDraw and label the bones of the foot\nRelate the structure of bones of the foot to their functions\nDefine a joint\nList the three types of joints\nDescribe the types of joints\nList examples of movable joints, hinge joints and bell and socket joints\nDefine Immovable joints\nName Immovable joints\nDefine muscles\nExplain the differences between the three types of muscles\nIdentifying biceps and triceps in the arm movement\nGenetics\nIntroduction\nGenetics is the study of inheritance. The fact that the offspring of any species resemble the parents indicates that the characters in the parents are passed on to the offspring. Factors that determine characters (genes) are passed on from parent to offspring through gametes or sex cells. In fertilisation the nucleus of the male gamete fuses with the nucleus of the female gamete.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.936407437836348, "ocr_used": false, "chunk_length": 7455, "token_count": 1542}}
{"text": "The fact that the offspring of any species resemble the parents indicates that the characters in the parents are passed on to the offspring. Factors that determine characters (genes) are passed on from parent to offspring through gametes or sex cells. In fertilisation the nucleus of the male gamete fuses with the nucleus of the female gamete. The offspring show the characteristics of both the male and the female. Genetics is the study of how this heritable material operates in individuals and their offspring. Variations within Plant and Animal Species\nVariation\nThe term variation means to differ from a standard. Genetics also deals with the study of differences between organisms belonging to one species. Organisms belonging to higher taxonomic groups e.g. phyla or classes are clearly different. Although organisms belonging to the same species are similar, they show a number of differences or variations such that no two organisms are exactly the same in every respect. Even identical twins, though similar in many aspects, are seen to differ if they grow in different environments. Their differences are as a result of the environment which modifies the expression of their genetic make-up or genotype. The two causes of variations are the genes and the environment. Genes determine the character while the environment modifies the expression of that character. Continuous and Discontinuous Variation\nContinuous Variations\nThe differences between the individual are not clear-cut. There are intermediates or gradations between any two extremes. Continuous variations are due to action of many genes e.g. skin complexion in humans. In continuous variation, the environment has a modifying effect in that it may enhance or suppress the expressions of the genes. Continuous variation can be represented in form of a histogram. Example of continuous variation in humans is weight, height and skin complexion. Linear measurements:\nIn humans, height shows gradation from tall, to tallest. So does the length of mature leaves of a plant. In most cases, continuous variation is as a result of the environment. Discontinuous Variations\nThese are distinct and clear cut differences within a species. Examples include:\nAbility to roll the tongue. An individual can either roll the tongue or not. Ability to taste phenylthiourea (PTC); some individuals can taste this chemical others cannot. Blood groups - and individual has one of the four blood groups A, B AB or O. There are no intermediates. Albinism - one is either an albino or not. Discontinuous variations is determined by the action of a single gene present in an individual.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9247344461305008, "ocr_used": false, "chunk_length": 2636, "token_count": 501}}
{"text": "There are no intermediates. Albinism - one is either an albino or not. Discontinuous variations is determined by the action of a single gene present in an individual. Structure and Properties of Chromosomes\nThese are threadlike structures found in the nucleus. They are normally very thin and coiled and are not easily visible unless the cell is dividing. When a cell is about to divide, the chromosomes uncoil and thicken. Their structure, number and behaviour is clearly observed during the process of cell division. The number of chromosomes is the same in all the body cells of an organism. In the body cells, the chromosomes are found in pairs. Each pair is made up of two identical chromosomes that make up a homologous pair. However sex chromosomes in human male are an exception in that the Y-chromosome is smaller. Number of Chromosomes\nDiploid Number (2n)\nThis is the number of chromosomes found in somatic cells. For example, in human 2n = 46 or 22 pairs (44 chromosomes) are known as autosomes (body chromosomes\")\nwhile 1 pair is known as the sex chromosomes. In Drosophila melanogaster, 2n = 8. Chromosome Structure\nAll chromosomes are not of the same size or shape. In human beings; each of the twenty three pairs have unique size and structure . On this basis they have been numbered 1 to 23. The sex chromosomes formthe 23rd pair. Properties of Chromosomes\nChromosomes are very long and thin. They are greatly and loosely coiled and fit within the nucleus. During cell division they shorten, become thicker and are easily observable. Each consists of two chromatids. The two chromatids are held at same position along the length, at the centromere. Chromatids separate during cell division in mitosis and in the second stage of meiosis. Chromosomes take most dyes and stain darker than any other part of the cell. This property has earned them the name \"chromatin material\"\nEach chromosome is made up of the following components:\nDeoxyribonucleic acid (DNA) - this carries the genes. It is the major component of the genetic material. Protein e.g. histones. Ribonucleic acid (RNA) is present in very small amounts. Enzymes concerned with DNA and RNA replication - these are DNA and RNA polymerases and ligases.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9023291925465838, "ocr_used": false, "chunk_length": 2226, "token_count": 495}}
{"text": "histones. Ribonucleic acid (RNA) is present in very small amounts. Enzymes concerned with DNA and RNA replication - these are DNA and RNA polymerases and ligases. Structure of DNA\nThe structure of DNA was first explained in 1953 by Watson and Crick. DNA was shown to be a double helix that coils around itself. The two strands are parallel and the distance between the two is constant. Components of DNA\nDNA is made up of repeating units called nucleotides. Each nucleotide is composed of:\nA five-carbon sugar (deoxyribose). Phosphate molecule. Nitrogenous base, four types are available i.e,\nAdenine - (A)\nGuanine - (G)\nCytosine - (C)\nThymine - (T)\nThe bases are represented by their initials as A, G, C and T respectively. The sugar alternates with the phosphate, and the two form the backbone of the strands. The bases combine in a specific manner, such that Adenine pairs with Thymine and Guanine pairs with Cytosine. The bases are held together by hydrogen bonds. A gene is the basic unit of inheritance consisting of a number of bases in linear sequence on the DNA. Genes exert their effect through protein synthesis. The sequence of bases that make up a gene determine the arrangement of amino acids to make a particular protein. The proteins manufactured are used to make cellular structures as well as hormones and enzymes. The types of proteins an organism manufactures determines its characteristics. For example, albinism is due to failure of the cells of an organism to synthesise the enzyme tyrosine required for the formation of the pigment melanin. First Law of Heredity\nIt is also known as Law of Segregation (Mendel's First Law). The characters of an organism are controlled by genes occurring in pairs known as Alleles. By definition, an allele is an alternative form of a gene controlling a particular characteristic. Of a pair of such alleles, only one is carried in each gamete. This is explained by first meiotic anaphase stage, when the homologous chromosomes are separated so that each carries one of the allelic genes. Monohybrid Inheritance\nThis is the study of the inheritance of one character trait that is represented by a pair of genes on homologous chromosomes. Gregor Mendel (an Austrian monk) was the first person to show the nature of inheritance.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9154241431800361, "ocr_used": false, "chunk_length": 2282, "token_count": 507}}
{"text": "This is explained by first meiotic anaphase stage, when the homologous chromosomes are separated so that each carries one of the allelic genes. Monohybrid Inheritance\nThis is the study of the inheritance of one character trait that is represented by a pair of genes on homologous chromosomes. Gregor Mendel (an Austrian monk) was the first person to show the nature of inheritance. He did this through a series of experiments using the garden pea, Pisum sativum. As opposed to others before him, the success in his work lay in the fact that:\nHe chose to study first a single character at a time (monohybrid inheritance). He then proceeded to study two characters at time (dihybrid inheritance) . He quantified his results by counting the number of offspring bearing each trait. Each character he chose was expressed in two clearly contrasting forms. Examples\nStem length: some plants were tall while others were short. Colour of unripe pods: some were green, others yellow. There were no intermediates. Mendel's Procedure\nFor each character, Mendel chose a plant that bred true. A true or pure breed continues to show a particular trait in all the offspring in several successive generations of self-fertilisation. He made one plant to act as the female by removing the stamens before the ovary was mature and protecting (e.g. by wrapping with paper). The female plant from contact with any stray pollen. When the ovary was mature, he carefully dusted pollen from the anthers of the selected male plant and transferred it to the stigma of the female plant. Observations were then made on the resulting seeds or on the plants obtained when those seeds were planted. Results\nFor each pair of contrasting characters he studied, Mendel obtained the same results. For example, when he crossed pure breeding tall plants with pure breeding short plants, the first offspring, known as the first filial generation (FI) were all tall. When these were selfed i.e. self-fertilisation allowed to take place, the second generation offspring also know as the second filial generation or F2 occurred in the ratio of 3 tall: 1 short. The same ratio was obtained for each of the other characters studied. From this it is clear that one character i.e. tall is dominant over the short character. A dominant character is that which is expressed alone in the offspring even when the opposite character is represented in the genotype.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9161577338433415, "ocr_used": false, "chunk_length": 2411, "token_count": 502}}
{"text": "From this it is clear that one character i.e. tall is dominant over the short character. A dominant character is that which is expressed alone in the offspring even when the opposite character is represented in the genotype. The unexpressed character is said to be recessive. From these results and others obtained when he studied two characters at the same time, Mendel concluded that gametes carry factors that are expressed in the offspring. These factors are what we know today as genes. Mendel put forward the following laws of inheritance:\nOf a pair of contrasting characters, only one can be represented in a gamete. For two or more pairs of such contrasting characters, each factor (gene) in the gamete acts independently of the others and may combine randomly with either of the factors of another pair during fertilisation. Genetic experiments carried out to date confirm Mendel's Laws of inheritance e.g. T.H. Morgan's work on inheritance in the fruit fly Drosophila melanogaster. Terms used in Genetics\nGenotype:\nThe genes present in an individual. The genetic constitution of an individual. It is expressed in alphabetical notation.e.g TT,Tt\nPhenotype:\nThe observed character or appearance i.e. the expression of the genes in the structure and physiology of the organism. In some cases the phenotype is the product of the genotype and the environment. Phenotype is expressed in words.eg TALL,SHORT,RED WHITE .etc. Alleles:\nThese are alternative forms of the same gene that control a pair of contrasting characters e.g. tall and short. They are found at the same position or gene-locus on each chromosome in a homologous pair. Homozygous:\nThis is a state where the alleles in an individual are similar e.g. TT (for tall)\nHeterozygous:\nThis is a state where the alleles are dissimilar i.e. each of the two genes responsible for a pair of contrasting characters are present\ne.g. Tt. (T for tall; t for short)\nHybrid:\nThis is the offspring resulting from crossing of two individuals with contrasting characters. Hybrid vigour or Heterosis:\nThe hybrid develops the best characteristics from both parents\ni.e. it is stronger or healthier, or yields more than either parent. Use of Symbols\nTo represent genes in the chromosomes, letters are used. It is customary to use a capital letter for the dominant characteristic and small letter for the recessive one.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9181049069373943, "ocr_used": false, "chunk_length": 2364, "token_count": 499}}
{"text": "it is stronger or healthier, or yields more than either parent. Use of Symbols\nTo represent genes in the chromosomes, letters are used. It is customary to use a capital letter for the dominant characteristic and small letter for the recessive one. The gametes are encircled. For example,a cross between a tall and a short pea plant is illustrated as follows;\nLet –T- represent gene for tallness. Let -t- represent gene for shortness. Fertilization-using checker board or Punnet square\nF1 genotype Tt\nF1 Phenotypic ratio =All tall. F2 Genotype TT,2Tt,tt\nF2 Phenotypic ratio;3 Tall;1 short\nTest Cross or Back Cross\nThis is a eras made between the F 1 bearing the dominant trait with the homozygous recessive parent. It is called a back cross because of using the first parent. It is also a test cross because it tests the genotype of the individual. Complete Dominance\nMendel happened to choose characters that showed complete dominance,\ni.e. the dominant trait completely masked the recessive one in the F1 generation. In man, certain characters are inherited in the same way\ne.g. colour of the skin; normal colour is dominant to albinism (lack of skin pigment). The children are all normal but have the gene for albinism. Such individuals are referred to as carriers. Other characters that show complete dominance in humans are:\nAbility to roll the tongue. Polydactyly (having more than 5 digits in one limb). Brachydactyly - having short fingers. Achondroplasia - dwarf with bow legs. Incomplete Dominance\nIn this kind of inheritance there is no dominant or recessive gene but the two are expressed equally in the offspring,\nResulting in blending of the characters. The gene for red colour (R) in cattle and the gene for white colour(W) show incomplete dominance or co-dominance. The offspring are neither red nor write but are intermediate between the two. They are said to be roan. In humans, the sickle cell gene and the normal gene are co-dominant. Inheritance of ABO blood groups in humans\nBlood groups in human are determined by three alleles, A, B, and O. An individual can have only two of these genes. Genes A and Bare codominant, while gene 0 is recessive to A and B.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.900878221284411, "ocr_used": false, "chunk_length": 2178, "token_count": 511}}
{"text": "Inheritance of ABO blood groups in humans\nBlood groups in human are determined by three alleles, A, B, and O. An individual can have only two of these genes. Genes A and Bare codominant, while gene 0 is recessive to A and B. These are referred to as multiple alleles. The ABO Blood Group System\nRhesus Factor\nThe Rhesus factor is responsible for the presence of a protein (Antigen D) in the red blood cells. If blood from a Rhesus positive (Rh+) person is transferred into a person without the Rhesus factor (Rh-);\nThe recipients' body produces antibodies against the Rhesus factor. This causes agglutination of red blood cells which can be fatal if subsequent transfusion with Rh+ blood is done. Sex Determination in Humans\nXY type e.g. human male\nIn males, two types of sperms are produced. Half of then containing X chromosomes and half Y chromosomes. During fertilisation only one sperm fuses with the egg. If it is an X-carrying sperm then a female zygote is formed;\nIf it is a Y-carrying sperm then a male zygote is formed. It follows then that the chances of getting a boy or girl are half or fifty-fifty. Note also that it is essentially the type of sperm that fertilises the egg that determines the sex. Linkage\nThe term linkage describe the situation where genes or certain characters are located on the same chromosome. Offspring produced by sexual reproduction show only the parental characteristics and only sometimes few new recombinants. i.e. offspring with combinations of characteristics not found in either of the parents due to crossing over in first prophase of meiosis. Genes are said to be linked when they are located close together on the same chromosome such that they are always inherited together. Sex linked genes\nThese are genes that are located on the sex chromosomes. Sex-linkage - refers to carrying of the genes on the sex-chromosome. Gene for a trait may be present, yet offspring does not show the trait. This happens in human females (XX) where a gene for the trait is recessive. The female acts as a carrier. In human, sex linked characters found on the X chromosome include:\nHaemophilia:\nThis is a disease that affects the rate of clotting of blood, leading to excessive bleeding even from a minor cut. Haemophilia is more common in males than in females.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9150924552987353, "ocr_used": false, "chunk_length": 2293, "token_count": 508}}
{"text": "The female acts as a carrier. In human, sex linked characters found on the X chromosome include:\nHaemophilia:\nThis is a disease that affects the rate of clotting of blood, leading to excessive bleeding even from a minor cut. Haemophilia is more common in males than in females. A female my have the gene for haemophilia and not show the trait because the normal gene is dominant over the gene for haemophilia. Such females are referred to as carriers. If the carrier female offspring will be carriers while the other half will be normal. Half the males will be normal and the other heamophilic. Red-green colour-blindness\nRed-green colour-blindness is caused by a recessive gene found on the X chromosome. It is inherited in the same way as haemophilia. More males 1:10,000, less female 1: 100 million afflicted. It is the inability to distinguish between red and green colours in humans. Genes found on y-chromosome include:\nHairy pinna and hairy nose are carried on the Y - chromosome. Premature balding. Mutations\nMutations are sudden changes in the genotype that are inherited. Mutations are rare in nature and mutated genes are usually recessive to the normal (wild type) genes. Most mutations are generally harmful and some are lethal. A somatic mutation is a genetic change in somatic cells. Somatic mutations are only inherited if asexual reproduction takes place e.g. as in plants and unicellular animals. A gene mutation is a change in genes of reproductive cells and is always inherited. The resultant individual is called a mutant. The mutant has different characteristics from the rest of the population. Types of Mutations\nChromosomal mutations - are changes in number or structure of chromosomes. Gene mutations - also called point mutations - are changes in the chemical nature of the gene. Mutagens:\nThese are agents that cause mutations. The include ultra-violet light, Gamma rays., x-rays and cosmic rays. Certain chemicals e.g. mustard gas and colchicines also induce mutations. Causes and consequences of chromosomal mutations\nThere are three main types of chromosomal mutations. Changes in the diploid number of chromosomes (allopolyploidy). The diploid number changes to 3n (triploid) or 4n (tetraploid) and so on. This results from the doubling of the chromosome number in the gamete (2n).", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9075100356251862, "ocr_used": false, "chunk_length": 2313, "token_count": 499}}
{"text": "Changes in the diploid number of chromosomes (allopolyploidy). The diploid number changes to 3n (triploid) or 4n (tetraploid) and so on. This results from the doubling of the chromosome number in the gamete (2n). This is due to failure of the chromosome sets to separate during meiosis. The phenomenon is known as polyploidy. It is common in plant's and has been employed artificially to produce varieties of crops with hybrid vigour e.g. bread wheat is hexaploid (6n). This is allopolyploidy). Change in the total number of chromosomes involving the addition or loss of individual chromosomes (autopolyploidy). This is due to failure of individual chromosomes to separate during meiosis. One gamete gains an extra chromosome while the other loses a chromosome. The term non-disjunction is used to describe the failure of chromosomes to separate. Non-disjunction results in several disorders in humans:\nDown's syndrome\nThe individual has 47 chromosomes due to non-disjunction of chromosome 21. It is also known as trisomy 21. The individual has slanted eyes with flat and rounded face, mental retardation and large tongue and weak muscles. Turner's Syndrome\nThis brings about to a sterile and abnormally short female. It is due to loss of one of the sex chromosomes\ni.e. the individual has one X chromosome (44 + X) instead of two (44 + XX). Klinefelter's Syndrome\nThis results in a sterile male who may be mentally retarded. It is due to an additional X chromosome\ni.e. the individual i.e. 47 chromosomes (44 + XXY) instead of 46 (44 + XY). Changes in the structure of a chromosome during meiosis. A portion of a chromosome may break off and fail to unite again or it may be joined in the wrong way or to the wrong chromosome. These mutations are described as follows:\nDeletion:\nThis is the loss of a portion of a chromosome,\nDeletion results in individuals born with missing body parts . e.g. limbs in the extreme of cases. Inversion:\nA portion may break from a chromosome and then rejoin to it after turning though an angle-of 1800 . Translocation:\nThis is when a portion is joined to a non-homologous chromosome. Duplication:\nA certain section of an intact chromosome replicates such that the genes are repeated.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.893037170744752, "ocr_used": false, "chunk_length": 2216, "token_count": 502}}
{"text": "Inversion:\nA portion may break from a chromosome and then rejoin to it after turning though an angle-of 1800 . Translocation:\nThis is when a portion is joined to a non-homologous chromosome. Duplication:\nA certain section of an intact chromosome replicates such that the genes are repeated. Gene Mutations\nA gene mutation is a change in the structure of a gene. It may involve only a change in one base, e.g. adenine in place of thyamine yet the effect on the individual is profound e.g. sickle cell anemia . There are two main type of gene mutations:\nDue to insertion or deletion of one or more (base) pairs. Substitution of base pairs e.g. purine for pyrimidine. Genetically inherited disorders in humans\nAlbinism is a mutation that alters the gene responsible for synthesis of skin pigment (melanin). The gene for albinism is recessive. Sickle cell anemia is a common condition in Kenya. Individuals with the sickle-cell gene produce abnormal haemoglobin. It is due to gene mutation caused by substitution of the base adenine for thymine. The result is the inclusion of the amino acid valine (in place of glutamic acid) in the haemoglobin synthesised. As a result the red blood cells become sickle shaped when oxygen concentration becomes low i.e. inside tissues. This leads to blockage of capillaries. Tissues do not get sufficient oxygen. Homozygous individuals are seriously anaemic and die in early childhood. Heterozygous individuals have a mixed population of normal and sickled red blood cells. They are not seriously anaemic and can lead fairly normal lives. Haemophila (bleeder's diseases) is due to lack of gene for production of proteins responsible for blood clotting. Practical Applications of Genetics\nStudy of genetics has been put into a wide variety of uses en-compasing plants and animals and in particular humans. Blood transfusion\nBlood groups are genetically determined. As discussed earlier a person of blood group A can only get blood from another one of A or O. In case of emergencies and unavailability of blood, a patient may be given blood group A + when he/she is A-. First transfusion is fine since, by the time enough antibodies are produced most of the red blood cells of donor have completed their lifespan but a subsequent transfusion of A+ blood is fatal.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9173755979346951, "ocr_used": false, "chunk_length": 2292, "token_count": 500}}
{"text": "As discussed earlier a person of blood group A can only get blood from another one of A or O. In case of emergencies and unavailability of blood, a patient may be given blood group A + when he/she is A-. First transfusion is fine since, by the time enough antibodies are produced most of the red blood cells of donor have completed their lifespan but a subsequent transfusion of A+ blood is fatal. Plant and Animal breeding\nGenetics is applied mostly in plant and animal breeding in order to produce varieties that are most suitable to man's needs. This is done through artificial selection. Varieties are developed that are resistant to pests, diseases or harsh climatic conditions. Genetic counselling\nGenetic counselling involves advising about hereditary diseases and disorders so that they can make informed decisions. This is done through:\nTaking family history. Screening for genotypes e.g. through amniocentesis. In amniocentesis, cells are obtained from amniotic fluid during pregnancy. Conditions such as Down's syndrome can be detected using microscopy. Genetic Engineering\nThis is a technology that involves the manipulation of the genotype of an organism to get the desired trait. It also involves the transfer of gene coding for the desired trait from one organism to another. Application of Genetic Engineering\nPharmaceutical industries:\nMaking of hormones e.g. Human insulin and human growth hormone. Enzymes e.g. Alph-Anti-Trypsin (AAT) used to treat emphysema. (c) Proteins. Drugs and vaccines. Agricultural industries:\nTransgenic animals and plants are produced which are also called Genetically Modified Organisms (GMO's). A variety of tomato with improved paste and a longer shell life. Sheep for producing desired proteins in milk. Plants resistant to pests and diseases. Cloning\nThis is the making of identical copies of genes, DNA and whole organisms. Cloning is used in plants - that is tissue culture e.g. in development of various varieties of bananas and Eucalyptus trees. The first mammal to be cloned successfully was Dolly - the sheep. A nucleus from the cell obtained from the udder of the sheep was inserted in an unfertilised egg without a nucleus. This zygote was introduced into the uterus of a sheep and developed to full term. Gene therapy\nInvolves injecting genes into patients of certain diseases\ne.g. Parkinson's diseases.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.921032585696149, "ocr_used": false, "chunk_length": 2363, "token_count": 492}}
{"text": "This zygote was introduced into the uterus of a sheep and developed to full term. Gene therapy\nInvolves injecting genes into patients of certain diseases\ne.g. Parkinson's diseases. The injected gene alters metabolism to bring about the cure of the disease. Practical Activities\nTo demonstrate Continuous variations\nHeight of students\nStudents should work in pairs, use chalk and metre rule to mark level of top of head onto the wall\nOr door as one student stands straight without shoes, next to the wall or door. The height for each student is recorded on chalk board. The frequency distribution of height is recording as the height is grouped into various classes. A histogram to represent frequency against height is drawn. The normal bell shaped curve is observed. Discontinuous variations - ability to roll tongue\nThe number of students who can roll their tongue is recorded as well as the number of non-tongue rollers. The ratio of tongue-rollers to non tonguerollers is worked out. Gene for the ability to roll the tongue is dominant, therefore is expected more tongue rollers. Demonstration of Mitosis and Meisosis\nMitosis\nPlasticene is used to represent number and shapes of various chromosomes e.g. 8 in Drosophila melanogaster. Each stage of mitosis illustrated e.g. interphase,\nEach is rolled to appear long is and coiled, prophase is each made into a ball and then shaped to the appropriate length; and split into two to represent chromatids. Centromeres for different chromosomes can be illustrated in different positions. Each stage of mitosis is illustrated and telophase can be illustrated by surrounding the \"chromosomes\" with a long many drawn plasticene to represent cell membrane. It is manipulated to show how telophase takes place. Meiosis\nThe same procedure is followed. Plasticine with contrasting colours is used to show clearly gene mixing in crossing over. Each pair of homologous chromosomes is represented by plasticene with two different colours e.g. red (paternal) blue for maternal chromosome. All the steps in the two stages of meiosis are illustrated up to the production of four haploid gametes. Human Finger Prints\nThe finger prints for each student's thumb, forefinger and middle fingers of the left hand is imprinted on a white paper. A rubber stamp with ink is used to and each finger -tip phalange is rolled onto the inkpad. For best results students work in pairs.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9223163023163025, "ocr_used": false, "chunk_length": 2405, "token_count": 499}}
{"text": "Human Finger Prints\nThe finger prints for each student's thumb, forefinger and middle fingers of the left hand is imprinted on a white paper. A rubber stamp with ink is used to and each finger -tip phalange is rolled onto the inkpad. For best results students work in pairs. Observations are made at all forefingers, thumb prints and differences noted. The main patterns are noted. It is also noted that no two, fingerprints are exactly similar. end\nEVOLUTION\nMeaning of Evolution and Current Concepts\nEvolution is the development of organisms from pre-existing simple organisms over a long period of time. It is based on the similarities in structure and function that is observed in all organisms. All are made up of cells, and similar chemical compounds are present. This indicates that all organism may have had a common origin. Evolution seeks to explain the diversity of life and also to answer the question as to the origin of life, as well as its present state. The Origin of Life\nHuman beings have tried to explain how life began. Currently held views are listed below:\nSpecial creation -life was created by a supernatural being within a particular time. Spontaneous generation life originated from non-living matter all at once. e.g. maggots arise from decaying meat. Steady state - life has no origin. Cosmozoan - life on earth originate from elsewhere, outer space. Bio-chemical evolution-life originated according to chemical and physical laws. Only special creation and chemical evolution will be discussed. Special Creation\nThe earliest idea is that of special creation which is recorded in the old testament (Genesis 1: 1-26). It states that God created the world and all living things in six days. Some hold the six days literally, while others say it may represent thousands of years. According to his theory, the earth and all organisms were created mature. Similarities in structure and function denote the stamp of a \"common Designer\"\nEvidence for this view arises from observations of life itself. Faith explains it all. By faith we understand that the universe was created by the command of God. Several scientists hold this view and their research confirms accounts in the old testament of a universal flood explains the disappearance of dinosaurs as vegetation decreased.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9171382725105446, "ocr_used": false, "chunk_length": 2296, "token_count": 456}}
{"text": "Faith explains it all. By faith we understand that the universe was created by the command of God. Several scientists hold this view and their research confirms accounts in the old testament of a universal flood explains the disappearance of dinosaurs as vegetation decreased. Chemical Evolution\nThe following is the line of thought held in this view to explain origin of life:\nThe composition of atmospheric gases was different from what it is today:\nThere was less oxygen, more carbon (IV) oxide, hence no ozone layers to filter the ultra-violet light. The high solar energy reached the earth and brought together hydrogen, carbon (IV) oxide and nitrogen to make organic compounds. These were: hydrocarbons, amino acids, nucleic acids, sugars, amino acids and proteins. The proteins coalesced and formed colloids. Proteins and lipids formed a \"cell membrane\" that enclosed the organic compounds, to form a primitive cell. The cell was surrounded by organic molecules that it fed on\nheterotrophically. This took place in water. From this cell progressively autotrophs evolved. That were similar to blue-green algae. They produced oxygen and as more oxygen was evolved ozone layer formed an blocked ultra violet radiation. This allowed formation of present day photo-autotrophs. Evidence for Organic Evolution\nMost of the evidence for evolution is indirect . i.e. it is based on studies carried out on present-day animals and plants. Direct evidence is obtained from studying the remains of animals and plants of the past. Fossil Records\nThe study of fossils is called paleontology. Fossils are remains of organisms that lived in ancient times. Most fossils are remains of hard parts of the body such as bones, teeth, shells and exoskeletons. Some fossils are just impressions of the body parts, e.g. footprints, leaf-vennation patterns, etc. Fossils are usually found in sedimentary rocks which have been formed by deposition of sediments over millions of years. The deeper the layer of sediments, the older the fossils found in that layer. Modem man, Homo sapiens, evolved from ape-like creatures 25 million years ago. These evolved to upright, tool using creature called Australopithecus afarensis which had a cranial capacity of 400-500 cc. This evolved through several intermediates; Homo habilis and Homo erectus to modem day human. Homo sapiens has a cranial capacity of 1350 - 1450 cc.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9125653285285547, "ocr_used": false, "chunk_length": 2393, "token_count": 503}}
{"text": "These evolved to upright, tool using creature called Australopithecus afarensis which had a cranial capacity of 400-500 cc. This evolved through several intermediates; Homo habilis and Homo erectus to modem day human. Homo sapiens has a cranial capacity of 1350 - 1450 cc. Homo sapiens is more intelligent. Main features in human evolution include bipedal posture, is an omnivore and has an opposable thumb. Limitations of the Fossil Evidence\nOnly partial preservation was usually possible because softer parts decayed. The fossil records are therefore incomplete. Distortion - parts of organisms might have become flattened during sedimentation. Subsequent geological activities e.g. erosion, earthquakes, faulting and uplifting may have destroyed some fossils. Geographical Distribution\nUntil about 250 million years ago, all the land masses on earth formed a single land mass (Pangaea). This is thought to have undergone continental drift, splitting into different continents. Consequently, organisms in certain regions became geographically isolated and did not have a chance to interbreed with other organisms in other regions. Such organisms underwent evolution in isolation and have become characteristically different from organisms in other regions. For example, pouched mammals (e.g. kangaroo, wallaby, koala bear) are found almost exclusively in Australia. The opossum is the only surviving representative of the pouched mammals in North America. Comparative Embryology\nDuring the early stages of development, the embryos of different vertebrates are almost indistinguishable. Fish, amphibian, bird and mammalian embryos have similar, features, indicating that they arose from a common ancestor. Similarities include:\nVisceral clefts, segmental muscle blocks (myotomes) and a single circulation. Comparative Anatomy\nComparative anatomy is the study of organs in different species with the aim of establishing whether the organism are related. Organisms which have the same basic features are thought to have arisen from a common ancestor. The vertebrate pentadactyl limb evolved in different ways as an adaptation to different modes of life. e.g. as a flipper in whales, as a wing in bats and as a digging hand in moles. Such organs are said to be homologous, i.e. they have arisen from a common ancestor but they have assumed different functions. This is an example of divergent evolution . The wing of a butterfly and that of a bird are said to be analogous.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9170841982045491, "ocr_used": false, "chunk_length": 2471, "token_count": 505}}
{"text": "they have arisen from a common ancestor but they have assumed different functions. This is an example of divergent evolution . The wing of a butterfly and that of a bird are said to be analogous. i.e. they have originated from different ancestors but they perform the same function. This is an example of convergent evolution. Cell Biology\nAll eucaryotic cells have organelles such as mitochondria, membrane-bound nuclei, ribosomes, golgi bodies. Thus indicating that different organisms have a common ancestor. The presence of chloroplasts and cellulose cell walls indicates that green plants have a common ancestor. Blood pigments are conjugated proteins with a metal group. Similar pigments are found in different animal groups . e.g. haemoglobin is found in all vertebrates and in annelida (earthworm). This shows that all animals have a common origin. Mechanism of Evolution\nThe mechanism of evolution can be described as a process of natural selection acting on the heritable variations that occur among the members of a population. A population consists of a group of individuals of the same species. Each individual has a set of hereditary factors(genes). All the genes in a population constitute a gene pool. When reproduction takes place, genes pair with one another randomly. Genes which occur in great numbers in the gene pool, will occur in greater numbers in the next generation. Several theories have been proposed over the years to explain how evolution took place. Lamark’s theory\nLamark had observed that if a part of the body of an organism was used extensively, it became enlarged and more efficient;\nIf a part of the body was not fully used, it would degenerate. By use and disuse of various body parts, the organism would change and acquire certain characteristics. He suggested that these characteristics would them be passed on to the offspring(next generation). In 1809, lamark published his book ‘’Theory Of Evolution’’. He proposed that new life forms arise from use and disuse of parts of existing organisms and through the inheritance of acquired characteristics. Lamark’s theory has been disapproped in that although use and disuse of parts does lead to acquired characteristics, such characteristics are not inheritable since they are effects produced by the environment and not by genes. Evolution by natural selection\nIn 1859, charles Darwin published his theory of evolution’ in a book called origin of species by means of natural selection’.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9192439418416802, "ocr_used": false, "chunk_length": 2476, "token_count": 494}}
{"text": "He proposed that new life forms arise from use and disuse of parts of existing organisms and through the inheritance of acquired characteristics. Lamark’s theory has been disapproped in that although use and disuse of parts does lead to acquired characteristics, such characteristics are not inheritable since they are effects produced by the environment and not by genes. Evolution by natural selection\nIn 1859, charles Darwin published his theory of evolution’ in a book called origin of species by means of natural selection’. Darwin’s theory was based on the following evidence;the population of a given species remains constant over a long period of time. The number of young ones is more than the number of adults. More offsprings are produced than can possibly survive. Variation occurs withing a given population,i.e all members of the same species are not alike. On the basis of these observations. Darwin made the following conclusions;\nThere is a struggle for existence among individuals in a given population. Individuals who are not suitably adapted (e.i. who have unfavourable variations)are less able to pass their characteristics to the next generation. Natural selection operates on the population, selecting those individuals with favourable variations;\ni.e. environment favours individuals that are more adapted. They win competition e.g. for food and survive.i.e. ‘’survival of the fittest’’. They attain sexual maturity and pass on the characteristics to their offsprings. Natural selection\nPeppered moth (Industrial melanism)\nThe peppered moth, Biston betularia, exists in two distinct forms;\nA speckled white form(the normal form) and the melanic, dark form. The moths normally rest on the tree trunks and branches wherre they are camouflaged against predators. The first melanic moths were observed in 1848 around Manchester in Britain. Since that time, their numbers has increased tremendously, out-numbering the speckled white form. The increase in the population of the melanic form is correlated with environmental changes brought about by industrialization and pollution. Smoke and soot from factories have darkened the tree trunks over the years. This has resulted in the preservation of the mutation in Biston betularia leading to the evolution of the melanic form. This form is almost invisible against the dark background of the tree trunks and is less subject to predation than the speckled form.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9211265432098766, "ocr_used": false, "chunk_length": 2430, "token_count": 489}}
{"text": "Smoke and soot from factories have darkened the tree trunks over the years. This has resulted in the preservation of the mutation in Biston betularia leading to the evolution of the melanic form. This form is almost invisible against the dark background of the tree trunks and is less subject to predation than the speckled form. The peppered form is more abundant in areas away from the soot and smoke of factories. This is because it is well camouflaged by the lichen-covered tree trunks against which it rests and is therefore not easily detected by predators. The existence of two or more distinct forms within a species (as exemplified by Biston betularia) is called polymorphism. Resistance to Drugs\nCertain strains of organisms have developed resistance to drugs and antibiotics. Following continued use of such drugs and antibiotics, some of the individuals in a population of bacteria or other microorganisms survive and are able to pass their characteristics to the next generation. When a patient fails to take full dosage of the antibiotics prescribed the pathogen develops resistance to the drugs hence become difficult to control. Some mosquitoes have developed resistance to certain pesticides. Practical Activities\nComparison of Vertebrate Limbs\nLimbs of various vertebrates are provided:\ne.g. fish- Tilapia, amphibian-frog reptiles, lizard; bird - domestic fowl (chicken), mammal- rabbit. Their anatomy can be studied. The following can be noted:\nThat all limbs have five sets of bones;\nA single upper bone- the femur in hind limb and the humerus in fore limb\nTwo lower limb bones -i.e. the tibia & fibula in the hind limb & ulna & radius in the forelimb. Small bones - i.e. ankle (tarsals) and wrist bones (carpals)\nThe bones making the foot and hand are metatarsals and metacarpals respectively. The bones of toes and of fingers i.e. phalanges\nObserve the various modifications of these bones in the various animals. Limbs of different mammals e.g. rabbit, cow, donkey reveal that the anatomy is adapted to mode or type of movement . e.g. the horse has a single digit. An outdoor activity to observe various sty les of movement in different mammals can be studied. It is noted that some move on tips of toes (donkey) others on the whole leg (rabbit).", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9173204054649626, "ocr_used": false, "chunk_length": 2269, "token_count": 493}}
{"text": "the horse has a single digit. An outdoor activity to observe various sty les of movement in different mammals can be studied. It is noted that some move on tips of toes (donkey) others on the whole leg (rabbit). Comparision of Wings of bird-and insect\nWings of birds and insects (grasshopper, butterfly or moth) are obtained. A hand lens or a dissecting microscope is used to observe the specimens. The differences in their anatomy is noted. Insect wings are membranous while those of birds are made up of feathers that interlock. Education tour to Archeological site/local Museum\nVisits to the local museum yield important information that greatly supplement study of evolution. The National museum in Nairobi has many fossils. Visit to the various archeological sites that exist in Kenya is recommended. end\nRECEPTION, RESPONSE AND CO-ORDINATION\nIN PLANTS AND ANIMALS\nIntroduction\nThe structures involved in detecting the changes may be located far away from the ones that respond. There is need for a communication system within the body. The nervous system and the endocrine system perform this function,\ni.e. linking the parts of the body that detect changes to those that respond to them. Irritability\nLiving organisms are capable of detecting changes in their internal and external environments and responding to these changes in appropriate ways. This characteristic is called irritability, and is of great survival value to the organism. Stimuli\nA stimulus is a change in the internal or external environment to which an organism responds. Examples of stimuli include light, heat, sound, chemicals, pH, water, food, oxygen and other organisms. Response\nA response is any change shown by an organism in reaction to a stimulus. The response involves movements of the whole or part of the body either towards the stimulus or away from it. It also results in secretion of substances e.g. hormones or enzymes by glands. Co-ordination\nCo-ordination is the working together of all the parts of the body to bring about appropriate responses to change in the environment. Reception\nReception is the detection of changes in the environment through receptors. Irritability in Plants\nResponse in plants is not as pronounced as in animals. This does not in anyway diminish the importance of irritability in plants. It is as important to their survival as it is in animals. Plants respond to a variety of stimuli in their environment. These stimuli include light, moisture, gravity and chemicals.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9226816539542353, "ocr_used": false, "chunk_length": 2491, "token_count": 498}}
{"text": "It is as important to their survival as it is in animals. Plants respond to a variety of stimuli in their environment. These stimuli include light, moisture, gravity and chemicals. Some plants also show response to touch. Tropisms\nPlants often respond by growing in a particular direction. Such growth movements are called tropisms. They are the result of unequal growth in the part of the plant that responds. The stimulus cause unequal distribution of growth hormones (auxins) produced in the plant. One side grows more than the other resulting in a bend either towards the stimulus (positive tropism) or away from the stimulus (negative tropism). Phototropism\nIf seedlings are exposed to light from one direction, their shoots grow towards the light. This response is called phototropism. Shoots are said to be positively phototropic because they grow towards the light. The tip of the shoot receives the light stimulus from one direction (unilateral stimulus) but the response occurs below the tip. The response of the shoot is due to a hormone called auxin produced at the tip. It diffuses down the shoot to this zone of cell elongation where it causes the cells to elongate. Light causes auxin to migrate to the darker side. The auxin is more concentrated in the dark side than on the light side. The cells on the dark side grow faster than the ones on the light side. A growth curvature is therefore produced. Survival value:\nPositive phototropism by shoots ensure that sufficient light is absorbed by leaves for photosynthesis. Geotropism\nGeotropism is a growth response to gravity. Roots are positively geotropic because they grow down towards the direction of the force of gravity;\nshoots are negatively geotropic because they grow away from direction of force of gravity. If a seedling is kept in the dark with its plumule and radicle in a horizontal position, the plumule will eventually grow vertically upwards while the radicle will grow vertically downwards. The effect of gravity on roots and shoots can be explained as follows:\nWhen the seedling is placed in a horizontal position, more auxin settles on the lower side of the root and shoot due to the effect of gravity. Shoots respond to a higher concentration of auxin than roots. The lower side of the shoot grows faster than the upper side. Resulting in a growth curvature that makes the shoot grow vertically upwards. Root growth is inhibited by high concentrations of auxin.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9230077646097262, "ocr_used": false, "chunk_length": 2447, "token_count": 499}}
{"text": "The lower side of the shoot grows faster than the upper side. Resulting in a growth curvature that makes the shoot grow vertically upwards. Root growth is inhibited by high concentrations of auxin. Therefore, the lower side of the root grows at a slower rate than the upper side where there is less auxin concentration. This results in a growth curvature that makes the root grow vertically downwards. Survival Value:\nRoots in response to gravity grow downwards where they absorb water and get anchored in the soil. This results in absorption of nutrients needed for growth. Hydrotropism\nHydrotropism is the growth of roots towards water (moisture) . Survival Value\nIt ensures that plant roots grow towards moisture to obtain water needed for photosynthesis and transport of mineral salts. Chemotropism\nChemotropism is the response of parts of a plant towards chemical substances,\ne.g. the growth of the pollen tube towards the ovule in flowering plants is a chemotropic response. Survival Value\nThis ensures that fertilisation take place and the perpetuation of the species continues. Thigmotropism\nThigmotropism is a growth response to touch. e.g. tendrils of climbing plant bend around objects that they come in contact with. Survival Value\nThis provides support and the leaves stay in a position suitable for absorption of light and gaseous exchange for photosynthesis. Tactic Movements in Plants and other Organisms\nA tactic movement is one made by a whole organism or a motile part of an organisms (e.g. a gamete) in response to a stimulus. Tactic movements are named according to the nature of the stimulus that brings about the response. Phototaxis is movement in response to direction and intensity of light. Free-swimming algae such as Chlamydomonas usually tend to concentrate where light intensity is optimum and will respond to light by swimming towards it. This is an example of phototactic response. Osmotaxis is movement in response to changes in osmotic conditions e.g. freshwater amoeba. Survival Value\nEnsures favourable conditions for existence. Chemotaxis is movement in response to concentration of chemical substances. Survival Value\nIn bryophytes, antherozoids move towards archegonia to effect fertilisation\nSurvival Value of taxis:\nThese ensure conditions favourable for life bring maximum benefit to the organism.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9276068376068376, "ocr_used": false, "chunk_length": 2340, "token_count": 485}}
{"text": "Survival Value\nEnsures favourable conditions for existence. Chemotaxis is movement in response to concentration of chemical substances. Survival Value\nIn bryophytes, antherozoids move towards archegonia to effect fertilisation\nSurvival Value of taxis:\nThese ensure conditions favourable for life bring maximum benefit to the organism. Nastic Movements\nA nastic movement is one made by part of a plant in response to stimulus which is not coming from any particular direction. Nastic movements are also named according to the nature of the stimulus. Seismonasty/haptonasty - response to shock. The 'sensitive plant' Mimosa pudica responds to touch by folding up its leaves. This is an example of a seismonastic response. Production of auxins and their effects on plant growth\nAuxins are produced by plant apices, i.e. root apex and shoot apex. They bring about cell elongation resulting in growth. They are diffusible substances which effect growth when in very small amounts. Roots require lower concentrations than shoots. The effect of auxins on the growth of roots and shoots has already been discussed. Auxins also exert other effects on plant growth and development. There are various other chemical substances which have been shown to influence plant growth and development. Effects of Auxin on Plant Growth\nApical Dominance\nAuxins inhibit the growth of side branches. This is referred to as apical dominance. If the terminal bud is removed, side branches develop from the lateral buds. This knowledge is applied in pruning. As long as the main stem is allowed to remain intact, the development of side branches is suppressed. Pruning the terminal bud removes the main sources of auxin, thus allowing side branches to sprout. Growth of adventitious roots\nAdventitious roots develop from the stem. Auxins stimulate the growth of such roots. Parthenocarpy\nThis refers to the formation of fruits without fertilisation. This can be induced by treating unpollinated flowers with auxin. This phenomenon is applied in the development of seedless fruit varieties. Auxins, together with other plant hormones, are involved in secondary growth, falling of leaves and ripening of fruits. Reception, Responses and Coordination in Animals\nThe nervous and endocrine systems (together known as the neuro-endocrine system) act as a co-ordinating system. They linking the receptors to the effectors and regulating their activities. Receptors\nReceptors are cells that detect or receive stimuli.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9271261588069327, "ocr_used": false, "chunk_length": 2481, "token_count": 508}}
{"text": "Reception, Responses and Coordination in Animals\nThe nervous and endocrine systems (together known as the neuro-endocrine system) act as a co-ordinating system. They linking the receptors to the effectors and regulating their activities. Receptors\nReceptors are cells that detect or receive stimuli. They may be scattered more uniformly all over the body surface\ne.g. receptors for pain, touch, temperature; or they may be located in a special sense organ e.g. receptors for light, sound, taste and smell. Motor nerves link the Central Nervous System (CNS) to the effectors. Its cell body is located at one end of the axon. It transmits nerve impulses from the CNS to the effectors. Effectors\nThese are the cells, organs, or organelles which enable the organism to respond. They include muscles, glands, cilia and flagella. The Nervous System\nComponents of the nervous system in humans\nEvery organ is the human body is connected to nerves. The nervous system is made up of nerve cells (neurons) which transmit impulses from one part of the body to another. It consists of the following:\nThe Central Nervous System (CNS) is a concentrated mass of interconnected nerve cells which make up the brain and the spinal cord. The peripheral nervous system is made up of nerves which link the CNS to the receptors and the effectors. Sensory nerves link the sensory cells (receptors) to the central nervous system and transmit nerve impulses from a sense organ to the CNS. Structure and Functions of Neurons\nA nerve cell consists of a cell body (centron) where the nucleus is located, and projections called dendrites arise. One of the projections is drawn out into an axon i.e. the longest process. Each axon contains axoplasm which is continuous with the cytoplasm in the cell body. The axon is enclosed in a fatty myelin sheath which is secreted by Schwarm cell. The myelin sheath is interrupted at approximately 1 mm intervals by constrictions known as nodes of Ranvier. The myelin sheath is enclosed by a thin membrane called the neurilemma, which is part of the Schwann cell in contact with axon. The myelin sheath and nodes of Ranvier enhance transmission of the impulse. There are three types of neurons:\nSensory neurone\nAlso known as afferent neurone.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9178723253757737, "ocr_used": false, "chunk_length": 2250, "token_count": 502}}
{"text": "The myelin sheath is enclosed by a thin membrane called the neurilemma, which is part of the Schwann cell in contact with axon. The myelin sheath and nodes of Ranvier enhance transmission of the impulse. There are three types of neurons:\nSensory neurone\nAlso known as afferent neurone. Transmits impulses from sensory cells to the CNS. The cell body of a sensory nerve cell is located at some distance along the length of the axon outside the CNS. Motor neurone\nKnown as efferent or effector neurone\nTransmit impulses from the CNS to the effectors(muscles and glands)\nIts cell body is located inside the CNS. Intermediate or connector neurone\nAlso called relay neurone\nFound inside the CNS. The connect sensory and motor neurons with each other and with other nerve cells in the CNS. Functions of the neurone\nThe nerve impulse is electrical in nature. Its transmission depends on differences in electrical potential between the inside and the outside of the axion. The outside is positive while the inside is negative. The stimulus triggers a change that affects the permeability of neurone membrane. The result is a change in the composition of ions on either side of the membrane. The outside becomes negative as the inside becomes positive due to sodium ions rushing in. The above constitutes a nervous impulse which is transmitted along the sensory neurone to the CNS. The speed of transmission is very high. Certain mammalian axions transmit impulses at the rate of 100m/s. The dendrites of neurons do not connect directly to each other, but they leave a small gap called synapse. The transmission of an impulse from one cell to the next takes place through synapse. Synaptic knobs are structures found at the ends of dendrites. Thus the dendrites of one nerve cell make contact with the dendrites of the adjacent nerve cell through the synapses. Impulses are transmitted in the form of a chemical transmitter substance which crosses the gap between one dendrite and the next. The transmitter substance is found within synaptic vesicles. The chemical substance is either acetylcholine or noradrnaline. The synaptic vesicles burst and release the transmitter substance when an impulse arrives at the synaptic knob. Impulses in motor neurone s are trans mitted to effectors. The space between motor end dendrite and muscle is known as neuro-muscular Junction. Synaptic vesicles in the ends of the dendrites release the transmitter substance across the neural muscular junction.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9250655861139733, "ocr_used": false, "chunk_length": 2480, "token_count": 510}}
{"text": "Impulses in motor neurone s are trans mitted to effectors. The space between motor end dendrite and muscle is known as neuro-muscular Junction. Synaptic vesicles in the ends of the dendrites release the transmitter substance across the neural muscular junction. Functions of Major Parts of the Human Brain\nThe Central Nervous System (CNS) consists of the brain and the spinal cord. The CNS co-ordinates body activities by receiving impulses from sensory cells from different parts of the body. It then sends the impulses to the appropriate effectors. The brain is enclosed within the cranium or braincase. It is covered and protected by membranes known as meninges. When meninges are infected by bacterial or fungi they cause meningitis. The brain consist of the following parts:\nCerebrum. This is the largest part of the brain. It consists of two cerebral hemispheres. It is highly folded in order to increase the surface area. The cerebrum controls learning, intelligence, thought, imagination and reasoning. The medulla oblongata (brain stem). The medulla oblongata has centres which control breathing (ventilation) rate,\nheart beat rate (cardiac frequency),\nswallowing, salivation, blood pressure\ntemperature regulation, hearing, taste and touch. The cerebellum\nIs located in front of the medulla and is a folded dorsal expansion of the hindbrain. It controls posture movement and balance. The hypothalamus\nControls functions such as body temperature and osmoregulation. The pituitary gland\nIs an endocrine organ that secretes a number of hormones which control osmoregulation, growth, metabolism and sexual development. Optic lobes -control the sense of sight. Olfactory lobes -control the sense of smell. Spinal Cord\nThe spinal cord is located within the vertebral column and consist of the following:\nThe grey matter forms the central part of the spinal cord. It consists of nervecell bodies and intermediate nerve fibres. The white matter of the spinal cord carries sensory nerve fibers while the ventral root carries motor nerve fibers. Simple And Conditioned Reflex Actions\nSimple Reflex Action\nA simple reflex action is an automatic response to a stimulus. The route that is followed by impulses during a reflex action is called a reflex arc. A reflex action follows the following sequence:\nA receptor is stimulated and an impulse is transmitted along a sensory nerve fibre to the spinal cord.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9246256239600666, "ocr_used": false, "chunk_length": 2404, "token_count": 499}}
{"text": "Simple And Conditioned Reflex Actions\nSimple Reflex Action\nA simple reflex action is an automatic response to a stimulus. The route that is followed by impulses during a reflex action is called a reflex arc. A reflex action follows the following sequence:\nA receptor is stimulated and an impulse is transmitted along a sensory nerve fibre to the spinal cord. The impulse is picked up by an intermediate neurone within the CNS. The intermediate nerve fibre transmit the impulse to a motor nerve fibre which is connected to an effector. The effector responds. Examples of reflex action include:\nPulling the hand away from a hot object. The knee jerk. Sneezing. Conditioned Reflexes\nThese are learned responses. When two or more stimuli are presented to an animal at the same time and repeatedly, the animal eventually responds to either stimulus. For example, if a hungry animal is presented with food, it will respond by salivating. If a bell is rung at the same time as the food is presented to the animal, the animal will learn to associate the sound of the bell with food. Eventually, the animal can be made to salivate at the sound of the bell alone. This response is called conditioned reflex and is one of the ways by which animals learn. The Role of Endocrine System in Human Beings\nEndocrine system consists of glands that secrete hormones. The glands have no ducts and are known as endocrine glands. Other glands are known as exocrine glands because they have ducts. The pancreas has an outer exocrine portion and an inner endocrine portion. Hormones are chemical substances, protein in nature which are secreted at one part of the body and have effects on other parts not necessarily near the point of secretion. They are secreted directly into blood and transported by blood. Each hormone either has a generalised co-ordinating effect on the body or brings about a specific response in a particular target organ. Hormones produced in humans and the in effects on the body. Adrenaline\nEnhance activity of sympathetic nervous system. Over secretion\nIncreased heartbeat\nHigh blood pressure\nThin toneless muscles. Under secretion\nLow blood pressure\nInability to withstand stress\nMuscular weakness\nThyroxine\nOver secretion is termed hyperthyroidism this causes:\nIncreased Basal Metabolic Rate (BMR) hence increased temperature. Person becomes very angry, nervous and hands may shake. Increased heartbeat which lead to cardiac failure.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9249692496924969, "ocr_used": false, "chunk_length": 2439, "token_count": 496}}
{"text": "Under secretion\nLow blood pressure\nInability to withstand stress\nMuscular weakness\nThyroxine\nOver secretion is termed hyperthyroidism this causes:\nIncreased Basal Metabolic Rate (BMR) hence increased temperature. Person becomes very angry, nervous and hands may shake. Increased heartbeat which lead to cardiac failure. Under secretion is termed hypothyroidism:\nPoor growth and mental retardation (cretinism). Reduced metabolic rate hence decreased temperature. Person becomes inactive and slothful. Eyes and face become puffy as fluid gets stored under skin. In extreme cases the tongue is swollen and skin becomes rough. Enlarged thyroid gland. Comparison between endocrine and nervous system\nSimilarities\nBoth endocrine and nervous system are involved in the coordination of body functions. Both have target organs. Both are controlled via a negative feedback mechanism, i.e too high production results in a reduced production. Effects of drugs abuse on the human health. Drug abuse can be defined as misuse of drugs. Drugs are chemical compounds that affect the working of body or kill disease causing microorganisms. Prescription drugs\nAre drugs prescribed by a doctor. Prescribed drugs can be abused through taking overdose which may cause death. Over the counter drugs(OCD)\nAre self prescribed drugs. These have harmful effects and may lead to tolerance such that higher doses are needed. Below is a list of effects of hard drugs on human health\nLung cancer caused by nicotine. Emphysema. Liver cirrhosis -caused by alcohol. Interferes with vision - alcohol. Sterility - khat (rniraa). Sleeplessness - insomnia - khat (miraa). Hallucinations - Canabis sativa (Bang i). Digestive system is upset, nausea. Diarrhoea and vomiting. Headache and double vision. Skin tone changes - e.g. too dark. Appetite is extreme - very poor or very great. Weight loss. Personality changes e.g. irritable and confused. Convulsions, lethargy and depressions due to inhalation of solvents e.g. glue. Structure and Function of Parts of the Human Eye\nStructure\nThe human eye is spherical in shape and situated within a socket or orbit in the skull. It is attached to the skull by three pairs of muscle, which also control its movement. It is made up of three main layers; sclerotic layer, choroid and the light sensitive retina.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9196366782006921, "ocr_used": false, "chunk_length": 2312, "token_count": 494}}
{"text": "Structure and Function of Parts of the Human Eye\nStructure\nThe human eye is spherical in shape and situated within a socket or orbit in the skull. It is attached to the skull by three pairs of muscle, which also control its movement. It is made up of three main layers; sclerotic layer, choroid and the light sensitive retina. Sclerotic layer\nOutermost white part situated at the sides and back of the eye. Made up of collagen fibres. It protects the eye and gives its shape. Cornea\nThis is the transparent front part of the sclera that allows light to pass through. It is curved, bulging at the front. It thus reflects light rays hence helps to focus light rays onto the retina. Choroid\nThe second or middle layer. It has many blood vessels that supply nutrients to the eye and remove metabolic wastes from the eye. It has dark pigments to absorb stray light and prevent its reflection inside the eye. Ciliary body\nIs glandular and secretes aqueous humour. It has blood vessels for supplying of nutrients excretion and gaseous exchange. It has ciliary muscles - which contract and relax to change the shape of lens during accommodation. Suspensory ligaments\nAre inelastic and attach the lens onto the cilliary body holding it in position. Lens\nBiconvex in shape, to refract light. Crystalline and transparent to allow light to pass through and focus it on to the retina. Aqueous humour\nFound between lens and the cornea. Transparent to allow light to pass through it. It is watery thus helping in focusing. Helps maintain shape of eye ball. To convey nutrients and oxygen to cornea, and remove waste products. Iris\nThe coloured part of the eye has an opening - the pupil at the centre. Iris has circular and radial muscles which controls size of the pupil, hence the amount of light entering the eye through the pupil. Vitreous humour\nIt is a fluid. Found between lens and retina. Is viscous and gives eye the shape. It is transparent and refracts light. Retina\nRetina contains light sensitive cells and is situated at the back of the eye. There are two types of light sensitive cells in the retina:\nRods - are sensitive to low-intensity light and detect black and white. Nocturnal mammals have more rods. Cones - are sensitive to high intensity of light;\nThey detect bright colour.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9182654402102497, "ocr_used": false, "chunk_length": 2283, "token_count": 503}}
{"text": "There are two types of light sensitive cells in the retina:\nRods - are sensitive to low-intensity light and detect black and white. Nocturnal mammals have more rods. Cones - are sensitive to high intensity of light;\nThey detect bright colour. Diurnal mammals have more cones. Fovea centralis\nFovea centralis (yellow spot) is the most sensitive part of the retina. Consists mainly of cones for accurate vision (visual acuity). Optic nerve\nOptic nerve, has neurons for transmission of impulse to the brain for interpretation. Blind spot\nBlind spot is located at the point where the optic nerve leaves the eye on its way to the brain. It is not sensitive to light it has no rods or cones. Eye lid\nEye lid is a loose skin that covers the eye. It closes by reflex action. Protects it from mechanical damage and from too much light. Eyelashes\nPrevent dust and other particles from entering eye. Conjuctiva\nIt is transparent and thin and allows light to pass through. It is a tough layer that is continuous with the epithelium of the eye lids. It protects the cornea. Accommodation\nAccommodation refers to the change in the shape of the lens in order to focus images. Rays from a distant object would be focused at a point behind the retina if the lens were not adjusted appropriately. When the eye is focusing at a distant object, the cilliary muscles are relaxed and the suspensory ligament are stretched tight. The lens is pulled thin, thus allowing light rays from a distant object to be properly focused on to the retina. When the eye is looking at near object, the ciliary muscles contract and the suspensory ligament become slack. The lens becomes more convex. This allows light rays from near object to be focused onto the retina. Control of light intensity entering the eye\nIn bright light (high intensity) the circular muscles of the iris contract. The diameter of the pupil decreases and less light enters. This protects retina from damage by too much light. In dim light circular muscles of iris relax (radial ones contract). Pupil's size (diameter) increases, more light enters the eye. Image formation and Interpretation\nLight rays from an object enter the cornea and are directed onto the lens through the pupil. They are refracted by the cornea and the lens. The latter brings the rays into fine focus.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9179575941151018, "ocr_used": false, "chunk_length": 2311, "token_count": 499}}
{"text": "Image formation and Interpretation\nLight rays from an object enter the cornea and are directed onto the lens through the pupil. They are refracted by the cornea and the lens. The latter brings the rays into fine focus. It makes the light rays converge so that an image is focused at a point on the retina. The image on the retina is inverted. This stimulate, the rods and cones on the retina and impulses generated are transmitted through the optic nerve to the brain. The brain interprets the image as upright. Common Eye Defects and their Correction\nShort-sightedness (Myopia)\nA shortsighted person cannot focus distant objects properly. Light rays from a distant object fall at a point in front of the retina. This may be due to the eyeball being too long. This defect can be corrected using spectacles with concave lenses. The lenses make the light rays diverge before they reach the eye. Long-sightedness (Hypermetropia)\nA long-sighted person cannot focus near objects properly. Light rays from the object are not focused on the retina. This may be due to the eyeball being too short. This defect may be corrected by using spectacles with convex lenses which make light rays converge before they reach the eye. Astigmatism\nAstigmatism refers to a condition in which the cornea or the lens is uneven, so that images are not focused properly on the retina. This defect can be corrected by wearing spectacles with special cylindrical lenses. Presbyopia is a condition in which light rays from a near object are not focused on the retina. This is caused by hardening or loss of elasticity of lense due to old age. This defect is corrected by wearing convex (converging) lenses. Structure and Functions of Parts of Human Ear\nThe Mammalian Ear\nThe mammalian ear performs two major functions:\nhearing and detecting changes in the positions of the body to bring about balance and posture. The ear is divided into three sections. The Outer Ear\nThis consists of:\nAn outer flap, the pinna which is made up of cartilage. The function of the pinna is to catch and direct sounds. The external auditory canal is a tube through which sound travel. The lining of the tube secretes wax, which traps dust particles and microorganisms. The tympanum is a membrane stretching across the inner end of the external auditory canal. The tympanum vibrates when it is hit by sound waves.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9208121827411169, "ocr_used": false, "chunk_length": 2364, "token_count": 502}}
{"text": "The lining of the tube secretes wax, which traps dust particles and microorganisms. The tympanum is a membrane stretching across the inner end of the external auditory canal. The tympanum vibrates when it is hit by sound waves. The Middle Ear\nThis is a chamber containing three small bones called the ear ossic1es, the malleus, incus and stapes. The three ossic1es articulate with one another to amplify vibrations. The vibrations are transmitted from the tympanum to the oval window. At the end of the chamber is a membrane called the oval window. When the tympanum vibrates, it causes the ear ossic1es to move forwards and backwards. This causes the oval window to vibrate. The Eustachian tube connects the middle ear to the pharynx. It allows air to get in and out of the middle ear, thus equalising the pressure between the inside and the outside of the tympanum. The Inner Ear\nThis consists of a series of chambers filled with fluid. It comprises the cochlea and semicircular canals. Cochlea is a coiled tube that occupies a small space and accommodates a large number of sensory cells. The cells are connected to the brain through the auditory nerve. They detect vibrations which lead to hearing. Hearing\nThe sound waves set the tympanum vibrating and are transformed into vibrations. The vibrations are transmitted to the oval window by the three ossicles. Vibrations of the oval window cause the fluids inside the cochlea tube to vibrate. The membranes inside the cochlea have sensory cells which change the sound vibrations to nerve impulses. These are transmitted to the brain through the auditory nerve. Hearing is perceived in the brain. Balance and posture\nThe semi-circular canals\nThere are three semi-circular canals in each ear. They are situated at right angles to each other and each one is sensitive to movement in a different plane. They are filled with fluid and each has a swelling called the ampulla at one end. Inside the ampulla are sensory cells. Balance and posture are detected by these cells. Movement of the head in a given direction causes the fluid to move the hairs on sensory cells. This transmit impulses to the brain through the auditory nerve so that the movement is registered. Defects of the ear\nAcute labyrinthitis\nThis is an inflammation of the middle ear and cochlea. It may lead to deafness.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9193347355301082, "ocr_used": false, "chunk_length": 2334, "token_count": 499}}
{"text": "This transmit impulses to the brain through the auditory nerve so that the movement is registered. Defects of the ear\nAcute labyrinthitis\nThis is an inflammation of the middle ear and cochlea. It may lead to deafness. It can be treated by using certain drugs but sometimes an operation may be necessary. Tinnitis:\nThis is a sensation of noises in the ear. It is caused among others by accumulation of wax in the ear or use of certain drugs e.g. quinine. Treatment is by removal of wax, stopping use of the causative drug. Vertigo - Giddiness\nThis is disorientation of body in space - one of the causes is dilation of endolymph. Corrections: Use of appropriate drugs. Deafness. This is inability to hear. It is presented in various degrees in various individuals, some have partial hearing, others are completely deaf. This may be as a result of:\nChronic infection of cochlea. Lack of sensory cells. Excess wax in external auditory canal. Fusion of ear ossicles. Otitis Media\nThis is the inflammation of middle ear due to build-up of fluid. It is marked by the swelling of tissues surrounding the Eustachian tube due to infection or severe congestion. A strong negative pressure creates a vacuum in the middle ear. Treatment - use of antibiotics or surgery. Practical Activities\nTo investigate tactic response\nTactic response in fly maggots are investigated using choice chambers(s). Responses to various stimuli are observed e.g. to chemical substances - chemotaxis. On one side of choice chambers is placed beef/fish that has been dried in the sun. On the opposite chambers is placed rotting meat/fish. Ten maggots are placed at the center and choice chamber is covered. After 10 minutes the number of maggots at each end is counted. Most of the maggots have moved to the chamber with rotting meat. Tropisms\nMaize or been seeds are soaked and germinated, to the stage when radical and coleoptile/plumule just appear . (about 5 days for beans and seven days for maize). Seedlings with straight radic1es and plumules are used .. Geotropism\nThe seedlings are placed horizontally on the medium (Soil or vermiculite or saw dust or sand). Observations are done after three days and results recorded.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9134209900519203, "ocr_used": false, "chunk_length": 2194, "token_count": 485}}
{"text": "Seedlings with straight radic1es and plumules are used .. Geotropism\nThe seedlings are placed horizontally on the medium (Soil or vermiculite or saw dust or sand). Observations are done after three days and results recorded. Phototropism\nA potted plant or a young seedling planted in a beaker is kept next to a window which is the only source of light in the laboratory. Alternatively, a dark box may be used. Observations are made after 3-5 days and results recorded. The shoots grow bending towards the same light. Etiolation\nYoung seedlings are placed in a dark box. It is kept moist but not exposed to light. After two weeks the seedlings are removed and observations made to note the following:\nColour of leaves is yellow. Size of leaves is small\nLength of internodes is long\nLength of stem elongated long and thin. Other seedlings that were grown in light are observed (as control) and similar measurements taken. They are green in colour with larger leaves, shorter internodes and the stem is shorter and thicker. Those in the dark have smaller yellow leaves, long thing stems with long internodes. (etiolated). Experiment to Determine Distance of the Blind Spot\nStudents should work in pairs so that one takes measurements while the other observes. A cross and a dot are marked on a white paper . The two points are 6-9 cm apart. The paper is held 50 cm away from the face. Closing the left eye, the paper is slowly moved towards the face as the right eye is fixed on the cross. At 50 cm distance the cross and the dot are seen clearly. As-the paper is moved closer to the face, the dot disappears. The distance at which the dot disappears is measured. This is the distance of the blind spot. When the light rays from the dot are focused on the blind spot it disappears hence the dot is not seen. The Knee Jerk Experiment\nStudents work in pairs, one student sits on the table, high stool or bench with one leg crossed over the other. The other student chops the crossed knee just below the knee cap with the edge of palm or wooden ruler. It is observed that the crossed knee jerks. This is a spinal reflex. END\nSupport and Movement in Plants and Animals\nNecessity for support and movement\nMovement is a characteristic of all living organisms. It enables animals and plants to adjust to their environment.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9081527458557489, "ocr_used": false, "chunk_length": 2312, "token_count": 506}}
{"text": "This is a spinal reflex. END\nSupport and Movement in Plants and Animals\nNecessity for support and movement\nMovement is a characteristic of all living organisms. It enables animals and plants to adjust to their environment. Most animals move from place to place but some are sessile (i.e. fixed to the substratum). Majority of plants move only certain parts. However, though not easily observed all living protoplasm shows movement of one type or another. Necessity for support and movement in plants\nThey enable plants to be held upright to trap maximum light for photosynthesis and gaseous exchange. To hold flowers and fruits in appropriate position for pollination and dispersal respectively. To enable plants to grow to great heights and withstand forces of environment e.g. strong winds. Movement of male gametes to effect fertilisation and ensure perpetuation of a species. Plant parts move in response to certain stimuli in the environment of tropisms. Tissue distribution in Monocotyledonous and Dicotyledonous plants\nVascular bundles are the main support tissues in plants. In monocotyledonous stem they are scattered all over the stem. while in dicotyledonous stem they are found in a ring or rings. In monocots the xylem and phloem alternate around with pith in the centre. In dicots of the xylem forms a star in the centre - there is no pith. Phloem is found in between the arms of xylem. Dicotyledonous plants have cambium which brings about secondary growth resulting in thickening of the stem and root hence providing support. Secondary xylem becomes wood, providing more support to the plant. Role of support tissues in young and old plant\nPlants are held upright by strengthening tissues ;\nparenchyma,\ncollenchyma,\nsclerenchyma\nxylem tissue. Parenchyma and collenchyma are the main support tissues in young plants. Parenchyma –\nThey are found below the epidermis. They form the bulk of packing tissue within the plant between other tissues . They are tightly packed and turgid they provide support. Collenchyma –\nTheir cell walls have additional cellulose deposited in the corners. This provides them with extra mechanical strength. Sclerenchyma –\nTheir cells are dead due to large deposits of lignin on the primary cell wall.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.925100312082033, "ocr_used": false, "chunk_length": 2243, "token_count": 498}}
{"text": "Collenchyma –\nTheir cell walls have additional cellulose deposited in the corners. This provides them with extra mechanical strength. Sclerenchyma –\nTheir cells are dead due to large deposits of lignin on the primary cell wall. The lignified wall is thick and inner lumen is small, hence provide support. Sclerenchyma fibres are arranged in elongated and in longitudinal sheets giving extra support. They are found in mature plants. Xylem –\nHas two types of specialised cells. Vessels and tracheids. Vessels are thick-walled tubes with lignin deposited in them. They give support and strength to the plant. Tracheids are spindle-shaped cells arranged with ends overlapping. Their walls are lignified. They help to support and strengthen the plant. Plants with weak stems obtain their support in the following ways. Some use thorn or spines to adhere to other plants or objects. Some have twinning stems which grow around objects which they come into contact with. Others use tendrils for support. Tendrils are parts of a stem or leaf that have become modified for twinning around objects when they gain support. In passion fruit and pumpkin, parts of lateral branches are modified to form tendrils. In the morning glory, the leaf is modified into a tendril. Support and Movement in Animals\nNecessity for support and movement in animals. Animals move from place to place:\nIn search of food. To escape from predators. To escape from hostile environment. To look for mates and breeding grounds. The skeleton, which is a support structure helps to maintain the shape of the body. Movement is effected by action of muscles that are attached to the skeleton. Types and Functions of Skeletons\nTwo main types will be considered. These are exoskeleton and endoskeleton. Exoskeleton\nExoskeleton is hard outer covering of arthropods made up of mainly chitin. Which is secreted by epidermal cells and hardens on secretion. It is strengthened by addition of other substances e.g. tannins and proteins to become hard and rigid. On the joints such as those in the legs the exoskeleton is thin and flexible to allow for movement. Functions of Exoskeleton\nProvide support. Attachment of muscles for movement. Protection of delicate organs and tissues. Prevention of water loss. Endoskeleton:\nIt forms an internal body framework. This is a type of skeleton characteristic of all vertebrates.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9227981458069955, "ocr_used": false, "chunk_length": 2373, "token_count": 506}}
{"text": "Prevention of water loss. Endoskeleton:\nIt forms an internal body framework. This is a type of skeleton characteristic of all vertebrates. The endoskeleton is made of cartilage, bone or both. It is made up of living tissues and grows steadily as animal grows. Muscles are attached on the skeleton. The muscles are connected to bones by ligaments. Functions\nThe functions of endoskeleton include support, protection and movement. Locomotion in a finned fish e.g. tilapia. Most of the fishes are streamlined and have backward directed fins to reduce resistance due to water. External features-of Tilapia\nScales tapers towards the back and overlap forwards to provide a smooth surface for a streamlined body. The head is not flexible. This helps the fish to maintain forward thrust. Slimy mucous enables the fish to escape predators and protects the scales from getting wet. The pectoral and pelvic fins are used mainly for steering, ensuring that the fish is balanced. They assist the fish to change direction. The dorsal and anal fins keep the fish upright preventing it from rolling sideways. The caudal or tail fin has a large surface area, and displaces a lot of water when moved sideways creating forward movement of the fish. In order to change position in water the fish uses the swim bladder. When filled with air the relative density of the body is lowered and the fish moves up in the water. When air is expelled, the relative density rises and the fish sinks to a lower level. Swimming action in fish is brought about by contraction of muscle blocks (myotomes). These muscles are antagonistic when those on the left contract, those on the right relax. The muscles are attached to the transverse processes on the vertebra. The vertebra are flexible to allow sideways movement. Mammalian skeleton\nThe mammalian skeleton is divided into two:\nAxial and appendicular. Axial skeleton is made up of the skull and the vertebral column. Appendicular skeleton is made up of the pelvic and pectoral girdles and limbs (hind limb and forelimbs). The Axial Skeleton\nThis consists of the ;\nskull,\nthe sternum,\nribs,\nthe vertebral column. The Skull\nThe skull is made up of cranium and facial bones. The cranium; encloses and protects the brain. It is made up of many bones joined together by immovable joints. The facial bones consists of the upper and lower jaws.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9198133220195164, "ocr_used": false, "chunk_length": 2357, "token_count": 503}}
{"text": "The cranium; encloses and protects the brain. It is made up of many bones joined together by immovable joints. The facial bones consists of the upper and lower jaws. At the posterior end of the cranium are two smooth rounded protuberances, the occipital condyles. These condyles articulate with the atlas vertebra to form a hinge joint, which permits the nodding of the head. Sternum and ribs –\nThey form the rib-cage. The rib-cage encloses the thoracic cavity protecting delicate organs such as the heart and lungs. The ribs articulate with the vertebral column at the back and the sternum at the front. The Vertebral Column\nThe vertebral column is made up of bones called vertebrae placed end to end. The vertebrae articulate with one another at the articulating facets. In between one vertebra and another is the cartilaginous material called intervertebal disc. The discs act as shock absorbers and allow for slight movement. Each vertebra consists of a centrum and a neural arch which projects into a neural spine. The neural canal is the cavity enclosed by the centrum and the neural arch. The spinal cord is located inside the canal. The neural spine and other projections e.g. transverse processes serve as points of attachment of muscles. Type and number of vertebrae in human and rabbit\nCervical Vertebrae\nThese are found in the neck region of a mammal. The distinguishing feature is a pair of verte-braterial canals in the neural arch, through which the blood vessels of the neck pass. Another feature is the structure of the transverse processes. They are flattened out and are known as cervical ribs. The fIrst cervical vertebra is known as the Atlas. It has a large neural canal and no centrum. The second cervical vertebra, is called axis. The other five cervical vertebrae have no specific names. They have the same structure. The cervical vertebrae possess numerous processes for muscle attachment. Thoracic Vertebrae\nEach thoracic vertebra has a large centrum ,a large neural canal, neural arch and a long neural spine that projects upwards and backward. There is a pair of prezygapophyses and postzygapophyses for articulation with other vertebra . They have a pair of short transverse process. The thoracic vertebra also articulates with pair of ribs at tubercular and capitular facets.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9234171725932351, "ocr_used": false, "chunk_length": 2306, "token_count": 504}}
{"text": "There is a pair of prezygapophyses and postzygapophyses for articulation with other vertebra . They have a pair of short transverse process. The thoracic vertebra also articulates with pair of ribs at tubercular and capitular facets. Lumbar Vertebrae\nEach lumbar vertebra has a large, thick centrum for support of the body. It has a neural spine that projects upwards and forwards. There is a pair of large transverse process that are directed forwards. Above the prezygapophyses lies a pair of processes called metapophyses,\nBelow postzygapophyses lies the anapophyses. Metapophyses and anapophysis serve for attachment pf muscles of the abdomen. In some mammals, there may be another process on lower side of centrum called hypapophysis also for muscle attachment. Sacral Vertebrae\nThe sacral vertebrae are fused together to form a rigid bony structure, the sacrum. The centrum of each vertebra is large, but the neural canal is narrow. The neural spine is reduced to a small notch. The transverse processes of the first sacral vertebra are large and wing-like\nThey are firmly attached to the upper part of the pelvic girdle. Caudal Vertebrae\nHuman beings have only four of these vertebrae which are fused together to form coccyx. Animals with long tails have many caudal vertebrae. A typical caudal vertebra appears as a solid rectangular mass of bone. The entire bone consists of the centrum only. Appendicular Skeleton\nThe appendicular skeleton consist of the limbs and their girdles. Bones of Fore-limbs\nPectoral girdle\nPectoral girdle is made of scapula, coracoid and clavicle. A cavity known as glenoid cavity occurs at the apex of the scapula. The humerus of the fore limb fits into this cavity. The clavice is a curved bone connecting the scapular to the sternum. Humerus\nHumerus is found in the upper arm. It articulates with the scapula at the glenoid cavity of the pectoral girdle and forms a ball and socket joint. Ulna and radius\nThese are two bones found in the forearm.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.924773413897281, "ocr_used": false, "chunk_length": 1986, "token_count": 490}}
{"text": "Humerus\nHumerus is found in the upper arm. It articulates with the scapula at the glenoid cavity of the pectoral girdle and forms a ball and socket joint. Ulna and radius\nThese are two bones found in the forearm. The ulna has a projection called olecranon process and a sigmoid notch which articulates with the humerus. Bones of hind limb\nPelvic Girdle\nThe pelvic girdle consists of two halves fused at the pubic symphysis. Each half is made up of three fused bones:\nthe ilium,\nischium\npubis. Each half has cup-shaped cavity for the acetabulum for articulation with the head of the femur. Between the ischium and pubis is an opening obturator foramen where spinal nerves, blood vessels and a tough inflexible connective tissues pass. The ilium, ischium and pubis are fused to form the innominate bone. The Femur\nThe femur is the long bone joining the pelvic girdle and the knee. The head of the femur articulates with acetabulum forming the ball and socket joint at the hip. The femur has a long shaft. At the distal end it has condyles that articulate with the tibia to form a hinge joint at the knee. The patella covers the knee joint and prevents the upward movement of the lower leg. Tibia and Fibula\nThe tibia is a large bone, and the fibula a smaller bone is fused to it on the distal part. In humans the tibia and fibula are clearly distinguishable. Joints and Movement\nAjoint is a connection between two or more bones. Joints provide articulation between bones making movement possible. However some joints do not allow any movement e.g. the joints, between bones of the skull. Movable joints are of three main types:\nGliding joint\ne.g., joints which occur between the vertebrae wrists and ankles. The ends of the bones that make the joint are covered with cartilage. The bones are held together by tough ligaments. Synovial joint\nThe joint is enclosed by fibrous capsule lined by synovial membrane which secretes synovial fluid into the synovial cavity. The synovial fluid lubricates the joint. They are called synovial joints. They include hinge joint and ball and socket joint. Hinge joint\ne.g.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9197149643705463, "ocr_used": false, "chunk_length": 2105, "token_count": 508}}
{"text": "They are called synovial joints. They include hinge joint and ball and socket joint. Hinge joint\ne.g. knee joint. The joint allows movement in one plane. Ball and socket joint. e.g., hip joint. The joint allows rotation in all directions. Types, Locations and Function of Muscles\nThere are three types of muscles, located at various parts of the body. In order to function all use energy in form of ATP. These include smooth, skeletal and cardiac muscles. Smooth Muscle (Involuntary Muscles)\nThese are spindle-shaped and contain filaments with myofibrils. Each muscle is bound by plasma membrane. They are found lining internal organs such as alimentary canal, bladder, and blood vessels. They are controlled by involuntary part of the nervous system. They are concerned with movement of materials along the organs and tubes. They contract slowly and fatigue slowly . Skeletal Muscle (striated or voluntary muscle)\nSkeletal muscles are striated and have several nuclei. They are long fibres each containing myofibrils and many mitochondria. They have cross-striations or stripes. They are also called voluntary muscles because the contraction is controlled by voluntary nervous system. They are surrounded by connective tissue and are attached to bones by tendons. Their contraction brings about movement of bone, resulting in locomotion. They contract quickly and fatigue quickly. Cardiac Muscle\nConsist of a network of striated muscle fibres connected by bridges. Are short cells with numerous mitochondria and uninucleate. They are found exclusively in the heart. Contractions of cardiac muscles are generated from within the muscles and are rhythmic and continuous hence they are myogenic. They do not tire or fatigue. The rate can be modified by involuntary nervous system. Their contractions result in the heart pumping blood. Role of muscles in movement of the human arm\nMuscles that bring about movement are antagonistic, i.e. when one set contracts the other relaxes. Antagonistic muscles of human forelimb\nThe biceps muscles of the forelimb act as flexors while the triceps muscles act as extensors. The biceps has its point of origin on the scapula and the point of insertion on the radius. The triceps has its points of origin on the scapula and humerus and is inserted on the ulna. When the muscles contract, the limb acts as a lever with the pivot at the joint.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9221894736842106, "ocr_used": false, "chunk_length": 2375, "token_count": 496}}
{"text": "The biceps has its point of origin on the scapula and the point of insertion on the radius. The triceps has its points of origin on the scapula and humerus and is inserted on the ulna. When the muscles contract, the limb acts as a lever with the pivot at the joint. Contraction of biceps muscles bends (flexes) the arm while contractions of triceps extends the arm. Practical Activities\nTo observe prepared slides of transverse section of stems of herbaceous and woody plants. Permanent slides of transverse sections of:\nHerbaceous plant and Woody plant are obtained. The permanent slide of a herbaceous plant is placed onto the stage of the microscope. Observations under the low power and medium power objective is made. A plan diagram is drawn and labelled. The permanent slide of a woody plant is placed on the stage of the microscope. Observations under the low power and medium power objectives are made. A plan diagram is drawn and labelled. In both cases, support tissues such as parenchyma, collenchyma, sc1erenchyma and xylem are observed. To observe wilting in young herbaceous plants. A herbaceous potted plant e.g. bean plant is obtained. The plant is placed on the bench near a window and left for 3 days without watering on the third and subsequent day. The shoot droops due to fall in turgor pressure; caused by water loss. To examine the exoskeleton in an arthropod. Obtain a beetle and observe the external structure. The exoskeleton is on the outer surface with muscles attached on inner side. The exoskeleton is hardened by chitin. Movement is due to joints on the limbs. Also examine various shed cocoons of insects e.g., butterfly. To observe the external features of a finned fish. Fresh Tilapia is obtained and placed on a tray. Observations are made on the external features of the fish. A labelled drawing is made. Features like scales, fins a streamlined body and an operculum are seen. Opened operculum reveals the gills. To examine bones of the axial skeleton of a rabbit. Bones of the vertebra column are obtained. These are cervical, thoracic, lumbar and sacral. For each of the bones the distinguishing features are listed down. Labelled drawings of the anterior and lateral views is made. To observe bones of appendicular skeleton.", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9170753704697242, "ocr_used": false, "chunk_length": 2264, "token_count": 500}}
{"text": "For each of the bones the distinguishing features are listed down.\n\nLabelled drawings of the anterior and lateral views is made.\n\nTo observe bones of appendicular skeleton.\n\nBones of pectoral girdle and fore limb are obtained i.e., scapula, humerus, ulna and radius.\n\nLabelled drawing of each bone is made.\n\nObservations on how the bones articulate with one another is made.\n\nBones of pelvic girdle and hind limbs are obtained i.e., pelvic girdle, femur, tibia and fibula.\n\nLabelled drawings of each, bone is made.\n\nThe distinguishing features of each bone is noted.\n\nObservations on how the bones articulate with one another is made.\n\nEND\nBEST WISHES", "metadata": {"source": "FORM-4-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.910599078341014, "ocr_used": false, "chunk_length": 651, "token_count": 150}}
{"text": "FORM THREE BIOLOGY\nBy the end of form three work, the learner should be able to:\nClassify common organisms into their main taxonomic units\nWrite scientific names of organisms correctly\nList the kingdoms of organisms\nDescribe the general characteristics of Kingdom monera\nDescribe the general characteristics of Kingdom protoctista\nObserve, draw and name parts of spirogyra, amoeba, paramecium and euglena\nDescribe the general characteristics of Kingdom fungi\nList down all the members of kingdom fungi\nDraw and name parts of bread mold (mucor), yeast and mushrooms\nDescribe the main characteristics of kingdom plantae\nDescribe the main characteristics of bryophyta\nIdentify examples of hyophyta\nObserve draw and name parts of liverworts and moss plants\nIdentify examples of pleridophyta\nObserve draw and name parts of fern plant\nIdentify examples of division spermatophyta\nIdentify major sub-division of spermatophyta\nList main characteristics of angiospermae\nDifferentiate between angiospermae and gymnospermae\nState the characteristics of angiospermapyta\nIdentify and state major characteristics of classes of angiospermapytaegdicotyledonae&monocotyledonoe\ndescribe the general characteristics of kingdom animalia\ndescribe the general characteristics of Phylum arthropoda\nlist down the classes of the Phylum arthropoda\ndescribe the general characteristics of Class crustacean\ndescribe the general characteristics of Class insect\ndescribe the general characteristics of Class arachnida\nlist down the members of class arachnida and insect\nDescribe the general characteristics of Classeschilopoda and diplopoda\nList down the members of class chilopoda and diplopoda\nDescribe the general characteristics of Phylum chordate\ndescribe the general characteristics of Pisces and amphibian\ndescribe the general characteristics of reptilian\ndescribe the general characteristics of Class aves\nDescribe the general characteristics of Class Mammalia\nIdentify different types of members of Class Mammalia\nConstruct a simple dichotomous to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\ndraw and label organisms correctly\nDefine the term ecology and identify terms used in ecology\nDefine the term ecology and identify terms used in ecology\nIdentify the types of ecosystems\nState and explain how light determines distribution of organisms in an ecosystem\nIdentify and describe how temperature determines distribution of organisms in an ecosystem\nIdentify and describe how Rainfall and humidity determines distribution of organisms in an ecosystem\ndescribe how Wind and atmospheric pressure determines distribution of organisms in an ecosystem\nWrite down correct answers to questions asked in the test\ndescribe how salinity affects the distribution of organisms in aquatic ecosystems\ndescribe how waves, currents and tides affects the distribution of organisms in aquatic ecosystem\nDescribe how Edaphic factors affects the distribution of organisms in an ecosystem\nMeasure certain factors in samples of different soils\nDescribe how Geological factors affect the distribution of organisms in an ecosystem\nDescribe how Abiotic factors affect the distribution of organisms in an ecosystem\nDescribe how competition affects the distribution of organisms in an ecosystem\nDescribe how Predation and Symbiosis affects the distribution of organisms in an ecosystem\nDifferentiate between Parasitism and saprophytism\nDescribe how Parasitism and saprophytism influence the distribution of organisms in an ecosystem e.g.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9423428413181945, "ocr_used": false, "chunk_length": 3611, "token_count": 725}}
{"text": "FORM THREE BIOLOGY\nBy the end of form three work, the learner should be able to:\nClassify common organisms into their main taxonomic units\nWrite scientific names of organisms correctly\nList the kingdoms of organisms\nDescribe the general characteristics of Kingdom monera\nDescribe the general characteristics of Kingdom protoctista\nObserve, draw and name parts of spirogyra, amoeba, paramecium and euglena\nDescribe the general characteristics of Kingdom fungi\nList down all the members of kingdom fungi\nDraw and name parts of bread mold (mucor), yeast and mushrooms\nDescribe the main characteristics of kingdom plantae\nDescribe the main characteristics of bryophyta\nIdentify examples of hyophyta\nObserve draw and name parts of liverworts and moss plants\nIdentify examples of pleridophyta\nObserve draw and name parts of fern plant\nIdentify examples of division spermatophyta\nIdentify major sub-division of spermatophyta\nList main characteristics of angiospermae\nDifferentiate between angiospermae and gymnospermae\nState the characteristics of angiospermapyta\nIdentify and state major characteristics of classes of angiospermapytaegdicotyledonae&monocotyledonoe\ndescribe the general characteristics of kingdom animalia\ndescribe the general characteristics of Phylum arthropoda\nlist down the classes of the Phylum arthropoda\ndescribe the general characteristics of Class crustacean\ndescribe the general characteristics of Class insect\ndescribe the general characteristics of Class arachnida\nlist down the members of class arachnida and insect\nDescribe the general characteristics of Classeschilopoda and diplopoda\nList down the members of class chilopoda and diplopoda\nDescribe the general characteristics of Phylum chordate\ndescribe the general characteristics of Pisces and amphibian\ndescribe the general characteristics of reptilian\ndescribe the general characteristics of Class aves\nDescribe the general characteristics of Class Mammalia\nIdentify different types of members of Class Mammalia\nConstruct a simple dichotomous to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\ndraw and label organisms correctly\nDefine the term ecology and identify terms used in ecology\nDefine the term ecology and identify terms used in ecology\nIdentify the types of ecosystems\nState and explain how light determines distribution of organisms in an ecosystem\nIdentify and describe how temperature determines distribution of organisms in an ecosystem\nIdentify and describe how Rainfall and humidity determines distribution of organisms in an ecosystem\ndescribe how Wind and atmospheric pressure determines distribution of organisms in an ecosystem\nWrite down correct answers to questions asked in the test\ndescribe how salinity affects the distribution of organisms in aquatic ecosystems\ndescribe how waves, currents and tides affects the distribution of organisms in aquatic ecosystem\nDescribe how Edaphic factors affects the distribution of organisms in an ecosystem\nMeasure certain factors in samples of different soils\nDescribe how Geological factors affect the distribution of organisms in an ecosystem\nDescribe how Abiotic factors affect the distribution of organisms in an ecosystem\nDescribe how competition affects the distribution of organisms in an ecosystem\nDescribe how Predation and Symbiosis affects the distribution of organisms in an ecosystem\nDifferentiate between Parasitism and saprophytism\nDescribe how Parasitism and saprophytism influence the distribution of organisms in an ecosystem e.g. Tick and cattle\nDescribe the interaction between organisms in an ecosystem\nDescribe the role of decomposers in Nitrogen cycle & carbon cycle\nDefine the terms food chain and food web\nConstruct food chains and food webs\nDescribe energy flow in a local ecosystem and Construct food chains and food webs\nDefine population\nList down the characteristics of population\nExplain the use of quadrants and transects as methods of Population estimation\nExplain the capture –recapture method of population estimation\nUse quadrant method to estimate population of named organisms within the compound\nDescribe total count, aerial count and aerial photography and other methods of population estimation\nRelate to the adaptations of xerophytes to their habitats\nRelate to the adaptations of mesophytes to their habitats\nRelate to the adaptations of hydrophytes to their habitats\nObserve, draw and label parts of named hydrophytes, mesophytes and xerophyte plants\nRelate to the adaptations of halophytes to their habitats\nExplain pollution and give examples of pollutants\nDescribe the various air pollutants\nDiscuss the effects of air pollution on the environment\nSuggest methods of controlling air pollution\nDescribe various causes of Land/ soil pollution\nDiscuss the effects of Land/ soil pollution and human health in rural and urban centers\nSuggest methods of controlling Land/ soil pollution\nDescribe the causes of Water pollution\nIdentify other causes of environmental pollution in rural and urban centers\nDiscuss the effects of water pollution on human health in rural and urban centers and other organisms\nSuggest methods of controlling water pollution\nIdentify symptoms of cholera and typhoid fever\nState methods of transmission\nSuggest control measures\nIdentify the causes, symptoms and methods of transmission and control of malaria\nIdentify the causes, symptoms and methods of transmission of amoebic dysentery\nSuggest control methods of amoebic dysentery\nIdentify the causes, symptoms and methods of transmission of ascariosis\nIdentify the causes, symptoms and methods of transmission and control of schistomiasis\nDefine reproduction and state its importance\nDifferentiate between asexual and sexual reproduction\nDescribe the appearance and location of chromosomes\nDefine mitosis\nDescribe chromosomicmovement during mitosis\nDescribe e the movement of chromosomes in mitosis\nIdentify stages of mitosis\nIdentify and describe stages of mitosis\nState the significance of mitosis in reproduction\nDefine meiosis\nState the stages of meiosis\nDescribe the chromosome movement during meiosis\nObserve the stages of meiosis\nDescribe the movement of chromosomes during meiosis\nState the significance of meiosis in reproduction\nDifferentiate between mitosis and meiosis\nState and describe the importance of Binary fission\nObserve spore formation in bread mould (mucor) and binary fission in paramecium\nState and describing the importance of budding in reproduction\nObserving drawing and budding cells of yeast\nDescribe the external structure of a typical flower\nDescribe the internal structure of a typical flower\nObserve, describe and draw different types of pollen grains\nDescribe the structure of ovules\nDescribe other characteristics of flowers\nDescribe and compare adaptations of wind and insect pollinated flowers\nDescribe the features and mechanisms that hinder self-pollination and self-fertilization\nDescribe the process of fertilization in flowering plants\nDescribe and explain how embryo and seeds are formed in flowering plants\nDescribe how fruits are formed in flowering plants\nDifferentiate between a fruit and a seed\nDescribe and explain how different seeds and fruits are dispersed\nClassifying various types of fruits and describe their placentation\nDifferentiate between internal and external fertilization\nDescribe external fertilization in amphibians\nRelate the structure of mammalian male reproductive system to its functions\nRelate the structure of mammalian male reproductive organ and spermatozoa to its function\nRelate the structure of mammalian female reproductive system to its function\nRelate the structure of mammalian ovum to its function\nDescribe internal fertilization in mammals\nDescribe the fertilization process\nDescribe implantation and the role of the placenta in mammals\nDefine gestation in mammals\nIdentify different gestation periods in different mammals\nDescribe birth and explain parental care\nDescribe the role of hormones in reproduction of humans\nDescribe the role of hormones in the menstrual cycle\nIdentify symptoms and explain the methods of transmission and prevention of gonorrhea and herpes simplex\nIdentify symptoms and explain the methods of transmission and prevention of syphilis and trichomoniasis\nIdentify symptoms and explain the methods of transmission and prevention of candidiasis and hepatitis\nIdentify the causes and modes of transmission of HIV/AIDS and prevention of HIV and AIDS\nIdentify effects of HIV/AIDS in human economy\nIdentify the symptoms of HIV/AIDS and stages of HIV and AIDS\nExplain ways of preventing and controlling the spread of HIV/AIDS\nDiscuss the social effects of HIV/AIDS\nExplain the advantages and disadvantages of sexual and asexual reproduction\nDefine the terms growth and development\nDescribe the sigmoid growth curve\nDescribe the phases of sigmoid curve\nDescribe the intermittent growth curve\nAnalyze data on growth rate\nDraw growth curves\nDefine seed dormancy\nIdentify factors affecting viability and dormancy of seeds\nIdentify factors affecting seed dormancy\nDefine seed germination\nDifferentiate between types of seed germination\nIdentifying Conditions necessary for germination - oxygen\nInvestigate the necessity of water and warmth\nDescribe the region of growth in seedlings\nIdentify the regions of growth\nDetermine the regions of growth in seedlings\nMeasure the aspect of growth in a given seedling\nDescribe growth in plants I.e.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9404854470946727, "ocr_used": false, "chunk_length": 9517, "token_count": 1881}}
{"text": "FORM THREE BIOLOGY\nBy the end of form three work, the learner should be able to:\nClassify common organisms into their main taxonomic units\nWrite scientific names of organisms correctly\nList the kingdoms of organisms\nDescribe the general characteristics of Kingdom monera\nDescribe the general characteristics of Kingdom protoctista\nObserve, draw and name parts of spirogyra, amoeba, paramecium and euglena\nDescribe the general characteristics of Kingdom fungi\nList down all the members of kingdom fungi\nDraw and name parts of bread mold (mucor), yeast and mushrooms\nDescribe the main characteristics of kingdom plantae\nDescribe the main characteristics of bryophyta\nIdentify examples of hyophyta\nObserve draw and name parts of liverworts and moss plants\nIdentify examples of pleridophyta\nObserve draw and name parts of fern plant\nIdentify examples of division spermatophyta\nIdentify major sub-division of spermatophyta\nList main characteristics of angiospermae\nDifferentiate between angiospermae and gymnospermae\nState the characteristics of angiospermapyta\nIdentify and state major characteristics of classes of angiospermapytaegdicotyledonae&monocotyledonoe\ndescribe the general characteristics of kingdom animalia\ndescribe the general characteristics of Phylum arthropoda\nlist down the classes of the Phylum arthropoda\ndescribe the general characteristics of Class crustacean\ndescribe the general characteristics of Class insect\ndescribe the general characteristics of Class arachnida\nlist down the members of class arachnida and insect\nDescribe the general characteristics of Classeschilopoda and diplopoda\nList down the members of class chilopoda and diplopoda\nDescribe the general characteristics of Phylum chordate\ndescribe the general characteristics of Pisces and amphibian\ndescribe the general characteristics of reptilian\ndescribe the general characteristics of Class aves\nDescribe the general characteristics of Class Mammalia\nIdentify different types of members of Class Mammalia\nConstruct a simple dichotomous to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\ndraw and label organisms correctly\nDefine the term ecology and identify terms used in ecology\nDefine the term ecology and identify terms used in ecology\nIdentify the types of ecosystems\nState and explain how light determines distribution of organisms in an ecosystem\nIdentify and describe how temperature determines distribution of organisms in an ecosystem\nIdentify and describe how Rainfall and humidity determines distribution of organisms in an ecosystem\ndescribe how Wind and atmospheric pressure determines distribution of organisms in an ecosystem\nWrite down correct answers to questions asked in the test\ndescribe how salinity affects the distribution of organisms in aquatic ecosystems\ndescribe how waves, currents and tides affects the distribution of organisms in aquatic ecosystem\nDescribe how Edaphic factors affects the distribution of organisms in an ecosystem\nMeasure certain factors in samples of different soils\nDescribe how Geological factors affect the distribution of organisms in an ecosystem\nDescribe how Abiotic factors affect the distribution of organisms in an ecosystem\nDescribe how competition affects the distribution of organisms in an ecosystem\nDescribe how Predation and Symbiosis affects the distribution of organisms in an ecosystem\nDifferentiate between Parasitism and saprophytism\nDescribe how Parasitism and saprophytism influence the distribution of organisms in an ecosystem e.g. Tick and cattle\nDescribe the interaction between organisms in an ecosystem\nDescribe the role of decomposers in Nitrogen cycle & carbon cycle\nDefine the terms food chain and food web\nConstruct food chains and food webs\nDescribe energy flow in a local ecosystem and Construct food chains and food webs\nDefine population\nList down the characteristics of population\nExplain the use of quadrants and transects as methods of Population estimation\nExplain the capture –recapture method of population estimation\nUse quadrant method to estimate population of named organisms within the compound\nDescribe total count, aerial count and aerial photography and other methods of population estimation\nRelate to the adaptations of xerophytes to their habitats\nRelate to the adaptations of mesophytes to their habitats\nRelate to the adaptations of hydrophytes to their habitats\nObserve, draw and label parts of named hydrophytes, mesophytes and xerophyte plants\nRelate to the adaptations of halophytes to their habitats\nExplain pollution and give examples of pollutants\nDescribe the various air pollutants\nDiscuss the effects of air pollution on the environment\nSuggest methods of controlling air pollution\nDescribe various causes of Land/ soil pollution\nDiscuss the effects of Land/ soil pollution and human health in rural and urban centers\nSuggest methods of controlling Land/ soil pollution\nDescribe the causes of Water pollution\nIdentify other causes of environmental pollution in rural and urban centers\nDiscuss the effects of water pollution on human health in rural and urban centers and other organisms\nSuggest methods of controlling water pollution\nIdentify symptoms of cholera and typhoid fever\nState methods of transmission\nSuggest control measures\nIdentify the causes, symptoms and methods of transmission and control of malaria\nIdentify the causes, symptoms and methods of transmission of amoebic dysentery\nSuggest control methods of amoebic dysentery\nIdentify the causes, symptoms and methods of transmission of ascariosis\nIdentify the causes, symptoms and methods of transmission and control of schistomiasis\nDefine reproduction and state its importance\nDifferentiate between asexual and sexual reproduction\nDescribe the appearance and location of chromosomes\nDefine mitosis\nDescribe chromosomicmovement during mitosis\nDescribe e the movement of chromosomes in mitosis\nIdentify stages of mitosis\nIdentify and describe stages of mitosis\nState the significance of mitosis in reproduction\nDefine meiosis\nState the stages of meiosis\nDescribe the chromosome movement during meiosis\nObserve the stages of meiosis\nDescribe the movement of chromosomes during meiosis\nState the significance of meiosis in reproduction\nDifferentiate between mitosis and meiosis\nState and describe the importance of Binary fission\nObserve spore formation in bread mould (mucor) and binary fission in paramecium\nState and describing the importance of budding in reproduction\nObserving drawing and budding cells of yeast\nDescribe the external structure of a typical flower\nDescribe the internal structure of a typical flower\nObserve, describe and draw different types of pollen grains\nDescribe the structure of ovules\nDescribe other characteristics of flowers\nDescribe and compare adaptations of wind and insect pollinated flowers\nDescribe the features and mechanisms that hinder self-pollination and self-fertilization\nDescribe the process of fertilization in flowering plants\nDescribe and explain how embryo and seeds are formed in flowering plants\nDescribe how fruits are formed in flowering plants\nDifferentiate between a fruit and a seed\nDescribe and explain how different seeds and fruits are dispersed\nClassifying various types of fruits and describe their placentation\nDifferentiate between internal and external fertilization\nDescribe external fertilization in amphibians\nRelate the structure of mammalian male reproductive system to its functions\nRelate the structure of mammalian male reproductive organ and spermatozoa to its function\nRelate the structure of mammalian female reproductive system to its function\nRelate the structure of mammalian ovum to its function\nDescribe internal fertilization in mammals\nDescribe the fertilization process\nDescribe implantation and the role of the placenta in mammals\nDefine gestation in mammals\nIdentify different gestation periods in different mammals\nDescribe birth and explain parental care\nDescribe the role of hormones in reproduction of humans\nDescribe the role of hormones in the menstrual cycle\nIdentify symptoms and explain the methods of transmission and prevention of gonorrhea and herpes simplex\nIdentify symptoms and explain the methods of transmission and prevention of syphilis and trichomoniasis\nIdentify symptoms and explain the methods of transmission and prevention of candidiasis and hepatitis\nIdentify the causes and modes of transmission of HIV/AIDS and prevention of HIV and AIDS\nIdentify effects of HIV/AIDS in human economy\nIdentify the symptoms of HIV/AIDS and stages of HIV and AIDS\nExplain ways of preventing and controlling the spread of HIV/AIDS\nDiscuss the social effects of HIV/AIDS\nExplain the advantages and disadvantages of sexual and asexual reproduction\nDefine the terms growth and development\nDescribe the sigmoid growth curve\nDescribe the phases of sigmoid curve\nDescribe the intermittent growth curve\nAnalyze data on growth rate\nDraw growth curves\nDefine seed dormancy\nIdentify factors affecting viability and dormancy of seeds\nIdentify factors affecting seed dormancy\nDefine seed germination\nDifferentiate between types of seed germination\nIdentifying Conditions necessary for germination - oxygen\nInvestigate the necessity of water and warmth\nDescribe the region of growth in seedlings\nIdentify the regions of growth\nDetermine the regions of growth in seedlings\nMeasure the aspect of growth in a given seedling\nDescribe growth in plants I.e. Primary and secondary growths\ninvestigate primary and secondary growth in a seedling\nExplain the role of hormones in regulation of growth and development in plants\nExplain Apical dominance in plants\nDefine metamorphosis\nDistinguish between complete and incomplete metamorphosis\nDescribe complete metamorphosis in housefly and anopheles mosquito\nDescribe incomplete metamorphosis in a cockroach\nDescribe and explain the Role of growth hormones in metamorphosis in insects\nObserve metamorphosis in some insects\nClassification II\nGeneral Principles of Classification\nClassification is the science that puts organisms into distinct groups to make their study easy and systematic.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9407632689100365, "ocr_used": false, "chunk_length": 10193, "token_count": 2010}}
{"text": "FORM THREE BIOLOGY\nBy the end of form three work, the learner should be able to:\nClassify common organisms into their main taxonomic units\nWrite scientific names of organisms correctly\nList the kingdoms of organisms\nDescribe the general characteristics of Kingdom monera\nDescribe the general characteristics of Kingdom protoctista\nObserve, draw and name parts of spirogyra, amoeba, paramecium and euglena\nDescribe the general characteristics of Kingdom fungi\nList down all the members of kingdom fungi\nDraw and name parts of bread mold (mucor), yeast and mushrooms\nDescribe the main characteristics of kingdom plantae\nDescribe the main characteristics of bryophyta\nIdentify examples of hyophyta\nObserve draw and name parts of liverworts and moss plants\nIdentify examples of pleridophyta\nObserve draw and name parts of fern plant\nIdentify examples of division spermatophyta\nIdentify major sub-division of spermatophyta\nList main characteristics of angiospermae\nDifferentiate between angiospermae and gymnospermae\nState the characteristics of angiospermapyta\nIdentify and state major characteristics of classes of angiospermapytaegdicotyledonae&monocotyledonoe\ndescribe the general characteristics of kingdom animalia\ndescribe the general characteristics of Phylum arthropoda\nlist down the classes of the Phylum arthropoda\ndescribe the general characteristics of Class crustacean\ndescribe the general characteristics of Class insect\ndescribe the general characteristics of Class arachnida\nlist down the members of class arachnida and insect\nDescribe the general characteristics of Classeschilopoda and diplopoda\nList down the members of class chilopoda and diplopoda\nDescribe the general characteristics of Phylum chordate\ndescribe the general characteristics of Pisces and amphibian\ndescribe the general characteristics of reptilian\ndescribe the general characteristics of Class aves\nDescribe the general characteristics of Class Mammalia\nIdentify different types of members of Class Mammalia\nConstruct a simple dichotomous to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\nUse an already constructed dichotomous key to identify given organisms\ndraw and label organisms correctly\nDefine the term ecology and identify terms used in ecology\nDefine the term ecology and identify terms used in ecology\nIdentify the types of ecosystems\nState and explain how light determines distribution of organisms in an ecosystem\nIdentify and describe how temperature determines distribution of organisms in an ecosystem\nIdentify and describe how Rainfall and humidity determines distribution of organisms in an ecosystem\ndescribe how Wind and atmospheric pressure determines distribution of organisms in an ecosystem\nWrite down correct answers to questions asked in the test\ndescribe how salinity affects the distribution of organisms in aquatic ecosystems\ndescribe how waves, currents and tides affects the distribution of organisms in aquatic ecosystem\nDescribe how Edaphic factors affects the distribution of organisms in an ecosystem\nMeasure certain factors in samples of different soils\nDescribe how Geological factors affect the distribution of organisms in an ecosystem\nDescribe how Abiotic factors affect the distribution of organisms in an ecosystem\nDescribe how competition affects the distribution of organisms in an ecosystem\nDescribe how Predation and Symbiosis affects the distribution of organisms in an ecosystem\nDifferentiate between Parasitism and saprophytism\nDescribe how Parasitism and saprophytism influence the distribution of organisms in an ecosystem e.g. Tick and cattle\nDescribe the interaction between organisms in an ecosystem\nDescribe the role of decomposers in Nitrogen cycle & carbon cycle\nDefine the terms food chain and food web\nConstruct food chains and food webs\nDescribe energy flow in a local ecosystem and Construct food chains and food webs\nDefine population\nList down the characteristics of population\nExplain the use of quadrants and transects as methods of Population estimation\nExplain the capture –recapture method of population estimation\nUse quadrant method to estimate population of named organisms within the compound\nDescribe total count, aerial count and aerial photography and other methods of population estimation\nRelate to the adaptations of xerophytes to their habitats\nRelate to the adaptations of mesophytes to their habitats\nRelate to the adaptations of hydrophytes to their habitats\nObserve, draw and label parts of named hydrophytes, mesophytes and xerophyte plants\nRelate to the adaptations of halophytes to their habitats\nExplain pollution and give examples of pollutants\nDescribe the various air pollutants\nDiscuss the effects of air pollution on the environment\nSuggest methods of controlling air pollution\nDescribe various causes of Land/ soil pollution\nDiscuss the effects of Land/ soil pollution and human health in rural and urban centers\nSuggest methods of controlling Land/ soil pollution\nDescribe the causes of Water pollution\nIdentify other causes of environmental pollution in rural and urban centers\nDiscuss the effects of water pollution on human health in rural and urban centers and other organisms\nSuggest methods of controlling water pollution\nIdentify symptoms of cholera and typhoid fever\nState methods of transmission\nSuggest control measures\nIdentify the causes, symptoms and methods of transmission and control of malaria\nIdentify the causes, symptoms and methods of transmission of amoebic dysentery\nSuggest control methods of amoebic dysentery\nIdentify the causes, symptoms and methods of transmission of ascariosis\nIdentify the causes, symptoms and methods of transmission and control of schistomiasis\nDefine reproduction and state its importance\nDifferentiate between asexual and sexual reproduction\nDescribe the appearance and location of chromosomes\nDefine mitosis\nDescribe chromosomicmovement during mitosis\nDescribe e the movement of chromosomes in mitosis\nIdentify stages of mitosis\nIdentify and describe stages of mitosis\nState the significance of mitosis in reproduction\nDefine meiosis\nState the stages of meiosis\nDescribe the chromosome movement during meiosis\nObserve the stages of meiosis\nDescribe the movement of chromosomes during meiosis\nState the significance of meiosis in reproduction\nDifferentiate between mitosis and meiosis\nState and describe the importance of Binary fission\nObserve spore formation in bread mould (mucor) and binary fission in paramecium\nState and describing the importance of budding in reproduction\nObserving drawing and budding cells of yeast\nDescribe the external structure of a typical flower\nDescribe the internal structure of a typical flower\nObserve, describe and draw different types of pollen grains\nDescribe the structure of ovules\nDescribe other characteristics of flowers\nDescribe and compare adaptations of wind and insect pollinated flowers\nDescribe the features and mechanisms that hinder self-pollination and self-fertilization\nDescribe the process of fertilization in flowering plants\nDescribe and explain how embryo and seeds are formed in flowering plants\nDescribe how fruits are formed in flowering plants\nDifferentiate between a fruit and a seed\nDescribe and explain how different seeds and fruits are dispersed\nClassifying various types of fruits and describe their placentation\nDifferentiate between internal and external fertilization\nDescribe external fertilization in amphibians\nRelate the structure of mammalian male reproductive system to its functions\nRelate the structure of mammalian male reproductive organ and spermatozoa to its function\nRelate the structure of mammalian female reproductive system to its function\nRelate the structure of mammalian ovum to its function\nDescribe internal fertilization in mammals\nDescribe the fertilization process\nDescribe implantation and the role of the placenta in mammals\nDefine gestation in mammals\nIdentify different gestation periods in different mammals\nDescribe birth and explain parental care\nDescribe the role of hormones in reproduction of humans\nDescribe the role of hormones in the menstrual cycle\nIdentify symptoms and explain the methods of transmission and prevention of gonorrhea and herpes simplex\nIdentify symptoms and explain the methods of transmission and prevention of syphilis and trichomoniasis\nIdentify symptoms and explain the methods of transmission and prevention of candidiasis and hepatitis\nIdentify the causes and modes of transmission of HIV/AIDS and prevention of HIV and AIDS\nIdentify effects of HIV/AIDS in human economy\nIdentify the symptoms of HIV/AIDS and stages of HIV and AIDS\nExplain ways of preventing and controlling the spread of HIV/AIDS\nDiscuss the social effects of HIV/AIDS\nExplain the advantages and disadvantages of sexual and asexual reproduction\nDefine the terms growth and development\nDescribe the sigmoid growth curve\nDescribe the phases of sigmoid curve\nDescribe the intermittent growth curve\nAnalyze data on growth rate\nDraw growth curves\nDefine seed dormancy\nIdentify factors affecting viability and dormancy of seeds\nIdentify factors affecting seed dormancy\nDefine seed germination\nDifferentiate between types of seed germination\nIdentifying Conditions necessary for germination - oxygen\nInvestigate the necessity of water and warmth\nDescribe the region of growth in seedlings\nIdentify the regions of growth\nDetermine the regions of growth in seedlings\nMeasure the aspect of growth in a given seedling\nDescribe growth in plants I.e. Primary and secondary growths\ninvestigate primary and secondary growth in a seedling\nExplain the role of hormones in regulation of growth and development in plants\nExplain Apical dominance in plants\nDefine metamorphosis\nDistinguish between complete and incomplete metamorphosis\nDescribe complete metamorphosis in housefly and anopheles mosquito\nDescribe incomplete metamorphosis in a cockroach\nDescribe and explain the Role of growth hormones in metamorphosis in insects\nObserve metamorphosis in some insects\nClassification II\nGeneral Principles of Classification\nClassification is the science that puts organisms into distinct groups to make their study easy and systematic. Modern scientific classification is based on structure and functions.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9407580629445581, "ocr_used": false, "chunk_length": 10263, "token_count": 2020}}
{"text": "Tick and cattle\nDescribe the interaction between organisms in an ecosystem\nDescribe the role of decomposers in Nitrogen cycle & carbon cycle\nDefine the terms food chain and food web\nConstruct food chains and food webs\nDescribe energy flow in a local ecosystem and Construct food chains and food webs\nDefine population\nList down the characteristics of population\nExplain the use of quadrants and transects as methods of Population estimation\nExplain the capture –recapture method of population estimation\nUse quadrant method to estimate population of named organisms within the compound\nDescribe total count, aerial count and aerial photography and other methods of population estimation\nRelate to the adaptations of xerophytes to their habitats\nRelate to the adaptations of mesophytes to their habitats\nRelate to the adaptations of hydrophytes to their habitats\nObserve, draw and label parts of named hydrophytes, mesophytes and xerophyte plants\nRelate to the adaptations of halophytes to their habitats\nExplain pollution and give examples of pollutants\nDescribe the various air pollutants\nDiscuss the effects of air pollution on the environment\nSuggest methods of controlling air pollution\nDescribe various causes of Land/ soil pollution\nDiscuss the effects of Land/ soil pollution and human health in rural and urban centers\nSuggest methods of controlling Land/ soil pollution\nDescribe the causes of Water pollution\nIdentify other causes of environmental pollution in rural and urban centers\nDiscuss the effects of water pollution on human health in rural and urban centers and other organisms\nSuggest methods of controlling water pollution\nIdentify symptoms of cholera and typhoid fever\nState methods of transmission\nSuggest control measures\nIdentify the causes, symptoms and methods of transmission and control of malaria\nIdentify the causes, symptoms and methods of transmission of amoebic dysentery\nSuggest control methods of amoebic dysentery\nIdentify the causes, symptoms and methods of transmission of ascariosis\nIdentify the causes, symptoms and methods of transmission and control of schistomiasis\nDefine reproduction and state its importance\nDifferentiate between asexual and sexual reproduction\nDescribe the appearance and location of chromosomes\nDefine mitosis\nDescribe chromosomicmovement during mitosis\nDescribe e the movement of chromosomes in mitosis\nIdentify stages of mitosis\nIdentify and describe stages of mitosis\nState the significance of mitosis in reproduction\nDefine meiosis\nState the stages of meiosis\nDescribe the chromosome movement during meiosis\nObserve the stages of meiosis\nDescribe the movement of chromosomes during meiosis\nState the significance of meiosis in reproduction\nDifferentiate between mitosis and meiosis\nState and describe the importance of Binary fission\nObserve spore formation in bread mould (mucor) and binary fission in paramecium\nState and describing the importance of budding in reproduction\nObserving drawing and budding cells of yeast\nDescribe the external structure of a typical flower\nDescribe the internal structure of a typical flower\nObserve, describe and draw different types of pollen grains\nDescribe the structure of ovules\nDescribe other characteristics of flowers\nDescribe and compare adaptations of wind and insect pollinated flowers\nDescribe the features and mechanisms that hinder self-pollination and self-fertilization\nDescribe the process of fertilization in flowering plants\nDescribe and explain how embryo and seeds are formed in flowering plants\nDescribe how fruits are formed in flowering plants\nDifferentiate between a fruit and a seed\nDescribe and explain how different seeds and fruits are dispersed\nClassifying various types of fruits and describe their placentation\nDifferentiate between internal and external fertilization\nDescribe external fertilization in amphibians\nRelate the structure of mammalian male reproductive system to its functions\nRelate the structure of mammalian male reproductive organ and spermatozoa to its function\nRelate the structure of mammalian female reproductive system to its function\nRelate the structure of mammalian ovum to its function\nDescribe internal fertilization in mammals\nDescribe the fertilization process\nDescribe implantation and the role of the placenta in mammals\nDefine gestation in mammals\nIdentify different gestation periods in different mammals\nDescribe birth and explain parental care\nDescribe the role of hormones in reproduction of humans\nDescribe the role of hormones in the menstrual cycle\nIdentify symptoms and explain the methods of transmission and prevention of gonorrhea and herpes simplex\nIdentify symptoms and explain the methods of transmission and prevention of syphilis and trichomoniasis\nIdentify symptoms and explain the methods of transmission and prevention of candidiasis and hepatitis\nIdentify the causes and modes of transmission of HIV/AIDS and prevention of HIV and AIDS\nIdentify effects of HIV/AIDS in human economy\nIdentify the symptoms of HIV/AIDS and stages of HIV and AIDS\nExplain ways of preventing and controlling the spread of HIV/AIDS\nDiscuss the social effects of HIV/AIDS\nExplain the advantages and disadvantages of sexual and asexual reproduction\nDefine the terms growth and development\nDescribe the sigmoid growth curve\nDescribe the phases of sigmoid curve\nDescribe the intermittent growth curve\nAnalyze data on growth rate\nDraw growth curves\nDefine seed dormancy\nIdentify factors affecting viability and dormancy of seeds\nIdentify factors affecting seed dormancy\nDefine seed germination\nDifferentiate between types of seed germination\nIdentifying Conditions necessary for germination - oxygen\nInvestigate the necessity of water and warmth\nDescribe the region of growth in seedlings\nIdentify the regions of growth\nDetermine the regions of growth in seedlings\nMeasure the aspect of growth in a given seedling\nDescribe growth in plants I.e. Primary and secondary growths\ninvestigate primary and secondary growth in a seedling\nExplain the role of hormones in regulation of growth and development in plants\nExplain Apical dominance in plants\nDefine metamorphosis\nDistinguish between complete and incomplete metamorphosis\nDescribe complete metamorphosis in housefly and anopheles mosquito\nDescribe incomplete metamorphosis in a cockroach\nDescribe and explain the Role of growth hormones in metamorphosis in insects\nObserve metamorphosis in some insects\nClassification II\nGeneral Principles of Classification\nClassification is the science that puts organisms into distinct groups to make their study easy and systematic. Modern scientific classification is based on structure and functions. Organisms with similar anatomical and morphological characteristics are placed in one group while those with different structures are grouped separately.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9400440852314476, "ocr_used": false, "chunk_length": 6805, "token_count": 1319}}
{"text": "Primary and secondary growths\ninvestigate primary and secondary growth in a seedling\nExplain the role of hormones in regulation of growth and development in plants\nExplain Apical dominance in plants\nDefine metamorphosis\nDistinguish between complete and incomplete metamorphosis\nDescribe complete metamorphosis in housefly and anopheles mosquito\nDescribe incomplete metamorphosis in a cockroach\nDescribe and explain the Role of growth hormones in metamorphosis in insects\nObserve metamorphosis in some insects\nClassification II\nGeneral Principles of Classification\nClassification is the science that puts organisms into distinct groups to make their study easy and systematic. Modern scientific classification is based on structure and functions. Organisms with similar anatomical and morphological characteristics are placed in one group while those with different structures are grouped separately. Modern studies in genetics and cell biochemistry are used to give additional help in classifying organisms. There are seven major taxonomic groups. The kingdom is the largest group. Others are phylum (division for plants) class, order, family, genus and species, the smallest. Binomial Nomenclature\nLiving organisms are named using Latin or Latinised names. Every organism has two names. This double naming is called binomial nomenclature. This system of naming was devised by Carolus Linnaeus in the 18th Century. The first name is the generic name - the name of the genus. The second name is the name of the species. The generic name starts with a capital letter while that of the species starts with a small letter. The names are written in italics or are underlined in manuscripts. Examples:\nBean =Phaseolus vulgaris. Phaseolus is the generic name,\nvulgaris is specific name. Dog =Canis familiaris. Canis is the generic name\n,familiaris the specific name. General Characteristics of Kingdoms\nOrganisms are classified into five kingdoms. Monera,\nProtoctista,\nFungi,\nPlantae\nAnimalia. Viruses do not fit neatly into any of the above kingdoms. They are simple and not cellular. They are metabolically inactive outside the host cell. Most of them can be crystallised like chemical molecules. Therefore they do not exhibit the characteristics of living organisms.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9286838562700632, "ocr_used": false, "chunk_length": 2262, "token_count": 459}}
{"text": "They are metabolically inactive outside the host cell. Most of them can be crystallised like chemical molecules. Therefore they do not exhibit the characteristics of living organisms. Examples of Organisms in Each Kingdom and Their Economic Importance\nKingdom Monera\nGeneral Characteristics\nUnicellular and microscopic\nSome single cells ,others colonial\nNuclear material not enclosed within nuclear membrane-prokaryotic\nHave cell wall but not of cellulose. Have few organelles which are not membrane bound\nMitochondria absent\nMostly heterotrophic, feeding saprotrophically or parasitically,some are autotrophic. Reproduction mostly asexual through binary fission\nMost of them are anaerobes but others are aerobes\nMost move by flagella\nExamples include Escherichia coli, Vibrio cholerae and Clostridium tetani. Spherical known as Cocci. Rod shaped - e.g. Clostridium tetani\nSpiral shaped e.g. sprilla\nComa shaped- Vibrios -e.g., Vibrio cholerae. Economic importance of bacteria Benefits to man include:\nThey are used in food processing e.g., Lactobacillus used in processing of cheese, yoghurt. Involved in synthesis of vitamin Band K, in humans and breakdown of cellulose in herbivores. Genetic Engineering\nBacteria are easily cultured and are being used for making antibiotics, aminoacids and enzymes e.g. amylase, and invertase e.g., Escherichia coli. Nutrient cycling:\nSaprophytes\nThey are involved in decomposition of dead organic matter. They are useful in the nitrogen cycle. Nitrogen fixing and nitrifying bacteria. They increase soil fertility. Modem sewage works use bacteria in treatment of sewage. Cleaning oil spills in oceans and lakes. Harmful Effects\nBacteria cause disease:\nTo humans (e.g. Cholera). To animals (e.g. Anthrax). Bacteria cause food spoilage. Others cause food poisoning e.g. Salmonella. Denitrifying bacteria reduce soil fertility e.g., Pseudomonas denitrificans. Kingdom Protoctista\nExamples include ;\nAlgae such as spirogyra, Chlamydomonas, euglena, Sargassum\nAnd protozoa such as amoeba, paramecium and Trypanosoma.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9239629087359689, "ocr_used": false, "chunk_length": 2049, "token_count": 492}}
{"text": "Salmonella. Denitrifying bacteria reduce soil fertility e.g., Pseudomonas denitrificans. Kingdom Protoctista\nExamples include ;\nAlgae such as spirogyra, Chlamydomonas, euglena, Sargassum\nAnd protozoa such as amoeba, paramecium and Trypanosoma. General Characteristics\nThey are said to be eukaryotic since their nucleus is bound by a membrane\nMost are mobile, and use flagella, cilia and pseudopodia. Some are sessile. They reproduce mainly asexually, by binary fission, fragmentation and sporulation. Some reproduce sexually by conjugation. Some are heterotrophic e.g. paramecium. Others are autotrophic e.g. spirogyra. Economic importance of protoctista\nAlgae are the primary producers in aquatic food chains. They release a lot of oxygen to the atmosphere. Some cause human diseases like malaria and amoebic dysentry ,sleeping sickness\nSome are source of food for humans e.g. sargassum is a source of iodine\nSkeletons of diatoms used in paint making. Spirogyra: They have spiral chloroplast. They are green, thread-like filaments\nChlamydomonas:\nThis is a unicellular green algae and has a cup shaped chloroplast. They move towards light using the flagella\nCilia assist the organism to move. The shape is due to the presence of a thin flexible pellicle. Kingdom Fungi\nMulticellular fungi are made of thread-like structures called hyphae (singular hyphae) that form a mycelium. .e.g.Saccharomyces cereviseae(bread yeast). Others include Penicillium, Rhizopus, and edible mushroom\nEconomic Importance of Fungi\nBeneficial Effects\nSome fungi are used as food e.g. mushrooms. Some are decomposers which enhance decay to improve soil fertility - recycling of nutrients e.g., toadstools. Some are useful in brewing and bread making e.g., yeast. Yeast is used as food - a rich source of Vitamin B. Some are useful in production of antibiotics e.g., Penicillium griseofulvin. Used in sewage treatment e.g., Fusarium spp. Harmful Effects\nSome cause food poisoning by producing toxic compounds e.g.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9190140845070424, "ocr_used": false, "chunk_length": 1988, "token_count": 496}}
{"text": "Some are useful in production of antibiotics e.g., Penicillium griseofulvin. Used in sewage treatment e.g., Fusarium spp. Harmful Effects\nSome cause food poisoning by producing toxic compounds e.g. Aspergillus flavus which produces aflatoxins. Some cause food spoilage, fabric and wood spoilage through decomposition. Some cause diseases to humans e.g., athlete's foot and ringworms. Others cause diseases to plants e.g., potato blight (Irish potatoes) rust in tomatoes and smuts in cereals. Kingdom Plantae\nGeneral Characteristics\nThey are multicellular and eukaryotic. They are photosynthetic and have a pigment chlorophyll. Their cells have cellulose cell walls. They reproduce sexually, others asexually. Kingdom Plantae has three major divisions:\nBryophyta,\nPteridophyta\nSpermatophyta. Division Bryophyta\nThese include mosses and liverworts. Plant body is not differentiated into root, stem and leaves. They have simple structures which resemble leaves and stems. They have rhizoids for absorbing water and anchoring the plant to substratum. Life cycle consists of two morphologically different plants, the gametophyte and sporophyte. The two alternate. They show alternation of generations. The gamete producing gametophyte is the persistent plant. The sporophyte is attached to the gametophyte and is nutritionally dependent on it. They lack vascular system. Sexual reproduction is dependent on water. Division Pteridophyta:\nThese include ferns and horsetails. General Characteristics\nThey have root and shoot system. Leaves are compound known as fronds, they have a vascular system. They show alternation of generations whereby the spore bearing sporophyte is the main plant. Spores are borne in clusters on the underside of leaves making sari. The gametophyte is an independent minute structure called prothallus which is short lived. Sexual reproduction is dependent on water. Division Spermatophyta\nThese are the seed bearing plants. General Characteristics\nPlant body is differentiated into root, stem and leaves. Vascular tissue consists of xylem and phloem. Sexual reproduction is independent of water. Male gametophyte (pollen grain) germinates and grows to reach female gametophyte.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.926842584167425, "ocr_used": false, "chunk_length": 2198, "token_count": 494}}
{"text": "Vascular tissue consists of xylem and phloem. Sexual reproduction is independent of water. Male gametophyte (pollen grain) germinates and grows to reach female gametophyte. They are divided into two sub-divisions:\nGymnosperms\nAngiosperms. Gymnosperms\nThese are cone-bearing plants. Naked seeds. They are trees and shrubs. Xylem consists of tracheids only. Examples; pine, cypress and spruce. They show xerophytic characteristics like having needle-like leaves. Angiosperms\nSeeds are enclosed within a fruit. They comprise trees, shrubs and herbs. Xylem consists of vessels of tracheids. These are the most advanced plants. Angiosperms has two classes;\nMonocotyledonae\nDicotyledonae. Comparison of Dicotyledonae and Monocotyledonae\nEconomic Importance of Spermatophyta\nThey are a source of food for humans and other animals. Source of fue1- wood fuel and charcoal. Source of timber for building and for paper. Ornamental plants. Useful in textile industry. Kingdom Animalia\nMost animals move from place to place in search of food. Major phyla are:\nPlatyhelminthes (Tapeworm). Nematoda (Ascaris). Annelida (Earthworm). Mollusca (Snails). Arthropoda\nchordata\nPhylum Arthropoda\nDistinguishing Characteristics\nThey have jointed appendages, which are specialised for various functions. Their body is covered by a hardened exoskeleton made of chitin. It is shed at intervals to allow for growth. They have jointed body parts. Most are divided into head, thorax and abdomen. Some have two body parts,\nGeneral Characteristics\nBody is segmented. They have bilateral symmetry. Gaseous exchange is through tracheal system, book lungs or gills which opens to the outside through spiracles. Aquatic forms use gills. Reproduction is mainly sexual. They have an open circulatory system. Phylum Arthropoda divided into five classes;\nCrustacea,\nArachnida,\nChilopoda,\nDiplopoda\nInsecta\nThis division is based on:\nThe number of limbs. Presence and number of antennae. Number of body parts.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9207855564059829, "ocr_used": false, "chunk_length": 1969, "token_count": 489}}
{"text": "Phylum Arthropoda divided into five classes;\nCrustacea,\nArachnida,\nChilopoda,\nDiplopoda\nInsecta\nThis division is based on:\nThe number of limbs. Presence and number of antennae. Number of body parts. Class Crustacea\nMost of them are aquatic, a few are terrestrial found in moist places e.g., woodlouse. Distinguishing Characteristics\nTwo body parts head and thorax are fused to form cephalothorax and an abdomen . They have two pairs of antennae; one is small and branched, the other is long. They have five or more parts of limbs. Some of these are modified for other functions e.g., locomotion, feeding and defence. Exoskeleton hardened with deposits of calcium carbonate i.e. carapace. Other Characteristics\nMouthparts include a pair of mandibles and two pairs of maxillae. Gaseous exchange is through gills. They have a pair of compound eyes. Most crustaceans are free-living but a few are parasitic e.g., barnacles. Examples are cray-fish and crab. Class Arachnida\nMembers are carnivorous and paralyse prey using poison produced from poison claws. Distinguishing Characteristics\nThe body has two parts: cephalothorax and abdomen. Cephalothorax is head fused to thorax. A pair of chelicerae, on ventral side of cephalothorax. They have four pairs of walking legs. They have no antennae. Instead they have a pair of short pedipalps which are sensitive to touch. Most arachnids use book lungs for gaseous exchange. Other characteristics include simple eyes. Examples include garden spider, ticks, scorpions. Class Chilopoda\ne.g. Centipede\nDistinguishing Characteristics\nThe body has 2 body parts, a head and trunk. The body is elongate, and has 15 or more segments. Has a pair of legs on each segment. The body is dorso-ventrally flattened. Other characteristics include:\nHead has a pair of antennae. Gaseous exchange through tracheal system. Are carnivorous. Class Diplopoda e.g. Millipede\nDistinguishing Characteristics\nHas two parts: head, short thorax and a trunk . Body elongate with 9-100 segments. Has two pairs of legs on each segment.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9113869484347452, "ocr_used": false, "chunk_length": 2044, "token_count": 509}}
{"text": "Millipede\nDistinguishing Characteristics\nHas two parts: head, short thorax and a trunk . Body elongate with 9-100 segments. Has two pairs of legs on each segment. They have a cylindrical body. Gaseous exchange is by tracheal system. Other characteristics:\nHead has a pair of antennae. Are herbivorous. Class Insecta\nDistinguishing Characteristics\nBody is divided into three body parts head, thorax and abdomen. They have three pairs of legs .. Most insects have a pair or two of wings. Other characteristics include:\nA pair of antennae. They breathe through spiracles, and gaseous exchange is through tracheal system. The class is divided into several orders based on:\nMouth parts- - type e.g. biting or piercing. Position of mouthparts - ventral or anterior. Wings - presence or absence; number of wing types, structure, texture. Size of legs. Order Orthoptera\nHave biting and chewing mouthparts. Hind legs longer than other legs e.g. fore wings, leathery and longer than hind legs . e.g. locusts and grasshoppers . Swarming - locusts are a menace to farmers and the environment as they destroy crops and vegetation. Order Diptera –\nTrue flies e.g. houseflies, and mosquitoes have sucking and piercing mouthparts, 1 pair of wings. The second pair is vestigial- acts as balancer. Mouthparts are ventral. These are disease vectors e.g., female anopheles mosquito transmits malaria. Order Lepidoptera –\nButterflies and moths have sucking mouthparts,\nTwo pairs of wings covered by scales. This group is important to farmers in pollination. Order Hymenoptera –\nBees ,wasps, ants. They have sucking mouthparts, two pairs of wings which are membranous. Some are non-winged e.g. some ants. Bees are important in pollination i.e. in production of honey. Order Isoptera - Termites\nThey have biting mouthparts which are anterior. Most are wingless,\nThose with wings they are membranous and of the same size. They are important in nutrient cycling as they feed on cellulose. Order Coleoptera - Beetles\nHave biting mouthparts,\nTwo pairs of wings,\nFore wing hardened enclosing membranous wings.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.908550020467368, "ocr_used": false, "chunk_length": 2081, "token_count": 495}}
{"text": "Most are wingless,\nThose with wings they are membranous and of the same size. They are important in nutrient cycling as they feed on cellulose. Order Coleoptera - Beetles\nHave biting mouthparts,\nTwo pairs of wings,\nFore wing hardened enclosing membranous wings. Destruction of stored grains and legumes (pulses)\nPhylum Chordata\nThis name is derived from the term notochord. This is a long flexible rod-like structure. The more familiar chordates are known as vertebrates. In vertebrates the notochord exists only in embryonic stages of development which in later stages is replaced by a vertebral column. Main Characteristics of Vertebrates\nMembers of the phylum have a notochord in early stages of development. They have visceral clefts - which are slits perforating the body wall at the pharynx. In fish these slits become gills while in higher chordates these slits are only present in embryo. They have a dorsal, hollow nerve cord. It develops into a brain at the anterior and spinal cord at the posterior end. The spinal cord is enclosed within the vertebral column. They have segmented muscle blocks known as myotomes on either side of the body. They possess a post-anal tail although rudimentary in some. They have a closed circulatory system. The heart is ventrally located. They possess an internal skeleton. The main classes of phylum chordata are;\nPisces,\nAmphibia,\nReptilia,\nAves\nMammalia. Class Pisces\nThese are the fishes. Some fish have a skeleton made of cartilage e.g. the shark. Others like Tilapia have a bony skeleton. Distinguishing Characteristics\nThey are aquatic. Movement is by means of fins. They have a streamlined body. They have a lateral line for sensitivity. Their heart has two chambers, the auricle and ventricle - simple circulatory system. Other Characteristics\nTheir body temperature changes according to the temperature of the environment. They are ectothermic (poikilothermic). Body covered with scales. They have gills for gaseous exchange. Exhibit external fertilisation. Class Amphibia\nLarval forms are aquatic while adults are terrestrial. Adults return to water for breeding e.g. frogs, toads, newts, salamanders. Distinguishing Characteristics\nSkin is soft and without scales. They have four well developed limbs.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9219317678334072, "ocr_used": false, "chunk_length": 2257, "token_count": 503}}
{"text": "frogs, toads, newts, salamanders. Distinguishing Characteristics\nSkin is soft and without scales. They have four well developed limbs. The hind limbs are longer and more muscular than forelimbs. The limb can be used for walking, jumping and swimming\nGaseous exchange is through the skin, gills and lungs. Middle ear is present. Other Characteristics\nThey have a three-chambered heart with two atria and one ventricle. Fertilisation is external. They are ectothermic (poikilotherms). Class Reptilia\nExamples are snakes, crocodiles, lizards, chameleons, tortoises and turtles. Distinguishing Characteristics\nThe skin is dry and is covered by horny scales. Fertilisation is internal. Some species eggs contain a lot of yolk and have either leathery or calcareous shells. They have a double circulatory system. The heart has three chambers - two atria and a partly divided ventricle. However crocodiles have a four chamber heart. Other Characteristics\nThey are ectothermic (poikilothermic). Have 2 pairs of limbs. They use lungs for gaseous exchange. Class Aves\nThese are birds. They are terrestrial and arboreal and others are aquatic\ne.g. flamingo, goose, ostrich, penguin, hawk, dove. Distinguishing Characteristics\nBody is covered by feathers and legs with horny scales. They have two pairs of limbs. Fore limbs modified to form wings for flight. Hind limbs are for walking or swimming. The mouth is a protruding beak. They have hollow bones. They have double circulation with a four-chambered heart (2 atria, 2 ventricles). They have lungs for gaseous exchange. Lungs are connected to air sacs in bones. Fertilisation is internal. They lay eggs with calcareous brittle shell. They have constant body temperatures hence are homoiotherms (endothermic ). Class Mammalia\nThey are arboreal e.g. tree-squirrels,\nOthers terrestrial e.g. humans\nOthers are aquatic e.g. dolphins and whales. Distinguishing Characteristics\nThey have mammary glands hence name of the class. Body is covered with fur or hair. Their teeth are differentiated into four types (heterodont dentition). They have external ear-pinna. Most have sweat glands.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9139211784829129, "ocr_used": false, "chunk_length": 2122, "token_count": 501}}
{"text": "Their teeth are differentiated into four types (heterodont dentition). They have external ear-pinna. Most have sweat glands. They have a diaphragm that separates the body cavity into thoracic and abdominal. Other Characteristics\nInternal fertilisation - most give birth. They have a double circulatory system with a four-chambered heart. They are endothermic (homoiotherms) . Eg Duck-billed Platypus (egg-laying mammal)\nEg.Kangaroo (pouched mammal)\nThe young are born immature and are nourished in a pouch with milk from mammary glands. Placental Mammals\nThey give birth to fully developed young ones which are fed on milk from mammary glands. Some are aquatic. e.g. dolphins, whale,\nOthers are flying e.g, bat;\nMost are terrestrial e.g. rabbits, elephants, buffalo, giraffe, antelope, cow, human being. Placental mammals are divided into various orders:\nRodentia: e.g. rats, mice - have one pair 9f upper incisors. Insectivora: e.g. mole-they are like rodents:\nCarnivora: e.g. dog; lion - flesh eaters, they have long pointed canines. Cetacea: e.g. whales and dolphins Aquatic mammals. Forelimbs are flippers. Chiroptera: e.g. bats - Forelimbs form wings. Artiodactyla: e.g. antelopes, cattle - they are even toed with split hooves. Perissodactyla: e.g. horse, donkey - they are odd toed with hooves. Proboscidea: e.g. elephant - upper lip and nose elongated to form trunk. Lagomorpha: e.g. rabbit, hare - mammals with upper and lower incisors. Have larger hind legs than forelegs. Primata: e.g. gorilla, orang utang, chimpanzee, monkeys - some are arboreal, with hand and foot for grasping. Human - Homo sapiens - upright gait, opposable thumb hence use of tools. Construction and Use of Dichotomous Keys\nBiological keys are sets of statements that act as clues leading to the identification of an organism. By following the keys we can be able to place an organism in its group. The most common key is the dichotomous key.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9025889143536202, "ocr_used": false, "chunk_length": 1925, "token_count": 499}}
{"text": "Construction and Use of Dichotomous Keys\nBiological keys are sets of statements that act as clues leading to the identification of an organism. By following the keys we can be able to place an organism in its group. The most common key is the dichotomous key. This is a biological tool for identification of unknown organisms. The word dichotomous means branching into two. A single characteristic is considered at a time. Two contrasting statements are put forward to describe the characteristics in such a way as to separate the organisms. This continues until all the organisms have been identified. Rules Used to Construct a Dichotomous Key\nUse morphological characteristics as far as possible e.g. type of leaf - simple or compound. Select a single characteristic at a time and identify it by number. 1. Type of leaf. . Use identical forms of words for two contrasting statements e.g.:\nFlowers scented. Flowers not scented. Start with a major characteristic that divide the organisms into two large groups then proceed to lesser variations that would separate the organisms further into smaller groups. Use positive statements especially the first one. Avoid generalizations e.g. short plants. Be specific in your description e.g.:\nplants above 1m tall. plants below 1m tall. Some Common Features Used for Identification\nIn Plants\nLeaves\nType of leaf Leaf\n(a) Compound leaves. (b) Type of venation. Simple leaf\nTrifoliate\nPinnate\nType ofleaf margin. Type ofleaf arrangement on stem. The colour of leaf. The texture ofleaf; whether hairy or smooth. Shape of the leaf e.g. palmate. Stem\nType of stem - woody or herbaceous. Shape of stem - cylindrical or rectangular. Texture of stem smooth or spiny. Infloresence\nAre flowers terminal or lateral\nFor each flower:\nIs the flower regular or irregular? Number of floral parts for each whorl. Are floral parts free or fused? Roots\nType of root system- Taproot or fibrous? Function of the root. In Animals\nFeatures used to identify animals:\nType of mouthparts. Type of skeleton. Presence or absence of antennae. Body segmentation. Body covering: scales, fur, hair or feathers. Number of body parts. Locomotory structures: legs, wings and fins. Presence or absence of vertebral column. Presence and type of eves.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9127070119749212, "ocr_used": false, "chunk_length": 2257, "token_count": 496}}
{"text": "Locomotory structures: legs, wings and fins. Presence or absence of vertebral column. Presence and type of eves. Practical Activities\nTo examine Bryophyta\nA mature moss plant is obtained. The specimen is observed using a hand -lens. A labelled drawing showing structures is made: rhizoids, set a capsule, gametophyte, sporophyte .. To examine Pteridophyta\nA mature fern plant is obtained. It is observed using a hand lens. Sori can be seen on the lower side of fronds. A labelled drawing showing: frond, pinna, sorus, rhizome and adventitious roots. To examine Spermatophyta\nA mature twig of either cypress or pinus with cones is obtained. Observation of Male and female is made using a hand-lens. The naked seeds are noted. The leaves show xerophytic characteristics e.g. they are rolled, or needle-like. A mature bean plant with pods is obtained,\nObservation of the leaves, stem and roots is made. Leaves are compound, broad arid have network of veins. The Ieaf-has a leaf stalk. They have a tap root system. Floral parts are in five e.g. 5 petals. A bean seed has two cotyledons. A mature maize plant is obtained. Observation of the leaves, stems and roots is made. Leaves are simple, narrow and long with parallel veins .. The petiole is modified to form a leaf sheath. They have a-fibrous root system. Floral parts are in threes. A maize gram has one cotyledon,\nExamination of Arthropoda\nSpecimens of crayfish, millipede, centipede grasshopper and spider are obtained. Where specimens are not available photographs are used. External features of the specimens are observed. The differences in the following are noted:\nBody parts. Antennae. Other appendages. Eyes. Examination of Chordata\nThe following specimens are obtained:\nTilapia, frog, Lizard, bird and rabbit. Using observable features each specimen is placed into its class. Features used include:\nBody covering. Limbs. Type of teeth. END\nECOLOGY\nIntroduction\nEcology is the study of organisms and their environment. All organisms show interdependence on one another. Organisms are affected by their environment, and they in turn affect the environment.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9133354254273102, "ocr_used": false, "chunk_length": 2115, "token_count": 498}}
{"text": "END\nECOLOGY\nIntroduction\nEcology is the study of organisms and their environment. All organisms show interdependence on one another. Organisms are affected by their environment, and they in turn affect the environment. Green plants manufacture food by photosynthesis which other organisms obtain directly or indirectly. Growth of plants is mainly affected by environmental factors such as soil and climatic factors. On the other hand, organisms modify the environment through various activities. This interrelationship comprises the study of ecology. The study of ecology is important in several fields of study such as agriculture and environmental studies. Concepts and Terms Used in Ecology\nHabitat:\nThis is the place or \"home\" that an organism lives or is found,\ne.g., forest or grassland. Niche:\nA niche is the functional unit in the habitat. It includes not only the specific place in which an organism lives but also how the organism functions. To avoid or reduce competition, organisms are separated or segregated by their niches,\nfor example, different species of birds make their nest on one tree, some at tips of terminal branches, and others feed on leaves, some on flowers and yet others on fruits of the same tree, i.e., food niche. Yet others feed on same food, e.g., worms in the same place but at different times - time niche. Population:\nThe term population refers to the total number of individuals of a species living in a given area at a particular time. Density is the number of individuals of a population found in a unit area, i.e.,\nDispersion:\nThis is the distribution of individuals in the available space. Dispersion may be uniform as in maize plants in a plantation;\nrandom as in cactus plants in the savannah ecosystem or clumped together as in human population in cities. Community:\nThis is the term used to describe all the organisms living together in an area. During the development of an ecosystem, the species composition of a community changes progressively through stages. Finally a steady state is reached and this is described as the climax community. This development of an ecosystem is termed succession. Each stage in development of an ecosystem is a sere. Succession is primary when it starts with bare ground, and secondary when it starts in a previously inhabited area e.g. after clearing a forest. The Ecosystem:\nThe community and the abiotic or non-living environment together make up an ecosystem or ecological system.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9200324412003245, "ocr_used": false, "chunk_length": 2466, "token_count": 485}}
{"text": "Succession is primary when it starts with bare ground, and secondary when it starts in a previously inhabited area e.g. after clearing a forest. The Ecosystem:\nThe community and the abiotic or non-living environment together make up an ecosystem or ecological system. In this system energy flow is clearly defined from producers to consumers and nutrient cycling takes place in paths that links all the organisms and the non-living environment. Biomass:\nThis is the mass of all the organisms in a given area. Ideally, it is the dry mass that should be compared. Carrying capacity:\nThis is the maximum sustainable density in a given area e.g. the number of herbivores a given area can support without overgrazing. Factors in an Ecosystem\nAbiotic factors (environmental factors)\nTemperature\nIs the hotness or coldness of an area or habitat. It directly affects the distribution and productivity (yield) of populations and communities. Most organisms are found in areas where temperature is moderate. However, certain plants and animals have adaptations that enable them to live in areas where temperatures are in the extremes such as the hot deserts and the cold polar regions. Temperatures not only influence distribution of organisms but also determine the activities of animals. High temperature usually accelerates the rates of photosynthesis, transpiration, evaporation and the decomposition and recycling of organic matter in the ecosystem. Light –\nLight is required by green plants for photosynthesis. Light intensity, duration and quality affect organisms in one way or another. Atmospheric Pressure\nThe force per unit area of atmospheric air that is exerted on organisms at different altitudes. Growth of plants and activity of animals is affected by atmospheric pressure\ne.g., rate of transpiration in plants and breathing in animals. Salinity\nThis is the salt content of soil or water. Animals and plants living in saline conditions have special adaptations. Humidity\nThis describes the amount of moisture (water vapour) in the air. Humidity affects the rate of transpiration in plants and evaporation in animals. pH\nIs the measure of acidity or alkalinity of soil solution or water. pH is very important to organisms living in water and soil. Most prefer a neutral pH. Wind:\nIs moving air currents and it influences the dispersion of certain plants by effecting the dispersal of spores, seeds and fruits. Air currents also modify the temperature and humidity of the surroundings. Topography:\nThese are surface features of a place.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9245669291338583, "ocr_used": false, "chunk_length": 2540, "token_count": 496}}
{"text": "Wind:\nIs moving air currents and it influences the dispersion of certain plants by effecting the dispersal of spores, seeds and fruits. Air currents also modify the temperature and humidity of the surroundings. Topography:\nThese are surface features of a place. The topographical factors considered include altitudes, gradient (slope), depressions and hills. All these characteristics affect the distribution of organisms in an area\ne.g., the leeward and windward sides of a hill. Biotic factors:\nThese are the living components in an ecosystem,\ncompetition\npredation,\nsymbiosis,\nparasitism,\nhuman activities. Inter-relationships Between Organisms\nThe relationships between organisms in a given ecosystem is primarily a feeding one. Organisms in a particular habitat have different feeding levels referred to as trophic levels. There are two main trophic levels:\nProducers:\nThese organisms that occupy the first trophic level. They manufacture their own food hence are autotrophic. Consumers:\nThese are the organisms that feed on organic substances manufactured by green plants. They occupy different trophic levels as follows:\nPrimary consumers:\nThese are herbivores and feed on green plants. Secondary consumers:\nThese are carnivores and feed on flesh. First order carnivores feed on herbivores while second order carnivores feed on other carnivores, i.e., tertiary consumers. Omnivores:\nThese are animals that feed on both plant and animal material. They can be primary, secondary or tertiary consumers. Competition:\nThis describes the situation where two or more organisms in the same habitat require or depend on the same resources. Organisms in an ecosystem compete for resources like food, space, light, water and mineral nutrients. Competition takes place when the environmental resource is not adequate for all. Intraspecific competition. This is competition between organisms of the same species. For example, maize plants in a field compete for water and nutrients among themselves. Interspecific competition. This refers to competition between organisms of different species, e.g., different species of predators can compete for water and prey among themselves. Predation\nIt is a relationship whereby one animal (the predator) feeds on another (the prey). Saprophytism\nSaprophytism is the mode of nutrition common in certain species of fungi and bacteria. Such organisms feed on dead organic material and release nutrients through the process of decomposition or decay. Saprophytes produce enzymes, which digest the substrates externally. The simpler substances are then absorbed.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9256944444444445, "ocr_used": false, "chunk_length": 2592, "token_count": 503}}
{"text": "Such organisms feed on dead organic material and release nutrients through the process of decomposition or decay. Saprophytes produce enzymes, which digest the substrates externally. The simpler substances are then absorbed. Saprophytes help in reducing the accumulation of dead bodies of plants and animals. Harmful saprophytes cause rapid decay of foods such as fruits, vegetables, milk and meat. Others damage buildings by causing wood rot. Some fungi produce poisonous substances called aflatoxins. These substances are associated with cereal crops which are stored under warm, moist conditions. If the infected grain is eaten, it may cause serious illness, and death. Parasitism\nThis is an association between members of different species. The parasite lives on or in the body of another organism, the host. The parasite derives benefits such as food and shelter from the host but the heist suffers harm as a result. Symbiosis\nThis is an association in which organisms of different species derive mutual benefit from one another. Some symbiotic associations are loose and the two partners gain very little from each other. Other symbiotic associations are more intimate and the organisms show a high degree of interdependence. Nitrogen cycle –\nIs the interdependence of organisms on one another and the physical environment as nitrogen is traced from and back into the atmosphere\nAlthough nitrogen is abundant in the atmosphere, most organisms are not able to utilise it directly. Some bacteria are capable of converting atmospheric nitrogen into forms which can be used by other living organisms. These bacteria are referred to as nitrogen fIxing bacteria. Symbiotic nitrogen fixing bacteria live in the root nodules of leguminous plants such as beans and peas. Non-symbiotic nitrogen fixing bacteria live in the soil. Nitrifying' bacteria convert ammonia into nitrites and nitrates. Denitrifying bacteria convert nitrates into atmospheric nitrogen. Energy Flow in an Ecosystem\nMost of the energy used in an ecosystem is derived from the sun. Solar energy is trapped by photosynthetic plants. It flows through different trophic levels . At each level energy is lost as heat to space and also through respiration. Besides animals lose energy through excretion and defecation. The amount of energy passed on as food from one trophic level to another decreases progressively. The energy in the organisms is recycled back to plants through the various nutrient or material cycles. Food Chains\nA food chain is a linear relationship between producers and consumers.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.928654970760234, "ocr_used": false, "chunk_length": 2565, "token_count": 492}}
{"text": "The amount of energy passed on as food from one trophic level to another decreases progressively. The energy in the organisms is recycled back to plants through the various nutrient or material cycles. Food Chains\nA food chain is a linear relationship between producers and consumers. It represents the transfer of food energy from green plants through repeated stages of eating and being eaten. Types of Food Chain\nGrazing food chain - starts with green plants. Detritus food chain - starts with dead organic material (debris or detritus). Detritivores:\nDetritivores feed on organic wastes and dead matter derived from the grazing food chain. Many different types of organisms feed on detritus. They include fungi, protozoa, insects, mites annelids and nematodes. Examples of Food Chains\nGreen plants~ aphids ~ lady-bird beetle\nGreen plants ~antelope -lion\nAlgae ~Tilapia ~ kingfisher\nPlant debris ~bacteria -eprotozoa ~ mosquito larva\nPhytoplankron-eZooplankton ~ Tilapia\n~ Nile perch ~ Human\nFood Web\nIn a natural community, several food chains are interlinked to form a food web. Several herbivores may feed on one plant . Similarly, a given herbivore may feed on different plants and may in turn be eaten by different carnivores. Decomposers\nThese are mainly bacteria and fungi. These organisms feed on dead organic matter thereby causing decomposition and decay and releasing nutrients for plants. They form a link between the biotic and the abiotic components. Pyramid of Numbers\nRefers to the number of organisms in each trophic level presented in a graphic form and a pyramid shape is obtained. The length of each bar is drawn proportional to the number of organisms represented at that level. This is because a herbivore feeds on many green plants. One carnivore also feeds on many herbivores. In a forest the shape of the pyramid is not perfect. This is because very many small animals such as insects, rodents and birds feed on one tree. Pyramid of Biomass\nThis is the mass of the producers and consumers at each trophic level drawn graphically. Population Estimation Methods\nIt is important to find or estimate the sizes of the different populations in a habitat. Direct counting or head count which involves the counting of every individual, is not always applicable for all organisms .", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9222608695652175, "ocr_used": false, "chunk_length": 2300, "token_count": 497}}
{"text": "Pyramid of Biomass\nThis is the mass of the producers and consumers at each trophic level drawn graphically. Population Estimation Methods\nIt is important to find or estimate the sizes of the different populations in a habitat. Direct counting or head count which involves the counting of every individual, is not always applicable for all organisms . e.g., it is impossible to count directly the numbers of grasshoppers in an area. Different sampling methods are thus used. A sample acts as a representative of the whole population. . Sampling Methods\nQuadrat Method\nA Quadrat is a square, made of woos metal/hard plastic. It can also be established on the ground using pegs, rope/permanent coloured ink, using metre rule or measuring tape. The size is usually one square metre (1M2), in grassland. In wooded or forest habitat it is usually larger, and can reach upto 20 m2 depending on particular species under investigation. The number of each species found within the quadrat is counted and recorded. Total number of organisms is then calculated by, finding the average quadrats and multiplying it with the total area of the whole habitat. The number of quadrats and their positions is determined by the type of vegetation studied. In a grassland, the quadrat frame can be thrown at random. In other habitats of forest, random numbers that determine the locus at which to establish a quadrat are used. Line Transect\nA line transect is a string or rope that is stretched along across the area in which all the plants that are touched are counted. It is tied on to a pole or tent peg. It is particularly useful where there is change of populations traversing through grassland, to woodland to forest land. This method can also be used in studying the changes in growth patterns in plants over a period of time. Belt Transect\nTwo line transects are set parallel to each other to enclose a strip through the habitat to be studied. The width is determined by the type of habitat, i.e., grass or forest and by the nature of investigation. In grassland it can be 0.5 m or 1 m. Sometimes it can be 20 metres or more especially when counting large herbivores. The number of organisms within the belt is counted and recorded. Capture-recapture method\nThis is used for animals such as fish, rodents, arthropods and birds. The animals are caught, marked, counted and released.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.909656055070951, "ocr_used": false, "chunk_length": 2367, "token_count": 503}}
{"text": "The number of organisms within the belt is counted and recorded. Capture-recapture method\nThis is used for animals such as fish, rodents, arthropods and birds. The animals are caught, marked, counted and released. For example, grasshoppers can be caught with a net and marked using permanent ink. After sometime, the same area is sampled again, i.e., the grasshoppers are caught again. The total number caught during the second catch is recorded. The number of marked ones is also recorded:\nLet the number caught and marked be a. The total number in the second catch be b. The number of marked ones in the second catch be c. The total number of grasshoppers in the area be T. The total number T can be estimated using the following formula:\nTotal Number =\nThe following assumptions are made:\nNo migration, i.e., no movement in and out of the study area. There is even distribution of the organisms in the study area. There is random distribution of the organisms after the first capture. No births or deaths during the activity. After the estimation, the results can be used to show anyone of the following population characteristics:\nDensity:\nDensity is calculated by dividing the number of organisms by the size of the area studied. Frequency:\nFrequency is the number of times that a species occurs in the area being studied. Percentage Cover:\nThis is the proportion of the area covered by a particular species. For example, a given plant species may cover the whole. of a given area. In this case the plant is said to have 100% cover. Dominance:\nThis is the term used to describe a species that exerts the most effect on others. The dominance may be in terms of high frequency or high density. Adaptations of Plants to Various Habitats\nOrganisms have developed structural features that enable them to live successfully in their particular habitats. Plants found beneath the canopies of trees are adapted to low light intensities by having broad leaves. Xerophytes\nThese are plants that grow in dry habitats,\ni.e., in deserts and semi-deserts. They have adaptations to reduce the rate of transpiration in order to save on water consumption. Others have water storage structures. Adaptations include:\nReduction of leaf surface area by having needle-like leaves, rolling up of leaves and shedding of leaves during drought to reduce water loss or transpiration.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.915861900513997, "ocr_used": false, "chunk_length": 2360, "token_count": 483}}
{"text": "They have adaptations to reduce the rate of transpiration in order to save on water consumption. Others have water storage structures. Adaptations include:\nReduction of leaf surface area by having needle-like leaves, rolling up of leaves and shedding of leaves during drought to reduce water loss or transpiration. Thick cuticle; epidermis consisting of several layers of cells;\nleaves covered with wax or resin to reduce evaporation. Sunken stomata, creating spaces with humid still air to reduce water holes. Few, small stomata, on lower epidermis to reduce water loss. Stomata open at night (reversed stomatal rhythm) to reduce water loss . Deep and extensive root systems for absorption of water. Development of flattened shoots and succulent tissue for water storage e.g. Opuntia. Mesophytes\nThese are the ordinary land plants which grow in well-watered habitats. They have no special adaptations. Stomata are found on both upper and lower leaf surfaces for efficient gaseous exchange and transpiration. However, those found in constantly wet places e.g. tropical rain forests, have features that increase transpiration. These plants are called hygrophytes. The leaves are broad to increase surface areas for transpiration and thin to ensure short distance for carbon (IV) oxide to reach photosynthetic cells and for light penetration. The stomata are raised above the epidermis to increase the rate of transpiration. They have grandular hairs or byhathodes that expel water into the saturated atmosphere. This phenomenon is called guttation. Hydrophytes (Water plants)\nWater plants are either submerged, emergent or floating. Submerged Plants\nThe leaves have an epidermis with very thin walls and a delicate cuticle. They have no stomata. Water is excreted from special glands and pores at the tips. Other adaptations include the following:\nPresence of large air spaces and canals (aerenchyma) for gaseous exchange and buoyancy. Some plants have filamentous leaves In order to increase the surface area for absorption of light, gases and mineral salts. Some plants are rootless, hence support provided by water. Mineral salts and water absorbed by all plant surfaces. In some plants, the stem and leaves are covered with a waxy substance to reduce absorption of water. e.g. Ceratophyllum and Elodea sp. Floating Plants\nTheir structure is similar to that of mesophytes.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9239258635214828, "ocr_used": false, "chunk_length": 2374, "token_count": 503}}
{"text": "e.g. Ceratophyllum and Elodea sp. Floating Plants\nTheir structure is similar to that of mesophytes. The leaves are broad to increase the surface area for water loss. They have more stomata on the upper surface than on the lower surface to increase rate of water loss. Examples are Pistia sp. (water lettuce), Salvinia and Nymphea. Halophytes (Salt plants)\nThese are plants that grow in salt marshes and on coastlines. They have root cells that concentrate salts and enable them to take in water by osmosis. They have salt glands which excrete salts. Fruits have large aerenchymatous tissues for air storage that makes them float. Some have shiny leaves to reduce water loss. The mangrove plants have roots that spread horizontally, and send some branches into the air. These aerial roots are known as breathing roots or pneumatophores. They have lenticel-Iike openings called pneumatothodes through which gaseous exchange takes place. Pollution\nEffect of Pollution on Human Beings and other Organisms\nPollution\nThis is the introduction of foreign material, poisonous compounds and excess nutrients or energy to the environment in harmful proportions. Any such substance is called a pollutant. Effects and Control of causes of Pollutants in Air, Water and Soil\nIndustrialisation and urbanisation are the main causes of pollution. As human beings exploit natural resources the delicate balance in the biosphere gets disturbed. The disturbance leads to the creation of conditions that are un-favourable to humans and other organisms. Sources of Pollutants\nMotor vehicles release carbon (II) oxide, sulphur (IV) oxide, and nitrogen oxides and hydrocarbons. Agricultural chemicals, fertilisers and pesticides. Factories, manufacturing and metal processing industries. They release toxic substances and gases as well as synthetic compounds that are bio-undegradable. They release solid particles or droplets of poisonous substances e.g. arsenic, beryllium, lead and cadmium. Radioactive waste: Leakages from nuclear power stations and testing sites release radioactive elements like strontium-90 which can eventually reach man through the food chain. Domestic waste and sewage are released raw into water bodies. Oil spills from accidents in the seas and leakage of oil tankers as well as from offshore drilling and storage and processing. Water Pollution. In most cases, chex,pical wastes from industries are discharged into water.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9243958642113848, "ocr_used": false, "chunk_length": 2426, "token_count": 505}}
{"text": "Oil spills from accidents in the seas and leakage of oil tankers as well as from offshore drilling and storage and processing. Water Pollution. In most cases, chex,pical wastes from industries are discharged into water. Toxic chemicals such as mercury compounds may be ingested by organisms. Insecticides like DDT, and weedkillers eventually get into the water and contaminate it. Oil and detergents also pollute water. Excess nitrates and phosphates from sewage and fertilisers cause overgrowth of algae and bacteria in water. This is called eutrophication. As a result there is insufficient oxygen which causes the deaths of animals in the water. Air pollution:\nSmoke from industries and motor vehicles contains poisonous chemicals like carbon (II) oxide, carbon (IV) oxide, sulphur (IV) oxide and oxides of nitrogen. When sulphur (IV) oxide and oxides of nitrogen dissolve in rain, they fall as acid rain. Accumulation of carbon (IV) oxide in the atmosphere causes the infrared light to be confined within the atmosphere, the earth's temperature rises. This is called the greenhouse effect. Carbon particles in smoke coat the leaves of plants and hinder gaseous exchange and photosynthesis. The particles also form smog in the air. Lead compounds are from vehicle exhaust pipes. All these have negative effects on man and the environment. Soil/Land pollution:\nPlastics and other man-made materials are biologically non-degradable i.e they are not acted upon by micro-organisms. Scrap metal and slag from mines also pollute land. Failure to rehabilitate mines and quarries also pollute land. Effects of Pollutants to Humans and other organisms\nChemical pollutants e.g. nitrogen oxides, fluorides, mercury and lead cause physiological and metabolic disorders to humans and domestic animals. Some hydrocarbons as well as radioactive pollutants acts as mutagens (cause mutations) and carcinogens induce cancer. Radioactive pollutants like strontium, caesium and lithium are absorbed into body surface and cause harm to bone marrow and the thyroid gland. Communicable diseases like cholera are spread through water polluted with sewage. Thermal pollution result in death of some fish due to decreased oxygen in the water. Oil spills disrupt normal functioning of coastal ecosystems. Birds that eat fish die due to inability to fly as feathers get covered by oil. Molluscs and crustaceans on rocky shores also die.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9240149315636667, "ocr_used": false, "chunk_length": 2411, "token_count": 491}}
{"text": "Oil spills disrupt normal functioning of coastal ecosystems. Birds that eat fish die due to inability to fly as feathers get covered by oil. Molluscs and crustaceans on rocky shores also die. Control of Air Pollution\nUse of lead-free petrol and low sulphur diesel in vehicles. Use of smokeless fuels e.g electricity or solar. Filtration of waste gases to remove harmful gases. Liquid dissolution of waste gases. In Kenya, factories are subjected to thorough audits to ensure that they do not pollute the environment. Factories should be erected far away from residential areas. Reduce volume or intensity of sound. Use of ear muffs. Vehicle exhaust systems should be fitted with catalytic oxidisers. Regular servicing of vehicles to ensure complete combustion of fuel. Water Pollution\nTreatment of sewage. Treatment of industrial waste before discharge into water. Use of controlled amounts of agrochemicals. Organic farming and biological control. Avoid spillage of oils and other chemicals into water. Good water management. Stiff penalties for oil spillage. Use of Pseudomonas bacteria that naturally feed on oil and break it up. Soil Pollution\nAddition of lime to farms to counteract the effect of agrochemicals. Recycling of solid waste. Compacting and incineration of solid waste. Use of biodegradable materials and chemicals. Good soil management to avoid soil erosion. Human Diseases\nThe term disease denotes any condition or disorder that disrupts the steady state of well being of the body. Health is a state of physical, mental and emotional well being in the internal environment of the body. Some of the causes of diseases are due to entry of pathogens and parasites. Pathogens include bacteria, viruses, protozoa and fungi. Parasites are organisms which live on or in the body of another organisms. Vectors are animals that carry the pathogen from are person to another. Most are ectoparasites that transmit the disease as they feed. Bacterial Diseases\nCholera\nCausative agent a bacterium Vibrio cholerae. Transmission - It is spread through water and food contaminated by human faeces containing the bacteria. The bacteria produce a powerful toxin, enterotoxin, that causes inflammation of the wall of the intestine leading to:\nSevere diarrhoea that leads to excessive water loss from body. Abdominal pain\nVomiting\nDehydration which may lead to death.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9258664412510567, "ocr_used": false, "chunk_length": 2366, "token_count": 487}}
{"text": "Transmission - It is spread through water and food contaminated by human faeces containing the bacteria. The bacteria produce a powerful toxin, enterotoxin, that causes inflammation of the wall of the intestine leading to:\nSevere diarrhoea that leads to excessive water loss from body. Abdominal pain\nVomiting\nDehydration which may lead to death. Prevention and Control\nAdequate sanitation such as water purification sewage treatment and proper disposal of human faeces. Public and personal hygiene e.g washing hands before meals and washing fruits and vegetables, boiling drinking water. Vaccination\nCarriers should be identified, isolated and treated during outbreaks. Treatment\nUse of appropriate antibiotics. Correcting fluid loss by injecting fluids or by administration of oral rehydration solutions. Typhoid\nCausative agent. The disease is caused by Salmonella typhi. Transmission is through contaminated water and food. It is also transmitted by certain 'e.g foods, e.g. oysters, mussels and shell fish. Symptoms\nFever\nMuscle pains\nHeadache\nSpots on the trunk of the body\nDiarrhoea\nIn severe cases mental confusion may result and death. Prevention\nBoil drinking water. Proper sewage treatnient. Proper disposal of faeces, if not flushed use deep pit latrines. Observe personal hygiene e.g. washing hands before meals. Washing fruits and vegetables. Treatment\nUse of appropriate antibiotics. Protozoa\nMalaria\nMalaria is caused by the protozoan plasmodium. The most common species of plasmodium are P. falciparum, P. vivax, P. rnalariae and P. ovale with varying degree of severity. Transmission\nIs by female anopheles mosquito as it gets a blood meal. Symptoms\nHeadache, sweating, shivering, high temperature (40-41 0C) chills and joint pains. The abdomen becomes tender due to destruction of red blood cells by the parasites . Prevention\nDestroy breeding grounds for mosquitoes by clearing bushes and draining stagnant water. Kill mosquito larvae by spraying water surfaces with oil. Use insecticides to kill adult mosquitoes\nSleeping under a mosquito net. Take preventive drugs. Treatment\nUse appropriate anti-malarial drugs. Amoebic dysentry (Amoebiasis)\nCause\nThis disease is caused by Entamoeba histolytica.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9207218149243233, "ocr_used": false, "chunk_length": 2219, "token_count": 492}}
{"text": "Take preventive drugs. Treatment\nUse appropriate anti-malarial drugs. Amoebic dysentry (Amoebiasis)\nCause\nThis disease is caused by Entamoeba histolytica. The parasites live in the intestinal tract but may occasionally spread to the liver. Transmission - They are transmitted through contaminated water and food especially salads. Symptoms –\nAbdominal pain, nausea and diarrhoea. The parasites cause ulceration of the intestinal tract, which results in diarrhoea. Prevention and control\nProper disposal of human faeces. Boiling water before drinking. Personal hygiene e.g. washing hands before meals. Washing vegetables and steaming particularly salads and fruits before eating. Treatment\nTreatment of infected people with appropriate drugs. Parasitic Diseases\nAscaris lumbricoides\nAscaris lumbricoides lives in the intestines of a man or pig, feeding on the digested food of the host. The body of the worm is tapered at both ends. The female is longer than the male. Mode of transmission\nThe host eats food contaminated with the eggs, the embryo worms hatch out in the intestine. The embryo worms then bore into the blood vessels of the intestine. They are carried in the bloodstream to the heart and then into the lungs. As they travel through the bloodstream, they grow in size. After sometime, the worms are coughed out from the air passages and into the oesophagus. They are then swallowed, eventually finding their way into the intestines where they grow into mature worms. Effects of Ascaris lumbricoides on the host\nThe parasites feed on the host's digested food. This results in malnutrition especially in children. If the worms are too many, they may block the intestine and interfere with digestion. The worms sometimes wander along the alimentary canal and may pass through the nose or mouth. In this way, they interfere with breathing and may cause serious illness. The larvae may cause severe internal bleeding as they penetrate the wall of the intestine. Adaptive Characteristics\nThe female lays as many as 25 million eggs. This ensures the continuation of the species. Eggs are covered by a protective cuticle that prevents them from dehydration. The adult worms tolerate low oxygen concentration. Have mouth parts for sucking food and other fluids in the intestines. Has a thick cuticle or pellicle to protect it from digestive enzymes produced by the host. Control and Prevention\nPersonal hygiene e.g.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9232163569467006, "ocr_used": false, "chunk_length": 2419, "token_count": 501}}
{"text": "Have mouth parts for sucking food and other fluids in the intestines. Has a thick cuticle or pellicle to protect it from digestive enzymes produced by the host. Control and Prevention\nPersonal hygiene e.g. washing hands before eating. Proper disposal of faeces. Washing of fruits and vegetables. Treatment\nDeworm using appropriate drugs ant-helmintics. Schistosoma\nSchistosoma or bilharzia worm is a flat worm, parasitic on human beings and fresh water snails. (Biomphalaria and Bulinus.)\nThe snail act as intermediate host. Mode of Transmission\nSchistosomiasis also known as a bilharsiasis is caused by several species of the genus schistosoma. Schistosoma haematobium infects the urinary system mainly the bladder\nS. japonicum and S. mansoni both infect the intestines. Schistosoma haemotobium is common in East Africa where irrigation is practised and where slow moving fresh water streams harbour snails. It is spread through contamination of water by faeces and urine from infected persons. The embryo (miracidium) that hatch in water penetrates into snails of the species Biompharahia and Bulinus. Inside the snail's body, the miracidium undergoes development and multiple fission to produce rediae. The rediae are released into the water and develop to form cercariae which infect human through:\nDrinking the water\nWading in water;\nBathing in snail-infested water. The cercaria burrows through the skin and enters blood vessel. Effects on the host\nInflammation of tissues where egg lodge. Ulceration where eggs calcify. Egg block small arteries in lungs leading to less aeration of blood. The body turns blue - a condition known as cyanosis. If eggs lodge in heart or brain, lesions formed can lead to death. Bleeding occurs as the worms burrow into blood vessels (faeces or urine has blood). Pain and difficulty in passing out urine. Nausea and vomiting. When eggs lodge in liver ulceration results in liver cirrhosis. Death eventually occurs. Adaptive Characteristics\nThe female has a thin body and fits into small blood vessels to lay eggs. Eggs are able to burrow out of blood vessel into intestine lumen. Many eggs are laid to ensure the survival of the parasite.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9239540229885058, "ocr_used": false, "chunk_length": 2175, "token_count": 496}}
{"text": "Adaptive Characteristics\nThe female has a thin body and fits into small blood vessels to lay eggs. Eggs are able to burrow out of blood vessel into intestine lumen. Many eggs are laid to ensure the survival of the parasite. Large numbers of cercariae are released by snail. The miracidia and cercariae larvae have glands that secrete lytic enzymes which soften the tissue to allow for penetration into host. The male has a gynecophoric canal that carries the female to ensure that eggs are fertilised before being shed. Has suckers for attachment. Prevention and Control\nDrain all stagnant water\nBoil drinking water. Do not wade bare feet in water. Wear long rubber boots and gloves (for those who work in rice fields). Eliminate snails, by spraying with molluscides. Reporting to doctor early when symptoms appear for early treatment. Practical Activities\nEcology is best studied outdoors. Students identify a habitat within or near the school compound, e.g. a flower bed. The quadrat method is used. Observation and recording of the various animals as well as their feeding habits is done. Birds that feed on the plants or arthropods in the area studied are noted through observation of habitat at various times of the day. Food chains are constructed e.g green plants ~ caterpillar ~ lizard and many others involving all organisms in the area. The numbers of animals in 1 m2 is counted directly or estimated e.g small arthropods like black ants. The number of plants is easily counted and recorded and ratio of consumers to producers calculated. It will be noted that in terms of numbers where invertebrates are involved, there are very many consumers of one plant. Several other quadrats are established and studied and averages calculated. Adaptions to Habitat\nHydrophytes\nSpecimen of hydrophytes e.g water lily is observed. Students should note the poorly developed root systems and broad leaves. Stomata distribution on leaf surface is studied through microscopy or by emersing a leaf in hot water and counting number of bubbles evolved. Mesophytes –\nOrdinary plants e.g bean hibiscus and zebrina can be studied. Size of leaves is noted and stomata distribution studied. Xerophytes\nSpecimen include Euphorbia, cactus and sisal which are easily available. The root system e.g in sisal is noted as shallow but extensive.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9210830581661164, "ocr_used": false, "chunk_length": 2325, "token_count": 496}}
{"text": "Size of leaves is noted and stomata distribution studied. Xerophytes\nSpecimen include Euphorbia, cactus and sisal which are easily available. The root system e.g in sisal is noted as shallow but extensive. It will be noted that sisal has fleshy leaves and stem while cactus and Euphorbia have fleshy stem but leaves are reduced to small hair-like structures. Comparison of Root nodules from fertile and poor soils\nRoot nodules –\nAre swellings on roots of leguminous plants. Soil fertility determines number of root nodules per plant. Bean plants are best used in this study. One plot can be manured while the other is not. Similar seeds are planted in the two plots. The plants are uprooted when fully mature (vegetatively) i.e any time after flowering and before drying. The number of nodules per plant is counted. An average for each plot is calculated. It is noted that the beans from fertile soil have more and large nodules than those grown in poor soils. Estimation of Population using Sampling Methods\nThe number of organisms both producers and the various consumers is recorded in each area studied e.g. using a quadrat. The total area of the habitat studied is measured. The average number of organisms per quadrat (1 m2) is calculated after establishing as many quadrats as are necessary to cover the area adequately. Total population of organisms is calculated from the area. Abiotic environment is studied within the area sampled. Air temperature soil surface temperature are taken and recorded. This is best done at different times of day, i.e., morning afternoon and evening. Any variations are noted. pH of the soil is measured using pH distilled water to make a solution. Litmus papers can be used to indicate if soil is acidic or alkaline, but pH paper or meter gives more precise pH values. Humidity is measured using anhydrous blue cobalt chloride paper which gives a mere indication of level of humidity. A windsock is used to give an indication of direction of wind. As all the abiotic factors are recorded observations are made to find the relationships between behaviour of organism and the environmental factors for example:\nThe temperature affects the behaviour of animals. The direction of wind will affect growth of plants. The level of humidity determines the type, number and distribution of organisms in an area. REPRODUCTION IN PLANTS AND ANIMALS\nIntroduction\nThe process by which mature individuals produce offspring is called reproduction.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.920940449909401, "ocr_used": false, "chunk_length": 2472, "token_count": 509}}
{"text": "The direction of wind will affect growth of plants. The level of humidity determines the type, number and distribution of organisms in an area. REPRODUCTION IN PLANTS AND ANIMALS\nIntroduction\nThe process by which mature individuals produce offspring is called reproduction. Reproduction is a characteristic of all living organisms and prevents extinction of a species. There are two types of reproduction: sexual and asexual reproduction. Sexual reproduction involves the fusion of male and female gametes to form a zygote. Asexual reproduction does not involve gametes. Cell Division\nCell division starts with division of nucleus. In the nucleus are a number of thread-like structures called chromosomes, which occur in pairs known as homologous chromosomes. Each chromosome contains-genes that determine the characteristics of an organism. The cells in each organism contains a specific number of chromosomes. There are two types of cell division:\nMitosis –\nThis takes place in all body cells of an organism to bring about increase in number of cells, resulting in growth and repair. The number of chromosomes in daughter cells remain the same as that in the mother cell. Meiosis –\nThis type of cell division takes place in reproductive organs (gonads) to produce gametes. The number of chromosomes in the gamete is half that in the mother cell. Mitosis\nMitosis is divided into four main stages. Prophase, Metaphase, Anaphase and Telophase. These stages of cell division occur in a smooth and continuous pattern. Interphase\nThe term interphase is used to describe the state of the nucleus when the cell is just about to divide. During this time the following take place:\nReplication of genetic material so that daughter cells will have the same number of chromosomes as the parent cell. Division of cell organelles such as mitochondria, ribosomes and centrioles. Energy for cell division is synthesised and stored in form of Adenosine Triphosphate (ATP) to drive the cell through the entire process. During. interphase, the following observations can be made:\nChromosomes are seen as long, thin, coiled thread-like structures. Nuclear membrane and nucleolus are intact. Prophase\nThe chromosomes shorten and thicken. Each chromosome is seen to consist of a pair of chromatids joined at a point called centromere. Centrioles (in animal cells) separate and move to opposite poles of the cell. The centre of the nucleus is referred to as the equator.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9241013071895425, "ocr_used": false, "chunk_length": 2448, "token_count": 503}}
{"text": "Each chromosome is seen to consist of a pair of chromatids joined at a point called centromere. Centrioles (in animal cells) separate and move to opposite poles of the cell. The centre of the nucleus is referred to as the equator. Spindle fibres begin to form, and connect the centriole pairs to the opposite poles. The nucleolus and nuclear membrane disintegrate and disappear. Metaphase\nSpindle fibres lengthen. In animal cells they attach to the centrioles at both poles. Each chromosome moves to the equatorial plane and is attached to the spindle fibres by the centromeres. Chromatids begin to separate at the centromere. Anaphase\nChromatids separate and migrate to the opposite poles due to the shortening of spindle fibres . Chromatids becomes a chromosome. In animal cell, the cell membrane starts to constrict. Telophase\nThe cell divides into two. In animal cells it occurs through cleavage of cell membrane. In plants cells, it is due to deposition of cellulose along the equator of the cell.(Cell plate formation). A nuclear membrane forms around each set of chromosome. Chromosomes later become less distinct. Significance of Mitosis\nIt brings about the growth of an organism:\nIt brings about asexual reproduction. Ensures that the chromosome number is retained. Ensures that the chromosomal constitution of the offspring is the same as the parents. Meiosis\nMeiosis involves two divisions of the parental cell resulting into four daughter cells. The mother cell has the diploid number of chromosomes. The four cells (gametes) have half the number of chromosomes (haploid) that the mother cell had. In the first meiotic division there is a reduction in the chromosome number because homologous chromosomes and not chromatids separate. Each division has four stages Prophase, Metaphase, Anaphase and Telophase. Interphase\nAs in mitosis the cell prepares for division. This involves replication of chromosomes, organelles and build up of energy to be used during the meiotic division. First Meiotic division\nProphase I\nHomologous chromosomes lie side by side in the process of synapsis forming pairs called bivalents. Chromosomes shorten and thicken hence become more visible. Chromosomes may become coiled around each other and the chromatids may remain in contact at points called chiasmata (singular chiasma).", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9255493321844034, "ocr_used": false, "chunk_length": 2321, "token_count": 503}}
{"text": "First Meiotic division\nProphase I\nHomologous chromosomes lie side by side in the process of synapsis forming pairs called bivalents. Chromosomes shorten and thicken hence become more visible. Chromosomes may become coiled around each other and the chromatids may remain in contact at points called chiasmata (singular chiasma). Chromatids cross-over at the chiasmata exchanging chromatid portions. Important genetic changes usually result. Metaphase I\nSpindle fibres are fully formed and attached to the centromeres. The bivalents move to the equator of the spindles. Anaphase I\nHomologous chromosomes separate and migrate to opposite poles. This is brought about by shortening of spindle fibres hence pulling the chromosomes. The number of chromosomes at each pole is half the number in the mother cell. Telophase I\nCytoplasm divides to separate the two daughter cells. Second Meiotic Division\nUsually the two daughter cells go into a short resting stage (interphase)\nbut sometimes the chromosomes remain condensed and the daughter cells go straight into metaphase of second meiotic division. The second meiotic division takes place just like mitosis. Prophase II\nEach chromosome is seen as a pair of chromatids. Metaphase II\nSpindle forms and are attached to the chromatids at the centromeres. Chromatids move to the equator. Anaphase II\nSister chromatids separate from each other\nThen move to opposite poles, pulled by the shortening of the spindle fibres. Telophase II\nThe spindle apparatus disappears. The nucleolus reappears and nuclear membrane is formed around each set of chromatids. The chromatids become chromosomes. Cytoplasm divides and four daughter cells are formed. Each has a haploid number of chromosomes. Significance of Meiosis\nMeiosis brings about formation of gametes that contain half the number of chromosomes as the parent cells. It helps to restore the diploid chromosomal constitution in a species at fertilisation. It brings about new gene combinations that lead to genetic variation in the offsprings. Asexual Reproduction\nAsexual reproduction is the formation of offspring from a single parent. The offspring are identical to the parent. Types of asexual reproduction. Binary fission in amoeba. Spore formation in Rhizopus. Budding in yeast. Binary fission\nThis involves the division of the parent organism into two daughter cells.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9301990681914444, "ocr_used": false, "chunk_length": 2361, "token_count": 504}}
{"text": "Spore formation in Rhizopus. Budding in yeast. Binary fission\nThis involves the division of the parent organism into two daughter cells. The nucleus first divides into two and then the cytoplasm separates into two portions\nBinary fission also occurs in bacteria, Paramecium, Trypanosoma and Euglena. Spore formation in Rhizopus\nRhizopus is a saprophytic fungus which grows on various substrate such as bread, rotting fruits or other decaying organic matter. The vegetative body is called mycelium which has many branched threads called hyphae. Horizontal hyphae are called stolons. Vertical hyphae are called sporangiophore. The tips of sporangiophore become swollen to form sporangia, the spore bearing structure. Each sporangium contains many spores. As it matures and ripens, it turns black in colour. When fully mature the sporangium wall burst and release spores which are dispersed by wind or insects. When spores land on moist substratum, they germinate and grow into a new Rhizopus and start another generation. Spore formation in ferns\nThe fern plant is called a sporophyte. On the lower side of the mature leaves are sari (Singular: sorus) which bear spores. Budding in Yeast\nBudding involves the formation of a protrusion called a bud from the body of the organism. The bud separates from the parent cell, in yeast budding goes on so fast and the first bud starts to form another bud before the separation. A short chain or mass of cells is formed. Sexual Reproduction in Plants\nIn flowering plants, the flower is the reproductive organ which is a specialised shoot consisting of a modified stem and leaves. The stem-like part is the pedicel and receptacle, while modified leaves form corolla and calyx. Structure of a flower\nA typical flower consists of the following parts:\nCalyx –\nmade up of sepals. They enclose and protect the flower when it is in a bud. Some flowers have an outer whorl made of sepal-like structures called epicalyx. Corolla –\nconsists of petals. The petals are brightly coloured in insect - pollinated flowers. Androecium –\nIs the male part of the flower. It consists of stamens. Each stamen consists of a filament whose end has an anther.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9217111315547379, "ocr_used": false, "chunk_length": 2174, "token_count": 504}}
{"text": "Androecium –\nIs the male part of the flower. It consists of stamens. Each stamen consists of a filament whose end has an anther. Inside the anther are pollen sacs which contain pollen grains. Gynoecium (pistil) –\nIs the female part of the flower. It consists of one or more carpels. Each carpel consists of an ovary, a sty le and a stigma. The ovary contains ovules which become seeds after fertilisation. A monocarpous pistil has one carpel e.g. beans. A polycarpous pistil has many carpels. If the carpes are free, it is called apocarpous as in rose and Bryophyllum,\nIn carpels that are fused it is called syncarpous as in Hibiscus. A complete flower has all the four floral parts. A regular flower can be divided into two halves by any vertical section passing through the centre. e.g. morning glory. Irregular flower can be divided into two halves in only one plane e.g. crotalaria. Pollination\nThis is the transfer of pollen grains from the anther to the stigma. Types of pollination\nSelf pollination is the transfer of pollen grains from the anther of one flower to the stigma of the same flower. Cross-pollination is the transfer of pollen grains from the anther of one flower to the stigma of a different flower, of the same species. Agents of pollination\nAgents of pollination include wind, insects, birds and mammals. Insect pollinators include bees, butterflies and mosquitoes. Mechanisms that hinder self-pollination\nStamens ripen early and release their pollen grains before the stigma, mature. This is called protandry e.g. in sunflower. The stigma matures earlier and dries before the anthers release the pollen grains. This is called protogyny and is common in grasses. Self sterility or incompatibility\nPollen grains are sterile to the stigma of the same flower, e.g. in maize flower. Shorter stamens than pistils. Fertilisation in Plants\nThe pollen grain contains the generative nucleus and a tube nucleus. When the pollen grain lands on the stigma, it absorbs nutrient and germinates forming a pollen tube. This pollen tube grows through the style pushing its way between the cells. It gets nourishment from these cells.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9182795698924733, "ocr_used": false, "chunk_length": 2139, "token_count": 502}}
{"text": "When the pollen grain lands on the stigma, it absorbs nutrient and germinates forming a pollen tube. This pollen tube grows through the style pushing its way between the cells. It gets nourishment from these cells. The tube nucleus occupies the position at the tip of the growing pollen tube. The generative nucleus follows behind the tube nucleus, and divides to form two male gamete nuclei. The pollen tube enters the ovule through the micropyle. When the pollen tube penetrates the ovule disintegrates and the pollen tube bursts open leaving a clear way for the male nuclei. One male nucleus fuses with the egg cell nucleus to form a diploid zygote which develops into an embryo. The other male gamete nucleus fuses with the polar nucleus to form a triploid nucleus which forms the primary endosperm. This is called double fertilisation. After fertilisation the following changes take place in a flower:\nThe integuments develops into seed coat (testa). The zygote develops into an embryo. The triploid nucleus develops into an endosperm. The ovules become seeds. The ovary develops into a fruit. The ovary wall develops into pericarp. The style, dries up and falls off leaving a scar. The corolla, calyx and stamens dry up and fall off. In some the calyx persists. Fruit formation\nFruit development without fertilisation is called parthenocarpy\ne.g. as in pineapples and bananas. Such fruits do not have seeds. Classification of fruits\nFalse fruits develops from other parts such as calyx, corolla and receptacle,\ne.g. apple and pineapple which develops from an inflorescence. True fruits develop from the ovary, e.g. bean fruit (pod). True fruits can be divided into fleshy or succulent fruits e.g. berries and drupes and dry fruits. The dry ones can be divided into Dehiscent which split open to release seeds and indehiscent which do not open. Types of fruits\nPlacentation\nThis is the arrangement of the ovules in an ovary. Marginal placentation:\nThe placenta appears as one ridge on the ovary wall e.g. bean. Parietal placentation:\nThe placenta is on the ridges on ovary wall. Ovules are in them e.g. pawpaw. Axile placentation:\nThe placenta is in the centre.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9184672206832872, "ocr_used": false, "chunk_length": 2166, "token_count": 506}}
{"text": "Ovules are in them e.g. pawpaw. Axile placentation:\nThe placenta is in the centre. Ovary is divided into a number of loculi. e.g. orange. Basal placentation. The placenta is formed at the base of the ovary e.g. sunflower. Free Central placentation. Placenta is in the centre of the ovary. There are no loculi e.g. in primrose. Methods of fruit and seed dispersal\nAnimal dispersal\nFleshy fruits are eaten by animals. Animals are attracted to the fruits by the bright colour, scent or the fact that it is edible. The seeds pass through the digestive tract undamaged and are passed out with faeces. E.g. tomatoes and guavas. Such seeds have hard, resistant seed coats. Others have fruits with hooks or spines that stick on animal fur or on clothes. Later the seeds are brushed of or fall off on their own e.g. Bidens pilosa (Black jack). Wind dispersal\nFruits and seeds are small and light in order to be carried by air currents. A fruit that is a capsule e.g. tobacco split or has pores at the top e.g. Mexican poppy. The capsule is attached to along stalk when swayed by wind the seeds are released and scattered. Some seeds have hairy or feather-like structures which increase their surface area so that they can be blown off by the wind e.g. Sonchus. Others have wing-like structures e.g. Jacaranda and Nandi Flame. These extensions increase the surface area of fruits and seeds such that they are carried by the wind. Water dispersal\nFruits like coconut have fibrous mescocarp which is spongy to trap air, the trapped air make the fruit light and buoyant to float on water. Plants like water lily produce seeds whose seed coats trap air bubbles. The air bubbles make the seeds float on water and are carried away. The pericarp and seed coat are waterproof. Self dispersal (explosive) Mechanism\nThis is seen in pods like bean and pea. Pressure inside the pod forces it to open along lines of weakness throwing seeds away from parent plant. Reproduction in Animals\nSexual reproduction involves the fusion of gametes. In animals two individuals are involved, a male and a female. Special organs known as gonads produce gametes.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9138287864534338, "ocr_used": false, "chunk_length": 2126, "token_count": 498}}
{"text": "Reproduction in Animals\nSexual reproduction involves the fusion of gametes. In animals two individuals are involved, a male and a female. Special organs known as gonads produce gametes. In males testes produce sperms while in females ovaries produce ova. The fusion of male gamete and female gamete to form a zygote is called fertilisation. There are two types of fertilisation. External and internal. External fertillsation\nExample in amphibians takes place in water. The male mounts the female and shed sperms on the eggs as they are laid. Eggs are covered by slippery jelly-like substance which provides protection. Many eggs are released to increase the chances of survival. Internal fertilisation\nThis occurs in reptiles, birds and mammals. Fertilisation occurs within the body of the female. Fewer eggs are produced because there are higher chances of fertilisation since sperms are released into the female body. Reproduction in Humans\nStructure of female reproduction system\nThe female reproduction system consist of the following:\nOvaries\nAre two oval cream coloured structures found in lower abdomen below the kidneys. Oviducts. They produce the ova. Are tubes which conduct the ova produced by the ovaries to the uterus. Fertilisation occurs in the upper part of the oviduct. Uterus\nThe uterus is a hollow muscular organ found in the lower abdomen. The embryo develops inside the uterus. The inner lining endometrium supplies nutrients to embryo. The embryo is implanted into the inner uterine wall- the endometrium which nourishes the embryo. The thick muscles of the uterus assist in parturition. Cervix\nHas a ring of muscles that separates the uterus from the vagina. It forms the opening to the uterus\nVagina\nIs a tube that opens to the outside and it acts as the copulatory and birth canal through the vulva. Structure of male reproductive system\nThe male reproductive system consists of the following:\nTestis:\nEach testis is a mass of numerous coiled tubes called semniferous tubules. Each is enclosed within a scrotal sac that suspends them between the thighs. This ensures that sperms are maintained at a temperature lower than that of the main body. Seminiferous tubules\nThe lining of seminiferous tubules consists of actively dividing cells which give rise to sperms.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9268356643356644, "ocr_used": false, "chunk_length": 2288, "token_count": 488}}
{"text": "Each is enclosed within a scrotal sac that suspends them between the thighs. This ensures that sperms are maintained at a temperature lower than that of the main body. Seminiferous tubules\nThe lining of seminiferous tubules consists of actively dividing cells which give rise to sperms. Between the seminiferous tubules are interstitial cells which produce the male hormones called androgens e.g. testosterone. The seminiferous tubules unite to form the epididymis, which is a coiled tube where sperms are stored temporarily . Vas deferens (sperm duct) is the tube through which sperms are carried from testis to urethra. Seminal vesicle produces an alkaline secretion which nourishes the spermatozoa. Prostate gland\nProduces an alkaline secretion to neutralise vaginal fluids. Cowpers' gland\nSecretes an alkaline fluid. All these fluids together with spermatozoa form semen. Urethra\nIs a long tube through which the semen is conducted during copulation. It also removes urine from the bladder. Penis\nIs an intro-mittent organ which is inserted into the vagina during copulation . Fertilisation in Animals\nFertilisation is preceded by copulation in which the erect penis is inserted into the vagina. This leads to ejaculation of semen. The sperms swim through the female's genital tract to the upper part of the oviduct. The head of the sperm penetrates the egg after the acrosome_ releases lytic enzymes t dissolve the egg membrane. The tail is left behind. Sperm nucleus fuses with that of the ovum and a zygote is formed. A fertilisation membrane forms around the zygote which prevents other sperms from penetrating the zygote. Implantation:\nAfter fertilisation the zygote begins to divide mitoticaly as it moves towards the uterus. It becomes embedded in the wall of the uterus a process called implantation. By this time the zygote is a hollow ball of cells called blastocyst or embryo. In the uterus the embryo develops villi which project into uterus for nourishment later the villi and endometrium develop into placenta. Embryonic membranes\nEmbryonic membranes develop around the embryo. The outermost membrane is the chorion which forms the finger-like projections (chorionic villi) which supply nutrients to the embryo.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.926354989774026, "ocr_used": false, "chunk_length": 2229, "token_count": 499}}
{"text": "In the uterus the embryo develops villi which project into uterus for nourishment later the villi and endometrium develop into placenta. Embryonic membranes\nEmbryonic membranes develop around the embryo. The outermost membrane is the chorion which forms the finger-like projections (chorionic villi) which supply nutrients to the embryo. The amnion surrounds the embryo forming a fluid filled cavity within which the embryo lies. Amniotic cavity is filled with amniotic fluid. This fluid acts as a shock absorber and protects the foetus against mechanical injury. It also regutates temperature. The chorionic villi, allantois together with the endometrium from the placenta. The embryo is attached to the placenta by a tube called umbilical cord which has umbilical vein and artery. The maternal blood in the placenta flows in the spaces lacuna and surrounds capillaries from umbilical vein and artery. The umbilical cord increase in length as the embryo develops. Role of placenta\nProtection\nMaternal blood and foetal blood do not mix. This ensures that the pathogens and toxins from maternal blood do not reach the foetus. The placenta allows maternal antibodies to pass into the foetus, providing the foetus with immunity. Nutrition\nThe placenta facilitates the transfer of nutrients from maternal blood to foetus. Excretion\nPlacenta facilitates the removal of nitrogenous wastes from the foetus' blood to maternal blood. Gaseous exchange\nOxygen from the maternal blood diffuses into the foetal blood while carbon (IV) oxide from foetal blood diffuse into maternal blood. Production of hormones\nPlacenta produces progesterone and oestrogen. Gestation period\nThe period between conception and birth is called gestation. In humans gestation takes nine months (40 weeks). The embryo differentiates into tissues and organs during this period. Week 1 to 3:\nZygote divides to form blastocyst. Implantation takes place. The three germ layers form endoderm, mesoderm and ectoderm. Nervous system starts to form. Week 4 to 7:\nDevelopment of circulating and digestive systems. Further development of nervous system, formation of sensory organs,\nAll major internal organs are developed. At week 5, heartbeat starts . Week 8 to 24:\nAll organs well developed including sex organs. Hair, finger and toe nails grow. Foetus move and eyelids open.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9158290670737874, "ocr_used": false, "chunk_length": 2333, "token_count": 508}}
{"text": "Week 8 to 24:\nAll organs well developed including sex organs. Hair, finger and toe nails grow. Foetus move and eyelids open. Week 25- 30:\nThe fully developed foetus responds to touch and noises and moves vigorously. The head turns and faces downwards ready for birth. Week 31-40:\nFoetus increases in size. Birth occurs. Reproductive Hormones\nSecondary Sexual Characteristics\nMale\nTesterone is the main androgen that stimulates the development of secondary sexual characteristics. Broadening of the shoulders. Deepening of the voice due to enlargement of larynx. Hair at the pubic area, armpit and chin regions. Penis and testis enlarge and produce sperms. Body becomes more masculine. Female\nEnlargement of mammary glands. Hair grows around pubic and armpit regions. Widening of the hips. Ovaries mature and start producing ova. Menstruation starts. Oestrogen triggers the onset of secondary sexual characteristics. Sexually transmitted infections (STl)\nMenstrual Cycle\nThis is characterized by discharge of blood and tissue debris (menses) from the uterus every 28 days. This is due to the breakdown of the endometrium which occurs when the level of progesterone falls and the girl starts to menstruate. The follicle stimulating hormone (FSH) causes the Graafian follicle to develop and also stimulate the ovary to release oestrogen. Oestrogen hormone triggers the onset of secondary sexual characteristics. Luteinising hormone (L.H) causes the mature ovum to be released from the Graafian follicle - a process called ovulation. After ovulation progesterone hormone is produced. After menstruation, the anterior lobe of the pituitary gland starts secreting the follicle stimulating hormone (FS.H) which causes the Graafian follicle to develop in the ovary. It also stimulates the ovary tissues to secrete oestrogen. Oestrogen brings about the repair and healing of the inner lining of the uterus (endometrium) which had been destroyed during menstruation. Oestrogen level stimulates the pituitary gland to produce (Luteinising Hormone (L.H). This hormone makes the mature Graafian follicle to release the ovum into the funnel of oviduct, a process called ovulation.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9133209653781968, "ocr_used": false, "chunk_length": 2166, "token_count": 489}}
{"text": "Oestrogen brings about the repair and healing of the inner lining of the uterus (endometrium) which had been destroyed during menstruation. Oestrogen level stimulates the pituitary gland to produce (Luteinising Hormone (L.H). This hormone makes the mature Graafian follicle to release the ovum into the funnel of oviduct, a process called ovulation. After releasing the ovum, the Graafian follicle changes into a yellow body called corpus luteum. The luteinising hormone stimulates the corpus luteum to secrete a hormone called progesterone which stimulates the thickening and vascularisation of endometrium. This prepares the uterine wall for implantation of the blastocyst. If fertilisation takes place, the level of progesterone increases and thus inhibits FSH from stimulating the maturation of another Graafian follicle. If fertilisation does not occur, the corpus luteum disintegrates and the level of progesterone goes down. The endometrium, sloughs off and menstruation occurs. Advantages of Reproduction Asexual\nGood qualities from parents are retained in the offspring without variation. New individuals produced asexually mature faster. Process does not depend on external factors which may fail such as pollination. New individuals obtain nourishment from parent and so are able to survive temporarily under unsuitable conditions. No indiscriminate spreading of individuals which can result in wastage of offspring. Takes a shorter time and leads to rapid colonization. Disadvantages of asexual reproduction\nNew offspring may carry undesirable qualities from parents. Offspring may be unable to withstand changing environmental conditions. Faster maturity can cause overcrowding and stiff competition. Reduced strength and vigour of successive generations. Advantages of sexual reproduction\nLeads to variations. Variations which are desirable often show hybrid vigour. High adaptability of individuals to changing environmental conditions. Variations provide a basis for evolutionary changes. Disadvantages of sexual reproduction\nFusion is difficult if two individuals are isolated. Some variations may have undesirable qualities. Population growth is slow. Practical Activities\nExamining the stages of mitosis\nAbout 2 mm of a root tip of onion bulb is cut off and placed on a microscope slide. A stain e.g. aceto-orcein is added and the root tip macerated using a scapel. A cover slip is added and observations made. Different stages of mitosis can be observed.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9306562061596595, "ocr_used": false, "chunk_length": 2474, "token_count": 504}}
{"text": "aceto-orcein is added and the root tip macerated using a scapel. A cover slip is added and observations made. Different stages of mitosis can be observed. Examining the stages of meiosis\nAn unopened bud of Tradescantia is obtained\nThe anther is removed and placed on a microscope slide. A few drops of hydrochloric acid and acetic-orcein stain are added. A cover slip is placed on the anther. Pressing the cover slip gives a thin squash, which is observed under the microscope. Different stages of meiosis are observed. To observe the structure of Rhizopus\nRhizopus grow on moist bread left under suitable temperature\nA piece of moist bread is placed on a petri-dish or enclosed in a plastic bag and observe daily for four days. Under a low power microscope the sporangia and stolons can be observed. To examine spores on sori of ferns\nObtain the fern plant. Detach a frond from the plant and observe the under-side using a hand lens to see the raised brown patches - the sori. Open up the sorus to observe the sporangia. Examine insect and wind pollinated flowers\nObtain insect pollinated flowers e.g. crotalaria, hibiscus/Ipomea, Solanum, incunum. Note the scent, colour and nectar guides. A description of the calyx, corolla, androecium and gynoecium is made. Obtain a wmd pollinated flower e.g,' maize, star-grass, sugar-cane, Kikuyu grass. Observe the glumes, spikes and spikelet. Examine a single floret, and identify the androecium and gynoecium. Classifying fruits\nObtain different fruits - oranges, mangoes, maize, castor oil, bean pod, black jack . Observe the fruits, classify them into succulent, dry-dehiscent or indehiscent. Dissection of Fruits\nObtain an orange and a mango fruit. Make a transverse section. Observe the cut surface and draw and label the parts. Note that the fruit is differentiated into epicarp, mesocarp and endocarp. Obtain a pod of a legume. Open up the pod and observe the exposed surface. Draw and label the parts. Note that the fruit wall is not differentiated.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9156000000000001, "ocr_used": false, "chunk_length": 2000, "token_count": 503}}
{"text": "Open up the pod and observe the exposed surface. Draw and label the parts. Note that the fruit wall is not differentiated. Dispersal of fruits and seeds\nObtain animal dispersal fruits, like oranges, tomatoes, black jack, sodom apple. Identify the way by which each is adapted to dispersal by animals. Obtain wind dispersed fruit/seed\ne.g. Nandi flame, Jacaranda Sonchus, cotton seed, Tecoma. END\nGROWTH AND DEVELOPMENT\nConcept of Growth and Development\nGrowth is a characteristic feature of all living organisms. Most multicellular organisms start life as a single cell and gradually grow into complex organisms with many cells. This involves multiplication of cells through the process of cell division. This quantitative permanent increase in size of an organism is referred to as growth. For growth to take place the following aspects occur\nCells of organisms assimilate nutrients hence increase in mass. Cell division (mitosis) that lead to increase in the number of cells. Cell expansion that leads to enlargement an increase in the volume and size of the organism. It is therefore possible to measure growth using such parameters as mass, volume, length, height, surface area. On the other hand development is the qualitative aspect of growth which involves differentiation of cells and formation of various tissues in the body of the organism in order for tissues to be able to perform special functions. It is not possible to measure ac aspects of development quantitative. Therefore development can be assessed terms of increase in complexity of organism e.g. development of leaves, flowers and roots. A mature human being has millions of cells in the body yet he or she started from; single cell, that is, a fertilised egg. During sexual reproduction mammals an ovum fuses with a sperm form a zygote. The zygote divides rapidly without increasing in size, first into 2, 4, 8, 16,32, 64 and so on, till it forms a mass cells called morula. These first cell division is called cleavages. The morula develops a hollow part, resulting into a structure known as a blastula (blastocyst). Later, blastocyst cells differentiate into an inner layer (endoderm) and the outer layer (ectoderm). The two-layered embryo implants into the uterine wall and, by obtaining nutrients from the maternal blood, starts to grow and develop.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.911501667360219, "ocr_used": false, "chunk_length": 2327, "token_count": 490}}
{"text": "The morula develops a hollow part, resulting into a structure known as a blastula (blastocyst). Later, blastocyst cells differentiate into an inner layer (endoderm) and the outer layer (ectoderm). The two-layered embryo implants into the uterine wall and, by obtaining nutrients from the maternal blood, starts to grow and develop. BlastocoeJ\nFertilised egg 2-celled stage 4-celled stage\n{zygote) Morula Blastula\n(mass of cells)\nAs the embryo grows and develops, changes occur in cell sizes and cell -types. Such changes are referred to as growth and development respectively. These processes lead to morphological and physiological changes in the developing young\norganism resulting into an adult that is more complex and efficient. In the early stages, all the cells of the embryo look alike, but as the development process\ncontinues the cells begin to differentiate and become specialised into different tissues to\nperform different functions. Growth involves the synthesis of new material and protoplasm. This requires a continuous supply of food, oxygen, water, warmth and means of removing\nwaste products. In animals, growth takes place all over the body but the rates differ in the various parts of the body and at different times. In plants however, growth and cell division mostly take place at the root tip just behind the root cap and stem apex. This is referred to as apical growth which leads to the lengthening of the plant. However, plants do not only grow upwards and downwards but sideways as well. This growth leads to an increase in width (girth) by the activity of cambium cells. The increase in girth is termed as secondary growth. Study Question 1-State two major differences between growth and development\nMeasurement of growth\nGrowth can be estimated by measuring some aspect of the organism such as height, weight, volume and length over a specified period of time. The measurements so obtained if plotted against time result into a growth curve. Study Question 2\nThe following results were obtained from a study of germination and early growth of maize. The grains were sown in soil in a greenhouse and.at two-day intervals. Samples were taken, oven dried and weighed. See table . Table\nPlot a graph of dry mass of embryo against time after sowing. Describe the shape of the graph. For most organisms when the measurements are plotted they give an S-shaped graph called a sigmoid curve such as in figure .", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9153946434016671, "ocr_used": false, "chunk_length": 2431, "token_count": 507}}
{"text": "Table\nPlot a graph of dry mass of embryo against time after sowing. Describe the shape of the graph. For most organisms when the measurements are plotted they give an S-shaped graph called a sigmoid curve such as in figure . Time\nFig. 4.2: TSie sigmoid growth curve\nThis pattern is due to the fact that growth tends to be slow at first and then speeds up\nand finally slows down as adult size is reached. A sigmoid curve may therefore be divided into four parts. Lag phase (slow growth)\nThis is the initial phase during which little growth occurs. The growth rate is slow due to various factors namely:\n(i) The number of cells dividing are few. (ii) The cells have not yet adjusted to the surrounding environmental factors. Exponential phase (log phase)\nThis is the second phase during which growth is rapid or proceeds exponentially. During this phase the rate of growth is at its maximum and at any point, the rate of growth is proportional to the amount of material or numbers of cells of the organism already present. This rapid growth is due to:\n(i) An increase in number of cells dividing,2-4-8-16-32-64 following a geometric progression,\n(ii) Cells having adjusted to the new environment,\n(iii) Food and other factors are not limiting hence cells are not competing for resources,\n(iv) The rate of cell increase being higher than the rate of cell death. Decelerating Phase\nThis is the third phase during which time growth becomes limited as a result of the effect of some internal or external factors, or the interaction of both. The slow growth is due to: (\ni) The fact that most cells are fully differentiated. (ii) Fewer ceils still dividing,\n(iii) Environmental factors (external and internal) such as:\nshortage of oxygen and nutrients due to high demand by the increased number of cells. space is limited due to high number of cells. accumulation of metabolic waste products inhibits growth. limited acquisition of carbon (IV) oxide as in the case of plants. Plateau (stationary) phase\nThis is the phase which marks the period where overall growth has ceased and the\nparameters under consideration remain constant. This is due to the fact that:\nThe rate of cell division equals the rate of cell death. Nearly all cells and tissues are fully differentiated, therefore there is no further increase in the number of cells. The nature of the curve during this phase may vary depending on the nature of the parameter, the species and the interns!", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9045417585137072, "ocr_used": false, "chunk_length": 2451, "token_count": 513}}
{"text": "This is due to the fact that:\nThe rate of cell division equals the rate of cell death. Nearly all cells and tissues are fully differentiated, therefore there is no further increase in the number of cells. The nature of the curve during this phase may vary depending on the nature of the parameter, the species and the interns! factors. In some cases, the curve continue to increase slightly until organism dies as is the case monocotyledonous plants, man invertebrates, fish and certain reptiles. indicates positive growth. In some ot cases the curve flattens out indicating change in growth while other growth curv may tail off indicating a period of negat growth rate. This negative pattern characteristic of many mammals includi humans and is a sign of physical senesee associated with increasing age. Study Question 3\nWhat happens during the following; log and stationary phases of growth? However, the sigmoid curve does not to all organisms, for example, arthropods. I insects, growth takes place at intervals-volume changes are plotted against time., different curve is obtained. This is cal intermittent growth curve. See figure 43,\nThe intermittent growth in insects is due to the fact that they have an exoskeleton and hence growth is possible only when it is shed. This shedding process is known as moulting or ecdysis. However, cell division continues to take place during the inter-moult phase but the expansion of tissues is limited by the unshed exoskeleton. Practical Activity I: Project\nTo measure the growth of a plant\nRequirements\nSmall plots/boxes, meter rule and seeds of beans (or green grams, peas, maize),\nProcedure\n. Place some soil in the box or prepare a small plot outside the laboratory. Plant some seeds in the box and place it in a suitable place outside the laboratory (or plant the\nseeds in your plot). Water the seeds daily. Observe the box/plot daily and note the day the seedlings emerge out of the soil. .Measure the height of the shoot from the soil level up to the tip of the shoot. Repeat this with four other seedlings. Work out the average height of the shoots for this day. Repeat procedure 5 every three days for at least three weeks. Record the results in a table form. On the same seedlings measure the length of one leaf from each of the fiveseedlings (from leaf apex to itsattachment on the stem). Calculate the average length of the leaves and record in the table.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9142100546976678, "ocr_used": false, "chunk_length": 2413, "token_count": 509}}
{"text": "Record the results in a table form. On the same seedlings measure the length of one leaf from each of the fiveseedlings (from leaf apex to itsattachment on the stem). Calculate the average length of the leaves and record in the table. Plot a graph of the height of the shoot against time. On the same axes plotlength of leaf against time. Compare the two graphs drawn. 4.2 Growth and Development in Plants\nThe main growth and development phase in plants begins with the germination of the mature seed. Seeds are of two kinds depending on the number of cotyledons or embryo leaves. Practical Activity 2\nTo investigate structural differences between monocotyledonous and dicotyledonous seeds\nS '\n-Z 4.5\n£4.0 qa\n3 3.5\n3.0\n2.5\n2.0\n1.5\nAdult\nmoulting\n2ndinstar>/^grawth phase\n^^tf^_/ jntermoult phaseji—i—i—i—ii—l__\n246 8101214161820222426283032343638404244\nTime in days Fig. 4.3: Growth curve showing increase in length of the short homed grasshopper\nRequirements\nBean seeds and maize grains which have been soaked overnight. Scalpel or razor blades, iodine solution, Petri-dish and hand lens. Procedure\nUsing a scalpel or razor blade make longitudinal sections (LS) of both the bean seed and the maize grain. Observe the LS of the specimens using a hand lens. Note any structural difference between the specimens. Draw the LS of each specimen and label. Puta drop of iodine solution on the cut surfaces of both specimens. Note any differences in colouration with iodine on the surfaces of the two specimens. On your diagrams indicate the distribution of the stain. Account for the difference in distribution of the colouration with iodine in the two specimens. Structure of the Seed\nA typical seed consists of a seed coat enclosing an embryo. The seed coat is the outer covering which, in most seeds, is made\n-Remains of style\nPosition of plumule\nPosition of radicle Scutellum\nAttachment to External Parent plant\nPosition of radicle\nMicropyle Hilum\nTesta\nStructure of monocotyledonous seed (maize grain)\nExternal (bj Structure of dicotyledonous seed\nFig.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.8771928539018046, "ocr_used": false, "chunk_length": 2051, "token_count": 509}}
{"text": "Account for the difference in distribution of the colouration with iodine in the two specimens. Structure of the Seed\nA typical seed consists of a seed coat enclosing an embryo. The seed coat is the outer covering which, in most seeds, is made\n-Remains of style\nPosition of plumule\nPosition of radicle Scutellum\nAttachment to External Parent plant\nPosition of radicle\nMicropyle Hilum\nTesta\nStructure of monocotyledonous seed (maize grain)\nExternal (bj Structure of dicotyledonous seed\nFig. 4.4: Structure of seeds\nup of the two layers, an outer testa and inner one, the legmen. The testa is thick; the tegmen is a transparent membrane tissue. The two layers protect the seed bacteria, fungi and other organisms whk may damage it. There is a scar called hilurn on one part of the seed. This is point where the seed had been attached the seed stalk or funicle. Near one end of 1 hilum is a tiny pore, the micropyle. This allows water and air into the embryo, embryo is made up of one or two seed leavi or cotyledons, a plumule (embryonic sh( and a radicle (the embryonic root). The of the radicle is opposite the micropyle. In some seeds the cotyledons swollen as they contain stored food for growing plumule and radicle. Such seeds, called non-endospermic seeds. In ot cases, the seeds have their food stored in: endosperm. Such seeds are call endospermic seeds. Seeds with one cotyk are referred to as monocotyledonous wi those with two are referred to dicotyledonous. This is the major basis i differentiation between the two large cb of plants, the monocotyledonae aa dicotyledonae. Scutellum\nCotyledon\nCoieoptile\nPlumule\nRadicle\nColeorhiza\nInternal\nPlumule\nRadicle\nCotyledon\nInternal\nFused pericaT and testa\nDormancy in Seeds\nThe embiyo of a dry, fully developed seed usually passes through a period of rest after ripening period. During this time the seed performs all its life (physiological) processes very slowly and uses up little food. This is a period of dormancy.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9150169729137396, "ocr_used": false, "chunk_length": 1974, "token_count": 506}}
{"text": "Scutellum\nCotyledon\nCoieoptile\nPlumule\nRadicle\nColeorhiza\nInternal\nPlumule\nRadicle\nCotyledon\nInternal\nFused pericaT and testa\nDormancy in Seeds\nThe embiyo of a dry, fully developed seed usually passes through a period of rest after ripening period. During this time the seed performs all its life (physiological) processes very slowly and uses up little food. This is a period of dormancy. Even if all the favourable environmental conditions for germination are provided to the seed during this period of dormancy, the seed will not germinate. This is due to the fact that the seed embryo may need to undergo further development before germination. Some seeds can germinate immediately after being_shed from the parent plant (e.g. most tropical plants) while others must pass through dormancy period, lasting for weeks, months or even years before the seed can germinate. Dormancy provides the seeds with enough time for dispersal so that they can germinate in a suitable environment. It also enables seeds to survive during adverse environmental conditions without depleting their food reserves. The embryo has time to develop until favourable conditions are available e.g. availability of water. Factors that Cause Dormancy\nEmbryo may not yet be fully developed. Presence of chemical inhibitors that inhibit germination in seeds e.g.abscisic acid. Very low concentrations of hormones e.g. gibberellins and enzymes reduces the ability of seeds to germinate. Hard and impermeable seed coats prevent entry of air and water in some seeds e.g. wattle. In some seeds the absence of certain wavelengths of light make them remain dormant e.g. in some lettuce plants. Freezing of seeds during winter lowers their enzymatic activities rendering them dormant. Ways of Breaking Dormancy\nWhen the seed embryos are mature then the seed embryos can break dormancyand germinate. Increase in concentration of hormones e.g. cytokinins and gibberellins stimulate germination. Favourable environmental factors such as water, oxygen and suitable temperature. Some wavelengths of light trigger the production of hormones like gibberellins leading to breaking of dormancy. Scarification i.e. weakening of the testa is needed before seeds with hard impermeable seed coats can germinate. This is achieved naturally by saprophytic bacteria and fungi or by passing through the gut of animals.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.926751230877794, "ocr_used": false, "chunk_length": 2367, "token_count": 503}}
{"text": "Scarification i.e. weakening of the testa is needed before seeds with hard impermeable seed coats can germinate. This is achieved naturally by saprophytic bacteria and fungi or by passing through the gut of animals. In agriculture the seeds of some plants are weakened by boiling, roasting and cracking e.g. wattle. Seed Germination\nThe process by which the seed develops into a seedling is known as germination. It refers to all the changes that take place when a seed becomes a seedling. At the beginning of germination water is absorbed into the seed through the micropyle in a process known as imbibition and causes the seed to swell. The cells of the cotyledons become turgid and active. They begin to make use of the water to dissolve and break down the food substances stored in the cotyledons. The soluble food is transported to the growing plumule and radicle. The plumule grows into a shoot while the radicle grows into a root. The radical emerges from the seed through micropyle, bursting the seed coat as it does so. Conditions Necessary for Germination\nSeeds can easily be destroyed by unfavourable conditions such as excessive heat, cold or animals. Seeds need certain conditions to germinate and grow. Some of these conditions are external, for example water, oxygen and suitable temperature while others are internal such as enzymes, hormones and viability of the seeds themselves. Water\nA non-germinating seed contains very little water. Without water a seed cannot germinate. Water activates the enzymes and provides the medium for enzymes to act and break down the stored food into soluble form. Water hydrolyses and dissolves the food materials and is also the medium of transport of dissolved food substances through the various cells to the growing region of the radical and plumule. Besides, water softens the seed coat which can subsequently burst and facilitate the emergence of the radicle. Oxygen\nGerminating seeds require energy for cell division and growth. This energy is obtained from the oxidation of food substances stored in the seed through respiration thus making oxygen an important factor in seed germination. Seed in water logged soil or seed buried deep into the soil will not germinate due to lack of oxygen. Temperature\nMost seeds require suitable temperature before they can germinate. Seeds will not germinate below 0°C or above 47° C. The optimum temperature for seeds to germinate is 30°C.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9212859726010605, "ocr_used": false, "chunk_length": 2435, "token_count": 496}}
{"text": "Temperature\nMost seeds require suitable temperature before they can germinate. Seeds will not germinate below 0°C or above 47° C. The optimum temperature for seeds to germinate is 30°C. At higher temperature the protoplasm is killed and the enzymes in the seed are denatured. At very low temperatures the enzymes become inactive. Therefore, the protoplasm and the enzymes work best within the optimum temperature range. The rate of germination increases with temperature until it reaches an optimum. This varies from plant to plant. Enzymes\nEnzymes play a vital role during germination in the breakdown and subsequent oxidation of food. Food is stored in seeds in form of carbohydrates, fats and proteins which are in insoluble form. The insoluble food is converted into a soluble form by the enzymes. Carbohydrates are broken down into glucose by the diastase enzyme, fats into fatty acids and glycerol by lipase, and proteins into amino acids by protease. Enzymes are also necessary for the conversion of hydrolysed products to new plant tissues. Hormones\nSeveral hormones play a vital role in germination since they act as growth stimulators. These include gibberellins and cytokinins. These hormones also counteract the effect of germination inhibitors. Viability\nOnly seeds whose embryos are alive and healthy will be able to germinate and grow. Seeds stored for long periods usually lose their viability due to depletion of their food reserves and destruction of their embryo by pests and diseases. Study Question 4\nIn an experiment to investigate the effect of neat on germination of seeds, ten bags each containing 60 pea seeds were placed in a water-bath maintained at 85°C . After every two minutes a bag was removed and seeds contained in it planted. The number that germinated was recorded. The procedure used for pea seeds was repeated for wattle seeds. The results obtained were as shown in the table 4,2,\n(a)Using a suitable scale and on the same axes, draw graphs of time in hot water against number of seeds that germinated for each plant. Use horizontal axis for time and the verticalaxis for the seeds that germinated. Explain why the ability of pea seeds to germinate declined with time of exposure to heat. Explain why the ability of the wattle seeds to germinate improved with time of exposure to heat.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9136547671069177, "ocr_used": false, "chunk_length": 2324, "token_count": 485}}
{"text": "Use horizontal axis for time and the verticalaxis for the seeds that germinated. Explain why the ability of pea seeds to germinate declined with time of exposure to heat. Explain why the ability of the wattle seeds to germinate improved with time of exposure to heat. Practical Activity 3\nTo investigate conditions necessary for seed germination\nRequirements\nCotton wool, seeds, water, six fiat bottomed flasks, 2 corks, 2 test-tubes, blotting paper, incubator, refrigerator, thermometer, pyrogallic acid and sodium hydroxide. Procedure\nPrepare three set-ups as shown in figure 4.5. Leave the set-ups to stand for five days. Record all the observable changes that have taken place in the flasks hi each set up in a table form as shown\nStudy Question 5\nWhich condition was being investigated in set-up I, II and III? For each set-up explain the results obtained. What was the role of flask B in each set-up? Types of Germination\nThe nature of germination varies in different seeds. During germination the cotyledons may be brought above the soil surface. This type of germination is called epigeal germination. If during germination the cotyledons remain underground the type of germination is known as hypogeal. SET UP 1\nPyrogallic acid + NaOH\nCotton wool\nSeeds Moist\nWater\nSET UP 2\nMoist cotton wool\nDry cotton wool\nSETUPS\nFig. 4.5: Set-up for investigating conditions necessary for germination\nEpigeal Germination\nDuring the germination of a bean seed, the radicle grows out through the micropyle. It grows downwards into the soil as a primary root from which other roots arise. The part of the embryo between the cotyledon and the radicle is called the hypocotyl. This part curves and pushes upwards through the soil protecting the delicate shoot tip. The hypocotyls then straightens and elongates carrying with it the two cotyledons which turn green and leafy. They start manufacturing food for the growing seedling. The plumule which is lying between two cotyledons, begins to grow into first foliage leaves which start manufacturing food. Hyopgeal Germination\nIn maize, the endosperm provides food to the embryo which begins to grow.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9094404777083599, "ocr_used": false, "chunk_length": 2139, "token_count": 492}}
{"text": "They start manufacturing food for the growing seedling. The plumule which is lying between two cotyledons, begins to grow into first foliage leaves which start manufacturing food. Hyopgeal Germination\nIn maize, the endosperm provides food to the embryo which begins to grow. The radicle along with a protective covering(c(?/eorfci2a) grows out of the seed. The epicotyl is the part of the embryo between the cotyledon and the plumule. The epicotyl elongates and the plumule grows out of the coleoptile and forms the first foliage leaves. The seedling now begins to produce its own food and the endosperm soon shrivels. This type of germination in which the cotyledon remains below the ground is known as hypogeal germination. Practical Activity 4\nTo investigate epigeal and hypogeal germination\nRequirements\nTin or box, soil, water, maize grains and bean seeds. Procedure\nPlace equal amounts of soil into two containers labelled A and B. In A, plant a few maize grains.In B, plant a few bean seeds. Water the seeds and continue watering daily until they germinate. Place your set-ups on the laboratory bench. Observe daily for germination. On the first day the seedlings emerge from the soil, observe them carefully with\nregard to the soil level. Carefully uproot one or two seedlings from each set. Observe and draw the seedlings from each set Label the parts and indicate the soil level on your diagram. On the fifth day since emergence, again uproot another seedling. Observe and draw. Indicate the soil level on your diagram.. Tabulate the differences between the two types of germination studied. Cotyledon Plumule\nRadicle\nRoot\nCotyledon Hypocotyf\nFig. 4.6 (a): Epigeal germination for castor seeds\nPlumule\nColeoptile\nScutellum\nColeorhiza\nPrimary and Secondary Growth\nThe region of growth in plants is found in localised areas called meristems as shown . A meristem is a group of undifferentiated cells in plants which are capable of continuous mitotic cell division.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9166024340770792, "ocr_used": false, "chunk_length": 1972, "token_count": 468}}
{"text": "Cotyledon Plumule\nRadicle\nRoot\nCotyledon Hypocotyf\nFig. 4.6 (a): Epigeal germination for castor seeds\nPlumule\nColeoptile\nScutellum\nColeorhiza\nPrimary and Secondary Growth\nThe region of growth in plants is found in localised areas called meristems as shown . A meristem is a group of undifferentiated cells in plants which are capable of continuous mitotic cell division. The main meristems in flowering plants are found at the tips of shoots and roots, in young leaves, at the bases of the inter-nodes, and in\nApicai meristem\nLeaf primordium\nVascular tissues\n\" beginning to form\nEpidermis xylem Phloem Pericycte\nCortex\nMedulla Cambium\n(c) Epidermis\nNode\nPhloem\nXylefti Epidermis\nCortege\nEndodermis\nRoot cap\nZone of cell division\nZone of cell elongation (expansion)\nZone Trf cell\ndifferentiation\nPermanent tissues\nFig. 4.7(a) and (b): Longitudinal section of the root tip and apex Fig. 4.7(c) and (d): Transverse section of the stem and rooi\nzone\nvascular cambium and cork cambium. T\nhe meristems at the tips of the shoots and the roots are known as apical meristems and are responsible for primary growth. The cambium meristems are responsible for secondary growth. Primary Growth\nPrimary growth occurs at the tips of roots and shoots due to the activity of apical meristems. These meristems originate from the embryonic tissues. In this growth there are three distinctive regions, the region of cell division, cell ejpngarion and eel] differentiation. See figure 4.7. The regipn of cell division is an area of actively dividing meristematic cells. These cells have thin cell walls, dense cytoplasm and no vacuoles. In the region of cell elongation, the cells become enlarged to their maximum size by the stretching of their walls. Vacuoles start forming and enlarging. In the region of ceH differentiation the cells attain their permanent size, have large vacuoles and thickened watt cells. The cells also differentiate into tissues specialised for specific functions.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9103777338556545, "ocr_used": false, "chunk_length": 1969, "token_count": 505}}
{"text": "Vacuoles start forming and enlarging. In the region of ceH differentiation the cells attain their permanent size, have large vacuoles and thickened watt cells. The cells also differentiate into tissues specialised for specific functions. Primary growth results into an increase in the length of shoots and roots. Study Question\n;:Mgure;4-S indkate the appearance of cells at different regions at the apical meristems.Nudeus -Cytoplasm\nFig. 4.8\nRearrange them into three regions:\nZone of cell division. Zone of cell elongation.; -,\nZone of cell differentiation. (jb) Name specialised tissues formed at tl\nzone of cell differentiation. Region of Growth in a root\nThis is determined by taking a young germinating seedling whose radicle is then marked with the Indian ink at intervals of 2 mm. The seedling is left to grow for sometime (about 24 hours or overnight) and then the ink marks are examined. When the distance between successive ink marks are measured, it is found that the first few ink marks, especially between the 2nd and 3\"1 mark above tip of root have increased significantly. This shows that growth has occurred in the region just behind the tip of the root. The difference between the length of each new interval and the initial interval of 2 mm gives the increase in the length of that interval during that period of time. From this the rate of growth of the root region can be calculated. See figure 4.9. „Increase in length . „_\nGrowth = ^ . .—^— X 100\nOriginal length\nPractical Activity 5 To determine the region of growth in roots\nRequirements\nGerminating bean seeds with radicle of about 1cm in length, cork, pin, beaker or gas jar, water, Indian ink, blotting paper or filter paper, marker and ruler marked in mm. Procedure\nBent wire\nRuler\nTake the germinating been seed, andusing a blotting paper, dry the radicaltaking care not to damage the root. Using a marker and ruler make light inkmarks 2mm apart along the length ofthe root. See figure 4.10(a). 4. 5\nMake a drawing of the marked root. Pin the seedling onto the cork and place it in the beaker containing a little water. See figure 4.10(b). Leave it overnight. Take out the seedling and examine the ink marks. 6.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.8827781639825436, "ocr_used": false, "chunk_length": 2192, "token_count": 517}}
{"text": "Leave it overnight. Take out the seedling and examine the ink marks. 6. 7. Measure the distances between the successive ink marks and record. Make a well labelled drawing of the seedling at the end of the experiment and compare with the drawing of the. seedling at the start of the experiment. Study Question 7\nWhat part of the radicle has the inkmarks moved further apart? Give an explanation for your answersin (a) above. What is the increase in length withineach interval? Work out the rate of growth for theroot\nSecondary Growth\nSecondary growth results in an increase in width or girth due to activity of the cambium. In secondary growth new tissues are formed by vascular cambium and cork cambium. In monocotyledons plants there are no cambium cell in the vascular bundles.The growth in diameter is due to the enlargement of the primary cells. Secondary growth in dicotyledonous pjants begins with the division of vascular cambium to produce new cambium cells between the vascular bundles. This forms a continuous cambium ring. These cambium cells divide to form the new cells that are added to the older ones. The cambium cells have now become meristematic. The new cells produced to the outer side of cambium differentiate to become secondary phloem and those to the inner side differentiate to become the secondary xylem. More secondary xylem is formed than secondary phloem. The interfascuiar cambium a/so cuts orTparenchymatous cells which form secondary medullary rays as seen in figure 4.11 (a), (b) and (c). As a result of the increase in the volume of the secondary tissues, pressure is exerted on the outer cells of the stem. This results in stretching and rupturing of the epidermal cells. In order to replace the protective outer layer of the stem, a new band of cambium cells are formed in the cortex. These cells, called cork cambium orphellogen originate from the cortical cells. The cork cambium divides to produce new cells on either side. The cells on the inner side of the cork cambium differentiate into secondary cortex and those produced on the outer side become cork cells. Cork cells are dead with thickened walls. Their walls become coated with a waterproof substance called suberin. The cork cells increase in number and become the bark of the stem. This prevents loss of water, infection from fungi and damage from insects.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9155055103428517, "ocr_used": false, "chunk_length": 2356, "token_count": 510}}
{"text": "Their walls become coated with a waterproof substance called suberin. The cork cells increase in number and become the bark of the stem. This prevents loss of water, infection from fungi and damage from insects. The corky bark is also resistant to fire and thus acts as an insulatory layer. The bark is normally impermeable to water and respiratory gases. Periodically the cork cells, instead of being tightly packed, they form a loose mass. This mass is known as Jenticel. The lenticles make it possible for\nTransverse section of dicotyledonous stem\nEpidermis Primary phloem\nSecondary phloem\nSecondary xylem\nPrimary xy/em\nCortex\nBeginning of secondary growth in dicot stem\nPrimary phloem\nSecondary ph/oem\nSecondary xylem Primary xylem\nEpidermis Medullary ray\nCambium ring\nEpidermis\nCork cell Cork cambium Secondary cortex Primary cortex\nFig. 4.12: Section through a lenticel\nThe rate of secondary growth in a stem varies with seasonal changes. During rainy season, xylem vessels and tracheids are formed In large numbers. These cells are large, have thin walls and the wood has a light texture. In the dry season, the xylem and trancheids formed are few in number. They are small, thick-walled and their wood has a dark texture. This leads to the development of two distinctive layers within the secondary xylem formed m a year, called annual rings. See figure 4.13. It is possible to determine the age of a tree by counting the number of annual rings. Furthermore climatic changes of the past years can be infered from the size of the ring. Primary phloem Secondary phloem\nCambium ring Medullary ray Pith\nPrimary xylem Cortex\nFig. 4.13: Annual rings\nCork\nRole of Growth Hormones in Plants\nPlant hormones are chemicals produced in very small amounts within the plant body, and play a very important part in regulating plant growth and development. Most growth hormones are produced at the tip of a shoot and transported downwards to the root. The root tip produces very small quantities of the hormones. There are many different types of plant hormones and one well-known group is the auxins. Indoie acetic acid (IAA) is one best known auxin. Auxins are produced at the shoot and root tips.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9138975780446988, "ocr_used": false, "chunk_length": 2191, "token_count": 508}}
{"text": "There are many different types of plant hormones and one well-known group is the auxins. Indoie acetic acid (IAA) is one best known auxin. Auxins are produced at the shoot and root tips. Maximum influence on growth in plants occurs when auxins are produced simultaneously with other plant hormones e.g. gibberellins. Maximum growth response in stems requires more IAA than tn roots. Auxins are known to have various effects on the growth and development in plants. They stimulate cell division and cell elongation in stems and roots leading to primary growth. Auxins cause tropic responses, which are growth responses in plants due to external stimuli acting from a given direction. On the other hand IAA stimulates the growth of adventitious roots which develop from the stem rather than tbe main root. Cuttings can be encouraged to develop roots with the help of IAA. If the cut end of a stem is dipped into IAA, root sprouting is faster. IAA is also used to induce parthenocarpy. This is the growth of an ovary into a fruit without fertilisation. This is commonly u^ed by horticulturalists to bring about a good crop of fruits particularly pineapples. Auxins are known . to inhibit development of side branches from lateral buds. They therefore enhance apical dominance. During secondary growth auxins Play an important role by initiating cell division in the cambium and differentiation of these cambium cells into vascular tissues. Auxins in association with other plant hormones such as the cytokinins induce the formation of callus tissue which causes the healing of wounds. When the concentration of auxins falls in the plant, it promotes formation of an abscission layer leading to leaf fall. A synthetic auxin, 2,4-dichlorophenoxyacetic acid (2,4-D) induces distorted growth and excessive respiration leading to death of the plant. Hence it can be used as a selective weed killer. GibbereHins are another important group of plant growth hormone. GibbereHins are a mixture of compounds and have a very high effect on growth. The most important in growth is gibberellic acid. Gibbereilins are distinguished from auxins by their stimulation of rapid cell division and cell elongation in dwarf varieties of certain plants. Dwarf conditions are thought to be caused by a shortage of gibberellins due to a genetic deficiency. Gibberellins are important in fruit formation.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9201080573173598, "ocr_used": false, "chunk_length": 2376, "token_count": 503}}
{"text": "Gibbereilins are distinguished from auxins by their stimulation of rapid cell division and cell elongation in dwarf varieties of certain plants. Dwarf conditions are thought to be caused by a shortage of gibberellins due to a genetic deficiency. Gibberellins are important in fruit formation. They induce the growth of ovaries into fruits after fertilisation. They also induce parthenocarpy. Gibberellins also promote formation of side branches from lateral buds and breaks dormancy in buds. This is common in species of temperate plants whose buds become dormant in winter. In addition, this hormone also inhibits sprouting of adventitious roots from stem cuttings, it retards formation of abscission layer hence reduces leaf fall. Gibberellins also break seed dormancy by activating the enzymes involved in the breakdown of food substances during germination. Cytokuuns also known as kinetins, are growth substances which promote growth in plants when they interact with auxins. In the presence of auxins, they stimulate cell division thereby bringing about growth of\nroots, leaves and buds. They also stimulate formation of the callus tissues in plants. The callus tissue is used in the repair of wounds in damaged parts of plants. Cytokinins promote flowering and breaking of seed dormancy in some plant species. They also promote formation of adventitious roots from stems and stimulate lateral bud development in shoots. When in high concentration cytokinins induce cell enlargement of leaves but in low concentration they encourage leaf senescence and hence leaf fall. Ethylene is a growth substance produced in plants in gaseous form. Its major effect in plants is that it causes ripening and falling of fruits. This is widely applied in horticultural farms in ripening and harvesting of fruits. It stimulates formation of abscission layer leading to leaf fall, induces thickening of stems by promoting cell division and differentiation at the cambium meristem. But it inhibits stem elongation. Ethylene promotes breaking of seed dormancy in some seeds and flower formation mostly in pineapples. Abscisic acid is a plant hormone whose effects are inhibitory in nature. It inhibits seed germination leading to seed dormancy, inhibits sprouting of buds from stems and retards stem elongation. In high concentration, abscisic acid causes closing of the stomata. This effect is important in that it enables plants to reduce water loss. It also promotes leaf and fruit fall.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9301574485264434, "ocr_used": false, "chunk_length": 2477, "token_count": 502}}
{"text": "In high concentration, abscisic acid causes closing of the stomata. This effect is important in that it enables plants to reduce water loss. It also promotes leaf and fruit fall. Another hormone, florigen is produced in plants where it promotes flowering. Apical Dominance\nAlthough auxins, particularly IAA are important stem and root elongation, they are known to exert profound effects on other aspects of plant growth and development. If an apical bud which normally contains high concentrations of auxins is removed, it is\nobserved that more lateral buds lower down the stem sprout, producing many branches. This shows that high concentrations of auxins have an inhibitor}' effect on sprouting of lateral buds and therefore hinders growth of many branches. This forms the basis of pruning in agriculture where more branches are required for increased harvest particularly on crops like coffee and tea. The failure of lateral buds to develop in the presence of an apical bud is due to the diffusion of auxins from the shoot apex downwards in concentrations higher than that promoting lateral bud development. Practical Activity 6\nTo investigate apical dominance in plants\nRequirements\nTomato seedlings growing in a tin. Procedure\nCut off the terminal buds from 3seedlings in the tin, leaving the otherseedlings with the terminal buds intact,\nLeave the seedlings to continuegrowing for five more days. Study Questions 8\nlist the differences noticed betweenthe two groups of seedlings? Explainhow the differences come about. From your observations, explain thebasis for pruning tea and coffee. Growth and Development in Animals\nIn higher animals, most cells with the exception of the nerve cells, retain their power of division. Thus, there is a continued breakdown and replacement of cells. Animal cells undergo rapid cell division and cell differentiation but, unlike plant cells, they undergo very little cell enlargement. In most animals growth occurs through: their life till they die. This type of growth called continuous growth. Arthropods e.g. insects show rapid growth immediately after moulting with periods when no growth increase occurs. This is called discontinuous growth. Insects exhibit two types of reproducti processes. In some insects, the ova in t female are fertilised by the spermatozoa frc the male. This is a typical example of sexi reproduction, common in butterflies ai moths.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9230063227953411, "ocr_used": false, "chunk_length": 2404, "token_count": 481}}
{"text": "Insects exhibit two types of reproducti processes. In some insects, the ova in t female are fertilised by the spermatozoa frc the male. This is a typical example of sexi reproduction, common in butterflies ai moths. In other insects like the black and t green aphids, the eggs are usually product without being fertilised and are able to --- into adult insects. This type of asexual reproduction is referred to ; parthenogenesis. Growth and Development in Insects\nMajority of insects lay eggs that hatch int larvae, which is an immature stage, usual! quite different from the adults in morpholog and behaviour. Depending on the insec species a larva is referred to as a grub, maggot or a caterpillar. Generally the larv eats a lot, grows rapidly and sheds its cuticl several times until it reaches full size t< become a pupa. The pupa is an inactive, non feeding stage during which extensivi breakdown and re-organisation of body tissui occur, eventually giving rise to the imago o adult form. Such changes, callec metamorphosis, do occur in butterflies moths, bees, wasps and flies. Insects which pass through these stages, namely, egg-larva-pupa, into imago/adult in their developmenl are said to undergo complete metamorphosis. Development in a Housefly(An example of complete metamorphosis)\nWhen the egg of a housefly is laid, it measures about 1mm in length. The eggs are laid in batches of between 100 to 150. The larvae which hatch from the eggs grow and feed on decaying matter. After several moults and increase in size, a Jarva reaches about 1cm in length. This takes about 5 days. After this, the larva changes into a pupa encased in a pupal case called die puparium, from which the adult fly later emerges. After emergence, the adult tgkes about two weeks of feeding and growing to attain sexual maturity, i.e. the males can mate and the females are able to lay eggs. Figure 4.14 summarises the life cycle of a housefly. Incomplete Metamorphosis\nDevelopment in some insects like the locust and cockroaches, involves the.egg hatching into a nymph which e!cie!y resembles the adult in every form, except for size and lack of sexual maturity.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9062710835971356, "ocr_used": false, "chunk_length": 2151, "token_count": 504}}
{"text": "the males can mate and the females are able to lay eggs.\n\nFigure 4.14 summarises the life cycle of a housefly.\n\nIncomplete Metamorphosis\nDevelopment in some insects like the locust and cockroaches, involves the.egg hatching into a nymph which e!cie!y resembles the adult in every form, except for size and lack of sexual maturity.\n\nPupa case Adult pushes out against the case\nFor such insects to reach the adult, stages, they undergo a series of moults.\n\nbefore fully acquiring the adult size and attaining the sexual maturity.\n\nThese insects are said to undergo incomplete metamorphosis.\n\nDevelopment in a Cockroach(An example of incomplete metamorphosis)\nCockroaches produce eggs enclosed in a case in groups of between 10 - 15.\n\nThe case known as ootheca is made up of cfaitm.\n\nThe ootheca is usually deposited in moist dark and warm places, for example in cracks of furniture or crevices in walls.\n\nIt takes about a month before the small wingless nymphs emerge.\n\nThe nymphs feed, and moult about ten times with the total nymphal period lasting about 16 days for all the adult structure to become fully developed.\n\nRole of Hormones in InsectMetamorphosis\nIn insects metamorphosis is controlled by hormones.\n\nThe hormones are produced in three glands namely;\nNeurosecretory cells in the brain ganglia, a pair of corpora allata (singular Corpus allatum) located in the mandibular segment and prothoracic glands in the thorax.\n\nDuring larval stages of the insect the corpora ailata produces juvenile hormone,\nThis leads to formation of larval cuticle., therefore moulting does not go beyond the larval stage.\n\nWhen the larva matures, the corpus allatum disintegrates-\nAt this time the neurosecretory cells stimulate the prothoracic glands to produce moulting hormone (ecdysone).\n\nEcdysone is responsible for moulting in insects leading to the laying of the adult cuticle.", "metadata": {"source": "FORM-3-BIOLOGY.docx", "file_type": "docx", "language": "en", "quality_score": 0.9123504273504273, "ocr_used": false, "chunk_length": 1872, "token_count": 427}}
{"text": "Study the equation below. Hydrated copper (II) sulphate white Solid T + Colourless Liquid H a) Identify the; i).White solid T ............................................................... ii). C olourless liquid H .......................................................{1 m ark} b) If H ydrated C opper (II) Sulphate had FIVE w ater m olecules, w rite the chemical equation for the above reaction. {1 m ark} .......................................................................................................................... .......................................................................................................... 10. During the 2012 London O lympic G ames, samples from four D ecathlon participants (Morgan, Bolton, Jimmy and Jade) w ere taken and tested for presence of two illegal steroids A and B. Paper chromatography w as used for the test. Steroid A Steroid B Morgan Bolton Jimmy Jade a) Which athlete(s) tested positive for the illegal steroid? {1 m ark} ........................................................................................................................... ............................................................................... b) On the filter paper representation below, draw the results for the Bolton.{2 marks} 1 1 . T h e c u r v e s b e l o w r e p r e s e n t t h e v a r i a t i o n o f t e m p e r a t u r e w i t h t i m e w h e n\npure and impure samples of a solid w ere heated. Solid K Temperature in ˚C Solid F Time in seconds a) Which of the two curves shows the variation in temperature for pure solid? Explain.{2 m arks} b) If 300 grams m ore of the pure substance w as added to the sample, show on the graph the time that the sample the pure substance w ill boil. {½ m ark}\nc) On the graph above, indicate the boiling point of the pure substance.{½ m ark} 12. Arnold, a student from Starehe Boys’ C entre Situated 3050m above the sea level boiled 100cm3 of pure w ater. Another student, Annette, from Mombasa 0 m etres above the sea level boiled the same volume of pure water. i. Which of the two students took the longest time to boil w ater? {1mark} .................................................................................................................. ii. Explain your answer in d, i) above? {1 m ark} ........................................................................................................................... ........................................... 13.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET-2.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6965043377674958, "ocr_used": false, "chunk_length": 2470, "token_count": 504}}
{"text": "Introduction​ ​to​ ​chemistry 1. Wooden​ ​splints​ ​F​ ​and​ ​G​ ​were​ ​placed​ ​in​ ​different​ ​zones​ ​of​ ​a​ ​Bunsen​ ​burner​ ​flame. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​The​ ​diagram​ ​below​ ​ ​gives​ ​the​ ​observations​ ​that​ ​were​ ​made (a)​ ​Explain​ ​the​ ​difference​ ​between​ ​F​ ​and​ ​G (b)​ ​Name​ ​the​ ​type​ ​of​ ​flame​ ​that​ ​was​ ​used​ ​in​ ​the​ ​above​ ​experiment 2. The diagrams​ ​below​ ​represent​ ​a​ ​list​ ​of​ ​apparatus​ ​which​ ​are​ ​commonly​ ​used​ ​in​ ​a​ ​chemistry laboratory:- (a)​ ​Give​ ​the​ ​correct​ ​order​ ​of​ ​the​ ​apparatus,​ ​using​ ​the​ ​letters​ ​only​,​ ​to​ ​show​ ​the​ ​correct arrangement ​ ​ ​ ​ ​ ​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prepare​ ​and​ ​investigate​ ​the​ ​nature​ ​of​ ​PH​ ​of​ ​a​ ​sample​ ​of​ ​onion solution (b)​ ​Name​ ​one​ ​chemical​ ​substance​ ​and​ ​apparatus​ ​that​ ​is​ ​needed​ ​in​ ​this​ ​experiment 3. (a)​ ​When​ ​the​ ​air-hole​ ​is​ ​fully​ ​opened,​ ​the​ ​bunsen​ ​burner​ ​produces​ ​a​ ​non-luminous​ ​flame. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain (b)​ ​Draw​ ​a​ ​labelled​ ​diagram​ ​of​ ​anon-luminous​ ​flame 4. (a)​ ​What​ ​is​ ​a​ ​drug?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7587756288217118, "ocr_used": false, "chunk_length": 1147, "token_count": 496}}
{"text": "(a)​ ​When​ ​the​ ​air-hole​ ​is​ ​fully​ ​opened,​ ​the​ ​bunsen​ ​burner​ ​produces​ ​a​ ​non-luminous​ ​flame. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain (b)​ ​Draw​ ​a​ ​labelled​ ​diagram​ ​of​ ​anon-luminous​ ​flame 4. (a)​ ​What​ ​is​ ​a​ ​drug? (b)​ ​Give​ ​two​ ​drugs​ ​that​ ​are​ ​commonly​ ​abused​ ​by​ ​the​ ​youth. 5. The​ ​diagram​ ​below​ ​shows​ ​three​ ​methods​ ​for​ ​collecting​ ​gases​ ​in​ ​the​ ​laboratory\n(a)​ ​Name​ ​the​ ​methods​ ​A​ ​and​ ​B (b)​ ​From​ ​the​ ​methods​ ​above,​ ​identify​ ​one​ ​that​ ​is​ ​suitable​ ​for​ ​collecting​ ​sulphur​ ​(IV) oxide. ​ ​ ​ ​ ​Explain 6. A​ ​mixture​ ​of​ ​hexane​ ​and​ ​water​ ​was​ ​shaken​ ​and​ ​left​ ​to​ ​separate​ ​as​ ​shown​ ​in​ ​the​ ​diagram below: State​ ​the​ ​identity​ ​of; (i)​ ​P​ ​………………………………..……..​ ​ ​ ​ ​ ​(ii)​ ​W​ ​………………………………….…. 7. The​ ​diagrams​ ​below​ ​are​ ​some​ ​common​ ​laboratory​ ​apparatus.​ ​Name​ ​each​ ​apparatus​ ​and ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​state​ ​its​ ​use Diagram Name Use\n(½mk​ ​) (½mk) (½mk) (½mk) 8.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7161605480676149, "ocr_used": false, "chunk_length": 1033, "token_count": 474}}
{"text": "A​ ​mixture​ ​of​ ​hexane​ ​and​ ​water​ ​was​ ​shaken​ ​and​ ​left​ ​to​ ​separate​ ​as​ ​shown​ ​in​ ​the​ ​diagram below: State​ ​the​ ​identity​ ​of; (i)​ ​P​ ​………………………………..……..​ ​ ​ ​ ​ ​(ii)​ ​W​ ​………………………………….…. 7. The​ ​diagrams​ ​below​ ​are​ ​some​ ​common​ ​laboratory​ ​apparatus.​ ​Name​ ​each​ ​apparatus​ ​and ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​state​ ​its​ ​use Diagram Name Use\n(½mk​ ​) (½mk) (½mk) (½mk) 8. The diagram​ ​below shows​ ​some parts​ ​of​ ​a​ ​Bunsen burner Explain​ ​how​ ​the​ ​parts​ ​labelled​ ​T​ ​and​ ​U​ ​are​ ​suited​ ​to​ ​their​ ​functions\n9. The​ ​diagram​ ​below​ ​shows​ ​the​ ​appearance​ ​of​ ​two​ ​pieces​ ​of​ ​paper​ ​placed​ ​in​ ​different​ ​parts of​ ​a non-luminous​ ​flame​ ​of​ ​a​ ​Bunsen​ ​burner​ ​and​ ​removed​ ​quickly​ ​before​ ​they​ ​caught​ ​fire. (a)​ ​What​ ​do​ ​the​ ​experiments​ ​show​ ​about​ ​the​ ​outer​ ​region​ ​of​ ​the​ ​flame? (b)​ ​From​ ​the​ ​above​ ​experiment,​ ​which​ ​part​ ​of​ ​the​ ​flame​ ​is​ ​better​ ​to​ ​use​ ​for​ ​heating?Give a​ ​reason 10.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7397459571372615, "ocr_used": false, "chunk_length": 1036, "token_count": 449}}
{"text": "The​ ​diagram​ ​below​ ​shows​ ​the​ ​appearance​ ​of​ ​two​ ​pieces​ ​of​ ​paper​ ​placed​ ​in​ ​different​ ​parts of​ ​a non-luminous​ ​flame​ ​of​ ​a​ ​Bunsen​ ​burner​ ​and​ ​removed​ ​quickly​ ​before​ ​they​ ​caught​ ​fire. (a)​ ​What​ ​do​ ​the​ ​experiments​ ​show​ ​about​ ​the​ ​outer​ ​region​ ​of​ ​the​ ​flame? (b)​ ​From​ ​the​ ​above​ ​experiment,​ ​which​ ​part​ ​of​ ​the​ ​flame​ ​is​ ​better​ ​to​ ​use​ ​for​ ​heating?Give a​ ​reason 10. A​ ​crystal​ ​of​ ​copper​ ​(II)​ ​sulphate​ ​was​ ​placed​ ​in​ ​a​ ​beaker​ ​of​ ​water.​ ​The​ ​beaker​ ​was​ ​left standing​ ​for ​ ​two​ ​days​ ​without​ ​shaking.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​that​ ​were​ ​made. 11. Study​ ​the​ ​information​ ​in​ ​the​ ​table​ ​below​ ​and​ ​answer​ ​questions​ ​that​ ​follow.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7807538016606027, "ocr_used": false, "chunk_length": 794, "token_count": 336}}
{"text": "A​ ​crystal​ ​of​ ​copper​ ​(II)​ ​sulphate​ ​was​ ​placed​ ​in​ ​a​ ​beaker​ ​of​ ​water.​ ​The​ ​beaker​ ​was​ ​left standing​ ​for ​ ​two​ ​days​ ​without​ ​shaking.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​that​ ​were​ ​made. 11. Study​ ​the​ ​information​ ​in​ ​the​ ​table​ ​below​ ​and​ ​answer​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(Letters​ ​given​ ​are​ ​not​ ​real​ ​symbols) Ions Electron​ ​arrangement Ionic​ ​radius​ ​(nm) A​+ B​+ C​2+ 2.8 2.8.8 2.8 0.95 0.133 0.065 Explain​ ​why​ ​the​ ​ionic​ ​radius​ ​of​ ​:- (a)​ ​B​+​​ ​is​ ​greater​ ​than​ ​that​ ​of​ ​A​+ ​ ​ ​(b)​ ​C​2+​​ ​is​ ​smaller​ ​than​ ​the​ ​of​ ​A​+ ​ ​Simple​ ​classification​ ​of​ ​substances 1. The​ ​diagram​ ​below​ ​shows​ ​the​ ​heating​ ​curve​ ​of​ ​a​ ​pure​ ​substance.​ ​Study​ ​it​ ​and​ ​answer​ ​the ​ ​questions​ ​that​ ​follow:\n(a)​ ​What​ ​physical​ ​changes​ ​are​ ​taking​ ​place​ ​at​ ​points​ ​X​ ​and​ ​Z​? (b)Explain​ ​what​ ​happens​ ​to​ ​the​ ​melting​ ​point​ ​of​ ​sodium​ ​chloride​ ​added​ ​to​ ​this​ ​substance 2.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.714832902912373, "ocr_used": false, "chunk_length": 1057, "token_count": 471}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(Letters​ ​given​ ​are​ ​not​ ​real​ ​symbols) Ions Electron​ ​arrangement Ionic​ ​radius​ ​(nm) A​+ B​+ C​2+ 2.8 2.8.8 2.8 0.95 0.133 0.065 Explain​ ​why​ ​the​ ​ionic​ ​radius​ ​of​ ​:- (a)​ ​B​+​​ ​is​ ​greater​ ​than​ ​that​ ​of​ ​A​+ ​ ​ ​(b)​ ​C​2+​​ ​is​ ​smaller​ ​than​ ​the​ ​of​ ​A​+ ​ ​Simple​ ​classification​ ​of​ ​substances 1. The​ ​diagram​ ​below​ ​shows​ ​the​ ​heating​ ​curve​ ​of​ ​a​ ​pure​ ​substance.​ ​Study​ ​it​ ​and​ ​answer​ ​the ​ ​questions​ ​that​ ​follow:\n(a)​ ​What​ ​physical​ ​changes​ ​are​ ​taking​ ​place​ ​at​ ​points​ ​X​ ​and​ ​Z​? (b)Explain​ ​what​ ​happens​ ​to​ ​the​ ​melting​ ​point​ ​of​ ​sodium​ ​chloride​ ​added​ ​to​ ​this​ ​substance 2. (a)​ ​State​ ​two​ ​differences​ ​between​ ​luminous​ ​flame​ ​and​ ​non-luminous​ ​flame (b)​ ​It​ ​is​ ​advisable​ ​to​ ​set​ ​a​ ​Bunsen​ ​burner​ ​to​ ​luminous​ ​flame​ ​prior​ ​to​ ​an​ ​experiment. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain 3.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6969437983447828, "ocr_used": false, "chunk_length": 973, "token_count": 450}}
{"text": "(b)Explain​ ​what​ ​happens​ ​to​ ​the​ ​melting​ ​point​ ​of​ ​sodium​ ​chloride​ ​added​ ​to​ ​this​ ​substance 2. (a)​ ​State​ ​two​ ​differences​ ​between​ ​luminous​ ​flame​ ​and​ ​non-luminous​ ​flame (b)​ ​It​ ​is​ ​advisable​ ​to​ ​set​ ​a​ ​Bunsen​ ​burner​ ​to​ ​luminous​ ​flame​ ​prior​ ​to​ ​an​ ​experiment. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain 3. The​ ​paper​ ​chromatography​ ​of​ ​a​ ​plant​ ​extract​ ​gave​ ​the​ ​following​ ​results: Solvent Number​ ​of​ ​spots X 6 Y 2 Z 3 (a)​ ​Which​ ​is​ ​the​ ​most​ ​suitable​ ​solvent​ ​for​ ​purifying​ ​the​ ​extract?​ ​Explain ​ ​(b)​ ​Ball​ ​pen​ ​cannot​ ​be​ ​used​ ​to​ ​mark​ ​solvent​ ​front​ ​in​ ​the​ ​above​ ​chromatography.​ ​Explain 4. Name​ ​the​ ​process​ ​which​ ​takes​ ​place​ ​when: ​ ​(a)​ ​Solid​ ​Carbon​ ​(Iv)​ ​Oxide​ ​(dry​ ​ice)​ ​changes​ ​directly​ ​into​ ​gas (b)​ ​A​ ​red​ ​litmus​ ​paper​ ​turns​ ​white​ ​when​ ​dropped​ ​into​ ​chlorine​ ​water (c)​ ​Propene​ ​gas​ ​molecules​ ​are​ ​converted​ ​into​ ​a​ ​giant​ ​molecule 5. A​ ​sample​ ​of​ ​copper​ ​turnings​ ​was​ ​found​ ​to​ ​be​ ​contaminated​ ​with​ ​copper​ ​(II)​ ​oxide.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7611729043063895, "ocr_used": false, "chunk_length": 1138, "token_count": 508}}
{"text": "The​ ​paper​ ​chromatography​ ​of​ ​a​ ​plant​ ​extract​ ​gave​ ​the​ ​following​ ​results: Solvent Number​ ​of​ ​spots X 6 Y 2 Z 3 (a)​ ​Which​ ​is​ ​the​ ​most​ ​suitable​ ​solvent​ ​for​ ​purifying​ ​the​ ​extract?​ ​Explain ​ ​(b)​ ​Ball​ ​pen​ ​cannot​ ​be​ ​used​ ​to​ ​mark​ ​solvent​ ​front​ ​in​ ​the​ ​above​ ​chromatography.​ ​Explain 4. Name​ ​the​ ​process​ ​which​ ​takes​ ​place​ ​when: ​ ​(a)​ ​Solid​ ​Carbon​ ​(Iv)​ ​Oxide​ ​(dry​ ​ice)​ ​changes​ ​directly​ ​into​ ​gas (b)​ ​A​ ​red​ ​litmus​ ​paper​ ​turns​ ​white​ ​when​ ​dropped​ ​into​ ​chlorine​ ​water (c)​ ​Propene​ ​gas​ ​molecules​ ​are​ ​converted​ ​into​ ​a​ ​giant​ ​molecule 5. A​ ​sample​ ​of​ ​copper​ ​turnings​ ​was​ ​found​ ​to​ ​be​ ​contaminated​ ​with​ ​copper​ ​(II)​ ​oxide. Describe how​ ​a​ ​sample​ ​of​ ​copper​ ​metal​ ​can​ ​be​ ​separated​ ​from​ ​the​ ​mixture 6. Copper​ ​(II)​ ​oxide​ ​and​ ​charcoal​ ​are​ ​black​ ​solids.​ ​How​ ​would​ ​you​ ​distinguish​ ​between​ ​the two​ ​solids? 7. a)​ ​What​ ​is​ ​chromatography? b)​ ​Give​ ​two​ ​applications​ ​of​ ​chromatography 8.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.768672597684881, "ocr_used": false, "chunk_length": 1084, "token_count": 469}}
{"text": "7. a)​ ​What​ ​is​ ​chromatography? b)​ ​Give​ ​two​ ​applications​ ​of​ ​chromatography 8. The​ ​two​ ​elements​ ​P​ ​and​ ​R​ ​were​ ​separately​ ​burned​ ​in​ ​air,​ ​the​ ​products​ ​gave​ ​the​ ​results ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​recorded​ ​in​ ​the​ ​table​ ​below: ELEMENTS​ ​ ​PHYSICAL STATE​ ​AT​ ​ROOM TEMPERATURE P​ ​SOLID R​ ​SOLID Physical​ ​states​ ​of​ ​products White​ ​solid​ ​powder only Colourless​ ​gases​ ​L​ ​and​ ​M Nature​ ​of​ ​solutions​ ​in​ ​water Basic L​ ​strongly​ ​acidic​ ​M​ ​slightly acidic (a)​ ​Suggest​ ​the​ ​identity​ ​of​ ​element​ ​R.​ ​……………………………………………..…….. (b)​ ​Describe​ ​how​ ​the​ ​nature​ ​of​ ​the​ ​solutions​ ​of​ ​the​ ​of​ ​the​ ​oxides​ ​were​ ​determined 9 The​ ​diagram​ ​below​ ​represents​ ​a​ ​paper​ ​chromatography​ ​for​ ​the​ ​three​ ​brands​ ​of​ ​soft​ ​drinks containing​ ​banned​ ​artificial​ ​food​ ​additives.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7788520012410798, "ocr_used": false, "chunk_length": 879, "token_count": 352}}
{"text": "b)​ ​Give​ ​two​ ​applications​ ​of​ ​chromatography 8. The​ ​two​ ​elements​ ​P​ ​and​ ​R​ ​were​ ​separately​ ​burned​ ​in​ ​air,​ ​the​ ​products​ ​gave​ ​the​ ​results ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​recorded​ ​in​ ​the​ ​table​ ​below: ELEMENTS​ ​ ​PHYSICAL STATE​ ​AT​ ​ROOM TEMPERATURE P​ ​SOLID R​ ​SOLID Physical​ ​states​ ​of​ ​products White​ ​solid​ ​powder only Colourless​ ​gases​ ​L​ ​and​ ​M Nature​ ​of​ ​solutions​ ​in​ ​water Basic L​ ​strongly​ ​acidic​ ​M​ ​slightly acidic (a)​ ​Suggest​ ​the​ ​identity​ ​of​ ​element​ ​R.​ ​……………………………………………..…….. (b)​ ​Describe​ ​how​ ​the​ ​nature​ ​of​ ​the​ ​solutions​ ​of​ ​the​ ​of​ ​the​ ​oxides​ ​were​ ​determined 9 The​ ​diagram​ ​below​ ​represents​ ​a​ ​paper​ ​chromatography​ ​for​ ​the​ ​three​ ​brands​ ​of​ ​soft​ ​drinks containing​ ​banned​ ​artificial​ ​food​ ​additives. 4 1 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​6 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​2 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​5 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​3 A B C BRANDS​ ​OF​ ​SOFT​ ​DRINKS A​ ​and​ ​C​ ​found​ ​to​ ​contain​ ​the​ ​banned​ ​artificial​ ​food​ ​additives.​ ​Which​ ​numbers​ ​indicate the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​banned​ ​artificial​ ​food​ ​additives?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6904027816240501, "ocr_used": false, "chunk_length": 1413, "token_count": 608}}
{"text": "The​ ​two​ ​elements​ ​P​ ​and​ ​R​ ​were​ ​separately​ ​burned​ ​in​ ​air,​ ​the​ ​products​ ​gave​ ​the​ ​results ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​recorded​ ​in​ ​the​ ​table​ ​below: ELEMENTS​ ​ ​PHYSICAL STATE​ ​AT​ ​ROOM TEMPERATURE P​ ​SOLID R​ ​SOLID Physical​ ​states​ ​of​ ​products White​ ​solid​ ​powder only Colourless​ ​gases​ ​L​ ​and​ ​M Nature​ ​of​ ​solutions​ ​in​ ​water Basic L​ ​strongly​ ​acidic​ ​M​ ​slightly acidic (a)​ ​Suggest​ ​the​ ​identity​ ​of​ ​element​ ​R.​ ​……………………………………………..…….. (b)​ ​Describe​ ​how​ ​the​ ​nature​ ​of​ ​the​ ​solutions​ ​of​ ​the​ ​of​ ​the​ ​oxides​ ​were​ ​determined 9 The​ ​diagram​ ​below​ ​represents​ ​a​ ​paper​ ​chromatography​ ​for​ ​the​ ​three​ ​brands​ ​of​ ​soft​ ​drinks containing​ ​banned​ ​artificial​ ​food​ ​additives. 4 1 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​6 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​2 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​5 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​3 A B C BRANDS​ ​OF​ ​SOFT​ ​DRINKS A​ ​and​ ​C​ ​found​ ​to​ ​contain​ ​the​ ​banned​ ​artificial​ ​food​ ​additives.​ ​Which​ ​numbers​ ​indicate the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​banned​ ​artificial​ ​food​ ​additives? 10.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6847037606038341, "ocr_used": false, "chunk_length": 1361, "token_count": 589}}
{"text": "(b)​ ​Describe​ ​how​ ​the​ ​nature​ ​of​ ​the​ ​solutions​ ​of​ ​the​ ​of​ ​the​ ​oxides​ ​were​ ​determined 9 The​ ​diagram​ ​below​ ​represents​ ​a​ ​paper​ ​chromatography​ ​for​ ​the​ ​three​ ​brands​ ​of​ ​soft​ ​drinks containing​ ​banned​ ​artificial​ ​food​ ​additives. 4 1 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​6 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​2 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​5 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​3 A B C BRANDS​ ​OF​ ​SOFT​ ​DRINKS A​ ​and​ ​C​ ​found​ ​to​ ​contain​ ​the​ ​banned​ ​artificial​ ​food​ ​additives.​ ​Which​ ​numbers​ ​indicate the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​banned​ ​artificial​ ​food​ ​additives? 10. Without​ ​using​ ​any​ ​laboratory​ ​chemical,​ ​describe​ ​a​ ​simple​ ​laboratory​ ​experiment​ ​to distinguish between​ ​calcium​ ​hydrogen​ ​carbonate​ ​and​ ​sodium​ ​hydrogen​ ​carbonate 11.​ ​Substance​ ​Q​ ​has​ ​a​ ​melting​ ​point​ ​of​ ​15​o​C​ ​and​ ​boiling​ ​point​ ​of​ ​70​o​C.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6631681101270571, "ocr_used": false, "chunk_length": 1146, "token_count": 505}}
{"text": "10. Without​ ​using​ ​any​ ​laboratory​ ​chemical,​ ​describe​ ​a​ ​simple​ ​laboratory​ ​experiment​ ​to distinguish between​ ​calcium​ ​hydrogen​ ​carbonate​ ​and​ ​sodium​ ​hydrogen​ ​carbonate 11.​ ​Substance​ ​Q​ ​has​ ​a​ ​melting​ ​point​ ​of​ ​15​o​C​ ​and​ ​boiling​ ​point​ ​of​ ​70​o​C. (a)​ ​On​ ​the​ ​same​ ​axes,​ ​draw​ ​the​ ​melting​ ​point​ ​and​ ​boiling​ ​point​ ​graph​ ​for​ ​Q​ ​and​ ​the​ ​room ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​temperature ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​State​ ​the​ ​physical​ ​state​ ​of​ ​substance​ ​Q​ ​at​ ​room​ ​temperature 12. Cooking​ ​oils​ ​comprise​ ​of​ ​a​ ​mixture​ ​of​ ​compounds​ ​which​ ​have​ ​a​ ​boiling​ ​point​ ​range ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​of​ ​23​o​C​ ​to​ ​27​o​C. (i)​ ​What​ ​evidence​ ​is​ ​then​ ​to​ ​support​ ​the​ ​statement​ ​that​ ​cooking​ ​oil​ ​is​ ​a​ ​mixture? (ii)Name​ ​another​ ​experimental​ ​technique​ ​that​ ​could​ ​be​ ​used​ ​to​ ​confirm​ ​your​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​part​ ​(i)​ ​above 13.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7269024533856723, "ocr_used": false, "chunk_length": 1019, "token_count": 446}}
{"text": "Cooking​ ​oils​ ​comprise​ ​of​ ​a​ ​mixture​ ​of​ ​compounds​ ​which​ ​have​ ​a​ ​boiling​ ​point​ ​range ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​of​ ​23​o​C​ ​to​ ​27​o​C. (i)​ ​What​ ​evidence​ ​is​ ​then​ ​to​ ​support​ ​the​ ​statement​ ​that​ ​cooking​ ​oil​ ​is​ ​a​ ​mixture? (ii)Name​ ​another​ ​experimental​ ​technique​ ​that​ ​could​ ​be​ ​used​ ​to​ ​confirm​ ​your​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​part​ ​(i)​ ​above 13. A​ ​form​ ​1​ ​student​ ​carried​ ​out​ ​the​ ​separation​ ​as​ ​shown​ ​in​ ​the​ ​set-up​ ​below:- ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​ ​Identify​ ​the​ ​method​ ​above................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Give​ ​one​ ​of​ ​its​ ​disadvantages ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Name​ ​a​ ​mixture​ ​which​ ​can​ ​be​ ​separated​ ​by​ ​the​ ​set-up​ ​above 14. What​ ​is​ ​meant​ ​by​ ​melting​ ​point​ ​and​ ​boiling​ ​point​ ​of​ ​a​ ​substance? 15.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6843178621659635, "ocr_used": false, "chunk_length": 948, "token_count": 400}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Give​ ​one​ ​of​ ​its​ ​disadvantages ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Name​ ​a​ ​mixture​ ​which​ ​can​ ​be​ ​separated​ ​by​ ​the​ ​set-up​ ​above 14. What​ ​is​ ​meant​ ​by​ ​melting​ ​point​ ​and​ ​boiling​ ​point​ ​of​ ​a​ ​substance? 15. The​ ​apparatus​ ​below​ ​were​ ​used​ ​by​ ​a​ ​student​ ​to​ ​study​ ​the​ ​effect​ ​of​ ​heat​ ​on​ ​hydrated ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​copper​ ​II​ ​sulphate\n(a)​ ​What​ ​is​ ​the​ ​role​ ​of​ ​the​ ​ice​ ​cold​ ​water…………… (b)​ ​Name​ ​liquid​ ​P………………………………………………………… (c)​ ​What​ ​observation​ ​is​ ​made​ ​in​ ​the​ ​boiling​ ​tube\n17. The​ ​diagram​ ​below​ ​shows​ ​chromatograms​ ​of​ ​blood​ ​samples​ ​obtained​ ​from​ ​three​ ​athletes. One athlete​ ​used illegal​ ​drug to improve performance in competition. (a)​ ​Name​ ​the​ ​line​ ​marked​ ​ ​M …………………………………………………. (b)Identify​ ​the​ ​athlete​ ​who​ ​used​ ​illegal​ ​drug ……………... ………………………. 18.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7280197802407615, "ocr_used": false, "chunk_length": 927, "token_count": 388}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​Displacement​ ​reaction​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​………………………………………… 19. Give​ ​two​ ​reasons​ ​why​ ​a​ ​luminous​ ​flame​ ​is​ ​not​ ​used​ ​for​ ​heating​ ​purposes 20. Classify​ ​the​ ​following​ ​processes​ ​as​ ​chemical​ ​changes​ ​or​ ​physical​ ​changes ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Process​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​physical​ ​or​ ​chemical ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Neutralization​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​……………………………………… ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sublimation​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​……………………………………… ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Fractional​ ​distillation​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​……………………………………….. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Displacement​ ​reaction​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​………………………………………… 21. Give​ ​two​ ​reasons​ ​why​ ​a​ ​luminous​ ​flame​ ​is​ ​not​ ​used​ ​for​ ​heating​ ​purposes 22. State​ ​two​ ​criteria​ ​for​ ​determining​ ​the​ ​purity​ ​of​ ​a​ ​substance Substance Water Concentrated sulphuric(VI)acid Concentrated sodium​ ​hydroxide\nEthene Slightly soluble Soluble Insoluble Ammonia Very​ ​ ​soluble Very​ ​ ​soluble Very​ ​ ​soluble Hydrogen Slightly​ ​soluble Insoluble Insoluble 23.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6836306122635559, "ocr_used": false, "chunk_length": 1229, "token_count": 484}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​Displacement​ ​reaction​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​………………………………………… 21. Give​ ​two​ ​reasons​ ​why​ ​a​ ​luminous​ ​flame​ ​is​ ​not​ ​used​ ​for​ ​heating​ ​purposes 22. State​ ​two​ ​criteria​ ​for​ ​determining​ ​the​ ​purity​ ​of​ ​a​ ​substance Substance Water Concentrated sulphuric(VI)acid Concentrated sodium​ ​hydroxide\nEthene Slightly soluble Soluble Insoluble Ammonia Very​ ​ ​soluble Very​ ​ ​soluble Very​ ​ ​soluble Hydrogen Slightly​ ​soluble Insoluble Insoluble 23. Study​ ​the​ ​information​ ​in​ ​the​ ​table​ ​below​ ​and​ ​answer​ ​the​ ​questions. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​i)​ ​A​ ​mixture​ ​contains​ ​ethene,​ ​Hydrogen​ ​and​ ​ammonia​ ​gases.​ ​Explain​ ​how​ ​a​ ​sample​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​ ​ ​gas​ ​can​ ​be​ ​obtained​ ​ ​from​ ​ ​this​ ​ ​mixture. ​ ​ ​24.a)i)​ ​The​ ​diagram​ ​below​ ​show​ ​spots​ ​of​ ​a​ ​pure​ ​substance​ ​A,​ ​B,​ ​and​ ​C​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​Spot​ ​D​ ​is​ ​that​ ​of​ ​a​ ​mixture ​ ​ ​ ​After​ ​development​ ​A,​ ​B​,​ ​and​ ​C​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​8cm,​ ​3cm​ ​and​ ​6cm​ ​respectively.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7325516887119558, "ocr_used": false, "chunk_length": 1198, "token_count": 505}}
{"text": "Study​ ​the​ ​information​ ​in​ ​the​ ​table​ ​below​ ​and​ ​answer​ ​the​ ​questions. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​i)​ ​A​ ​mixture​ ​contains​ ​ethene,​ ​Hydrogen​ ​and​ ​ammonia​ ​gases.​ ​Explain​ ​how​ ​a​ ​sample​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​ ​ ​gas​ ​can​ ​be​ ​obtained​ ​ ​from​ ​ ​this​ ​ ​mixture. ​ ​ ​24.a)i)​ ​The​ ​diagram​ ​below​ ​show​ ​spots​ ​of​ ​a​ ​pure​ ​substance​ ​A,​ ​B,​ ​and​ ​C​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​Spot​ ​D​ ​is​ ​that​ ​of​ ​a​ ​mixture ​ ​ ​ ​After​ ​development​ ​A,​ ​B​,​ ​and​ ​C​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​8cm,​ ​3cm​ ​and​ ​6cm​ ​respectively. D​ ​had​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​had​ ​moved​ ​6cm​ ​and​ ​8cm On​ ​the​ ​diagram​ ​above; ​ ​ ​I.​ ​Label​ ​the​ ​baseline​ ​(origin) ​ ​II.​ ​Show​ ​the​ ​positions​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development ​ ​ ​ii)​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​D b)​ ​Describe​ ​how​ ​solid​ ​ammonium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​solid​ ​mixture​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ammonium​ ​chloride​ ​and​ ​anhydrous​ ​calcium​ ​chloride c)​ ​The​ ​table​ ​below​ ​shows​ ​liquids​ ​that​ ​are​ ​miscible​ ​and​ ​those​ ​that​ ​are​ ​immiscible Liquid L​3 L​4 L​1 Miscible Miscible L​2 Miscible Immiscible Use​ ​the​ ​information​ ​given​ ​in​ ​the​ ​table​ ​to​ ​answer​ ​that​ ​questions​ ​that​ ​follow; i)​ ​Name​ ​the​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​separate​ ​L​1​​ ​and​ ​L​2​ ​from​ ​a​ ​mixture​ ​of​ ​the​ ​two ii)​ ​Describe​ ​how​ ​a​ ​mixture​ ​of​ ​L​2​​ ​and​ ​L​4​​ ​can​ ​be​ ​separated 25.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7208479905891935, "ocr_used": false, "chunk_length": 1676, "token_count": 751}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​i)​ ​A​ ​mixture​ ​contains​ ​ethene,​ ​Hydrogen​ ​and​ ​ammonia​ ​gases.​ ​Explain​ ​how​ ​a​ ​sample​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​ ​ ​gas​ ​can​ ​be​ ​obtained​ ​ ​from​ ​ ​this​ ​ ​mixture. ​ ​ ​24.a)i)​ ​The​ ​diagram​ ​below​ ​show​ ​spots​ ​of​ ​a​ ​pure​ ​substance​ ​A,​ ​B,​ ​and​ ​C​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​Spot​ ​D​ ​is​ ​that​ ​of​ ​a​ ​mixture ​ ​ ​ ​After​ ​development​ ​A,​ ​B​,​ ​and​ ​C​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​8cm,​ ​3cm​ ​and​ ​6cm​ ​respectively. D​ ​had​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​had​ ​moved​ ​6cm​ ​and​ ​8cm On​ ​the​ ​diagram​ ​above; ​ ​ ​I.​ ​Label​ ​the​ ​baseline​ ​(origin) ​ ​II.​ ​Show​ ​the​ ​positions​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development ​ ​ ​ii)​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​D b)​ ​Describe​ ​how​ ​solid​ ​ammonium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​solid​ ​mixture​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ammonium​ ​chloride​ ​and​ ​anhydrous​ ​calcium​ ​chloride c)​ ​The​ ​table​ ​below​ ​shows​ ​liquids​ ​that​ ​are​ ​miscible​ ​and​ ​those​ ​that​ ​are​ ​immiscible Liquid L​3 L​4 L​1 Miscible Miscible L​2 Miscible Immiscible Use​ ​the​ ​information​ ​given​ ​in​ ​the​ ​table​ ​to​ ​answer​ ​that​ ​questions​ ​that​ ​follow; i)​ ​Name​ ​the​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​separate​ ​L​1​​ ​and​ ​L​2​ ​from​ ​a​ ​mixture​ ​of​ ​the​ ​two ii)​ ​Describe​ ​how​ ​a​ ​mixture​ ​of​ ​L​2​​ ​and​ ​L​4​​ ​can​ ​be​ ​separated 25. A​ ​student​ ​left​ ​some​ ​crushed​ ​fruit​ ​mixture​ ​with​ ​water​ ​for​ ​some​ ​days.​ ​He​ ​found​ ​the​ ​mixture\nhad​ ​ ​fermented.​ ​He​ ​concluded​ ​that​ ​the​ ​mixture​ ​was​ ​contaminated​ ​with​ ​water​ ​and​ ​ethanol with boiling​ ​point​ ​of​ ​100​o​C​ ​and​ ​78​o​C​ ​respectively.​ ​The​ ​set-up​ ​of​ ​apparatus​ ​below​ ​are​ ​used​ ​to separate the​ ​mixture.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7286103345223176, "ocr_used": false, "chunk_length": 1968, "token_count": 879}}
{"text": "​ ​ ​24.a)i)​ ​The​ ​diagram​ ​below​ ​show​ ​spots​ ​of​ ​a​ ​pure​ ​substance​ ​A,​ ​B,​ ​and​ ​C​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​Spot​ ​D​ ​is​ ​that​ ​of​ ​a​ ​mixture ​ ​ ​ ​After​ ​development​ ​A,​ ​B​,​ ​and​ ​C​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​8cm,​ ​3cm​ ​and​ ​6cm​ ​respectively. D​ ​had​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​had​ ​moved​ ​6cm​ ​and​ ​8cm On​ ​the​ ​diagram​ ​above; ​ ​ ​I.​ ​Label​ ​the​ ​baseline​ ​(origin) ​ ​II.​ ​Show​ ​the​ ​positions​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development ​ ​ ​ii)​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​D b)​ ​Describe​ ​how​ ​solid​ ​ammonium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​solid​ ​mixture​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ammonium​ ​chloride​ ​and​ ​anhydrous​ ​calcium​ ​chloride c)​ ​The​ ​table​ ​below​ ​shows​ ​liquids​ ​that​ ​are​ ​miscible​ ​and​ ​those​ ​that​ ​are​ ​immiscible Liquid L​3 L​4 L​1 Miscible Miscible L​2 Miscible Immiscible Use​ ​the​ ​information​ ​given​ ​in​ ​the​ ​table​ ​to​ ​answer​ ​that​ ​questions​ ​that​ ​follow; i)​ ​Name​ ​the​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​separate​ ​L​1​​ ​and​ ​L​2​ ​from​ ​a​ ​mixture​ ​of​ ​the​ ​two ii)​ ​Describe​ ​how​ ​a​ ​mixture​ ​of​ ​L​2​​ ​and​ ​L​4​​ ​can​ ​be​ ​separated 25. A​ ​student​ ​left​ ​some​ ​crushed​ ​fruit​ ​mixture​ ​with​ ​water​ ​for​ ​some​ ​days.​ ​He​ ​found​ ​the​ ​mixture\nhad​ ​ ​fermented.​ ​He​ ​concluded​ ​that​ ​the​ ​mixture​ ​was​ ​contaminated​ ​with​ ​water​ ​and​ ​ethanol with boiling​ ​point​ ​of​ ​100​o​C​ ​and​ ​78​o​C​ ​respectively.​ ​The​ ​set-up​ ​of​ ​apparatus​ ​below​ ​are​ ​used​ ​to separate the​ ​mixture. (i)​ ​Name​ ​the​ ​piece​ ​of​ ​apparatus​ ​labelled​ ​W (ii)​ ​What​ ​is​ ​the​ ​purpose​ ​of​ ​the​ ​thermometer​ ​in​ ​the​ ​set-up?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7381304683756846, "ocr_used": false, "chunk_length": 1846, "token_count": 820}}
{"text": "D​ ​had​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​had​ ​moved​ ​6cm​ ​and​ ​8cm On​ ​the​ ​diagram​ ​above; ​ ​ ​I.​ ​Label​ ​the​ ​baseline​ ​(origin) ​ ​II.​ ​Show​ ​the​ ​positions​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development ​ ​ ​ii)​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​D b)​ ​Describe​ ​how​ ​solid​ ​ammonium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​solid​ ​mixture​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ammonium​ ​chloride​ ​and​ ​anhydrous​ ​calcium​ ​chloride c)​ ​The​ ​table​ ​below​ ​shows​ ​liquids​ ​that​ ​are​ ​miscible​ ​and​ ​those​ ​that​ ​are​ ​immiscible Liquid L​3 L​4 L​1 Miscible Miscible L​2 Miscible Immiscible Use​ ​the​ ​information​ ​given​ ​in​ ​the​ ​table​ ​to​ ​answer​ ​that​ ​questions​ ​that​ ​follow; i)​ ​Name​ ​the​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​separate​ ​L​1​​ ​and​ ​L​2​ ​from​ ​a​ ​mixture​ ​of​ ​the​ ​two ii)​ ​Describe​ ​how​ ​a​ ​mixture​ ​of​ ​L​2​​ ​and​ ​L​4​​ ​can​ ​be​ ​separated 25. A​ ​student​ ​left​ ​some​ ​crushed​ ​fruit​ ​mixture​ ​with​ ​water​ ​for​ ​some​ ​days.​ ​He​ ​found​ ​the​ ​mixture\nhad​ ​ ​fermented.​ ​He​ ​concluded​ ​that​ ​the​ ​mixture​ ​was​ ​contaminated​ ​with​ ​water​ ​and​ ​ethanol with boiling​ ​point​ ​of​ ​100​o​C​ ​and​ ​78​o​C​ ​respectively.​ ​The​ ​set-up​ ​of​ ​apparatus​ ​below​ ​are​ ​used​ ​to separate the​ ​mixture. (i)​ ​Name​ ​the​ ​piece​ ​of​ ​apparatus​ ​labelled​ ​W (ii)​ ​What​ ​is​ ​the​ ​purpose​ ​of​ ​the​ ​thermometer​ ​in​ ​the​ ​set-up? iii)​ ​At​ ​which​ ​end​ ​of​ ​the​ ​apparatus​ ​W​ ​should​ ​tap​ ​water​ ​be connected?…………………………… (iv)​ ​Which​ ​liquid​ ​was​ ​collected​ ​as​ ​the​ ​first​ ​distillate?​ ​Explain (v)​ ​What​ ​is​ ​the​ ​name​ ​given​ ​to​ ​the​ ​above​ ​method​ ​of​ ​separating​ ​mixture?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7561151611590127, "ocr_used": false, "chunk_length": 1772, "token_count": 771}}
{"text": "A​ ​student​ ​left​ ​some​ ​crushed​ ​fruit​ ​mixture​ ​with​ ​water​ ​for​ ​some​ ​days.​ ​He​ ​found​ ​the​ ​mixture\nhad​ ​ ​fermented.​ ​He​ ​concluded​ ​that​ ​the​ ​mixture​ ​was​ ​contaminated​ ​with​ ​water​ ​and​ ​ethanol with boiling​ ​point​ ​of​ ​100​o​C​ ​and​ ​78​o​C​ ​respectively.​ ​The​ ​set-up​ ​of​ ​apparatus​ ​below​ ​are​ ​used​ ​to separate the​ ​mixture. (i)​ ​Name​ ​the​ ​piece​ ​of​ ​apparatus​ ​labelled​ ​W (ii)​ ​What​ ​is​ ​the​ ​purpose​ ​of​ ​the​ ​thermometer​ ​in​ ​the​ ​set-up? iii)​ ​At​ ​which​ ​end​ ​of​ ​the​ ​apparatus​ ​W​ ​should​ ​tap​ ​water​ ​be connected?…………………………… (iv)​ ​Which​ ​liquid​ ​was​ ​collected​ ​as​ ​the​ ​first​ ​distillate?​ ​Explain (v)​ ​What​ ​is​ ​the​ ​name​ ​given​ ​to​ ​the​ ​above​ ​method​ ​of​ ​separating​ ​mixture? (vi)​ ​State​ ​two​ ​applications​ ​of​ ​the​ ​above​ ​method​ ​of​ ​separating​ ​mixtures (vi)​ ​What​ ​properties​ ​of​ ​the​ ​mixture​ ​makes​ ​it​ ​possible​ ​for​ ​the​ ​component​ ​to​ ​be​ ​separated ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​by​ ​ ​the​ ​above​ ​methods? 26.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7650734312416556, "ocr_used": false, "chunk_length": 1070, "token_count": 461}}
{"text": "iii)​ ​At​ ​which​ ​end​ ​of​ ​the​ ​apparatus​ ​W​ ​should​ ​tap​ ​water​ ​be connected?…………………………… (iv)​ ​Which​ ​liquid​ ​was​ ​collected​ ​as​ ​the​ ​first​ ​distillate?​ ​Explain (v)​ ​What​ ​is​ ​the​ ​name​ ​given​ ​to​ ​the​ ​above​ ​method​ ​of​ ​separating​ ​mixture? (vi)​ ​State​ ​two​ ​applications​ ​of​ ​the​ ​above​ ​method​ ​of​ ​separating​ ​mixtures (vi)​ ​What​ ​properties​ ​of​ ​the​ ​mixture​ ​makes​ ​it​ ​possible​ ​for​ ​the​ ​component​ ​to​ ​be​ ​separated ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​by​ ​ ​the​ ​above​ ​methods? 26. The​ ​set-up​ ​below​ ​was used​ ​to​ ​separate​ ​a mixture:- (a)​ ​Name​ ​the​ ​apparatus​ ​missing​ ​in​ ​the​ ​set-up\n(b)​ ​Give​ ​one​ ​example​ ​of​ ​mixture​ ​T (c)​ ​What​ ​is​ ​the​ ​name​ ​of​ ​this​ ​method​ ​of​ ​separation\n27.a)​ ​The​ ​diagram​ ​below​ ​shows​ ​a​ ​set​ ​–​ ​up​ ​used​ ​by​ ​a​ ​student​ ​to​ ​find​ ​out​ ​what​ ​happens ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​when​ ​Copper​ ​ ​(II)​ ​sulphate​ ​crystals​ ​are​ ​heated. ​ ​ ​ ​(i)​ ​State​ ​the​ ​observations​ ​made​ ​when​ ​the​ ​blue​ ​copper​ ​(II)​ ​sulphate​ ​crystals​ ​ ​are​ ​heated.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7497249217180854, "ocr_used": false, "chunk_length": 1129, "token_count": 503}}
{"text": "26. The​ ​set-up​ ​below​ ​was used​ ​to​ ​separate​ ​a mixture:- (a)​ ​Name​ ​the​ ​apparatus​ ​missing​ ​in​ ​the​ ​set-up\n(b)​ ​Give​ ​one​ ​example​ ​of​ ​mixture​ ​T (c)​ ​What​ ​is​ ​the​ ​name​ ​of​ ​this​ ​method​ ​of​ ​separation\n27.a)​ ​The​ ​diagram​ ​below​ ​shows​ ​a​ ​set​ ​–​ ​up​ ​used​ ​by​ ​a​ ​student​ ​to​ ​find​ ​out​ ​what​ ​happens ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​when​ ​Copper​ ​ ​(II)​ ​sulphate​ ​crystals​ ​are​ ​heated. ​ ​ ​ ​(i)​ ​State​ ​the​ ​observations​ ​made​ ​when​ ​the​ ​blue​ ​copper​ ​(II)​ ​sulphate​ ​crystals​ ​ ​are​ ​heated. ​ ​ ​ ​ ​(ii)​ ​Identify​ ​liquid​ ​Y​ ​and​ ​write​ ​an​ ​equation​ ​for​ ​its​ ​formation. b)​ ​Pellets​ ​of​ ​sodium​ ​hydrogen​ ​and​ ​anhydrous​ ​Copper​ ​(II)​ ​sulphate​ ​were​ ​put​ ​in​ ​separate Petri- ​ ​ ​ ​ ​dishes​ ​and​ ​left​ ​in​ ​the​ ​open​ ​for​ ​two​ ​hours.​ ​Explain​ ​the​ ​observation​ ​in​ ​each​ ​Petri-dish. 28.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7469497722753224, "ocr_used": false, "chunk_length": 917, "token_count": 424}}
{"text": "​ ​ ​ ​ ​(ii)​ ​Identify​ ​liquid​ ​Y​ ​and​ ​write​ ​an​ ​equation​ ​for​ ​its​ ​formation. b)​ ​Pellets​ ​of​ ​sodium​ ​hydrogen​ ​and​ ​anhydrous​ ​Copper​ ​(II)​ ​sulphate​ ​were​ ​put​ ​in​ ​separate Petri- ​ ​ ​ ​ ​dishes​ ​and​ ​left​ ​in​ ​the​ ​open​ ​for​ ​two​ ​hours.​ ​Explain​ ​the​ ​observation​ ​in​ ​each​ ​Petri-dish. 28. The​ ​chromatography​ ​below​ ​shows​ ​the​ ​constituents​ ​of​ ​a​ ​flower​ ​extract​ ​using​ ​an​ ​organic solvent:- (a)​ ​(i)​ ​Name​ ​a​ ​possible​ ​organic​ ​solvent​ ​you​ ​can​ ​use​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​one​ ​property​ ​that​ ​makes​ ​the​ ​red​ ​pigment​ ​to​ ​move​ ​the​ ​furthest​ ​distance​ ​from M ​ ​ ​ ​ ​ ​(iii)​ ​Describe​ ​how​ ​one​ ​could​ ​get​ ​a​ ​sample​ ​of​ ​yellow​ ​pigment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iv)​ ​On​ ​the​ ​diagram​ ​indicate​ ​solvent​ ​front (b)​ ​Describe​ ​how​ ​Aluminium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​mixture​ ​of​ ​aluminium chloride\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​and​ ​sodium​ ​chloride ​ ​ ​ ​ ​29.​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solid Cold​ ​water Hot​ ​water R Soluble Soluble V Insoluble Insoluble S Insoluble Insoluble ​ ​Describe​ ​how​ ​the​ ​mixture​ ​of​ ​solid​ ​R,​ ​S,​ ​and​ ​V​ ​can​ ​be​ ​separated 30.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7644245283018868, "ocr_used": false, "chunk_length": 1325, "token_count": 564}}
{"text": "b)​ ​Pellets​ ​of​ ​sodium​ ​hydrogen​ ​and​ ​anhydrous​ ​Copper​ ​(II)​ ​sulphate​ ​were​ ​put​ ​in​ ​separate Petri- ​ ​ ​ ​ ​dishes​ ​and​ ​left​ ​in​ ​the​ ​open​ ​for​ ​two​ ​hours.​ ​Explain​ ​the​ ​observation​ ​in​ ​each​ ​Petri-dish. 28. The​ ​chromatography​ ​below​ ​shows​ ​the​ ​constituents​ ​of​ ​a​ ​flower​ ​extract​ ​using​ ​an​ ​organic solvent:- (a)​ ​(i)​ ​Name​ ​a​ ​possible​ ​organic​ ​solvent​ ​you​ ​can​ ​use​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​one​ ​property​ ​that​ ​makes​ ​the​ ​red​ ​pigment​ ​to​ ​move​ ​the​ ​furthest​ ​distance​ ​from M ​ ​ ​ ​ ​ ​(iii)​ ​Describe​ ​how​ ​one​ ​could​ ​get​ ​a​ ​sample​ ​of​ ​yellow​ ​pigment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iv)​ ​On​ ​the​ ​diagram​ ​indicate​ ​solvent​ ​front (b)​ ​Describe​ ​how​ ​Aluminium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​mixture​ ​of​ ​aluminium chloride\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​and​ ​sodium​ ​chloride ​ ​ ​ ​ ​29.​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solid Cold​ ​water Hot​ ​water R Soluble Soluble V Insoluble Insoluble S Insoluble Insoluble ​ ​Describe​ ​how​ ​the​ ​mixture​ ​of​ ​solid​ ​R,​ ​S,​ ​and​ ​V​ ​can​ ​be​ ​separated 30. Given​ ​a​ ​mixture​ ​of​ ​lead​ ​(II)​ ​oxide,​ ​ammonium​ ​chloride​ ​and​ ​sodium​ ​chloride,​ ​describe how​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​mixture​ ​can​ ​ ​be​ ​separated​ ​to​ ​obtain​ ​a​ ​sample​ ​of​ ​each.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7633308776349248, "ocr_used": false, "chunk_length": 1451, "token_count": 621}}
{"text": "28. The​ ​chromatography​ ​below​ ​shows​ ​the​ ​constituents​ ​of​ ​a​ ​flower​ ​extract​ ​using​ ​an​ ​organic solvent:- (a)​ ​(i)​ ​Name​ ​a​ ​possible​ ​organic​ ​solvent​ ​you​ ​can​ ​use​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​one​ ​property​ ​that​ ​makes​ ​the​ ​red​ ​pigment​ ​to​ ​move​ ​the​ ​furthest​ ​distance​ ​from M ​ ​ ​ ​ ​ ​(iii)​ ​Describe​ ​how​ ​one​ ​could​ ​get​ ​a​ ​sample​ ​of​ ​yellow​ ​pigment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iv)​ ​On​ ​the​ ​diagram​ ​indicate​ ​solvent​ ​front (b)​ ​Describe​ ​how​ ​Aluminium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​mixture​ ​of​ ​aluminium chloride\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​and​ ​sodium​ ​chloride ​ ​ ​ ​ ​29.​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solid Cold​ ​water Hot​ ​water R Soluble Soluble V Insoluble Insoluble S Insoluble Insoluble ​ ​Describe​ ​how​ ​the​ ​mixture​ ​of​ ​solid​ ​R,​ ​S,​ ​and​ ​V​ ​can​ ​be​ ​separated 30. Given​ ​a​ ​mixture​ ​of​ ​lead​ ​(II)​ ​oxide,​ ​ammonium​ ​chloride​ ​and​ ​sodium​ ​chloride,​ ​describe how​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​mixture​ ​can​ ​ ​be​ ​separated​ ​to​ ​obtain​ ​a​ ​sample​ ​of​ ​each. 31.​ ​ ​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​separate​ ​two​ ​miscible​ ​liquids​ ​Q​ ​and​ ​T ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(Boling​ ​points;​ ​Q​ ​=98°​ ​C,​ ​T=78°C) (a)​ ​Identify​ ​the​ ​mistakes​ ​in​ ​the​ ​setup​ ​above (b)Identify​ ​Distillate​ ​X 32.Name​ ​the​ ​process​ ​which​ ​takes​ ​place​ ​when: a)​ ​Solid​ ​Carbon​ ​(IV)​ ​oxide​ ​(dry​ ​ice)​ ​changes​ ​directly​ ​into​ ​gas.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7462096417514374, "ocr_used": false, "chunk_length": 1615, "token_count": 698}}
{"text": "The​ ​chromatography​ ​below​ ​shows​ ​the​ ​constituents​ ​of​ ​a​ ​flower​ ​extract​ ​using​ ​an​ ​organic solvent:- (a)​ ​(i)​ ​Name​ ​a​ ​possible​ ​organic​ ​solvent​ ​you​ ​can​ ​use​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​one​ ​property​ ​that​ ​makes​ ​the​ ​red​ ​pigment​ ​to​ ​move​ ​the​ ​furthest​ ​distance​ ​from M ​ ​ ​ ​ ​ ​(iii)​ ​Describe​ ​how​ ​one​ ​could​ ​get​ ​a​ ​sample​ ​of​ ​yellow​ ​pigment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iv)​ ​On​ ​the​ ​diagram​ ​indicate​ ​solvent​ ​front (b)​ ​Describe​ ​how​ ​Aluminium​ ​chloride​ ​can​ ​be​ ​separated​ ​from​ ​a​ ​mixture​ ​of​ ​aluminium chloride\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​and​ ​sodium​ ​chloride ​ ​ ​ ​ ​29.​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solid Cold​ ​water Hot​ ​water R Soluble Soluble V Insoluble Insoluble S Insoluble Insoluble ​ ​Describe​ ​how​ ​the​ ​mixture​ ​of​ ​solid​ ​R,​ ​S,​ ​and​ ​V​ ​can​ ​be​ ​separated 30. Given​ ​a​ ​mixture​ ​of​ ​lead​ ​(II)​ ​oxide,​ ​ammonium​ ​chloride​ ​and​ ​sodium​ ​chloride,​ ​describe how​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​mixture​ ​can​ ​ ​be​ ​separated​ ​to​ ​obtain​ ​a​ ​sample​ ​of​ ​each. 31.​ ​ ​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​separate​ ​two​ ​miscible​ ​liquids​ ​Q​ ​and​ ​T ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(Boling​ ​points;​ ​Q​ ​=98°​ ​C,​ ​T=78°C) (a)​ ​Identify​ ​the​ ​mistakes​ ​in​ ​the​ ​setup​ ​above (b)Identify​ ​Distillate​ ​X 32.Name​ ​the​ ​process​ ​which​ ​takes​ ​place​ ​when: a)​ ​Solid​ ​Carbon​ ​(IV)​ ​oxide​ ​(dry​ ​ice)​ ​changes​ ​directly​ ​into​ ​gas. b)​ ​A​ ​red​ ​litmus​ ​paper​ ​turns​ ​white​ ​when​ ​dropped​ ​into​ ​chlorine​ ​water.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.750780870587634, "ocr_used": false, "chunk_length": 1701, "token_count": 736}}
{"text": "Given​ ​a​ ​mixture​ ​of​ ​lead​ ​(II)​ ​oxide,​ ​ammonium​ ​chloride​ ​and​ ​sodium​ ​chloride,​ ​describe how​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​mixture​ ​can​ ​ ​be​ ​separated​ ​to​ ​obtain​ ​a​ ​sample​ ​of​ ​each. 31.​ ​ ​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​separate​ ​two​ ​miscible​ ​liquids​ ​Q​ ​and​ ​T ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(Boling​ ​points;​ ​Q​ ​=98°​ ​C,​ ​T=78°C) (a)​ ​Identify​ ​the​ ​mistakes​ ​in​ ​the​ ​setup​ ​above (b)Identify​ ​Distillate​ ​X 32.Name​ ​the​ ​process​ ​which​ ​takes​ ​place​ ​when: a)​ ​Solid​ ​Carbon​ ​(IV)​ ​oxide​ ​(dry​ ​ice)​ ​changes​ ​directly​ ​into​ ​gas. b)​ ​A​ ​red​ ​litmus​ ​paper​ ​turns​ ​white​ ​when​ ​dropped​ ​into​ ​chlorine​ ​water. c)​ ​Propene​ ​gas​ ​molecules​ ​are​ ​converted​ ​into​ ​a​ ​giant​ ​molecule. 33. The​ ​following​ ​diagram​ ​shows​ ​a​ ​paper​ ​chromatogram​ ​of​ ​substances​ ​A,​ ​B,​ ​C,​ ​and​ ​D​ ​which ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​are​ ​coloured ​ ​ ​ ​ ​ ​(a)​ ​Indicate​ ​the​ ​solvent​ ​front​ ​on​ ​the​ ​chromatogram ​ ​ ​ ​ ​ ​(b)​ ​Which​ ​substance​ ​is​ ​pure? ………………………………………..", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7284771166649239, "ocr_used": false, "chunk_length": 1095, "token_count": 496}}
{"text": "33. The​ ​following​ ​diagram​ ​shows​ ​a​ ​paper​ ​chromatogram​ ​of​ ​substances​ ​A,​ ​B,​ ​C,​ ​and​ ​D​ ​which ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​are​ ​coloured ​ ​ ​ ​ ​ ​(a)​ ​Indicate​ ​the​ ​solvent​ ​front​ ​on​ ​the​ ​chromatogram ​ ​ ​ ​ ​ ​(b)​ ​Which​ ​substance​ ​is​ ​pure? ……………………………………….. ​ ​ ​ ​ ​ ​(c)​ ​Substance​ ​E​ ​is​ ​a​ ​mixture​ ​of​ ​C​ ​and​ ​D​.​ ​Indicate​ ​its​ ​chromatogram​ ​in​ ​the​ ​diagram 34.​ ​ ​ ​ ​ ​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​following​ ​questions.​ ​A​ ​mixture​ ​contains three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solids​ ​ ​ ​A,​ ​B​,​ ​and​ ​C​.​ ​The​ ​solubility​ ​of​ ​these​ ​solids​ ​in​ ​different​ ​liquids​ ​is​ ​as​ ​shown​ ​below:-\nSolid Water Alcohol Ether A Soluble Insoluble Insoluble B Insoluble Soluble Very​ ​soluble C Soluble Soluble Insoluble ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain​ ​how​ ​you​ ​will​ ​obtain​ ​sample​ ​C​ ​from​ ​the​ ​mixture 35. State​ ​and​ ​explain​ ​the​ ​observations​ ​made​ ​when​ ​iodine​ ​crystals​ ​is​ ​heated​ ​in​ ​a​ ​boiling​ ​tube? ​ ​Acids,​ ​bases​ ​and​ ​combustion 1.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7487050960735172, "ocr_used": false, "chunk_length": 1083, "token_count": 464}}
{"text": "​ ​ ​ ​ ​ ​(c)​ ​Substance​ ​E​ ​is​ ​a​ ​mixture​ ​of​ ​C​ ​and​ ​D​.​ ​Indicate​ ​its​ ​chromatogram​ ​in​ ​the​ ​diagram 34.​ ​ ​ ​ ​ ​ ​ ​Study​ ​the​ ​information​ ​below​ ​and​ ​answer​ ​the​ ​following​ ​questions.​ ​A​ ​mixture​ ​contains three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solids​ ​ ​ ​A,​ ​B​,​ ​and​ ​C​.​ ​The​ ​solubility​ ​of​ ​these​ ​solids​ ​in​ ​different​ ​liquids​ ​is​ ​as​ ​shown​ ​below:-\nSolid Water Alcohol Ether A Soluble Insoluble Insoluble B Insoluble Soluble Very​ ​soluble C Soluble Soluble Insoluble ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Explain​ ​how​ ​you​ ​will​ ​obtain​ ​sample​ ​C​ ​from​ ​the​ ​mixture 35. State​ ​and​ ​explain​ ​the​ ​observations​ ​made​ ​when​ ​iodine​ ​crystals​ ​is​ ​heated​ ​in​ ​a​ ​boiling​ ​tube? ​ ​Acids,​ ​bases​ ​and​ ​combustion 1. The​ ​table​ ​below​ ​shows​ ​solutions​ ​A,​ ​B​ ​and​ ​C​ ​are​ ​tested​ ​and​ ​observations​ ​records​ ​as shown: Solution Observations​ ​on​ ​indicator A Methyl​ ​orange​ ​turns​ ​yellow B Phenolphthalein​ ​turns​ ​colourless C Litmus​ ​turns​ ​purple (a)​ ​Using​ ​the​ ​table​ ​above,​ ​name​ ​an​ ​acid (b)​ ​How​ ​does​ ​the​ ​pH​ ​value​ ​of​ ​1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​1M​ ​aqueous​ ​ammonia?​ ​Explain 2.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7668620460450783, "ocr_used": false, "chunk_length": 1273, "token_count": 535}}
{"text": "State​ ​and​ ​explain​ ​the​ ​observations​ ​made​ ​when​ ​iodine​ ​crystals​ ​is​ ​heated​ ​in​ ​a​ ​boiling​ ​tube? ​ ​Acids,​ ​bases​ ​and​ ​combustion 1. The​ ​table​ ​below​ ​shows​ ​solutions​ ​A,​ ​B​ ​and​ ​C​ ​are​ ​tested​ ​and​ ​observations​ ​records​ ​as shown: Solution Observations​ ​on​ ​indicator A Methyl​ ​orange​ ​turns​ ​yellow B Phenolphthalein​ ​turns​ ​colourless C Litmus​ ​turns​ ​purple (a)​ ​Using​ ​the​ ​table​ ​above,​ ​name​ ​an​ ​acid (b)​ ​How​ ​does​ ​the​ ​pH​ ​value​ ​of​ ​1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​1M​ ​aqueous​ ​ammonia?​ ​Explain 2. The​ ​information​ ​below​ ​gives​ ​PH​ ​values​ ​of​ ​solutions​ ​V,​ ​W,​ ​X,​ ​Y​ ​Z Solution PH​ ​values V W X Y Z ​ ​ ​ ​ ​ ​ ​ ​ ​2 6.5 11 14 4.5 ​ ​(a)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Calcium​ ​hydroxide?……………………………………………….", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7278216897319985, "ocr_used": false, "chunk_length": 909, "token_count": 393}}
{"text": "​ ​Acids,​ ​bases​ ​and​ ​combustion 1. The​ ​table​ ​below​ ​shows​ ​solutions​ ​A,​ ​B​ ​and​ ​C​ ​are​ ​tested​ ​and​ ​observations​ ​records​ ​as shown: Solution Observations​ ​on​ ​indicator A Methyl​ ​orange​ ​turns​ ​yellow B Phenolphthalein​ ​turns​ ​colourless C Litmus​ ​turns​ ​purple (a)​ ​Using​ ​the​ ​table​ ​above,​ ​name​ ​an​ ​acid (b)​ ​How​ ​does​ ​the​ ​pH​ ​value​ ​of​ ​1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​1M​ ​aqueous​ ​ammonia?​ ​Explain 2. The​ ​information​ ​below​ ​gives​ ​PH​ ​values​ ​of​ ​solutions​ ​V,​ ​W,​ ​X,​ ​Y​ ​Z Solution PH​ ​values V W X Y Z ​ ​ ​ ​ ​ ​ ​ ​ ​2 6.5 11 14 4.5 ​ ​(a)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Calcium​ ​hydroxide?………………………………………………. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Rain​ ​water?……………………………………………………… (b)​ ​Which​ ​solution​ ​would​ ​react​ ​most​ ​vigorously​ ​with​ ​Zinc​ ​carbonate 3.a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​solutions.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7133865010972542, "ocr_used": false, "chunk_length": 1099, "token_count": 473}}
{"text": "The​ ​table​ ​below​ ​shows​ ​solutions​ ​A,​ ​B​ ​and​ ​C​ ​are​ ​tested​ ​and​ ​observations​ ​records​ ​as shown: Solution Observations​ ​on​ ​indicator A Methyl​ ​orange​ ​turns​ ​yellow B Phenolphthalein​ ​turns​ ​colourless C Litmus​ ​turns​ ​purple (a)​ ​Using​ ​the​ ​table​ ​above,​ ​name​ ​an​ ​acid (b)​ ​How​ ​does​ ​the​ ​pH​ ​value​ ​of​ ​1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​1M​ ​aqueous​ ​ammonia?​ ​Explain 2. The​ ​information​ ​below​ ​gives​ ​PH​ ​values​ ​of​ ​solutions​ ​V,​ ​W,​ ​X,​ ​Y​ ​Z Solution PH​ ​values V W X Y Z ​ ​ ​ ​ ​ ​ ​ ​ ​2 6.5 11 14 4.5 ​ ​(a)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Calcium​ ​hydroxide?………………………………………………. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Rain​ ​water?……………………………………………………… (b)​ ​Which​ ​solution​ ​would​ ​react​ ​most​ ​vigorously​ ​with​ ​Zinc​ ​carbonate 3.a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​solutions. Indicator Colour​ ​in Methyl​ ​Orange Acidic​ ​Solution Basic​ ​Solution Yellow Phenolphthalein Colourless b)​ ​How​ ​does​ ​the​ ​PH​ ​value​ ​of​ ​0.1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of 0.1M ​ ​ ​ ​ ​aqueous​ ​ammonia?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7259382756521332, "ocr_used": false, "chunk_length": 1310, "token_count": 549}}
{"text": "The​ ​information​ ​below​ ​gives​ ​PH​ ​values​ ​of​ ​solutions​ ​V,​ ​W,​ ​X,​ ​Y​ ​Z Solution PH​ ​values V W X Y Z ​ ​ ​ ​ ​ ​ ​ ​ ​2 6.5 11 14 4.5 ​ ​(a)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Calcium​ ​hydroxide?………………………………………………. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Rain​ ​water?……………………………………………………… (b)​ ​Which​ ​solution​ ​would​ ​react​ ​most​ ​vigorously​ ​with​ ​Zinc​ ​carbonate 3.a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​solutions. Indicator Colour​ ​in Methyl​ ​Orange Acidic​ ​Solution Basic​ ​Solution Yellow Phenolphthalein Colourless b)​ ​How​ ​does​ ​the​ ​PH​ ​value​ ​of​ ​0.1M​ ​potassium​ ​hydroxide​ ​solution​ ​compare​ ​with​ ​that​ ​of 0.1M ​ ​ ​ ​ ​aqueous​ ​ammonia? Explain. 4. Use​ ​the​ ​information​ ​given​ ​below​ ​to​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solution G H I J K pH 1.5 6.5 13.0 7.0 8.0 (a)​ ​Which​ ​of​ ​the​ ​solutions​ ​would​ ​be​ ​used​ ​to​ ​relieve​ ​a​ ​stomach​ ​upset​ ​caused​ ​by indigestion? (b)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Dilute​ ​sulphuric​ ​acid?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6967687810925336, "ocr_used": false, "chunk_length": 1177, "token_count": 512}}
{"text": "4. Use​ ​the​ ​information​ ​given​ ​below​ ​to​ ​answer​ ​the​ ​questions​ ​that​ ​follow: Solution G H I J K pH 1.5 6.5 13.0 7.0 8.0 (a)​ ​Which​ ​of​ ​the​ ​solutions​ ​would​ ​be​ ​used​ ​to​ ​relieve​ ​a​ ​stomach​ ​upset​ ​caused​ ​by indigestion? (b)​ ​Which​ ​solution​ ​is​ ​likely​ ​to​ ​be: ​ ​ ​ ​ ​ ​ ​(i)​ ​Dilute​ ​sulphuric​ ​acid? ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Sodium​ ​hydroxide​ ​solution? 5. Solid​ ​copper​ ​(II)​ ​oxide​ ​is​ ​a​ ​base​ ​although​ ​it​ ​does​ ​not​ ​turn​ ​litmus​ ​paper​ ​to​ ​blue.​ ​Explain ​ ​6. Below​ ​are​ ​the​ ​pH​ ​values​ ​of​ ​4​ ​types​ ​of​ ​medicine​ ​represented​ ​by​ ​letters​ ​P,​ ​Q,​ ​R​ ​and​ ​S MEDICINE pH VALUES P Q R S 7.0 5.0 8.0 6.0 ​ ​a)​ ​It​ ​is​ ​not​ ​advisable​ ​to​ ​use​ ​S​ ​when​ ​a​ ​patient​ ​has​ ​indigestion​ ​.Explain ​ ​b)​ ​What​ ​is​ ​the​ ​role​ ​of​ ​chemistry​ ​in​ ​drug​ ​manufacture 7.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6862099440280187, "ocr_used": false, "chunk_length": 892, "token_count": 425}}
{"text": "5. Solid​ ​copper​ ​(II)​ ​oxide​ ​is​ ​a​ ​base​ ​although​ ​it​ ​does​ ​not​ ​turn​ ​litmus​ ​paper​ ​to​ ​blue.​ ​Explain ​ ​6. Below​ ​are​ ​the​ ​pH​ ​values​ ​of​ ​4​ ​types​ ​of​ ​medicine​ ​represented​ ​by​ ​letters​ ​P,​ ​Q,​ ​R​ ​and​ ​S MEDICINE pH VALUES P Q R S 7.0 5.0 8.0 6.0 ​ ​a)​ ​It​ ​is​ ​not​ ​advisable​ ​to​ ​use​ ​S​ ​when​ ​a​ ​patient​ ​has​ ​indigestion​ ​.Explain ​ ​b)​ ​What​ ​is​ ​the​ ​role​ ​of​ ​chemistry​ ​in​ ​drug​ ​manufacture 7. Explain​ ​why​ ​very​ ​little​ ​Carbon​ ​(IV)​ ​oxide​ ​gas​ ​is​ ​evolved​ ​when​ ​dilute​ ​sulphuric​ ​(VI)​ ​acid is​ ​added​ ​to​ ​ ​ ​lead​ ​(II)​ ​carbonate 8​ ​.State​ ​one​ ​commercial​ ​use​ ​of​ ​Calcium​ ​Oxide 9. The​ ​following​ ​data​ ​gives​ ​the​ ​pH​ ​values​ ​of​ ​some​ ​solutions Solution pH P Q R 14.0 6.8 2.5 (a)​ ​What​ ​colour​ ​change​ ​would​ ​occur​ ​in​ ​solution​ ​P​ ​on​ ​addition​ ​of​ ​two​ ​drops​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​phenolphthalein​ ​ ​indicator?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7042953446511135, "ocr_used": false, "chunk_length": 977, "token_count": 456}}
{"text": "Below​ ​are​ ​the​ ​pH​ ​values​ ​of​ ​4​ ​types​ ​of​ ​medicine​ ​represented​ ​by​ ​letters​ ​P,​ ​Q,​ ​R​ ​and​ ​S MEDICINE pH VALUES P Q R S 7.0 5.0 8.0 6.0 ​ ​a)​ ​It​ ​is​ ​not​ ​advisable​ ​to​ ​use​ ​S​ ​when​ ​a​ ​patient​ ​has​ ​indigestion​ ​.Explain ​ ​b)​ ​What​ ​is​ ​the​ ​role​ ​of​ ​chemistry​ ​in​ ​drug​ ​manufacture 7. Explain​ ​why​ ​very​ ​little​ ​Carbon​ ​(IV)​ ​oxide​ ​gas​ ​is​ ​evolved​ ​when​ ​dilute​ ​sulphuric​ ​(VI)​ ​acid is​ ​added​ ​to​ ​ ​ ​lead​ ​(II)​ ​carbonate 8​ ​.State​ ​one​ ​commercial​ ​use​ ​of​ ​Calcium​ ​Oxide 9. The​ ​following​ ​data​ ​gives​ ​the​ ​pH​ ​values​ ​of​ ​some​ ​solutions Solution pH P Q R 14.0 6.8 2.5 (a)​ ​What​ ​colour​ ​change​ ​would​ ​occur​ ​in​ ​solution​ ​P​ ​on​ ​addition​ ​of​ ​two​ ​drops​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​phenolphthalein​ ​ ​indicator? (b)​ ​State​ ​the​ ​pH​ ​value​ ​of​ ​a​ ​resulting​ ​solution​ ​when​ ​equal​ ​moles​ ​of​ ​solution​ ​P​ ​and​ ​R​ ​react 10. In​ ​an​ ​experiment,​ ​ammonia​ ​gas​ ​was​ ​prepared​ ​by​ ​heating​ ​ammonium​ ​salt​ ​with​ ​an​ ​alkali.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7162233980830375, "ocr_used": false, "chunk_length": 1084, "token_count": 503}}
{"text": "The​ ​following​ ​data​ ​gives​ ​the​ ​pH​ ​values​ ​of​ ​some​ ​solutions Solution pH P Q R 14.0 6.8 2.5 (a)​ ​What​ ​colour​ ​change​ ​would​ ​occur​ ​in​ ​solution​ ​P​ ​on​ ​addition​ ​of​ ​two​ ​drops​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​phenolphthalein​ ​ ​indicator? (b)​ ​State​ ​the​ ​pH​ ​value​ ​of​ ​a​ ​resulting​ ​solution​ ​when​ ​equal​ ​moles​ ​of​ ​solution​ ​P​ ​and​ ​R​ ​react 10. In​ ​an​ ​experiment,​ ​ammonia​ ​gas​ ​was​ ​prepared​ ​by​ ​heating​ ​ammonium​ ​salt​ ​with​ ​an​ ​alkali. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​drying,​ ​ammonia​ ​gas​ ​was​ ​collected​ ​at​ ​room​ ​temperature​ ​and​ ​pressure. (a)​ ​What​ ​is​ ​meant​ ​by​ ​the​ ​term​ ​alkali? (b)​ ​Explain​ ​using​ ​physical​ ​properties​ ​of​ ​the​ ​gas​ ​why​ ​ammonia​ ​is​ ​not​ ​collected​ ​by downward ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​delivery 11. The​ ​table​ ​shows​ ​the​ ​colours​ ​obtained​ ​when​ ​some​ ​indicators​ ​are​ ​added​ ​to​ ​solutions:- Solution Blue​ ​litmus​ ​paper Indicator​ ​W Distilled​ ​water ………………….. Colourless Calcium​ ​hydroxide Blue Pink Nitric​ ​acid ………………………… Colourless (a)​ ​Complete​ ​the​ ​table​ ​by​ ​filling​ ​in​ ​the​ ​missing​ ​colours (b)​ ​Identify​ ​indicator​ ​W\n12.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7462569796130374, "ocr_used": false, "chunk_length": 1224, "token_count": 514}}
{"text": "(b)​ ​Explain​ ​using​ ​physical​ ​properties​ ​of​ ​the​ ​gas​ ​why​ ​ammonia​ ​is​ ​not​ ​collected​ ​by downward ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​delivery 11. The​ ​table​ ​shows​ ​the​ ​colours​ ​obtained​ ​when​ ​some​ ​indicators​ ​are​ ​added​ ​to​ ​solutions:- Solution Blue​ ​litmus​ ​paper Indicator​ ​W Distilled​ ​water ………………….. Colourless Calcium​ ​hydroxide Blue Pink Nitric​ ​acid ………………………… Colourless (a)​ ​Complete​ ​the​ ​table​ ​by​ ​filling​ ​in​ ​the​ ​missing​ ​colours (b)​ ​Identify​ ​indicator​ ​W\n12. (a)​ ​Flower​ ​extracts​ ​can​ ​be​ ​used​ ​as​ ​Acid-base​ ​indicators.​ ​Give​ ​two​ ​limitations​ ​of​ ​such ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​indicators (b)​ ​The​ ​diagram​ ​below​ ​shows​ ​spots​ ​of​ ​pure​ ​substances​ ​W,​ ​X​,​ ​and​ ​Y​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​ ​ ​Spot​ ​Z​ ​is​ ​that​ ​of​ ​a​ ​mixture After​ ​development​ ​W,​ ​X​,​ ​and​ ​Y​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​9cm​3​,​ ​4cm​3​​ ​and​ ​7cm​3 respectively.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.730742558063461, "ocr_used": false, "chunk_length": 1019, "token_count": 427}}
{"text": "The​ ​table​ ​shows​ ​the​ ​colours​ ​obtained​ ​when​ ​some​ ​indicators​ ​are​ ​added​ ​to​ ​solutions:- Solution Blue​ ​litmus​ ​paper Indicator​ ​W Distilled​ ​water ………………….. Colourless Calcium​ ​hydroxide Blue Pink Nitric​ ​acid ………………………… Colourless (a)​ ​Complete​ ​the​ ​table​ ​by​ ​filling​ ​in​ ​the​ ​missing​ ​colours (b)​ ​Identify​ ​indicator​ ​W\n12. (a)​ ​Flower​ ​extracts​ ​can​ ​be​ ​used​ ​as​ ​Acid-base​ ​indicators.​ ​Give​ ​two​ ​limitations​ ​of​ ​such ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​indicators (b)​ ​The​ ​diagram​ ​below​ ​shows​ ​spots​ ​of​ ​pure​ ​substances​ ​W,​ ​X​,​ ​and​ ​Y​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​ ​ ​Spot​ ​Z​ ​is​ ​that​ ​of​ ​a​ ​mixture After​ ​development​ ​W,​ ​X​,​ ​and​ ​Y​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​9cm​3​,​ ​4cm​3​​ ​and​ ​7cm​3 respectively. Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7324681152574398, "ocr_used": false, "chunk_length": 1160, "token_count": 490}}
{"text": "Colourless Calcium​ ​hydroxide Blue Pink Nitric​ ​acid ………………………… Colourless (a)​ ​Complete​ ​the​ ​table​ ​by​ ​filling​ ​in​ ​the​ ​missing​ ​colours (b)​ ​Identify​ ​indicator​ ​W\n12. (a)​ ​Flower​ ​extracts​ ​can​ ​be​ ​used​ ​as​ ​Acid-base​ ​indicators.​ ​Give​ ​two​ ​limitations​ ​of​ ​such ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​indicators (b)​ ​The​ ​diagram​ ​below​ ​shows​ ​spots​ ​of​ ​pure​ ​substances​ ​W,​ ​X​,​ ​and​ ​Y​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​ ​ ​Spot​ ​Z​ ​is​ ​that​ ​of​ ​a​ ​mixture After​ ​development​ ​W,​ ​X​,​ ​and​ ​Y​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​9cm​3​,​ ​4cm​3​​ ​and​ ​7cm​3 respectively. Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13. A​ ​beekeeper​ ​found​ ​that​ ​when​ ​stung​ ​by​ ​a​ ​bee,​ ​application​ ​of​ ​a​ ​little​ ​solution​ ​of​ ​sodium ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​carbonate​ ​helped​ ​to​ ​relieve​ ​the​ ​irritation​ ​of​ ​the​ ​affected​ ​area.​ ​Explain 14.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7259085629817337, "ocr_used": false, "chunk_length": 1230, "token_count": 533}}
{"text": "(a)​ ​Flower​ ​extracts​ ​can​ ​be​ ​used​ ​as​ ​Acid-base​ ​indicators.​ ​Give​ ​two​ ​limitations​ ​of​ ​such ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​indicators (b)​ ​The​ ​diagram​ ​below​ ​shows​ ​spots​ ​of​ ​pure​ ​substances​ ​W,​ ​X​,​ ​and​ ​Y​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​ ​ ​Spot​ ​Z​ ​is​ ​that​ ​of​ ​a​ ​mixture After​ ​development​ ​W,​ ​X​,​ ​and​ ​Y​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​9cm​3​,​ ​4cm​3​​ ​and​ ​7cm​3 respectively. Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13. A​ ​beekeeper​ ​found​ ​that​ ​when​ ​stung​ ​by​ ​a​ ​bee,​ ​application​ ​of​ ​a​ ​little​ ​solution​ ​of​ ​sodium ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​carbonate​ ​helped​ ​to​ ​relieve​ ​the​ ​irritation​ ​of​ ​the​ ​affected​ ​area.​ ​Explain 14. 10g​ ​of​ ​sodium​ ​hydrogen​ ​carbonate​ ​were​ ​dissolved​ ​in​ ​20cm​3​​ ​of​ ​water​ ​in​ ​a​ ​boiling​ ​tube.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7122061259737056, "ocr_used": false, "chunk_length": 1158, "token_count": 519}}
{"text": "Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13. A​ ​beekeeper​ ​found​ ​that​ ​when​ ​stung​ ​by​ ​a​ ​bee,​ ​application​ ​of​ ​a​ ​little​ ​solution​ ​of​ ​sodium ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​carbonate​ ​helped​ ​to​ ​relieve​ ​the​ ​irritation​ ​of​ ​the​ ​affected​ ​area.​ ​Explain 14. 10g​ ​of​ ​sodium​ ​hydrogen​ ​carbonate​ ​were​ ​dissolved​ ​in​ ​20cm​3​​ ​of​ ​water​ ​in​ ​a​ ​boiling​ ​tube. Lemon ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​juice​ ​was​ ​then​ ​added​ ​dropwise​ ​with​ ​shaking​ ​until​ ​there​ ​was​ ​no​ ​further​ ​change. (a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7352278545826934, "ocr_used": false, "chunk_length": 1116, "token_count": 487}}
{"text": "Lemon ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​juice​ ​was​ ​then​ ​added​ ​dropwise​ ​with​ ​shaking​ ​until​ ​there​ ​was​ ​no​ ​further​ ​change. (a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7258229674018539, "ocr_used": false, "chunk_length": 1378, "token_count": 591}}
{"text": "(a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7162759397790228, "ocr_used": false, "chunk_length": 1486, "token_count": 644}}
{"text": "15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7198731501057083, "ocr_used": false, "chunk_length": 1518, "token_count": 654}}
{"text": "(a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0\n​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer\n3.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7164869290271583, "ocr_used": false, "chunk_length": 1592, "token_count": 688}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer\n3. (a)​ ​What​ ​is​ ​rust?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.717555923777962, "ocr_used": false, "chunk_length": 680, "token_count": 303}}
{"text": "Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer\n3. (a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7060219138056976, "ocr_used": false, "chunk_length": 1369, "token_count": 620}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer\n3. (a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.691207806037792, "ocr_used": false, "chunk_length": 1355, "token_count": 621}}
{"text": "(a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7021970654714018, "ocr_used": false, "chunk_length": 1356, "token_count": 616}}
{"text": "(b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air. State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment\n7.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7066889966505023, "ocr_used": false, "chunk_length": 1446, "token_count": 649}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air. State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment\n7. A​ ​form​ ​two​ ​student​ ​in​ ​an​ ​attempt​ ​to​ ​stop​ ​rusting​ ​put​ ​copper​ ​and​ ​Zinc​ ​ ​in​ ​contact​ ​with​ ​iron ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​as​ ​shown:- ​ ​ ​ ​ ​ ​ ​(a)​ ​State​ ​whether​ ​rusting​ ​occurred​ ​after​ ​one​ ​week​ ​if​ ​the​ ​set-ups​ ​were​ ​left​ ​out ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Explain​ ​your​ ​answer​ ​in​ ​(a)​ ​above 8. In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7277562792775628, "ocr_used": false, "chunk_length": 1111, "token_count": 496}}
{"text": "State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment\n7. A​ ​form​ ​two​ ​student​ ​in​ ​an​ ​attempt​ ​to​ ​stop​ ​rusting​ ​put​ ​copper​ ​and​ ​Zinc​ ​ ​in​ ​contact​ ​with​ ​iron ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​as​ ​shown:- ​ ​ ​ ​ ​ ​ ​(a)​ ​State​ ​whether​ ​rusting​ ​occurred​ ​after​ ​one​ ​week​ ​if​ ​the​ ​set-ups​ ​were​ ​left​ ​out ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Explain​ ​your​ ​answer​ ​in​ ​(a)​ ​above 8. In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7346303133579476, "ocr_used": false, "chunk_length": 1006, "token_count": 450}}
{"text": "In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7499171926055235, "ocr_used": false, "chunk_length": 1354, "token_count": 589}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.75698552073036, "ocr_used": false, "chunk_length": 1186, "token_count": 506}}
{"text": "c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.762380555377413, "ocr_used": false, "chunk_length": 1006, "token_count": 426}}
{"text": "d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment. d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide\n(e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7561491130178128, "ocr_used": false, "chunk_length": 1163, "token_count": 500}}
{"text": "b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment. d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide\n(e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10. ​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7416425506436484, "ocr_used": false, "chunk_length": 911, "token_count": 402}}
{"text": "d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide\n(e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10. ​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7401732593164223, "ocr_used": false, "chunk_length": 1156, "token_count": 510}}
{"text": "​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up?", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7362035023799729, "ocr_used": false, "chunk_length": 1496, "token_count": 648}}
{"text": "c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7346448276858192, "ocr_used": false, "chunk_length": 1049, "token_count": 455}}
{"text": "d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air:\n(i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7530862897848897, "ocr_used": false, "chunk_length": 1357, "token_count": 570}}
{"text": "​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air:\n(i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous. ​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.752766669502822, "ocr_used": false, "chunk_length": 1314, "token_count": 550}}
{"text": "(iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air:\n(i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous. ​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7599322482871043, "ocr_used": false, "chunk_length": 1138, "token_count": 486}}
{"text": "​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish. ​ ​ ​ ​ ​ ​The​ ​ash​ ​was​ ​found​ ​to​ ​comprise​ ​of​ ​magnesium​ ​Oxide​ ​and​ ​Magnesium​ ​nitride ​ ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​leading​ ​to​ ​formation​ ​of​ ​the​ ​magnesium​ ​nitride ​ ​ ​ ​ ​ ​(ii)​ ​A​ ​little​ ​water​ ​was​ ​added​ ​to​ ​the​ ​products​ ​in​ ​the​ ​Petri​ ​dish.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​made.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7413577833209736, "ocr_used": false, "chunk_length": 1141, "token_count": 504}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish. ​ ​ ​ ​ ​ ​The​ ​ash​ ​was​ ​found​ ​to​ ​comprise​ ​of​ ​magnesium​ ​Oxide​ ​and​ ​Magnesium​ ​nitride ​ ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​leading​ ​to​ ​formation​ ​of​ ​the​ ​magnesium​ ​nitride ​ ​ ​ ​ ​ ​(ii)​ ​A​ ​little​ ​water​ ​was​ ​added​ ​to​ ​the​ ​products​ ​in​ ​the​ ​Petri​ ​dish.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​made. ​ ​ ​ ​ ​(iii)​ ​A​ ​piece​ ​of​ ​blue​ ​litmus​ ​paper​ ​was​ ​dipped​ ​into​ ​the​ ​solution​ ​formed​ ​in​ ​(b)​ ​above. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​State​ ​the​ ​ ​ ​observation​ ​made. 14.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7282306746665199, "ocr_used": false, "chunk_length": 982, "token_count": 445}}
{"text": "​ ​ ​ ​ ​(iii)​ ​A​ ​piece​ ​of​ ​blue​ ​litmus​ ​paper​ ​was​ ​dipped​ ​into​ ​the​ ​solution​ ​formed​ ​in​ ​(b)​ ​above. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​State​ ​the​ ​ ​ ​observation​ ​made. 14. A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M..................................................................................", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6918345420359926, "ocr_used": false, "chunk_length": 657, "token_count": 255}}
{"text": "14. A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air\n​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6895017502535414, "ocr_used": false, "chunk_length": 1329, "token_count": 518}}
{"text": "A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air\n​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6926821444324152, "ocr_used": false, "chunk_length": 1467, "token_count": 580}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air\n​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6913214990138067, "ocr_used": false, "chunk_length": 1690, "token_count": 696}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air\n​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7147562246670527, "ocr_used": false, "chunk_length": 1727, "token_count": 735}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7356139997988536, "ocr_used": false, "chunk_length": 1220, "token_count": 550}}
{"text": "A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid. 18.Below​ ​is​ ​a​ ​list​ ​of​ ​oxides.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7386125502404572, "ocr_used": false, "chunk_length": 1118, "token_count": 507}}
{"text": "Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid. 18.Below​ ​is​ ​a​ ​list​ ​of​ ​oxides. MgO,​ ​N​2​O​,​ ​K​2​O,​ ​CaO​ ​ans​ ​Al​2​O​3 Select:- a)​ ​A​ ​neutral​ ​oxide. b)​ ​A​ ​highly​ ​water​ ​soluble​ ​basic​ ​oxide. c)​ ​An​ ​oxide​ ​which​ ​can​ ​react​ ​with​ ​both​ ​sodium​ ​hydroxide​ ​solution​ ​and​ ​dilute​ ​hydrochloric acid. 19.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7389172592627944, "ocr_used": false, "chunk_length": 911, "token_count": 418}}
{"text": "b)​ ​A​ ​highly​ ​water​ ​soluble​ ​basic​ ​oxide. c)​ ​An​ ​oxide​ ​which​ ​can​ ​react​ ​with​ ​both​ ​sodium​ ​hydroxide​ ​solution​ ​and​ ​dilute​ ​hydrochloric acid. 19. The​ ​diagram​ ​below​ ​shows​ ​students​ ​set-up​ ​for​ ​the​ ​preparation​ ​and​ ​collection​ ​of​ ​oxygen gas\n(a)​ ​Name​ ​substance​ ​X​ ​used (b)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​the​ ​reaction​ ​of​ ​sodium​ ​peroxide​ ​with​ ​the​ ​substance​ ​named in​ ​1(a) 5.​ ​Water​ ​and​ ​hydrogen 1. (a)​ ​Hydrogen​ ​can​ ​reduce​ ​coppers​ ​Oxide​ ​but​ ​not​ ​alluminium​ ​oxide.​ ​Explain (b)​ ​When​ ​water​ ​reacts​ ​with​ ​potassium​ ​metal​ ​the​ ​hydrogen​ ​produced​ ​ignites​ ​explosively ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​on​ ​the​ ​surface​ ​of​ ​water. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​What​ ​causes​ ​this​ ​ignition? ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7389307083574599, "ocr_used": false, "chunk_length": 942, "token_count": 417}}
{"text": "(a)​ ​Hydrogen​ ​can​ ​reduce​ ​coppers​ ​Oxide​ ​but​ ​not​ ​alluminium​ ​oxide.​ ​Explain (b)​ ​When​ ​water​ ​reacts​ ​with​ ​potassium​ ​metal​ ​the​ ​hydrogen​ ​produced​ ​ignites​ ​explosively ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​on​ ​the​ ​surface​ ​of​ ​water. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​What​ ​causes​ ​this​ ​ignition? ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2. In​ ​an experiment,​ ​dry​ ​hydrogen​ ​gas​ ​was​ ​passed​ ​over​ ​hot​ ​copper​ ​(II)​ ​oxide​ ​in​ ​a​ ​combustion ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​tube​ ​as​ ​ ​shown​ ​in​ ​the​ ​diagram​ ​below:-\n(a)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​the​ ​other​ ​product,​ ​substance​ ​R​ ​could​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​laboratory. (b)​ ​Describe​ ​how​ ​copper​ ​could​ ​be​ ​obtained​ ​from​ ​the​ ​mixture​ ​containing​ ​copper​ ​(II)​ ​oxide\n3.​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​the​ ​reaction​ ​between​ ​metals​ ​and​ ​water.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7327316747370178, "ocr_used": false, "chunk_length": 1060, "token_count": 473}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2. In​ ​an experiment,​ ​dry​ ​hydrogen​ ​gas​ ​was​ ​passed​ ​over​ ​hot​ ​copper​ ​(II)​ ​oxide​ ​in​ ​a​ ​combustion ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​tube​ ​as​ ​ ​shown​ ​in​ ​the​ ​diagram​ ​below:-\n(a)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​the​ ​other​ ​product,​ ​substance​ ​R​ ​could​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​laboratory. (b)​ ​Describe​ ​how​ ​copper​ ​could​ ​be​ ​obtained​ ​from​ ​the​ ​mixture​ ​containing​ ​copper​ ​(II)​ ​oxide\n3.​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​the​ ​reaction​ ​between​ ​metals​ ​and​ ​water. ​ ​ ​ ​ ​(a)​ ​Identify​ ​solid​ ​X​ ​and​ ​state​ ​its​ ​purpose ​ ​ ​ ​ ​ ​ ​ ​ ​Solid​ ​X ………………..……………………………………………………………………….. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Purpose ……………………………………………………………………………………….. ​ ​ ​ ​ ​(b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​produces​ ​the​ ​flame. 4. Gas​ ​P​ ​was​ ​passed​ ​over​ ​heated​ ​magnesium​ ​ribbon​ ​and​ ​hydrogen​ ​gas​ ​was​ ​collected​ ​as shown in the​ ​diagram​ ​below:\n(i)​ ​Name​ ​gas​ ​P ...............................................................................................................", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6858709938948739, "ocr_used": false, "chunk_length": 1282, "token_count": 505}}
{"text": "​ ​ ​ ​ ​(b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​produces​ ​the​ ​flame. 4. Gas​ ​P​ ​was​ ​passed​ ​over​ ​heated​ ​magnesium​ ​ribbon​ ​and​ ​hydrogen​ ​gas​ ​was​ ​collected​ ​as shown in the​ ​diagram​ ​below:\n(i)​ ​Name​ ​gas​ ​P ............................................................................................................... (ii)​ ​Write​ ​an​ ​equation​ ​of​ ​the​ ​reaction​ ​that​ ​takes​ ​place​ ​in​ ​the​ ​combustion​ ​tube (iii)​ ​State​ ​one​ ​precaution​ ​necessary​ ​at​ ​the​ ​end​ ​of​ ​this​ ​experiment 5.​ ​ ​ ​ ​ ​ ​ ​When​ ​hydrogen​ ​is​ ​burnt​ ​and​ ​the​ ​product​ ​cooled,​ ​the​ ​following​ ​results​ ​are​ ​obtained​ ​as shown ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​diagram​ ​below: ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​formation​ ​of​ ​liquid​ ​Y ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Give​ ​a​ ​chemical​ ​test​ ​for​ ​liquid​ ​Y\n6. Jane​ ​set-up​ ​the​ ​experiment​ ​as​ ​shown​ ​below​ ​to​ ​collect​ ​a​ ​gas.​ ​The​ ​wet​ ​sand​ ​was​ ​heated before heating​ ​Zinc granules (a)​ ​Complete​ ​the​ ​diagram​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​the​ ​gas (b)​ ​Why​ ​was​ ​it​ ​necessary​ ​to​ ​heat​ ​wet​ ​sand​ ​before​ ​heating​ ​Zinc​ ​granules? 7.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7220028793299014, "ocr_used": false, "chunk_length": 1256, "token_count": 505}}
{"text": "(ii)​ ​Write​ ​an​ ​equation​ ​of​ ​the​ ​reaction​ ​that​ ​takes​ ​place​ ​in​ ​the​ ​combustion​ ​tube (iii)​ ​State​ ​one​ ​precaution​ ​necessary​ ​at​ ​the​ ​end​ ​of​ ​this​ ​experiment 5.​ ​ ​ ​ ​ ​ ​ ​When​ ​hydrogen​ ​is​ ​burnt​ ​and​ ​the​ ​product​ ​cooled,​ ​the​ ​following​ ​results​ ​are​ ​obtained​ ​as shown ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​diagram​ ​below: ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​formation​ ​of​ ​liquid​ ​Y ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Give​ ​a​ ​chemical​ ​test​ ​for​ ​liquid​ ​Y\n6. Jane​ ​set-up​ ​the​ ​experiment​ ​as​ ​shown​ ​below​ ​to​ ​collect​ ​a​ ​gas.​ ​The​ ​wet​ ​sand​ ​was​ ​heated before heating​ ​Zinc granules (a)​ ​Complete​ ​the​ ​diagram​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​the​ ​gas (b)​ ​Why​ ​was​ ​it​ ​necessary​ ​to​ ​heat​ ​wet​ ​sand​ ​before​ ​heating​ ​Zinc​ ​granules? 7. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Between​ ​N​ ​and​ ​M​ ​which​ ​part​ ​should​ ​be​ ​heated​ ​first?​ ​Explain (b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​occurring​ ​in​ ​the​ ​combustion​ ​tube. 8.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7469567553048616, "ocr_used": false, "chunk_length": 1095, "token_count": 483}}
{"text": "7. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Between​ ​N​ ​and​ ​M​ ​which​ ​part​ ​should​ ​be​ ​heated​ ​first?​ ​Explain (b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​occurring​ ​in​ ​the​ ​combustion​ ​tube. 8. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​electrolysis​ ​of​ ​a​ ​certain​ ​molten​ ​compound;- (a)​ ​Complete​ ​the​ ​circuit​ ​by​ ​drawing​ ​the​ ​cell​ ​in​ ​the​ ​gap​ ​left​ ​in​ ​the​ ​diagram (b)​ ​Write​ ​half-cell​ ​equation​ ​to​ ​show​ ​what​ ​happens​ ​at​ ​the​ ​cathode (c)​ ​Using​ ​an​ ​arrow​ ​show​ ​the​ ​direction​ ​of​ ​electron​ ​flow​ ​in​ ​the​ ​diagram​ ​above\n9.Hydrogen​ ​can​ ​be​ ​prepared​ ​by​ ​reacting​ ​zinc​ ​with​ ​dilute​ ​hydrochloric​ ​acid. a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Name​ ​an​ ​appropriate​ ​drying​ ​agent​ ​for​ ​hydrogen​ ​gas. c)​ ​Explain​ ​why​ ​copper​ ​metal​ ​cannot​ ​be​ ​used​ ​to​ ​prepare​ ​hydrogen​ ​gas. d)​ ​Hydrogen​ ​burns​ ​in​ ​oxygen​ ​to​ ​form​ ​an​ ​oxide. ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7610239739943112, "ocr_used": false, "chunk_length": 1070, "token_count": 477}}
{"text": "c)​ ​Explain​ ​why​ ​copper​ ​metal​ ​cannot​ ​be​ ​used​ ​to​ ​prepare​ ​hydrogen​ ​gas. d)​ ​Hydrogen​ ​burns​ ​in​ ​oxygen​ ​to​ ​form​ ​an​ ​oxide. ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction. ​ ​ ​ ​(ii)​ ​State​ ​two​ ​precautions​ ​that​ ​must​ ​be​ ​taken​ ​before​ ​the​ ​combustion​ ​begins​ ​and​ ​at​ ​the end​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​combustion. e)​ ​Give​ ​two​ ​uses​ ​of​ ​hydrogen​ ​gas. f)​ ​When​ ​zinc​ ​is​ ​heated​ ​to​ ​redness​ ​in​ ​a​ ​current​ ​of​ ​steam,​ ​hydrogen​ ​gas​ ​is​ ​obtained.​ ​Write an ​ ​ ​ ​ ​equation​ ​for​ ​the​ ​reaction. g)​ ​Element​ ​Q​ ​reacts​ ​with​ ​dilute​ ​acids​ ​but​ ​not​ ​with​ ​cold​ ​water.​ ​Element​ ​R​ ​does​ ​not​ ​react with ​ ​ ​ ​ ​dilute​ ​acids.​ ​Elements​ ​S​ ​displaces​ ​element​ ​P​ ​from​ ​its​ ​oxide.​ ​P​ ​reacts​ ​with​ ​cold​ ​water. Arrange ​ ​ ​ ​ ​the​ ​four​ ​elements​ ​in​ ​order​ ​of​ ​their​ ​reactivity,​ ​starting​ ​with​ ​the​ ​most​ ​reactive. h)​ ​Explain​ ​how​ ​hydrogen​ ​is​ ​used​ ​in​ ​the​ ​manufacture​ ​of​ ​margarine. 10. a)​ ​The​ ​set-up​ ​below​ ​ ​is​ ​used​ ​to​ ​investigate​ ​the​ ​properties​ ​ ​of​ ​ ​hydrogen.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7669308051507004, "ocr_used": false, "chunk_length": 1146, "token_count": 511}}
{"text": "h)​ ​Explain​ ​how​ ​hydrogen​ ​is​ ​used​ ​in​ ​the​ ​manufacture​ ​of​ ​margarine. 10. a)​ ​The​ ​set-up​ ​below​ ​ ​is​ ​used​ ​to​ ​investigate​ ​the​ ​properties​ ​ ​of​ ​ ​hydrogen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​i)​ ​On​ ​the​ ​diagram,​ ​indicate​ ​what​ ​should​ ​be​ ​done​ ​for​ ​the​ ​reaction​ ​to​ ​occur ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ii)​ ​Hydrogen​ ​gas​ ​is​ ​allowed​ ​to​ ​pass​ ​through​ ​the​ ​tube​ ​for​ ​some​ ​time​ ​before​ ​it​ ​is​ ​lit.​ ​Explain ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​iii)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​occurs​ ​in​ ​the​ ​combustion​ ​tube ​ ​ ​ ​ ​ ​ ​ ​ ​ ​iv)​ ​When​ ​the​ ​reaction​ ​is​ ​complete,​ ​hydrogen​ ​gas​ ​is​ ​passed​ ​through​ ​the​ ​apparatus​ ​until​ ​they ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​cool​ ​down​ ​.​ ​ ​ ​Explain ​ ​ ​ ​ ​ ​ ​ ​ ​ ​v)​ ​What​ ​property​ ​of​ ​hydrogen​ ​is​ ​being​ ​investigated? ​ ​ ​ ​ ​ ​ ​ ​ ​vi)​ ​What​ ​observation​ ​confirms​ ​the​ ​property​ ​stated​ ​in​ ​(​v)​ ​above? ​ ​ ​ ​ ​ ​ ​ ​ ​vii)​ ​Why​ ​is​ ​zinc​ ​oxide​ ​not​ ​used​ ​to​ ​investigate​ ​this​ ​property​ ​of​ ​hydrogen​ ​gas? 11.", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7320059412728035, "ocr_used": false, "chunk_length": 1085, "token_count": 485}}
{"text": "​ ​ ​ ​ ​ ​ ​ ​ ​vi)​ ​What​ ​observation​ ​confirms​ ​the​ ​property​ ​stated​ ​in​ ​(​v)​ ​above?\n\n​ ​ ​ ​ ​ ​ ​ ​ ​vii)​ ​Why​ ​is​ ​zinc​ ​oxide​ ​not​ ​used​ ​to​ ​investigate​ ​this​ ​property​ ​of​ ​hydrogen​ ​gas?\n\n11.\n\nThe​ ​set​ ​up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​K,​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​water​ ​and ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​calcium​ ​metal.\n\n(a)​ ​Name​ ​gas​ ​K……………………………………………………………..\n\n(b)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​beaker​ ​was​ ​found​ ​to​ ​be​ ​a​ ​weak​ ​base.\n\nExplain ​ ​ ​ ​ ​ ​ ​why​ ​the​ ​solution​ ​is​ ​a​ ​weak​ ​base", "metadata": {"source": "CHEMISTRY-FORM-1-REVISION-BOOKLET.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7177679324894516, "ocr_used": false, "chunk_length": 632, "token_count": 277}}
{"text": "ATOMIC STRUCTURE AND THE PERIODIC TABLE Preface The structure of the atom is extensively covered here. The periodic table is the arrangement of atoms of elements based on their atomic structure. Emphasis on the trends across and down the periodic table of atoms is important for the teacher facilitator. This work should be covered before “Chemical bonding and structure”. The two complements each other in understanding periodicity. Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element\n2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8906141658805748, "ocr_used": false, "chunk_length": 1671, "token_count": 376}}
{"text": "Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element\n2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20\n3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7452951432129515, "ocr_used": false, "chunk_length": 1825, "token_count": 575}}
{"text": "3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element\n2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20\n3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes. Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7245819641614193, "ocr_used": false, "chunk_length": 1516, "token_count": 495}}
{"text": "Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40\n4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) .", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6759328982299311, "ocr_used": false, "chunk_length": 1166, "token_count": 407}}
{"text": "i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40\n4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) . All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7123968693137914, "ocr_used": false, "chunk_length": 1548, "token_count": 505}}
{"text": "All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples. a) Chlorine occurs as 75% 3517Cl and 25% 3717Cl isotopes. Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;\n5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7715413335017247, "ocr_used": false, "chunk_length": 1678, "token_count": 472}}
{"text": "Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;\n5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.69559063098696, "ocr_used": false, "chunk_length": 1471, "token_count": 465}}
{"text": "Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;\n5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6939001506135651, "ocr_used": false, "chunk_length": 1604, "token_count": 516}}
{"text": "b) Calculate the relative atomic mass of potassium given that it exist as;\n5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms. The higher the relative atomic mass, the heavier the atom. Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7231564625850341, "ocr_used": false, "chunk_length": 1575, "token_count": 486}}
{"text": "Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons\n6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8916575812638805, "ocr_used": false, "chunk_length": 762, "token_count": 162}}
{"text": "An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons\n6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:\n7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)\n8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels\n9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels\n10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus\n11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2\n12 B.PERIODIC TABLE There are over 100 elements so far discovered.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8012630319683354, "ocr_used": false, "chunk_length": 6045, "token_count": 1623}}
{"text": "By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons\n6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:\n7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)\n8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels\n9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels\n10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus\n11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2\n12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8001261209179529, "ocr_used": false, "chunk_length": 5981, "token_count": 1607}}
{"text": "Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons\n6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:\n7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)\n8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels\n9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels\n10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus\n11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2\n12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8019196432255187, "ocr_used": false, "chunk_length": 5996, "token_count": 1602}}
{"text": "By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:\n7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)\n8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels\n9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels\n10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus\n11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2\n12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7964318782944484, "ocr_used": false, "chunk_length": 5649, "token_count": 1532}}
{"text": "Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev. The horizontal arrangement forms period. Atoms in the same period have the same the same number of energy levels in their electronic structure. i.e. The number of energy levels in the electronic configuration of an element determine the period to which the element is in the periodic table. e.g. Which period of the periodic table are the following isotopes/elements/atoms? a) 126C Electron structure 2:4 => 2 energy levels used thus Period 2 b) 2311Na Electron structure 2:8:1 => 3 energy levels used thus Period 3 c) 3919K Electron structure 2:8:8:1 => 4 energy levels used thus Period 4 d) 11H Electron structure 1: => 1 energy level used thus Period 1 The vertical arrangement of elements forms a group. Atoms in the same have the same the same group have the same number of outer energy level electrons as per their electronic structure. i.e. The number of electrons in the outer energy level an element determine the group to which the element is ,in the periodic table. 13 a) 126C Electron structure 2:4 => 4 electrons in outer energy level thus Group IV b) 2311C Electron structure 2:8:1 => 1 electron in outer energy level thus Group I c) 3919K Electron structure 2:8:8:1=>1 electron in outer energy level thus Group I d) 11H Electron structure 1: => 1 electron in outer energy level thus Group I By convention; (i)Periods are named using English numerals 1,2,3,4,… (ii)Groups are named using Roman numerals I,II,III,IV,… There are eighteen groups in a standard periodic table. There are seven periods in a standard periodic table. When an atom has maximum number of electrons in its outer energy level, it is said to be stable. When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.846643119781195, "ocr_used": false, "chunk_length": 2151, "token_count": 508}}
{"text": "When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable. THE STANDARD PERIODIC TABLE OF ELEMENTS\n14 All unstable atoms/isotopes try to be stable through chemical reactions. A chemical reaction involves gaining or losing outer electrons (electron transfer) .When electron transfer take place, an ion is formed. An ion is formed when an unstable atom gain or donate electrons in its outer energy level inorder to be stable. Whether an atom gain or donate electrons depend on the relative energy required to donate or gain extra electrons i.e. Examples 1. 199 F has electronic structure/configuration 2:7. It can donate the seven outer electrons to have stable electronic structure/configuration 2:. It can gain one extra electron to have stable electronic structure/configuration 2:8. Gaining requires less energy, and thus Fluorine reacts by gaining one extra electrons. 2. 2313 Al has electronic structure/configuration 2:8:3 It can donate the three outer electrons to have stable electronic structure/configuration 2:8. It can gain five extra electrons to have stable electronic structure/configuration 2:8:8. Donating requires less energy, and thus Aluminium reacts by donating its three outer electrons. Elements with less than four electrons in the outer energy level donates /lose the outer electrons to be stable and form a positively charged ion called cation. A cation therefore has more protons(positive charge) than electrons(negative charge) Generally metals usually form cation Elements with more than four electrons in the outer energy level gain /acquire extra electrons in the outer energy level to be stable and form a negatively charged ion called anion. An anion therefore has less protons(positive charge) than electrons(negative charge) Generally non metals usually form anion. Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains)\n15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8706718492237465, "ocr_used": false, "chunk_length": 2329, "token_count": 512}}
{"text": "Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains)\n15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3. 2311 Na Na -> Na+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 4. 2412Mg Mg -> Mg2+ + 2e (atom) (cation) ( 2 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 5. 168O O + 2e -> O2- (atom) ( 2 electrons gained/acquired) (anion) Electron 2:6 2:8 structure (unstable) (stable) 6. 147N N + 3e -> N3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 7. 3115P P + 3e -> P3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 8. 199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl\n16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6791221082658176, "ocr_used": false, "chunk_length": 1414, "token_count": 517}}
{"text": "199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl\n16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation. When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4\n17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7146555764929423, "ocr_used": false, "chunk_length": 1319, "token_count": 438}}
{"text": "When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4\n17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.773085231589768, "ocr_used": false, "chunk_length": 1561, "token_count": 538}}
{"text": "Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4\n17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7635328075374799, "ocr_used": false, "chunk_length": 1459, "token_count": 517}}
{"text": "Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae. Elements combine together to form a compound depending on their combining power. The combining power of atoms in an element is called Valency.Valency of an element is equal to the number of: (i)hydrogen atoms that an atom of element can combine with or displace. (ii)electrons gained /acquired in outer energy level by non metals to be stable/attain duplet/octet. (iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions\n18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8761575009221367, "ocr_used": false, "chunk_length": 1576, "token_count": 450}}
{"text": "(iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions\n18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals. Radical name Chemical formulae Combining power / Valency Ammonium NH4 + 1 Hydroxide OH- 1 Nitrate(V) NO3 - 1 Hydrogen carbonate HCO3- 1 Hydrogen sulphate(VI) HSO4- 1 Hydrogen sulphate(IV) HSO3- 1 Manganate(VII) MnO4- 1 Chromate(VI) CrO42- 2 Dichromate(VI) Cr2O72- 2 Sulphate(VI) SO42- 2 Sulphate(IV) SO32- 2 Carbonate(IV) CO32- 2 Phosphate(V) PO42- 3 Table showing the valency of some common metal and non metals Element/metal Valency Element/non metal Valency Hydrogen 1 Florine 1 Lithium 1 Chlorine 1 Beryllium 2 Bromine 1 Boron 3 Iodine 1 Sodium 1 Carbon 4 Magnesium 2 Nitrogen 3 Aluminium 3 Oxygen 2 Potassium 1 Phosphorus 3 Calcium 2 Zinc 2 Barium 2 Mercury 2 Iron 2 and 3 Copper 1 and 2 Manganese 2 and 4 Lead 2 and 4\n19 From the valency of elements , the chemical formular of a compound can be derived using the following procedure: (i)Identify the elements and radicals making the compound (ii)Write the symbol/formular of the elements making the compound starting with the metallic element (iii)Assign the valency of each element /radical as superscript. (iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8144664924888632, "ocr_used": false, "chunk_length": 1607, "token_count": 467}}
{"text": "(iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula. Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1\n20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8646801440044309, "ocr_used": false, "chunk_length": 2041, "token_count": 475}}
{"text": "Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1\n20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8605332154956549, "ocr_used": false, "chunk_length": 2190, "token_count": 514}}
{"text": "(d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1\n21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.841410381848953, "ocr_used": false, "chunk_length": 1856, "token_count": 487}}
{"text": "(e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1\n21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8322464225078825, "ocr_used": false, "chunk_length": 1550, "token_count": 417}}
{"text": "(f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1\n21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8265131351878491, "ocr_used": false, "chunk_length": 1593, "token_count": 431}}
{"text": "(g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -\n22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8321361815443561, "ocr_used": false, "chunk_length": 1609, "token_count": 432}}
{"text": "(h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -\n22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.834534231200898, "ocr_used": false, "chunk_length": 1620, "token_count": 434}}
{"text": "(i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -\n22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8393811228659905, "ocr_used": false, "chunk_length": 1639, "token_count": 425}}
{"text": "(j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -\n22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical. (l)Magnesium hydrogen sulphate(VI) Elements making compound Magnesium Hydrogen sulphate(VI) Symbol of elements/radicals in compound Mg HSO4 Assign valencies as superscript Mg 2 HSO4 1 Exchange/Interchange the valencies as subscript Mg1 HSO4 2 Divide by two to get smallest whole number ratio - - Chemical formula of Magnesium hydrogen sulphate(VI) is thus: Mg(HSO4)2 This means:1 atom of Magnesium combine with 2 hydrogen sulphate(VI) radical. Compounds are formed from chemical reactions. A chemical reaction is formed when atoms of the reactants break free to bond again and form products. A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:\n23 1. Write the word equation 2.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8513708927346193, "ocr_used": false, "chunk_length": 2018, "token_count": 506}}
{"text": "A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:\n23 1. Write the word equation 2. Write the correct chemical formula for each of the reactants and products 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. This is called balancing. Do not change the chemical formula of the products/reactants. 5. Assign in brackets, the physical state/state symbols of the reactants and products after each chemical formula as: (i) (s) for solids (ii) (l) for liquids (iii) (g) for gas (iv) (aq) for aqueous/dissolved in water to make a solution. Practice examples Write a balanced chemical equation for the following (a) Hydrogen gas is prepared from reacting Zinc granules with dilute hydrochloric acid. Procedure 1. Write the word equation Zinc + Hydrochloric acid -> Zinc chloride + hydrogen gas 2. Write the correct chemical formula for each of the reactants and products Zn + HCl -> ZnCl2 + H2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of Zn on the reactant side is equal to product side One atom of H in HCl on the reactant side is not equal to two atoms in H2 on product side. One atom of Cl in HCl on the reactant side is not equal to two atoms in ZnCl2 on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “2” to get “2” Hydrogen and “2” Chlorine on product and reactant side. Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)\n24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8703174097956543, "ocr_used": false, "chunk_length": 2069, "token_count": 507}}
{"text": "Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)\n24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1. Write the word equation Hydrogen peroxide -> Water + oxygen gas 2. Write the correct chemical formula for each of the reactants and products H2O2 -> H2O + O2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of H on the reactant side is equal to product side Two atom of O in H2O2 on the reactant side is not equal to three atoms (one in H2O and two in O2) on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply H2O2 by “2” to get “4” Hydrogen and “4” Oxygen on reactants Multiply H2O by “2” to get “4” Hydrogen and “2” Oxygen on product side When the “2” Oxygen in O2 and the“2” in H2O are added on product side they are equal to the“4” Oxygen on reactants side. 2H2O2 -> 2H2O + O2 5. Assign in brackets, the physical state/state symbols . 2H2O2(aq) -> 2H2O(l) + O2(g) (c) Chlorine gas is prepared from Potassium manganate(VII) reacting with hydrochloric acid to form potassium chloride solution, manganese(II) chloride solution,water and chlorine gas. Procedure 1. Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8248862372332897, "ocr_used": false, "chunk_length": 1798, "token_count": 505}}
{"text": "Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 25 Number of atoms of K and Mn on the reactant side is equal to product side Two atom of H in H2O on the product side is not equal to one atom on reactant side. Four atom of O in KMnO4 is not equal to one in H2O One atom of Cl in HCl on reactant side is not equal to three (one in H2O and two in Cl2) 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “16” to get “16” Hydrogen and “16” Chlorine on reactants Multiply KMnO4 by “2” to get “2” Potassium and “2” manganese, “2 x4 =8” Oxygen on reactant side. Balance the product side to get: 2 KMnO4 +16 HCl -> 2 KCl + 2 MnCl2 +8 H2O + 5 Cl2 5. Assign in brackets, the physical state/state symbols . 2KMnO4(s) +16 HCl(aq)-> 2 KCl (aq) + 2MnCl2(aq)+8 H2O(l)+5 Cl2(g) (d)Carbon(IV)oxide gas is prepared from Calcium carbonate reacting with hydrochloric acid to form calcium chloride solution, water and carbon(IV)oxide gas. Procedure 1. Write the word equation Calcium carbonate + Hydrochloric acid -> calcium chloride solution+ water +carbon(IV)oxide 2. Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8180846123067249, "ocr_used": false, "chunk_length": 1744, "token_count": 498}}
{"text": "Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. 5. Assign in brackets, the physical state/state symbols . CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) (d)Sodium hydroxide solution neutralizes hydrochloric acid to form salt and water. 26 NaOH(aq) + HCl(aq) -> NaCl (aq) + H2O(l) (e)Sodium reacts with water to form sodium hydroxide and hydrogen gas. 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) (f)Calcium reacts withwater to form calcium hydroxide and hydrogen gas Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g) (g)Copper(II)Oxide solid reacts with dilute hydrochloric acid to form copper(II)chloride and water. CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l) (h)Hydrogen sulphide reacts with Oxygen to form sulphur(IV)Oxide and water. 2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.797181090958851, "ocr_used": false, "chunk_length": 1381, "token_count": 498}}
{"text": "2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water. C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) (l)Ethyne(C2H2) gas burns in air to form Carbon(IV)Oxide and water. 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l)\n27 C.PERIODICITY OF CHEMICAL FAMILES/DOWN THE GROUP. The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table.i.e The number of occupied energy levels determine the Period and the valence electrons determine the Group. Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7725012926062921, "ocr_used": false, "chunk_length": 1479, "token_count": 514}}
{"text": "Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M+ The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels). Atomic and ionic radius The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size. The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also\n28 measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion. Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g. The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.853874988790243, "ocr_used": false, "chunk_length": 2065, "token_count": 509}}
{"text": "Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level. Table showing the atomic and ionic radius of some alkali metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Lithium Li 3 0.133 0.060 Sodium Na 11 0.157 0.095 Potassium K 19 0.203 0.133 The atomic radius of sodium is 0.157nM .The ionic radius of Na+ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius. Ionization energy The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1st ionization energy decrease down the group as\n29 the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8861804663996782, "ocr_used": false, "chunk_length": 2269, "token_count": 494}}
{"text": "e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium. Physical properties Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen. Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure. Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure. Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8841741898984908, "ocr_used": false, "chunk_length": 2089, "token_count": 485}}
{"text": "Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate. 2NaOH(aq) + CO2(g) -> Na2CO3(g) + H2O(l) (ii)Burning in air/oxygen Lithium burns in air with a crimson/deep red flame to form Lithium oxide 4Li (s) + O2(g) -> 2Li2O(s) Sodium burns in air with a yellow flame to form sodium oxide 4Na (s) + O2(g) -> 2Na2O(s) Sodium burns in oxygen with a yellow flame to form sodium peroxide 2Na (s) + O2(g) -> Na2O2 (s) Potassium burns in air with a lilac/purple flame to form potassium oxide 4K (s) + O2(g) -> 2K2O (s) (iii) Reaction with water: Experiment Measure 500 cm3 of water into a beaker. Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8078144718193644, "ocr_used": false, "chunk_length": 1062, "token_count": 336}}
{"text": "Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium. Caution: Keep a distance Observations Alkali metal Observations Comparative speed/rate of the reaction Lithium -Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Moderately vigorous Sodium -Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn Very vigorous\n31 phenolphthalein indicator pink -pH of solution = 12/13/14 Potassium -Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Explosive/burst into flames Explanation Alkali metals are less dense than water. They therefore float in water.They react with water to form a strongly alkaline solution of their hydroxides and producing hydrogen gas. The rate of this reaction increase down the group. i.e. Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8913817515512432, "ocr_used": false, "chunk_length": 1628, "token_count": 383}}
{"text": "Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions. Chemical equations 2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g) 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) 2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g) 2Cs(s) + 2H2O(l) -> 2CsOH(aq) + H2(g) 2Fr(s) + 2H2O(l) -> 2FrOH(aq) + H2(g) Reactivity increase down the group (iv) Reaction with chlorine: Experiment Cut about 0.5g of sodium into a deflagrating spoon with a lid cover. Introduce it on a Bunsen flame until it catches fire. Quickly and carefully lower it into a gas jar containing dry chlorine to cover the gas jar. Repeat with about 0.5g of Lithium. Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic. Observation Sodium metal continues to burn with a yellow flame forming white solid/fumes. 32 Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8374601524601525, "ocr_used": false, "chunk_length": 1443, "token_count": 409}}
{"text": "32 Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus. Chemical equations 2Li(s) + Cl2(g) -> 2LiCl(s) 2Na(s) + Cl2(g) -> 2NaCl(s) 2K(s) + Cl2(g) -> 2KCl(s) 2Rb(s) + Cl2(g) -> 2RbCl(s) 2Cs(s) + Cl2(g) -> 2CsCl(s) 2Fr(s) + Cl2(g) -> 2FrCl(s) Reactivity increase down the group The table below shows some compounds of the 1st three alkali metals Some uses of alkali metals include: (i)Sodium is used in making sodium cyanide for extracting gold from gold ore. (ii)Sodium chloride is used in seasoning food. (iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors. (iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8399721797190152, "ocr_used": false, "chunk_length": 1027, "token_count": 289}}
{"text": "(ii)Sodium chloride is used in seasoning food. (iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors. (iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents. Lithium sodium Potassium Hydroxide LiOH NaOH KOH Oxide Li2O Na2O K2O Sulphide Li2S Na2S K2S Chloride LiCl NaCl KCl Carbonate Li2CO3 Na2CO3 K2CO3 Nitrate(V) LiNO3 NaNO3 KNO3 Nitrate(III) - NaNO2 KNO2 Sulphate(VI) Li2SO4 Na2SO4 K2SO4 Sulphate(IV) - Na2SO3 K2SO3 Hydrogen carbonate - NaHCO3 KHCO3 Hydrogen sulphate(VI) - NaHSO4 KHSO4 Hydrogen sulphate(IV) - NaHSO3 KHSO3 Phosphate - Na3PO4 K3PO4 Manganate(VI) - NaMnO4 KMnO4 Dichromate(VI) - Na2Cr2O7 K2Cr2O7 Chromate(VI) - Na2CrO4 K2CrO4\n33 (v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride. (vi)Lithium is used in making special high strength glasses (vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops. Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals . The alkaline earth metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Beryllium Be 4 2:2 Be2+ 2 Magnesium Mg 12 2:8:2 Mg2+ 2 Calcium Ca 20 2:8:8:2 Ca2+ 2 Strontium Sr 38 2:8:18:8:2 Sr2+ 2 Barium Ba 56 2:8:18:18:8:2 Ba2+ 2 Radium Ra 88 2:8:18:32:18:8:2 Ra2+ 2 All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7514626787354062, "ocr_used": false, "chunk_length": 1452, "token_count": 512}}
{"text": "Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals . The alkaline earth metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Beryllium Be 4 2:2 Be2+ 2 Magnesium Mg 12 2:8:2 Mg2+ 2 Calcium Ca 20 2:8:8:2 Ca2+ 2 Strontium Sr 38 2:8:18:8:2 Sr2+ 2 Barium Ba 56 2:8:18:18:8:2 Ba2+ 2 Radium Ra 88 2:8:18:32:18:8:2 Ra2+ 2 All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent. They donate /lose the two outer electrons to have oxidation state M2+ The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Calcium is bigger/larger than that of Magnesium because Calcium has more/4 energy levels than Magnesium (3 energy levels). Atomic radius and ionic radius of alkaline earth metals increase down the group as the number of energy levels increases. The atomic radius of alkaline earth metals is bigger than the ionic radius. This is because they react by losing/donating the two outer electrons and hence lose the outer energy level. Table showing the atomic and ionic radius of the 1st three alkaline earth metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Beryllium Be 4 0.089 0.031 Magnesium Mg 12 0.136 0.065\n34 Calcium Ca 20 0.174 0.099 The atomic radius of Magnesium is 0.136nM .The ionic radius of Mg2+ is 0.065nM. This is because Magnesium reacts by donating/losing the two outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8102613957517348, "ocr_used": false, "chunk_length": 1899, "token_count": 502}}
{"text": "The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size. Calcium is more electropositive than Magnesium. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The two outer electrons in calcium experience less nuclear attraction and can be lost/ donated easily/with ease because of the higher/bigger atomic radius. Ionization energy For alkaline earth metals the 1st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of Magnesium is 900 kJmole-1 while that of Calcium is 590 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from magnesium to calcium. It requires therefore less energy to donate/lose outer electron in calcium than in magnesium. The minimum amount of energy required to remove a second electron from an ion of an element in its gaseous state is called the 2nd ionization energy. The 2nd ionization energy is always higher /bigger than the 1st ionization energy. This because once an electron is donated /lost form an atom, the overall effective nuclear attraction on the remaining electrons/energy level increase. Removing a second electron from the ion require therefore more energy than the first electron. The atomic radius of alkali metals is higher/bigger than that of alkaline earth metals.This is because across/along the period from left to right there is an increase in nuclear charge from additional number of protons and still additional number of electrons entering the same energy level. Increase in nuclear charge increases the effective nuclear attraction on the outer energy level which pulls it closer to the nucleus. e.g. Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on\n35 the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus. Physical properties Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9093921022586041, "ocr_used": false, "chunk_length": 2447, "token_count": 489}}
{"text": "Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on\n35 the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus. Physical properties Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals. This is because of the decrease in atomic radius of corresponding alkaline earth metal, increases the strength of metallic bond and the packing of the metallic structure. Alkaline earth metals are (i)ductile(able to form wire/thin long rods) (ii)malleable(able to be hammered into sheet/long thin plates) (iii)have high tensile strength(able to be coiled without breaking/ not brittle/withstand stress) Appearance: Alkali earth metals have a shiny grey metallic luster when their surface is freshly polished /scrubbed. The surface slowly tarnishes on exposure to air. This is because the metal surface slowly undergoes oxidation to form an oxide. This oxide layer should be removed before using the alkaline earth metals. Melting and boiling points: Alkaline earth metals have a relatively high melting/ boiling point than alkali metals. This is because alkali metals use only one delocalized electron to form a weaker metallic bond/structure. Alkaline earth metals use two delocalized electrons to form a stronger metallic bond /structure. The melting and boiling points decrease down the group as the atomic radius/size increase reducing the strength of metallic bond and packing of the metallic structure. e.g. Beryllium has a melting point of 1280oC. Magnesium has a melting point of 650oC.Beryllium has a smaller atomic radius/size than magnesium .The strength of metallic bond and packing of the metallic structure is thus stronger in beryllium. Electrical/thermal conductivity: Alkaline earth metals are good thermal and electrical conductors. The two delocalized valence electrons move randomly within the metallic structure. Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g. Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9149946561968734, "ocr_used": false, "chunk_length": 2395, "token_count": 506}}
{"text": "Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g. Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor. 36 Calcium is a better conductor than magnesium. Calcium has bigger/larger atomic radius than magnesium because the delocalized electrons are less attracted to the nucleus of calcium and thus more free /mobile and thus better the electrical conductor Summary of some physical properties of the 1st three alkaline earth metals Alkaline earth metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conduct- ivity 1st ionization energy 2nd ionization energy Beryllium Shiny grey Not easy 1280 3450 Good 900 1800 Magnesium Shiny grey Not Easy 650 1110 Good 736 1450 calcium Shiny grey Not easy 850 1140 Good 590 970 Chemical properties (i)Reaction with air/oxygen On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air. Example On exposure to air, the surface of magnesium ribbon is oxidized to form a thin film of Magnesium oxide . 2Mg(s) + O2(g) -> 2MgO(s) (ii)Burning in air/oxygen Experiment Hold a about 2cm length of Magnesium ribbon on a Bunsen flame. Stop heating when it catches fire/start burning. Caution: Do not look directly at the flame Put the products of burning into 100cm3 beaker. Add about 5cm3 of distilled water. Swirl. Test the mixture using litmus papers. Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8864697530692354, "ocr_used": false, "chunk_length": 2113, "token_count": 486}}
{"text": "Swirl. Test the mixture using litmus papers. Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride. 37 2Mg (s) + O2(g) -> 2MgO(s) 3Mg (s) + N2 (g) -> Mg3N2 (s) Magnesium oxide dissolves in water to form magnesium hydroxide. MgO(s) + H2O (l) -> Mg(OH)2(aq) Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas. Mg3N2 (s) + 6H2O(l) -> 3Mg(OH)2(aq) + 2NH3 (g) Magnesium hydroxide and ammonia are weakly alkaline with pH 8/9/10/11 and turns red litmus paper blue. Calcium burns in air with faint orange/red flame to form a mixture of both Calcium oxide and calcium nitride. 2Ca (s) + O2(g) -> 2CaO(s) 3Ca (s) + N2 (g) -> Ca3N2 (s) Calcium oxide dissolves in water to form calcium hydroxide. CaO(s) + H2O(l) -> Ca(OH)2(aq) Calcium nitride dissolves in water to form calcium hydroxide and produce ammonia gas. Ca3N2 (s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2NH3 (g) Calcium hydroxide is also weakly alkaline solution with pH 8/9/10/11 and turns red litmus paper blue. (iii)Reaction with water Experiment Measure 50 cm3 of distilled water into a beaker. Scrub/polish with sand paper 1cm length of Magnesium ribbon Place it in the water. Test the product-mixture with blue and red litmus papers. Repeat with Calcium metal. Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8262297662895094, "ocr_used": false, "chunk_length": 1682, "token_count": 508}}
{"text": "Repeat with Calcium metal. Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed. -Effervescence/bubbles/fizzing takes place in Calcium. -Red litmus paper turns blue. -Blue litmus paper remains blue. Explanations Magnesium slowly reacts with cold water to form Magnesium hydroxide and bubbles of Hydrogen gas that stick on the surface of the ribbon. Mg(s) + 2H2O (l) -> Mg(OH)2(aq) + H2 (g)\n38 Calcium moderately reacts with cold water to form Calcium hydroxide and produce a steady stream of Hydrogen gas. Ca(s) + 2H2O (l) -> Ca(OH)2(aq) + H2 (g) (iv)Reaction with water vapour/steam Experiment Put some cotton wool soaked in water/wet sand in a long boiling tube. Coil a well polished magnesium ribbon into the boiling tube. Ensure the coil touches the side of the boiling tube. Heat the cotton wool/sand slightly then strongly heat the Magnesium ribbon . Set up of apparatus Observations -Magnesium glows red hot then burns with a blindening flame. -Magnesium continues to glow/burning even without more heating. -White solid/residue. -colourless gas collected over water. Explanation On heating wet sand, steam is generated which drives out the air that would otherwise react with /oxidize the ribbon. Magnesium burns in steam/water vapour generating enough heat that ensures the reaction goes to completion even without further heating. White Magnesium oxide is formed and hydrogen gas is evolved. To prevent suck back, the delivery tube should be removed from the water before heating is stopped at the end of the experiment. Mg(s) + H2O (l) -> MgO(s) + H2 (g)\n39 (v)Reaction with chlorine gas. Experiment Lower slowly a burning magnesium ribbon/shavings into a gas jar containing Chlorine gas. Repeat with a hot piece of calcium metal. Observation -Magnesium continues to burn in chlorine with a bright blindening flame. -Calcium continues to burn for a short time. -White solid formed . -Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8908620364038944, "ocr_used": false, "chunk_length": 2175, "token_count": 504}}
{"text": "-Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid. Calcium oxide formed coat unreacted Calcium stopping further reaction Ca(s) + Cl2 (g) -> CaCl2 (s) (v)Reaction with dilute acids. Experiment Place about 4.0cm3 of 0.1M dilute sulphuric(VI)acid into a test tube. Add about 1.0cm length of magnesium ribbon into the test tube. Cover the mouth of the test tube using a thumb. Release the gas and test the gas using a burning splint. Repeat with about 4.0cm3 of 0.1M dilute hydrochloric/nitric(V) acid. Repeat with 0.1g of Calcium in a beaker with all the above acid Caution: Keep distance when using calcium Observation -Effervescence/fizzing/bubbles with dilute sulphuric(VI) and nitric(V) acids -Little Effervescence/fizzing/bubbles with calcium and dilute sulphuric(VI) acid. -Colourless gas produced that extinguishes a burning splint with an explosion/ “pop” sound. -No gas is produced with Nitric(V)acid. -Colourless solution is formed. Explanation Dilute acids react with alkaline earth metals to form a salt and produce hydrogen gas. Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8765246385589994, "ocr_used": false, "chunk_length": 1423, "token_count": 364}}
{"text": "Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used. Chemical equations\n40 Mg(s) + H2SO4 (aq) -> MgSO4(aq) + H2 (g) Mg(s) + 2HNO3 (aq) -> Mg(NO3)2(aq) + H2 (g) Mg(s) + 2HCl (aq) -> MgCl2(aq) + H2 (g) Ca(s) + H2SO4 (aq) -> CaSO4(s) + H2 (g) (insoluble CaSO4(s) coat/cover Ca(s)) Ca(s) + 2HNO3 (aq) -> Ca(NO3)2(aq) + H2 (g) Ca(s) + 2HCl (aq) -> CaCl2(aq) + H2 (g) Ba(s) + H2SO4 (aq) -> BaSO4(s) + H2 (g) (insoluble BaSO4(s) coat/cover Ba(s)) Ba(s) + 2HNO3 (aq) -> Ba(NO3)2(aq) + H2 (g) Ba(s) + 2HCl (aq) -> BaCl2(aq) + H2 (g) The table below shows some compounds of some alkaline earth metals Some uses of alkaline earth metals include: (i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity. (ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light. (iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone. (iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building. (iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms. (v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8084845673932883, "ocr_used": false, "chunk_length": 1502, "token_count": 492}}
{"text": "(iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building. (iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms. (v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth. Beryllium Magnesium Calcium Barium Hydroxide Be(OH)2 Mg(OH)2 Ca(OH)2 Ba(OH)2 Oxide BeO MgO CaO BaO Sulphide - MgS CaS BaS Chloride BeCl2 MgCl2 CaCl2 BaCl2 Carbonate BeCO3 MgCO3 CaCO3 BaCO3 Nitrate(V) Be(NO3)2 Mg(NO3)2 Ca(NO3)2 Ba(NO3)2 Sulphate(VI) BeSO4 MgSO4 CaSO4 BaSO4 Sulphate(IV) - - CaSO3 BaSO3 Hydrogen carbonate - Mg(HCO3)2 Ca(HCO3)2 - Hydrogen sulphate(VI) - Mg(HSO4)2 Ca(HSO4)2 -\n41 (vi)In the blast furnace-Limestone is added to the blast furnace to produce more reducing agent and remove slag in the blast furnace for extraction of Iron. (c)Group VII elements: Halogens Group VII elements are called Halogens. They are all non metals. They include: Element Symbol Atomic number Electronicc configuration Charge of ion Valency State at Room Temperature Fluorine Chlorine Bromine Iodine Astatine F Cl Br I At 9 17 35 53 85 2:7 2:8:7 2:8:18:7 2:8:18:18:7 2:8:18:32:18:7 F- Cl- Br- I- At- 1 1 1 1 1 Pale yellow gas Pale green gas Red liquid Grey Solid Radioactive All halogen atoms have seven electrons in the outer energy level. They acquire/gain one electron in the outer energy level to be stable. They therefore are therefore monovalent .They exist in oxidation state X- The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8056300454248115, "ocr_used": false, "chunk_length": 1721, "token_count": 511}}
{"text": "They therefore are therefore monovalent .They exist in oxidation state X- The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Chlorine is bigger/larger than that of Fluorine because Chlorine has more/3 energy levels than Fluorine (2 energy levels). Atomic radius and ionic radius of Halogens increase down the group as the number of energy levels increases. The atomic radius of Halogens is smaller than the ionic radius. This is because they react by gaining/acquiring extra one electron in the outer energy level. The effective nuclear attraction on the more/extra electrons decreases. The incoming extra electron is also repelled causing the outer energy level to expand to reduce the repulsion and accommodate more electrons. Table showing the atomic and ionic radius of four Halogens\n42 Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Fluorine F 9 0.064 0.136 Chlorine Cl 17 0.099 0.181 Bromine Br 35 0.114 0.195 Iodine I 53 0.133 0.216 The atomic radius of Chlorine is 0.099nM .The ionic radius of Cl- is 0.181nM. This is because Chlorine atom/molecule reacts by gaining/acquiring extra one electrons. The more/extra electrons/energy level experience less effective nuclear attraction /pull towards the nucleus .The outer enegy level expand/increase to reduce the repulsion of the existing and incoming gained /acquired electrons. Electronegativity The ease of gaining/acquiring extra electrons is called electronegativity. All halogens are electronegative. Electronegativity decreases as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and thus ease of gaining/acquiring extra electrons decrease. It is measured using Pauling’s scale. Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron. Table showing the electronegativity of the halogens.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8802411393474173, "ocr_used": false, "chunk_length": 2118, "token_count": 475}}
{"text": "It is measured using Pauling’s scale. Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron. Table showing the electronegativity of the halogens. Halogen F Cl Br I At Electronegativity (Pauling scale) 4.0 3.0 2.8 2.5 2.2 The electronegativity of the halogens decrease down the group from fluorine to Astatine. This is because atomic radius increases down the group and thus decrease electron – attracting power down the group from fluorine to astatine. Fluorine is the most electronegative element in the periodic table because it has the small atomic radius. Electron affinity The minimum amount of energy required to gain/acquire an extra electron by an atom of element in its gaseous state is called 1st electron affinity. The SI unit of electron affinity is kilojoules per mole/kJmole-1 . Electron affinity depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer energy level electrons and thus the lower the electron affinity. For halogens the 1st electron affinity decrease down the group as the\n43 atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. Due to its small size/atomic radius Fluorine shows exceptionally low electron affinity. This is because a lot of energy is required to overcome the high repulsion of the existing and incoming electrons. Table showing the election affinity of halogens for the process X + e -> X- The higher the electron affinity the more stable theion.i.e Cl- is a more stable ion than Br- because it has a more negative / exothermic electron affinity than Br- Electron affinity is different from: (i) Ionization energy. Ionization energy is the energy required to lose/donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state. (ii) Electronegativity. -Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process: X(g) + e -> X-(g) Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8992760895170788, "ocr_used": false, "chunk_length": 2264, "token_count": 504}}
{"text": "(ii) Electronegativity. -Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process: X(g) + e -> X-(g) Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions. It does not involve use of energy but theoretical arbitrary Pauling’ scale of measurements. Physical properties State at room temperature Fluorine and Chlorine are gases, Bromine is a liquid and Iodine is a solid. Astatine is radioactive . All halogens exist as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces. Melting/Boiling point The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius. Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid. Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour. Halogen F Cl Br I Electron affinity kJmole-1 -333 -364 -342 -295\n44 This is because Iodine molecules are held together by weak van-derwaals/intermolecular forces which require little heat energy to break. Electrical conductivity All Halogens are poor conductors of electricity because they have no free delocalized electrons. Solubility in polar and non-polar solvents All halogens are soluble in water(polar solvent). When a boiling tube containing either chlorine gas or bromine vapour is separately inverted in a beaker containing distilled water and tetrachloromethane (non-polar solvent), the level of solution in boiling tube rises in both water and tetrachloromethane. This is because halogen are soluble in both polar and non-polar solvents. Solubility of halogens in water/polar solvents decrease down the group. Solubility of halogens in non-polar solvent increase down the group. The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9121726738651662, "ocr_used": false, "chunk_length": 2139, "token_count": 499}}
{"text": "Solubility of halogens in non-polar solvent increase down the group. The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous. Table showing the physical properties of Halogens Halogen Formula of molecule Electrical conductivity Solubility in water Melting point(oC) Boiling point(oC) Fluorine F2 Poor Insoluble/soluble in tetrachloromethane -238 -188 Chlorine Cl2 Poor Insoluble/soluble in tetrachloromethane -101 -35 Bromine Br2 Poor Insoluble/soluble in tetrachloromethane 7 59 Iodine I2 Poor Insoluble/soluble in tetrachloromethane 114 sublimes Chemical properties (i)Displacement Experiment Place separately in test tubes about 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions. Add 5 drops of chlorine water to each test tube: Repeat with 5 drops of bromine water instead of chlorine water\n45 Observation Using Chlorine water -Yellow colour of chlorine water fades in all test tubes except with sodium chloride. -Coloured Solution formed. Using Bromine water Yellow colour of bromine water fades in test tubes containing sodium iodide. -Coloured Solution formed. Explanation The halogens displace each other from their solution. The more electronegative displace the less electronegative from their solution. Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine. Bromine is more electronegative than iodide but less 6than chlorine. On adding Bromine water, iodine is displaced from its solution but not chlorine.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8973411764705883, "ocr_used": false, "chunk_length": 1700, "token_count": 412}}
{"text": "On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine. Bromine is more electronegative than iodide but less 6than chlorine. On adding Bromine water, iodine is displaced from its solution but not chlorine. Table showing the displacement of the halogens (V) means there is displacement (x ) means there is no displacement Chemical /ionic equations With Fluorine F2(g) + 2NaCl-(aq) -> 2NaF(aq) + Cl2(aq) F2(g) + 2Cl-(aq) -> 2F-(aq) + Cl2(aq) F2(g) + 2NaBr-(aq) -> 2NaF(aq) + Br2(aq) F2(g) + 2Br-(aq) -> 2F-(aq) + Br2(aq) F2(g) + 2NaI-(aq) -> 2NaF(aq) + I2(aq) F2(g) + 2I-(aq) -> 2F-(aq) + I2(aq) With chlorine Cl2(g) + 2NaCl-(aq) -> 2NaCl(aq) + Br2(aq) Cl2(g) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq) Halogen ion in solution Halogen F- Cl- Br- I- F2 X Cl2 X X Br2 X X X I2 X X X X\n46 Cl2(g) + 2NaI-(aq) -> 2NaCl(aq) + I2(aq) Cl2(g) + 2I-(aq) -> 2Cl-(aq) + I2(aq) With Bromine Br2(g) + 2NaI-(aq) -> 2NaBr(aq) + I2(aq) Br2(g) + 2I-(aq) -> 2Br-(aq) + I2(aq) Uses of halogens (i) Florine – manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber. - Reduce tooth decay when added in small amounts/quantities in tooth paste. NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis. - Hydrogen fluoride is used to engrave words /pictures in glass.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7355881702757514, "ocr_used": false, "chunk_length": 1317, "token_count": 508}}
{"text": "- Reduce tooth decay when added in small amounts/quantities in tooth paste. NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis. - Hydrogen fluoride is used to engrave words /pictures in glass. (ii) Bromine - Silver bromide is used to make light sensitive photographic paper/films. (iii) Iodide – Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine. The table below to show some compounds of halogens. (i) Below is the table showing the bond energy of four halogens. Bond Bond energy k J mole-1 Cl-Cl 242 Br-Br 193 I-I 151 I. What do you understand by the term “bond energy” Element Halogen H Na Mg Al Si C P F HF NaF MgH2 AlF3 SiF4 CF4 PF3 Cl HCl NaCl MgCl AlCl3 SiCl3 CCl4 PCl3 Br HBr NaBr MgBr2 AlBr3 SiBr4 CBr4 PBr3 I Hl Nal Mgl2 All3 SiI4 Cl2 PBr3\n47 Bond energy is the energy required to break/ form one mole of chemical bond II. Explain the trend in bond Energy of the halogens above: -Decrease down the group from chlorine to Iodine -Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear @ charge an outer energy level that take part in bonding. (c)Group VIII elements: Noble gases Group VIII elements are called Noble gases. They are all non metals. Noble gases occupy about 1.0% of the atmosphere as colourless gaseous mixture. Argon is the most abundant with 0.9%. They exists as monatomic molecules with very weak van-der-waals /intermolecular forces holding the molecules. They include: All noble gas atoms have a stable duplet(two electrons in the 1st energy level) or octet(eight electrons in other outer energy level)in the outer energy level. They therefore do not acquire/gain extra electron in the outer energy level or donate/lose. They therefore are therefore zerovalent . The number of energy levels increases down the group from Helium to Randon.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8717963789062443, "ocr_used": false, "chunk_length": 1971, "token_count": 500}}
{"text": "They therefore do not acquire/gain extra electron in the outer energy level or donate/lose. They therefore are therefore zerovalent . The number of energy levels increases down the group from Helium to Randon. The more the number of energy levels the bigger/larger the atomic size/radius. e.g. The atomic size/radius of Argon is bigger/larger than that of Neon because Argon has more/3 energy levels than Neon (2 energy levels). Atomic radius noble gases increase down the group as the number of energy levels increases. The effective nuclear attraction on the outer electrons thus decrease down the group. The noble gases are generally unreactive because the outer energy level has the stable octet/duplet. The stable octet/duplet in noble gas atoms lead to a Element Symbol Atomic number Electron structure State at room temperature Helium He 2 2: Colourless gas Neon Ne 10 2:8 Colourless gas Argon Ar 18 2:8:8 Colourless gas Krypton Kr 36 2:8:18:8 Colourless gas Xenon Xe 54 2:8:18:18:8 Colourless gas Radon Rn 86 2:8:18:32:18:8 Radioctive\n48 comparatively very high 1st ionization energy. This is because losing /donating an electron from the stable atom require a lot of energy to lose/donate and make it unstable. As atomic radius increase down the group and the 1st ionization energy decrease, very electronegative elements like Oxygen and Fluorine are able to react and bond with lower members of the noble gases.e.g Xenon reacts with Fluorine to form a covalent compound XeF6.This is because the outer electrons/energy level if Xenon is far from the nucleus and thus experience less effective nuclear attraction. Noble gases have low melting and boiling points. This is because they exist as monatomic molecules joined by very weak intermolecular/van-der-waals forces that require very little energy to weaken and form liquid and break to form a gas. The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group. Noble gases are insoluble in water and are poor conductors of electricity.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.876254525722114, "ocr_used": false, "chunk_length": 2121, "token_count": 497}}
{"text": "The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group. Noble gases are insoluble in water and are poor conductors of electricity. Element Formula of molecule Electrical conductivity Solubility in water Atomic radius(nM) 1st ionization energy Melting point(0C) Boiling point(0C) Helium He Poor Insoluble 0.128 2372 -270 -269 Neon Ne Poor Insoluble 0.160 2080 -249 -246 Argon Ar Poor Insoluble 0.192 1520 -189 -186 Krypton Kr Poor Insoluble 0.197 1350 -157 -152 Xenon Xe Poor Insoluble 0.217 1170 -112 -108 Radon Rn Poor Insoluble 0.221 1134 -104 -93 Uses of noble gases Argon is used in light bulbs to provide an inert environment to prevent oxidation of the bulb filament Argon is used in arch welding as an insulator. Neon is used in street and advertisement light Helium is mixed with Oxygen during deep sea diving and mountaineering. Helium is used in weather balloon for meteorological research instead of Hydrogen because it is unreactive/inert.Hydrogen when impure can ignite with an explosion. Helium is used in making thermometers for measuring very low temperatures. 49 C. PERIODICITY OF ACROSS THE PERIOD. (See Chemical bonding and Structure)\nPERIODICITY OF CHEMICAL FAMILES (Patterns down the group) The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table. i.e The number of occupied energy levels determine the Period and the valence electrons determine the Group. Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8608336757180743, "ocr_used": false, "chunk_length": 1959, "token_count": 468}}
{"text": "The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M+ The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels). Atomic and ionic radius The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size. 2 The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion. Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g. The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8535331194251478, "ocr_used": false, "chunk_length": 2049, "token_count": 504}}
{"text": "The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level. Table showing the atomic and ionic radius of some alkali metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Lithium Li 3 0.133 0.060 Sodium Na 11 0.157 0.095 Potassium K 19 0.203 0.133 The atomic radius of sodium is 0.157nM .The ionic radius of Na+ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius. Ionization energy\n3 The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8840729591988794, "ocr_used": false, "chunk_length": 2314, "token_count": 506}}
{"text": "For alkali metals the 1st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium. Physical properties Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen. Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure. Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure. Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419\n4 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8911842982493872, "ocr_used": false, "chunk_length": 2144, "token_count": 478}}
{"text": "Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419\n4 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate. 2NaOH(aq) + CO2(g) -> Na2CO3(g) + H2O(l) (ii)Burning in air/oxygen Lithium burns in air with a crimson/deep red flame to form Lithium oxide 4Li (s) + O2(g) -> 2Li2O(s) Sodium burns in air with a yellow flame to form sodium oxide 4Na (s) + O2(g) -> 2Na2O(s) Sodium burns in oxygen with a yellow flame to form sodium peroxide 2Na (s) + O2(g) -> Na2O2 (s) Potassium burns in air with a lilac/purple flame to form potassium oxide 4K (s) + O2(g) -> 2K2O (s) (iii) Reaction with water: Experiment Measure 500 cm3 of water into a beaker. Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.816195307704834, "ocr_used": false, "chunk_length": 1495, "token_count": 447}}
{"text": "Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium. Caution: Keep a distance Observations\n5 Alkali metal Observations Comparative speed/rate of the reaction Lithium -Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Moderately vigorous Sodium -Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Very vigorous Potassium -Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Explosive/burst into flames Explanation Alkali metals are less dense than water. They therefore float in water.They react with water to form a strongly alkaline solution of their hydroxides and producing hydrogen gas. The rate of this reaction increase down the group. i.e. Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8915754273749128, "ocr_used": false, "chunk_length": 1627, "token_count": 383}}
{"text": "Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions. 6 Chemical equations 2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g) 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) 2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g) 2Cs(s) + 2H2O(l) -> 2CsOH(aq) + H2(g) 2Fr(s) + 2H2O(l) -> 2FrOH(aq) + H2(g) Reactivity increase down the group (iv) Reaction with chlorine: Experiment Cut about 0.5g of sodium into a deflagrating spoon with a lid cover. Introduce it on a Bunsen flame until it catches fire. Quickly and carefully lower it into a gas jar containing dry chlorine to cover the gas jar. Repeat with about 0.5g of Lithium. Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic. Observation Sodium metal continues to burn with a yellow flame forming white solid/fumes. Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8376583448913546, "ocr_used": false, "chunk_length": 1442, "token_count": 409}}
{"text": "Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus. Chemical equations 2Li(s) + Cl2(g) -> 2LiCl(s) 2Na(s) + Cl2(g) -> 2NaCl(s) 2K(s) + Cl2(g) -> 2KCl(s) 2Rb(s) + Cl2(g) -> 2RbCl(s) 2Cs(s) + Cl2(g) -> 2CsCl(s) 2Fr(s) + Cl2(g) -> 2FrCl(s) Reactivity increase down the group The table below shows some compounds of the 1st three alkali metals\n7 Some uses of alkali metals include: (i)Sodium is used in making sodium cyanide for extracting gold from gold ore. (ii)Sodium chloride is used in seasoning food. (iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors. (iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents. (v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride. (vi)Lithium is used in making special high strength glasses (vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops. Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals .", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8591408591408591, "ocr_used": false, "chunk_length": 1365, "token_count": 368}}
{"text": "(v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride. (vi)Lithium is used in making special high strength glasses (vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops. Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals . The alkaline earth metals include: Lithium sodium Potassium Hydroxide LiOH NaOH KOH Oxide Li2O Na2O K2O Sulphide Li2S Na2S K2S Chloride LiCl NaCl KCl Carbonate Li2CO3 Na2CO3 K2CO3 Nitrate(V) LiNO3 NaNO3 KNO3 Nitrate(III) - NaNO2 KNO2 Sulphate(VI) Li2SO4 Na2SO4 K2SO4 Sulphate(IV) - Na2SO3 K2SO3 Hydrogen carbonate - NaHCO3 KHCO3 Hydrogen sulphate(VI) - NaHSO4 KHSO4 Hydrogen sulphate(IV) - NaHSO3 KHSO3 Phosphate - Na3PO4 K3PO4 Manganate(VI) - NaMnO4 KMnO4 Dichromate(VI) - Na2Cr2O7 K2Cr2O7 Chromate(VI) - Na2CrO4 K2CrO4\n8 Element Symbol Atomic number Electron structure Oxidation state Valency Beryllium Be 4 2:2 Be2+ 2 Magnesium Mg 12 2:8:2 Mg2+ 2 Calcium Ca 20 2:8:8:2 Ca2+ 2 Strontium Sr 38 2:8:18:8:2 Sr2+ 2 Barium Ba 56 2:8:18:18:8:2 Ba2+ 2 Radium Ra 88 2:8:18:32:18:8:2 Ra2+ 2 All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent. They donate /lose the two outer electrons to have oxidation state M2+ The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7512635865900693, "ocr_used": false, "chunk_length": 1454, "token_count": 507}}
{"text": "They donate /lose the two outer electrons to have oxidation state M2+ The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Calcium is bigger/larger than that of Magnesium because Calcium has more/4 energy levels than Magnesium (3 energy levels). Atomic radius and ionic radius of alkaline earth metals increase down the group as the number of energy levels increases. The atomic radius of alkaline earth metals is bigger than the ionic radius. This is because they react by losing/donating the two outer electrons and hence lose the outer energy level. Table showing the atomic and ionic radius of the 1st three alkaline earth metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Beryllium Be 4 0.089 0.031 Magnesium Mg 12 0.136 0.065 Calcium Ca 20 0.174 0.099 The atomic radius of Magnesium is 0.136nM .The ionic radius of Mg2+ is 0.065nM. This is because Magnesium reacts by donating/losing the two outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. 9 Electropositivity All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size. Calcium is more electropositive than Magnesium. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The two outer electrons in calcium experience less nuclear attraction and can be lost/ donated easily/with ease because of the higher/bigger atomic radius. Ionization energy For alkaline earth metals the 1st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of Magnesium is 900 kJmole-1 while that of Calcium is 590 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from magnesium to calcium. It requires therefore less energy to donate/lose outer electron in calcium than in magnesium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8816314553990612, "ocr_used": false, "chunk_length": 2272, "token_count": 497}}
{"text": "e.g. The 1st ionization energy of Magnesium is 900 kJmole-1 while that of Calcium is 590 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from magnesium to calcium. It requires therefore less energy to donate/lose outer electron in calcium than in magnesium. The minimum amount of energy required to remove a second electron from an ion of an element in its gaseous state is called the 2nd ionization energy. The 2nd ionization energy is always higher /bigger than the 1st ionization energy. This because once an electron is donated /lost form an atom, the overall effective nuclear attraction on the remaining electrons/energy level increase. Removing a second electron from the ion require therefore more energy than the first electron. The atomic radius of alkali metals is higher/bigger than that of alkaline earth metals.This is because across/along the period from left to right there is an increase in nuclear charge from additional number of protons and still additional number of electrons entering the same energy level. Increase in nuclear charge increases the effective nuclear attraction on the outer energy level which pulls it closer to the nucleus. e.g. Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus. Physical properties Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals. This is because of the decrease in atomic radius of\n10 corresponding alkaline earth metal, increases the strength of metallic bond and the packing of the metallic structure. Alkaline earth metals are (i)ductile(able to form wire/thin long rods) (ii)malleable(able to be hammered into sheet/long thin plates) (iii)have high tensile strength(able to be coiled without breaking/ not brittle/withstand stress) Appearance: Alkali earth metals have a shiny grey metallic luster when their surface is freshly polished /scrubbed. The surface slowly tarnishes on exposure to air. This is because the metal surface slowly undergoes oxidation to form an oxide. This oxide layer should be removed before using the alkaline earth metals.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9057124549599962, "ocr_used": false, "chunk_length": 2349, "token_count": 495}}
{"text": "The surface slowly tarnishes on exposure to air. This is because the metal surface slowly undergoes oxidation to form an oxide. This oxide layer should be removed before using the alkaline earth metals. Melting and boiling points: Alkaline earth metals have a relatively high melting/ boiling point than alkali metals. This is because alkali metals use only one delocalized electron to form a weaker metallic bond/structure. Alkaline earth metals use two delocalized electrons to form a stronger metallic bond /structure. The melting and boiling points decrease down the group as the atomic radius/size increase reducing the strength of metallic bond and packing of the metallic structure. e.g. Beryllium has a melting point of 1280oC. Magnesium has a melting point of 650oC.Beryllium has a smaller atomic radius/size than magnesium .The strength of metallic bond and packing of the metallic structure is thus stronger in beryllium. Electrical/thermal conductivity: Alkaline earth metals are good thermal and electrical conductors. The two delocalized valence electrons move randomly within the metallic structure. Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g. Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor. Calcium is a better conductor than magnesium. Calcium has bigger/larger atomic radius than magnesium because the delocalized electrons are less attracted to the nucleus of calcium and thus more free /mobile and thus better the electrical conductor Summary of some physical properties of the 1st three alkaline earth metals\n11 Alkaline earth metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conduct- ivity 1st ionization energy 2nd ionization energy Beryllium Shiny grey Not easy 1280 3450 Good 900 1800 Magnesium Shiny grey Not Easy 650 1110 Good 736 1450 calcium Shiny grey Not easy 850 1140 Good 590 970 Chemical properties (i)Reaction with air/oxygen On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9002879089615933, "ocr_used": false, "chunk_length": 2375, "token_count": 507}}
{"text": "The more delocalized electrons the better the electrical conductor. Calcium is a better conductor than magnesium. Calcium has bigger/larger atomic radius than magnesium because the delocalized electrons are less attracted to the nucleus of calcium and thus more free /mobile and thus better the electrical conductor Summary of some physical properties of the 1st three alkaline earth metals\n11 Alkaline earth metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conduct- ivity 1st ionization energy 2nd ionization energy Beryllium Shiny grey Not easy 1280 3450 Good 900 1800 Magnesium Shiny grey Not Easy 650 1110 Good 736 1450 calcium Shiny grey Not easy 850 1140 Good 590 970 Chemical properties (i)Reaction with air/oxygen On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air. Example On exposure to air, the surface of magnesium ribbon is oxidized to form a thin film of Magnesium oxide . 2Mg(s) + O2(g) -> 2MgO(s) (ii)Burning in air/oxygen Experiment Hold a about 2cm length of Magnesium ribbon on a Bunsen flame. Stop heating when it catches fire/start burning. Caution: Do not look directly at the flame Put the products of burning into 100cm3 beaker. Add about 5cm3 of distilled water. Swirl. Test the mixture using litmus papers. Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride. 2Mg (s) + O2(g) -> 2MgO(s) 3Mg (s) + N2 (g) -> Mg3N2 (s) Magnesium oxide dissolves in water to form magnesium hydroxide. 12 MgO(s) + H2O (l) -> Mg(OH)2(aq) Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8602495577338346, "ocr_used": false, "chunk_length": 1961, "token_count": 505}}
{"text": "Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride. 2Mg (s) + O2(g) -> 2MgO(s) 3Mg (s) + N2 (g) -> Mg3N2 (s) Magnesium oxide dissolves in water to form magnesium hydroxide. 12 MgO(s) + H2O (l) -> Mg(OH)2(aq) Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas. Mg3N2 (s) + 6H2O(l) -> 3Mg(OH)2(aq) + 2NH3 (g) Magnesium hydroxide and ammonia are weakly alkaline with pH 8/9/10/11 and turns red litmus paper blue. Calcium burns in air with faint orange/red flame to form a mixture of both Calcium oxide and calcium nitride. 2Ca (s) + O2(g) -> 2CaO(s) 3Ca (s) + N2 (g) -> Ca3N2 (s) Calcium oxide dissolves in water to form calcium hydroxide. CaO(s) + H2O(l) -> Ca(OH)2(aq) Calcium nitride dissolves in water to form calcium hydroxide and produce ammonia gas. Ca3N2 (s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2NH3 (g) Calcium hydroxide is also weakly alkaline solution with pH 8/9/10/11 and turns red litmus paper blue. (iii)Reaction with water Experiment Measure 50 cm3 of distilled water into a beaker. Scrub/polish with sand paper 1cm length of Magnesium ribbon Place it in the water. Test the product-mixture with blue and red litmus papers. Repeat with Calcium metal. Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed. -Effervescence/bubbles/fizzing takes place in Calcium.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.826707358258993, "ocr_used": false, "chunk_length": 1692, "token_count": 512}}
{"text": "Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed. -Effervescence/bubbles/fizzing takes place in Calcium. -Red litmus paper turns blue. -Blue litmus paper remains blue. Explanations Magnesium slowly reacts with cold water to form Magnesium hydroxide and bubbles of Hydrogen gas that stick on the surface of the ribbon. Mg(s) + 2H2O (l) -> Mg(OH)2(aq) + H2 (g)\n13 Calcium moderately reacts with cold water to form Calcium hydroxide and produce a steady stream of Hydrogen gas. Ca(s) + 2H2O (l) -> Ca(OH)2(aq) + H2 (g) (iv)Reaction with water vapour/steam Experiment Put some cotton wool soaked in water/wet sand in a long boiling tube. Coil a well polished magnesium ribbon into the boiling tube. Ensure the coil touches the side of the boiling tube. Heat the cotton wool/sand slightly then strongly heat the Magnesium ribbon . Set up of apparatus Observations -Magnesium glows red hot then burns with a blindening flame. -Magnesium continues to glow/burning even without more heating. -White solid/residue. -colourless gas collected over water. Explanation On heating wet sand, steam is generated which drives out the air that would otherwise react with /oxidize the ribbon. Magnesium burns in steam/water vapour generating enough heat that ensures the reaction goes to completion even without further heating. White Magnesium oxide is formed and hydrogen gas is evolved. To prevent suck back, the delivery tube should be removed from the water before heating is stopped at the end of the experiment. 14 Mg(s) + H2O (l) -> MgO(s) + H2 (g) (v)Reaction with chlorine gas. Experiment Lower slowly a burning magnesium ribbon/shavings into a gas jar containing Chlorine gas. Repeat with a hot piece of calcium metal. Observation -Magnesium continues to burn in chlorine with a bright blindening flame. -Calcium continues to burn for a short time. -White solid formed . -Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8905294723418333, "ocr_used": false, "chunk_length": 2148, "token_count": 500}}
{"text": "-Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid. Calcium oxide formed coat unreacted Calcium stopping further reaction Ca(s) + Cl2 (g) -> CaCl2 (s) (v)Reaction with dilute acids. Experiment Place about 4.0cm3 of 0.1M dilute sulphuric(VI)acid into a test tube. Add about 1.0cm length of magnesium ribbon into the test tube. Cover the mouth of the test tube using a thumb. Release the gas and test the gas using a burning splint. Repeat with about 4.0cm3 of 0.1M dilute hydrochloric/nitric(V) acid. Repeat with 0.1g of Calcium in a beaker with all the above acid Caution: Keep distance when using calcium Observation -Effervescence/fizzing/bubbles with dilute sulphuric(VI) and nitric(V) acids -Little Effervescence/fizzing/bubbles with calcium and dilute sulphuric(VI) acid. -Colourless gas produced that extinguishes a burning splint with an explosion/ “pop” sound. -No gas is produced with Nitric(V)acid. -Colourless solution is formed. Explanation Dilute acids react with alkaline earth metals to form a salt and produce hydrogen gas. 15 Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8743314633006078, "ocr_used": false, "chunk_length": 1426, "token_count": 366}}
{"text": "15 Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used. Chemical equations Mg(s) + H2SO4 (aq) -> MgSO4(aq) + H2 (g) Mg(s) + 2HNO3 (aq) -> Mg(NO3)2(aq) + H2 (g) Mg(s) + 2HCl (aq) -> MgCl2(aq) + H2 (g) Ca(s) + H2SO4 (aq) -> CaSO4(s) + H2 (g) (insoluble CaSO4(s) coat/cover Ca(s)) Ca(s) + 2HNO3 (aq) -> Ca(NO3)2(aq) + H2 (g) Ca(s) + 2HCl (aq) -> CaCl2(aq) + H2 (g) Ba(s) + H2SO4 (aq) -> BaSO4(s) + H2 (g) (insoluble BaSO4(s) coat/cover Ba(s)) Ba(s) + 2HNO3 (aq) -> Ba(NO3)2(aq) + H2 (g) Ba(s) + 2HCl (aq) -> BaCl2(aq) + H2 (g) The table below shows some compounds of some alkaline earth metals Some uses of alkaline earth metals include: (i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity. (ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7608227926587302, "ocr_used": false, "chunk_length": 1024, "token_count": 383}}
{"text": "Chemical equations Mg(s) + H2SO4 (aq) -> MgSO4(aq) + H2 (g) Mg(s) + 2HNO3 (aq) -> Mg(NO3)2(aq) + H2 (g) Mg(s) + 2HCl (aq) -> MgCl2(aq) + H2 (g) Ca(s) + H2SO4 (aq) -> CaSO4(s) + H2 (g) (insoluble CaSO4(s) coat/cover Ca(s)) Ca(s) + 2HNO3 (aq) -> Ca(NO3)2(aq) + H2 (g) Ca(s) + 2HCl (aq) -> CaCl2(aq) + H2 (g) Ba(s) + H2SO4 (aq) -> BaSO4(s) + H2 (g) (insoluble BaSO4(s) coat/cover Ba(s)) Ba(s) + 2HNO3 (aq) -> Ba(NO3)2(aq) + H2 (g) Ba(s) + 2HCl (aq) -> BaCl2(aq) + H2 (g) The table below shows some compounds of some alkaline earth metals Some uses of alkaline earth metals include: (i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity. (ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light. Beryllium Magnesium Calcium Barium Hydroxide Be(OH)2 Mg(OH)2 Ca(OH)2 Ba(OH)2 Oxide BeO MgO CaO BaO Sulphide - MgS CaS BaS Chloride BeCl2 MgCl2 CaCl2 BaCl2 Carbonate BeCO3 MgCO3 CaCO3 BaCO3 Nitrate(V) Be(NO3)2 Mg(NO3)2 Ca(NO3)2 Ba(NO3)2 Sulphate(VI) BeSO4 MgSO4 CaSO4 BaSO4 Sulphate(IV) - - CaSO3 BaSO3 Hydrogen carbonate - Mg(HCO3)2 Ca(HCO3)2 - Hydrogen sulphate(VI) - Mg(HSO4)2 Ca(HSO4)2 -\n16 (iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7191315634947616, "ocr_used": false, "chunk_length": 1310, "token_count": 538}}
{"text": "(ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light. Beryllium Magnesium Calcium Barium Hydroxide Be(OH)2 Mg(OH)2 Ca(OH)2 Ba(OH)2 Oxide BeO MgO CaO BaO Sulphide - MgS CaS BaS Chloride BeCl2 MgCl2 CaCl2 BaCl2 Carbonate BeCO3 MgCO3 CaCO3 BaCO3 Nitrate(V) Be(NO3)2 Mg(NO3)2 Ca(NO3)2 Ba(NO3)2 Sulphate(VI) BeSO4 MgSO4 CaSO4 BaSO4 Sulphate(IV) - - CaSO3 BaSO3 Hydrogen carbonate - Mg(HCO3)2 Ca(HCO3)2 - Hydrogen sulphate(VI) - Mg(HSO4)2 Ca(HSO4)2 -\n16 (iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone. (iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building. (iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms. (v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth. (vi)In the blast furnace-Limestone is added to the blast furnace to produce more reducing agent and remove slag in the blast furnace for extraction of Iron. (c)Group VII elements: Halogens Group VII elements are called Halogens. They are all non metals. They include: Element Symbol Atomic number Electronicc configuration Charge of ion Valency State at Room Temperature Fluorine Chlorine Bromine Iodine Astatine F Cl Br I At 9 17 35 53 85 2:7 2:8:7 2:8:18:7 2:8:18:18:7 2:8:18:32:18:7 F- Cl- Br- I- At- 1 1 1 1 1 Pale yellow gas Pale green gas Red liquid Grey Solid Radioactive All halogen atoms have seven electrons in the outer energy level. They acquire/gain one electron in the outer energy level to be stable.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8041701171181829, "ocr_used": false, "chunk_length": 1712, "token_count": 515}}
{"text": "They are all non metals. They include: Element Symbol Atomic number Electronicc configuration Charge of ion Valency State at Room Temperature Fluorine Chlorine Bromine Iodine Astatine F Cl Br I At 9 17 35 53 85 2:7 2:8:7 2:8:18:7 2:8:18:18:7 2:8:18:32:18:7 F- Cl- Br- I- At- 1 1 1 1 1 Pale yellow gas Pale green gas Red liquid Grey Solid Radioactive All halogen atoms have seven electrons in the outer energy level. They acquire/gain one electron in the outer energy level to be stable. They therefore are therefore monovalent .They exist in oxidation state X- The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Chlorine is bigger/larger than that of Fluorine because Chlorine has more/3 energy levels than Fluorine (2 energy levels). 17 Atomic radius and ionic radius of Halogens increase down the group as the number of energy levels increases. The atomic radius of Halogens is smaller than the ionic radius. This is because they react by gaining/acquiring extra one electron in the outer energy level. The effective nuclear attraction on the more/extra electrons decreases. The incoming extra electron is also repelled causing the outer energy level to expand to reduce the repulsion and accommodate more electrons. Table showing the atomic and ionic radius of four Halogens Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Fluorine F 9 0.064 0.136 Chlorine Cl 17 0.099 0.181 Bromine Br 35 0.114 0.195 Iodine I 53 0.133 0.216 The atomic radius of Chlorine is 0.099nM .The ionic radius of Cl- is 0.181nM. This is because Chlorine atom/molecule reacts by gaining/acquiring extra one electrons. The more/extra electrons/energy level experience less effective nuclear attraction /pull towards the nucleus .The outer enegy level expand/increase to reduce the repulsion of the existing and incoming gained /acquired electrons.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8316544196948563, "ocr_used": false, "chunk_length": 1970, "token_count": 502}}
{"text": "Table showing the atomic and ionic radius of four Halogens Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Fluorine F 9 0.064 0.136 Chlorine Cl 17 0.099 0.181 Bromine Br 35 0.114 0.195 Iodine I 53 0.133 0.216 The atomic radius of Chlorine is 0.099nM .The ionic radius of Cl- is 0.181nM. This is because Chlorine atom/molecule reacts by gaining/acquiring extra one electrons. The more/extra electrons/energy level experience less effective nuclear attraction /pull towards the nucleus .The outer enegy level expand/increase to reduce the repulsion of the existing and incoming gained /acquired electrons. Electronegativity The ease of gaining/acquiring extra electrons is called electronegativity. All halogens are electronegative. Electronegativity decreases as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and thus ease of gaining/acquiring extra electrons decrease. It is measured using Pauling’s scale. Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron. Table showing the electronegativity of the halogens. Halogen F Cl Br I At Electronegativity (Pauling scale) 4.0 3.0 2.8 2.5 2.2\n18 The electronegativity of the halogens decrease down the group from fluorine to Astatine. This is because atomic radius increases down the group and thus decrease electron – attracting power down the group from fluorine to astatine. Fluorine is the most electronegative element in the periodic table because it has the small atomic radius. Electron affinity The minimum amount of energy required to gain/acquire an extra electron by an atom of element in its gaseous state is called 1st electron affinity. The SI unit of electron affinity is kilojoules per mole/kJmole-1 . Electron affinity depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer energy level electrons and thus the lower the electron affinity.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8657277760602569, "ocr_used": false, "chunk_length": 2093, "token_count": 491}}
{"text": "The SI unit of electron affinity is kilojoules per mole/kJmole-1 . Electron affinity depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer energy level electrons and thus the lower the electron affinity. For halogens the 1st electron affinity decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. Due to its small size/atomic radius Fluorine shows exceptionally low electron affinity. This is because a lot of energy is required to overcome the high repulsion of the existing and incoming electrons. Table showing the election affinity of halogens for the process X + e -> X- The higher the electron affinity the more stable theion.i.e Cl- is a more stable ion than Br- because it has a more negative / exothermic electron affinity than Br- Electron affinity is different from: (i) Ionization energy. Ionization energy is the energy required to lose/donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state. (ii) Electronegativity. -Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process: X(g) + e -> X-(g) Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions. Halogen F Cl Br I Electron affinity kJmole-1 -333 -364 -342 -295\n19 It does not involve use of energy but theoretical arbitrary Pauling’ scale of measurements. Physical properties State at room temperature Fluorine and Chlorine are gases, Bromine is a liquid and Iodine is a solid. Astatine is radioactive . All halogens exist as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces. Melting/Boiling point The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius. Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid. Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9069951153048949, "ocr_used": false, "chunk_length": 2262, "token_count": 500}}
{"text": "Melting/Boiling point The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius. Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid. Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour. This is because Iodine molecules are held together by weak van-derwaals/intermolecular forces which require little heat energy to break. Electrical conductivity All Halogens are poor conductors of electricity because they have no free delocalized electrons. Solubility in polar and non-polar solvents All halogens are soluble in water(polar solvent). When a boiling tube containing either chlorine gas or bromine vapour is separately inverted in a beaker containing distilled water and tetrachloromethane (non-polar solvent), the level of solution in boiling tube rises in both water and tetrachloromethane. This is because halogen are soluble in both polar and non-polar solvents. Solubility of halogens in water/polar solvents decrease down the group. Solubility of halogens in non-polar solvent increase down the group. The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous. 20 Table showing the physical properties of Halogens Halogen Formula of molecule Electrical conductivity Solubility in water Melting point(oC) Boiling point(oC) Fluorine F2 Poor Insoluble/soluble in tetrachloromethane -238 -188 Chlorine Cl2 Poor Insoluble/soluble in tetrachloromethane -101 -35 Bromine Br2 Poor Insoluble/soluble in tetrachloromethane 7 59 Iodine I2 Poor Insoluble/soluble in tetrachloromethane 114 sublimes Chemical properties (i)Displacement Experiment Place separately in test tubes about 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9064318605223383, "ocr_used": false, "chunk_length": 1962, "token_count": 479}}
{"text": "The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous. 20 Table showing the physical properties of Halogens Halogen Formula of molecule Electrical conductivity Solubility in water Melting point(oC) Boiling point(oC) Fluorine F2 Poor Insoluble/soluble in tetrachloromethane -238 -188 Chlorine Cl2 Poor Insoluble/soluble in tetrachloromethane -101 -35 Bromine Br2 Poor Insoluble/soluble in tetrachloromethane 7 59 Iodine I2 Poor Insoluble/soluble in tetrachloromethane 114 sublimes Chemical properties (i)Displacement Experiment Place separately in test tubes about 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions. Add 5 drops of chlorine water to each test tube: Repeat with 5 drops of bromine water instead of chlorine water Observation Using Chlorine water -Yellow colour of chlorine water fades in all test tubes except with sodium chloride. -Coloured Solution formed. Using Bromine water Yellow colour of bromine water fades in test tubes containing sodium iodide. -Coloured Solution formed. Explanation The halogens displace each other from their solution. The more electronegative displace the less electronegative from their solution. Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine. Bromine is more electronegative than iodide but less 6than chlorine. 21 On adding Bromine water, iodine is displaced from its solution but not chlorine.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8939432884536925, "ocr_used": false, "chunk_length": 1634, "token_count": 398}}
{"text": "On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine. Bromine is more electronegative than iodide but less 6than chlorine. 21 On adding Bromine water, iodine is displaced from its solution but not chlorine. Table showing the displacement of the halogens (V) means there is displacement (x ) means there is no displacement Chemical /ionic equations With Fluorine F2(g) + 2NaCl-(aq) -> 2NaF(aq) + Cl2(aq) F2(g) + 2Cl-(aq) -> 2F-(aq) + Cl2(aq) F2(g) + 2NaBr-(aq) -> 2NaF(aq) + Br2(aq) F2(g) + 2Br-(aq) -> 2F-(aq) + Br2(aq) F2(g) + 2NaI-(aq) -> 2NaF(aq) + I2(aq) F2(g) + 2I-(aq) -> 2F-(aq) + I2(aq) With chlorine Cl2(g) + 2NaCl-(aq) -> 2NaCl(aq) + Br2(aq) Cl2(g) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq) Cl2(g) + 2NaI-(aq) -> 2NaCl(aq) + I2(aq) Cl2(g) + 2I-(aq) -> 2Cl-(aq) + I2(aq) With Bromine Br2(g) + 2NaI-(aq) -> 2NaBr(aq) + I2(aq) Br2(g) + 2I-(aq) -> 2Br-(aq) + I2(aq) Halogen ion in solution Halogen F- Cl- Br- I- F2 X Cl2 X X Br2 X X X I2 X X X X\n22 Uses of halogens (i) Florine – manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber. - Reduce tooth decay when added in small amounts/quantities in tooth paste. NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis. - Hydrogen fluoride is used to engrave words /pictures in glass.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7337751464222053, "ocr_used": false, "chunk_length": 1320, "token_count": 510}}
{"text": "- Reduce tooth decay when added in small amounts/quantities in tooth paste. NB –large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis. - Hydrogen fluoride is used to engrave words /pictures in glass. (ii) Bromine - Silver bromide is used to make light sensitive photographic paper/films. (iii) Iodide – Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine. The table below to show some compounds of halogens. (i) Below is the table showing the bond energy of four halogens. Bond Bond energy k J mole-1 Cl-Cl 242 Br-Br 193 I-I 151 I. What do you understand by the term “bond energy” Bond energy is the energy required to break/ form one mole of chemical bond II. Explain the trend in bond Energy of the halogens above: -Decrease down the group from chlorine to Iodine Element Halogen H Na Mg Al Si C P F HF NaF MgF2 AlF3 SiF4 CF4 PF3 Cl HCl NaCl MgCl2 AlCl3 SiCl 4 CCl4 PCl3 Br HBr NaBr MgBr2 AlBr3 SiBr4 CBr4 PBr3 I Hl Nal Mgl2 All3 SiI4 C l 4 PBr3\n23 -Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear @ charge an outer energy level that take part in bonding. (c)Group VIII elements: Noble gases Group VIII elements are called Noble gases. They are all non metals. Noble gases occupy about 1.0% of the atmosphere as colourless gaseous mixture. Argon is the most abundant with 0.9%. They exists as monatomic molecules with very weak van-der-waals /intermolecular forces holding the molecules. They include: All noble gas atoms have a stable duplet(two electrons in the 1st energy level) or octet(eight electrons in other outer energy level)in the outer energy level. They therefore do not acquire/gain extra electron in the outer energy level or donate/lose. They therefore are therefore zerovalent . The number of energy levels increases down the group from Helium to Randon.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8703218806509947, "ocr_used": false, "chunk_length": 1975, "token_count": 504}}
{"text": "They therefore do not acquire/gain extra electron in the outer energy level or donate/lose. They therefore are therefore zerovalent . The number of energy levels increases down the group from Helium to Randon. The more the number of energy levels the bigger/larger the atomic size/radius. e.g. The atomic size/radius of Argon is bigger/larger than that of Neon because Argon has more/3 energy levels than Neon (2 energy levels). Atomic radius noble gases increase down the group as the number of energy levels increases. The effective nuclear attraction on the outer electrons thus decrease down the group. The noble gases are generally unreactive because the outer energy level has the stable octet/duplet. The stable octet/duplet in noble gas atoms lead to a comparatively very high 1st ionization energy. This is because losing /donating an electron from the stable atom require a lot of energy to lose/donate and make it unstable. Element Symbol Atomic number Electron structure State at room temperature Helium He 2 2: Colourless gas Neon Ne 10 2:8 Colourless gas Argon Ar 18 2:8:8 Colourless gas Krypton Kr 36 2:8:18:8 Colourless gas Xenon Xe 54 2:8:18:18:8 Colourless gas Radon Rn 86 2:8:18:32:18:8 Radioctive\n24 As atomic radius increase down the group and the 1st ionization energy decrease, very electronegative elements like Oxygen and Fluorine are able to react and bond with lower members of the noble gases.e.g Xenon reacts with Fluorine to form a covalent compound XeF6.This is because the outer electrons/energy level if Xenon is far from the nucleus and thus experience less effective nuclear attraction. Noble gases have low melting and boiling points. This is because they exist as monatomic molecules joined by very weak intermolecular/van-der-waals forces that require very little energy to weaken and form liquid and break to form a gas. The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group. Noble gases are insoluble in water and are poor conductors of electricity.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.876254525722114, "ocr_used": false, "chunk_length": 2121, "token_count": 497}}
{"text": "The intermolecular/van-der-waals forces increase down the group as the atomic radius/size increase from Helium to Radon. The melting and boiling points thus increase also down the group. Noble gases are insoluble in water and are poor conductors of electricity. Element Formula of molecule Electrical conductivity Solubility in water Atomic radius(nM) 1st ionization energy Melting point(0C) Boiling point(0C) Helium He Poor Insoluble 0.128 2372 -270 -269 Neon Ne Poor Insoluble 0.160 2080 -249 -246 Argon Ar Poor Insoluble 0.192 1520 -189 -186 Krypton Kr Poor Insoluble 0.197 1350 -157 -152 Xenon Xe Poor Insoluble 0.217 1170 -112 -108 Radon Rn Poor Insoluble 0.221 1134 -104 -93 Uses of noble gases Argon is used in light bulbs to provide an inert environment to prevent oxidation of the bulb filament Argon is used in arch welding as an insulator. Neon is used in street and advertisement light Helium is mixed with Oxygen during deep sea diving and mountaineering. Helium is used in weather balloon for meteorological research instead of Hydrogen because it is unreactive/inert. Hydrogen when impure can ignite with an explosion. Helium is used in making thermometers for measuring very low temperatures. STRUCTURE AND BONDING IONIC (ELECTROVALENT) BONDING Noble gases like neon or argon have eight electrons in their outer shells (or two in the case of helium). These noble gas structures are thought of as being in some way a \"desirable\" thing for an atom to have. When other atoms react, they try to organise electrons such that their outer shells are either completely full or completely empty. Chemical reactions occur so that atoms attain inert gas configuration by either losing valency electrons as in the case of metals, or gaining electrons as in the case of non metals. Ionic bonding in sodium chloride Sodium (2,8,1) has 1 electron more than a stable noble gas structure (2,8). If it gave away that electron it would become more stable. Chlorine (2,8,7) has 1 electron short of a stable noble gas structure (2,8,8).", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8434916907590467, "ocr_used": false, "chunk_length": 2031, "token_count": 501}}
{"text": "Ionic bonding in sodium chloride Sodium (2,8,1) has 1 electron more than a stable noble gas structure (2,8). If it gave away that electron it would become more stable. Chlorine (2,8,7) has 1 electron short of a stable noble gas structure (2,8,8). If it could gain an electron from somewhere it too would become more stable. If a sodium atom gives an electron to a chlorine atom, both become more stable. The sodium has lost an electron, so it no longer has equal numbers of electrons and protons. Because it has one more proton than electron, it has a charge of 1+. If electrons are lost from an atom, positive ions are formed. Positive ions are sometimes called cations because they move to the cathode during electrolysis. The chlorine has gained an electron, so it now has one more electron than proton. It therefore has a charge of 1-. If electrons are gained by an atom, negative ions are formed. A negative ion is sometimes called an anion since it drifts to the anode during electrolysis. The nature of ionic bond The sodium ions and chloride ions are held together by the strong electrostatic attractions between the positive and negative charges. You need one sodium atom to provide the extra electron for one chlorine atom, so they combine together 1:1. The formula is therefore NaCl. Properties of ionic compounds  All compounds with ionic bonding produce giant ionic structures.  Consist of oppositely charged ions arranged in an ionic lattice, the ions are held together by strong ionic bonds. e.g. NaCl is composed of Na+ ions and Cl- ions.  These bonds are hard to break, therefore ionic substances have very high melting and boiling points.  All exist as solids.  They conduct electricity when molten, because the ions are free to move, but do not conduct when solid.  They conduct electricity in the aqueous state because the ions are free to move.  Most ionic substances are soluble in water because the polar water molecules can accommodate the charged ions. COVALENT BONDING - SINGLE BONDS As well as achieving noble gas structures by transferring electrons from one atom to another as in ionic bonding, it is also possible for atoms to reach these stable structures by sharing electrons to give covalent bonds.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.892749818259271, "ocr_used": false, "chunk_length": 2238, "token_count": 502}}
{"text": " They conduct electricity in the aqueous state because the ions are free to move.  Most ionic substances are soluble in water because the polar water molecules can accommodate the charged ions. COVALENT BONDING - SINGLE BONDS As well as achieving noble gas structures by transferring electrons from one atom to another as in ionic bonding, it is also possible for atoms to reach these stable structures by sharing electrons to give covalent bonds. Depending on the number of electron pairs shared between atoms which participate in bonding, covalent bonds are classified as follows: Some simple covalent molecules Chlorine For example, two chlorine atoms could both achieve stable structures by sharing their single unpaired electron as in the diagram. The fact that one chlorine has been drawn with electrons marked as crosses and the other as dots is simply to show where all the electrons come from. In reality there is no difference between them. The two chlorine atoms are said to be joined by a covalent bond. The reason that the two chlorine atoms stick together is that the shared pair of electrons is attracted to the nucleus of both chlorine atoms. Hydrogen Hydrogen atoms only need two electrons in their outer level to reach the noble gas structure of helium. Once again, the covalent bond holds the two atoms together because the pair of electrons is attracted to both nuclei. This is another single bond. Hydrogen chloride\nThe hydrogen has a helium structure, and the chlorine an argon structure. Water Oxygen atom has six electrons in the outer shell, while each of the two hydrogen atoms has one each. After bonding, oxygen has 8 electrons while each hydrogen atom has two as shown by the molecule. NITROGEN GAS Each nitrogen atom has five electrons in the outer shell. Each needs 3 electrons to complete the outer shell. In the formation of the molecule, each nitrogen atom contributes three electrons and a triple bond is formed Characteristics of Covalent Compounds 1) Covalent compounds consist of molecules and not ions. The molecules do not have any electric charge on them. The molecules are held together by weak forces called Van der Waal's forces. 2) Covalent compounds are gases, volatile liquids or soft solids. As there are weak, Van der Waal's forces between the molecules, they are not held in rigid position. The state depends on the bond energy. If the bond energy is very low, they stay as gases, if it is appreciable they are volatile liquids.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9176146824475904, "ocr_used": false, "chunk_length": 2480, "token_count": 507}}
{"text": "As there are weak, Van der Waal's forces between the molecules, they are not held in rigid position. The state depends on the bond energy. If the bond energy is very low, they stay as gases, if it is appreciable they are volatile liquids. If very high, they exist as soft solids. 3) Covalent compounds generally have low melting and boiling points. As Van der Waal's forces are weak, a very small amount of energy is required to break the bond between the molecules corresponding to low melting point and boiling point. 4) Covalent compounds dissolve in organic solvents. As they do not contain ions, solvation does not take place when water is added to the compound. Hence they do not dissolve in water. 5) Covalent compounds are bad conductors of electricity. They do not contain ions in the fused state, nor do ions migrate on application of an electric potential. Hence, there is no conduction of current. 6) Covalent compounds are less dense when compared to water. Very weak Van der Waal's forces hold the molecules together, hence there are large inter molecular spaces. Consequently less number of molecules per unit volume, which means mass per unit volume is also less. Hence they have a low density. Exceptions  Diamond and graphite, the allotropes of carbon have high melting point.  Hydrogen chloride in the aqueous state conducts electricity. The giant covalent structure of diamond Carbon has an electronic arrangement of 2, 4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds. In the diagram some carbon atoms only seem to be forming two bonds (or even one bond), but that's not really the case. We are only showing a small bit of the whole structure. This is a giant covalent structure - it continues on and on in three dimensions. It is not a molecule, because the number of atoms joined up in a real diamond is completely variable - depending on the size of the crystal. The physical properties of diamond Diamond  Has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent bonds have to be broken throughout the structure before melting occurs.  Is very hard. This is again due to the need to break very strong covalent bonds operating in 3dimensions.  Doesnಬt conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9039814880125828, "ocr_used": false, "chunk_length": 2357, "token_count": 511}}
{"text": "This is again due to the need to break very strong covalent bonds operating in 3dimensions.  Doesnಬt conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move.  Is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and carbon atoms which could outweigh the attractions between the covalently bound carbon atoms. The giant covalent structure of graphite Graphite has a layer structure which is quite difficult to draw convincingly in three dimensions. The diagram below shows the arrangement of the atoms in each layer, and the way the layers are spaced. The bonding in graphite Each carbon atom uses three of its electrons to form simple bonds to its three close neighbours. That leaves a fourth electron in the bonding level. These \"spare\" electrons in each carbon atom become delocalised over the whole of the sheet of atoms in one layer. They are no longer associated directly with any particular atom or pair of atoms, but are free to wander throughout the whole sheet. The important thing is that the delocalised electrons are free to move anywhere within the sheet - each electron is no longer fixed to a particular carbon atom. There is, however, no direct contact between the delocalised electrons in one sheet and those in the neighbouring sheets. The atoms within a sheet are held together by strong covalent bonds - stronger, in fact, than in diamond because of the additional bonding caused by the delocalised electrons. So what holds the sheets together? In graphite you have the ultimate example of van der Waals dispersion forces. As the delocalised electrons move around in the sheet, very large temporary dipoles can be set up which will induce opposite dipoles in the sheets above and below - and so on throughout the whole graphite crystal. The physical properties of graphite Graphite  Has a high melting point, similar to that of diamond. In order to melt graphite, it isn't enough to loosen one sheet from another. You have to break the covalent bonding throughout the whole structure.  Has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like locks. You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper.  Has a lower density than diamond.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9177222319031174, "ocr_used": false, "chunk_length": 2479, "token_count": 511}}
{"text": "You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper.  Has a lower density than diamond. This is because of the relatively large amount of space that is \"wasted\" between the sheets.  Is insoluble in water and organic solvents - for the same reason that diamond is insoluble. Attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds in graphite.  Conducts electricity. The delocalised electrons are free to move throughout the sheets. If a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet and be replaced with new ones at the other end. The structure of silicon dioxide, SiO2 Silicon dioxide is also known as silicon (IV) oxide. The giant covalent structure of silicon dioxide There are three different crystal forms of silicon dioxide. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms. Notice that each silicon atom is bridged to its neighbours by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending on all 3 dimensions. The physical properties of silicon dioxide Silicon dioxide  Has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs.  Is hard. This is due to the need to break the very strong covalent bonds.  Doesnಬt conduct electricity. There aren't any delocalised electrons. All the electrons are held tightly between the atoms, and aren't free to move.  Is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure. Uses of Silica i) Quartz glass is used for manufacturing optical instruments. ii) Colored quartz is used for manufacturing gems. iii) Sand is used in manufacture of glass, porcelain, sand paper and mortar etc.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9130627332917479, "ocr_used": false, "chunk_length": 2438, "token_count": 509}}
{"text": "Uses of Silica i) Quartz glass is used for manufacturing optical instruments. ii) Colored quartz is used for manufacturing gems. iii) Sand is used in manufacture of glass, porcelain, sand paper and mortar etc. iv) Sand stone is used as a building material. CO-ORDINATE (DATIVE COVALENT) BONDING Co-ordinate (dative covalent) bonding A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the electron pair is attracted by both of the nuclei. In the formation of a simple covalent bond, each atom supplies one electron to the bond - but that doesn't have to be the case. A co-ordinate bond (also called a\ndative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom. The reaction between ammonia and hydrogen chloride If these colourless gases are allowed to mix, a thick white smoke of solid ammonium chloride is formed. Ammonium ions, NH4+, are formed by the transfer of a hydrogen ion from the hydrogen chloride to the lone pair of electrons on the ammonia molecule. When the ammonium ion, NH4+, is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen's nucleus is transferred from the chlorine to the nitrogen. The hydrogen's electron is left behind on the chlorine to form a negative chloride ion. Once the ammonium ion has been formed it is impossible to tell any difference between the dative covalent and the ordinary covalent bonds. Although the electrons are shown differently in the diagram, there is no difference between them in reality. INTERMOLECULAR BONDING - VAN DER WAALS FORCES (a) VAN DER WAALS FORCES Intermolecular attractions are attractions between one molecule and a neighbouring molecule. The forces of attraction which hold an individual molecule together (for example, the covalent bonds) are known as\nintramolecular attractions. All molecules experience intermolecular attractions, although in some cases those attractions are very weak. Even in a gas like hydrogen, H2, if you slow the molecules down by cooling the gas, the attractions are large enough for the molecules to stick together eventually to form a liquid and then a solid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9145493657740524, "ocr_used": false, "chunk_length": 2210, "token_count": 479}}
{"text": "The forces of attraction which hold an individual molecule together (for example, the covalent bonds) are known as\nintramolecular attractions. All molecules experience intermolecular attractions, although in some cases those attractions are very weak. Even in a gas like hydrogen, H2, if you slow the molecules down by cooling the gas, the attractions are large enough for the molecules to stick together eventually to form a liquid and then a solid. In hydrogen's case the attractions are so weak that the molecules have to be cooled to (-252°C) before the attractions are enough to condense the hydrogen as a liquid. Helium's intermolecular attractions are even weaker - the molecules won't stick together to form a liquid until the temperature drops to (-269°C). HYDROGEN BONDING Polar molecules, such as water molecules, have a weak, partial negative charge at one region of the molecule (the oxygen atom in water) and a partial positive charge elsewhere (the hydrogen atoms in water). Thus when water molecules are close together, their positive and negative regions are attracted to the oppositely-charged regions of nearby molecules. The force of attraction, shown here as a dotted line, is called a hydrogen bond. Each water molecule is hydrogen bonded to four others. The hydrogen bonds that form between water molecules account for some of the essential — and unique — properties of water.  The attraction created by hydrogen bonds keeps water liquid over a wider range of temperature than is found for any other molecule its size.  The energy required to break multiple hydrogen bonds causes water to have a high heat of vaporization; that is; a large amount of energy is needed to convert liquid water, where the molecules are attracted through their hydrogen bonds, to water vapor, where they are not. Liquid Water and Hydrogen Bonding Why water is a liquid? In many ways, water is a miracle liquid. Since the hydrogen and oxygen atoms in the molecule carry opposite (though partial) charges, nearby water molecules are attracted to each other like tiny little magnets. Hydrogen bonding makes water molecules \"stick\" together. This makes water have high melting and boiling points compared to other covalent compounds such as ammonia (NH3) which have similar molecular mass but are gases Ice and Hydrogen Bonding The structure that forms in the solid ice crystal actually has large holes in it. Therefore, in a given volume of ice, there are fewer water molecules than in the same volume of liquid water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9136328551915852, "ocr_used": false, "chunk_length": 2519, "token_count": 501}}
{"text": "Hydrogen bonding makes water molecules \"stick\" together. This makes water have high melting and boiling points compared to other covalent compounds such as ammonia (NH3) which have similar molecular mass but are gases Ice and Hydrogen Bonding The structure that forms in the solid ice crystal actually has large holes in it. Therefore, in a given volume of ice, there are fewer water molecules than in the same volume of liquid water. In other words, ice is less dense than liquid water and will float on the surface of the liquid. Surface Tension and hydrogen bonding As we just discussed, neighboring water molecules are attracted to one another. Molecules at the surface of liquid water have fewer neighbors and, as a result, have a greater attraction to the few water molecules that are nearby. This enhanced attraction is called surface tension. It makes the surface of the liquid slightly more difficult to break through than the interior. Water as a Solvent The partial charge that develops across the water molecule helps make it an excellent solvent. Water dissolves many substances by surrounding charged particles and \"pulling\" them into solution. For example, common table salt, sodium chloride, is an ionic substance that contains alternating sodium and chlorine ions. When table salt is added to water, the partial charges on the water molecule are attracted to the Na+ and Cl- ions. Why does ethanol have a higher boiling point than methoxymethane? Ethanol, CH3CH2-O-H, and methoxymethane, CH3-O-CH3, both have the same molecular formula, C2H6O. They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same. However, ethanol has a hydrogen atom attached directly to oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge. In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.913193970271548, "ocr_used": false, "chunk_length": 2410, "token_count": 501}}
{"text": "The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge. In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:\nethanol (with hydrogen bonding) 78.5°C methoxymethane (without hydrogen bonding) -24.8°C The hydrogen bonding in the ethanol has lifted its boiling point about 100°C.It is important to realise that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding. 4. BONDING IN METALS Bonding in metals Metal atoms have relatively few electrons in their outer shells. When they are packed together, each metal atom loses its outer electrons into a ಫseaಬ of free electrons (or mobile electrons). Having lost electrons, the atoms are no longer electrically neutral. They become positive ions because they have lost electrons but the number of protons in the nucleus has remained unchanged. Therefore the structure of a metal is made up of positive ions packed together. These ions are surrounded by electrons, which can move freely between the ions.  An ion is a charged particle made from an atom by the loss or gain of electrons.  Metal atoms most easily lose electrons, so they become positive ions. In doing so they achieve a more stable electron arrangement, usually that of the nearest noble gas. These free electrons are delocalized (not restricted to orbiting one positive ion) and form a kind of electrostatic ಫglueಬ holding the structure together. In an electrical circuit, metals can conduct electricity because the mobile electrons can move through the structure carrying charge. His type of bonding (called metallic boding) is present in alloys as well. Alloys, for example solder and brass, will conduct electricity. The physical properties of metals: This strong bonding generally results in dense, strong materials with high melting and boiling points.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9117460251502418, "ocr_used": false, "chunk_length": 2437, "token_count": 502}}
{"text": "His type of bonding (called metallic boding) is present in alloys as well. Alloys, for example solder and brass, will conduct electricity. The physical properties of metals: This strong bonding generally results in dense, strong materials with high melting and boiling points. Usually a relatively large amount of energy is needed to melt or boil metals. a. Metals are good conductors of electricity because these 'free' electrons carry the charge of an electric current when a potential difference (voltage!) is applied across a piece of metal. b. Metals are also good conductors of heat. This is also due to the free moving electrons. Nonmetallic solids conduct heat energy by hotter more strongly vibrating atoms, knocking against cooler less strongly vibrating atoms to pass the particle kinetic energy on. In metals, as well as this effect, the 'hot' high kinetic energy electrons move around freely to transfer the particle kinetic energy more efficiently to 'cooler' atoms. c. Typical metals also have a silvery surface but remember this may be easily tarnished by corrosive oxidation in air and water. d. Unlike ionic solids, metals are very malleable, they can be readily bent, pressed or hammered into shape. INTRODUCTION TO SALTS 1.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid. A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical. (b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid. Some acids are therefore: (i)monobasic acids generally denoted HX e.g. HCl, HNO3,HCOOH,CH3COOH. (ii)dibasic acids ; generally denoted H2X e.g. H2SO4, H2SO3, H2CO3,HOOCOOH. (iii)tribasic acids ; generally denoted H3X e.g. H3PO4. (c) Some salts are normal salts while other are acid salts. (i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9002893009363615, "ocr_used": false, "chunk_length": 2179, "token_count": 510}}
{"text": "(c) Some salts are normal salts while other are acid salts. (i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-)\nCarbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) acid H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8471644397869182, "ocr_used": false, "chunk_length": 1041, "token_count": 317}}
{"text": "(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-)\nCarbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) acid H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI)\nFe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7721563967238708, "ocr_used": false, "chunk_length": 1734, "token_count": 627}}
{"text": "(ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-)\nCarbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) acid H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI)\nFe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.769527400764635, "ocr_used": false, "chunk_length": 1706, "token_count": 619}}
{"text": "Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-)\nCarbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) acid H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI)\nFe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution. Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7681698631816649, "ocr_used": false, "chunk_length": 1705, "token_count": 628}}
{"text": "Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI)\nFe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution. Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt. (ii)Deliquescent salts /compounds are those that absorb water from the atmosphere and form a solution. Some salts which are deliquescent include: Sodium nitrate(V),Calcium chloride, Sodium hydroxide, Iron(II)chloride, Magnesium chloride. (iii)Efflorescent salts/compounds are those that lose their water of crystallization to the atmosphere.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7813685819620538, "ocr_used": false, "chunk_length": 1348, "token_count": 473}}
{"text": "(ii)Deliquescent salts /compounds are those that absorb water from the atmosphere and form a solution. Some salts which are deliquescent include: Sodium nitrate(V),Calcium chloride, Sodium hydroxide, Iron(II)chloride, Magnesium chloride. (iii)Efflorescent salts/compounds are those that lose their water of crystallization to the atmosphere. Some salts which effloresces include: sodium carbonate decahydrate, Iron(II)sulphate(VI)heptahydrate, sodium sulphate (VI)decahydrate. (e)Some salts contain water of crystallization.They are hydrated.Others do not contain water of crystallization. They are anhydrous. Table showing some hydrated salts. Name of hydrated salt Chemical formula Copper(II)sulphate(VI)pentahydrate CuSO4.5H2O Aluminium(III)sulphate(VI)hexahydrate Al2 (SO4) 3.6H2O Zinc(II)sulphate(VI)heptahydrate ZnSO4.7H2O Iron(II)sulphate(VI)heptahydrate FeSO4.7H2O Calcium(II)sulphate(VI)heptahydrate CaSO4.7H2O Magnesium(II)sulphate(VI)heptahydrate MgSO4.7H2O Sodium sulphate(VI)decahydrate Na2SO4.10H2O Sodium carbonate(IV)decahydrate Na2CO3.10H2O Potassium carbonate(IV)decahydrate K2CO3.10H2O Potassium sulphate(VI)decahydrate K2SO4.10H2O (f)Some salts exist as a simple salt while some as complex salts. Below are some complex salts.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8592961959143576, "ocr_used": false, "chunk_length": 1246, "token_count": 407}}
{"text": "Table showing some hydrated salts. Name of hydrated salt Chemical formula Copper(II)sulphate(VI)pentahydrate CuSO4.5H2O Aluminium(III)sulphate(VI)hexahydrate Al2 (SO4) 3.6H2O Zinc(II)sulphate(VI)heptahydrate ZnSO4.7H2O Iron(II)sulphate(VI)heptahydrate FeSO4.7H2O Calcium(II)sulphate(VI)heptahydrate CaSO4.7H2O Magnesium(II)sulphate(VI)heptahydrate MgSO4.7H2O Sodium sulphate(VI)decahydrate Na2SO4.10H2O Sodium carbonate(IV)decahydrate Na2CO3.10H2O Potassium carbonate(IV)decahydrate K2CO3.10H2O Potassium sulphate(VI)decahydrate K2SO4.10H2O (f)Some salts exist as a simple salt while some as complex salts. Below are some complex salts. Table of some complex salts Name of complex salt Chemical formula Colour of the complex salt Tetraamminecopper(II)sulphate(VI) Cu(NH3) 4 SO4 H2O Royal/deep blue solution\nTetraamminezinc(II)nitrate(V) Zn(NH3) 4 (NO3 )2 Colourless solution Tetraamminecopper(II) nitrate(V) Cu(NH3) 4 (NO3 )2 Royal/deep blue solution Tetraamminezinc(II)sulphate(VI) Zn(NH3) 4 SO4 Colourless solution (g)Some salts exist as two salts in one. They are called double salts.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8076748194440039, "ocr_used": false, "chunk_length": 1087, "token_count": 410}}
{"text": "Below are some complex salts. Table of some complex salts Name of complex salt Chemical formula Colour of the complex salt Tetraamminecopper(II)sulphate(VI) Cu(NH3) 4 SO4 H2O Royal/deep blue solution\nTetraamminezinc(II)nitrate(V) Zn(NH3) 4 (NO3 )2 Colourless solution Tetraamminecopper(II) nitrate(V) Cu(NH3) 4 (NO3 )2 Royal/deep blue solution Tetraamminezinc(II)sulphate(VI) Zn(NH3) 4 SO4 Colourless solution (g)Some salts exist as two salts in one. They are called double salts. Table of some double salts Name of double salts Chemical formula Trona(sodium sesquicarbonate) Na2CO3 NaHCO3.2H2O Ammonium iron(II)sulphate(VI) FeSO4(NH4) 2SO4.2H2O Ammonium aluminium(III)sulphate(VI) Al2(SO4) 3(NH4) 2SO4.H2O (h)Some salts dissolve in water to form a solution. They are said to be soluble. Others do not dissolve in water. They form a suspension/precipitate in water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8140353069832839, "ocr_used": false, "chunk_length": 865, "token_count": 293}}
{"text": "They are said to be soluble. Others do not dissolve in water. They form a suspension/precipitate in water. Table of solubility of salts Soluble salts Insoluble salts All nitrate(V)salts All sulphate(VI)/SO42- salts except Barium(II) sulphate(VI)/BaSO4 Calcium(II) sulphate(VI)/CaSO4 Lead(II) sulphate(VI)/PbSO4 All sulphate(IV)/SO32- salts except Barium(II) sulphate(IV)/BaSO3 Calcium(II) sulphate(IV)/CaSO3 Lead(II) sulphate(IV)/PbSO3 All chlorides/Cl- except Silver chloride/AgCl Lead(II)chloride/PbCl2(dissolves in hot water) All phosphate(V)/PO43- All sodium,potassium and ammonium salts All hydrogen carbonates/HCO3- All hydrogen sulphate(VI)/ HSO4- Sodium carbonate/Na2CO3, potassium carbonate/ K2CO3, ammonium carbonate (NH4) 2CO3 except All carbonates All alkalis(KOH,NaOH, NH4OH) except All bases 13 Salts can be prepared in a school laboratory by a method that uses its solubility in water. (a) Soluble salts may be prepared by using any of the following methods: (i)Direct displacement/reaction of a metal with an acid. By reacting a metal higher in the reactivity series than hydrogen with a dilute acid,a salt is formed and hydrogen gas is evolved. Excess of the metal must be used to ensure all the acid has reacted. When effervescence/bubbling /fizzing has stopped ,excess metal is filtered. The filtrate is heated to concentrate then allowed to crystallize. Washing with distilled water then drying between filter papers produces a sample crystal of the salt. i.e.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8723533289386948, "ocr_used": false, "chunk_length": 1480, "token_count": 406}}
{"text": "The filtrate is heated to concentrate then allowed to crystallize. Washing with distilled water then drying between filter papers produces a sample crystal of the salt. i.e. M(s) + H2X -> MX(aq) + H2(g) Examples Mg(s) + H2SO4(aq) -> MgSO4 (aq) + H2(g) Zn(s) + H2SO4(aq) -> ZnSO4 (aq) + H2(g) Pb(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2(g) Ca(s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2(g) Mg(s) + 2HNO3(aq) -> Mg(NO3) 2(aq) + H2(g) Mg(s) + 2HCl(aq) -> MgCl 2(aq) + H2(g) Zn(s) + 2HCl(aq) -> ZnCl 2(aq) + H2(g) (ii)Reaction of an insoluble base with an acid By adding an insoluble base (oxide/hydroxide )to a dilute acid until no more dissolves, in the acid,a salt and water are formed. Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7689734666607934, "ocr_used": false, "chunk_length": 846, "token_count": 310}}
{"text": "M(s) + H2X -> MX(aq) + H2(g) Examples Mg(s) + H2SO4(aq) -> MgSO4 (aq) + H2(g) Zn(s) + H2SO4(aq) -> ZnSO4 (aq) + H2(g) Pb(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2(g) Ca(s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2(g) Mg(s) + 2HNO3(aq) -> Mg(NO3) 2(aq) + H2(g) Mg(s) + 2HCl(aq) -> MgCl 2(aq) + H2(g) Zn(s) + 2HCl(aq) -> ZnCl 2(aq) + H2(g) (ii)Reaction of an insoluble base with an acid By adding an insoluble base (oxide/hydroxide )to a dilute acid until no more dissolves, in the acid,a salt and water are formed. Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g. PbO(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2O (l) Pb(OH)2(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + 2H2O (l) CaO (s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2O (l) MgO (s) + 2HNO3(aq) -> Mg(NO3) 2(aq) + H2O (l) MgO (s) + 2HCl(aq) -> MgCl 2(aq) + H2O (l) ZnO (s) + 2HCl(aq) -> ZnCl 2(aq) + H2O (l) Zn(OH)2(s) + 2HNO3(aq) -> Zn(NO3) 2(aq) + 2H2O (l) CuO (s) + 2HCl(aq) -> CuCl 2(aq) + H2O (l) CuO (s) + H2SO4(aq) -> CuSO4(aq) + H2O (l) Ag2O(s) + 2HNO3(aq) -> 2AgNO3(aq) + H2O (l) Na2O(s) + 2HNO3(aq) -> 2NaNO3(aq) + H2O (l) (iii)reaction of insoluble /soluble carbonate /hydrogen carbonate with an acid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6638093042024558, "ocr_used": false, "chunk_length": 1254, "token_count": 633}}
{"text": "Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g. PbO(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2O (l) Pb(OH)2(s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + 2H2O (l) CaO (s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2O (l) MgO (s) + 2HNO3(aq) -> Mg(NO3) 2(aq) + H2O (l) MgO (s) + 2HCl(aq) -> MgCl 2(aq) + H2O (l) ZnO (s) + 2HCl(aq) -> ZnCl 2(aq) + H2O (l) Zn(OH)2(s) + 2HNO3(aq) -> Zn(NO3) 2(aq) + 2H2O (l) CuO (s) + 2HCl(aq) -> CuCl 2(aq) + H2O (l) CuO (s) + H2SO4(aq) -> CuSO4(aq) + H2O (l) Ag2O(s) + 2HNO3(aq) -> 2AgNO3(aq) + H2O (l) Na2O(s) + 2HNO3(aq) -> 2NaNO3(aq) + H2O (l) (iii)reaction of insoluble /soluble carbonate /hydrogen carbonate with an acid. By adding an excess of a soluble /insoluble carbonate or hydrogen carbonate to adilute acid, effervescence /fizzing/bubbling out of carbon(IV)oxide gas shows the\nreaction is taking place. When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7506166660149377, "ocr_used": false, "chunk_length": 1195, "token_count": 493}}
{"text": "By adding an excess of a soluble /insoluble carbonate or hydrogen carbonate to adilute acid, effervescence /fizzing/bubbling out of carbon(IV)oxide gas shows the\nreaction is taking place. When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g. PbCO3 (s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2O (l)+ CO2(g) ZnCO3 (s) + 2HNO3(aq) -> Zn(NO3) 2(aq) + H2O (l)+ CO2(g) CaCO3 (s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2O (l)+ CO2(g) MgCO3 (s) + H2SO4(aq) -> MgSO4(aq) + H2O (l)+ CO2(g) Cu CO3 (s) + H2SO4(aq) -> CuSO4(aq) + H2O (l) + CO2(g) Ag2CO3 (s) + 2HNO3(aq) -> 2AgNO3(aq) + H2O (l) + CO2(g) Na2CO3 (s) + 2HNO3(aq) -> 2NaNO3(aq) + H2O (l) + CO2(g) K2CO3 (s) + 2HCl(aq) -> 2KCl(aq) + H2O (l) + CO2(g) NaHCO3 (s) + HNO3(aq) -> NaNO3(aq) + H2O (l) + CO2(g) KHCO3 (s) + HCl(aq) -> KCl(aq) + H2O (l) + CO2(g) (iv)neutralization/reaction of soluble base/alkali with dilute acid By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7315681838874376, "ocr_used": false, "chunk_length": 1258, "token_count": 514}}
{"text": "When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g. PbCO3 (s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2O (l)+ CO2(g) ZnCO3 (s) + 2HNO3(aq) -> Zn(NO3) 2(aq) + H2O (l)+ CO2(g) CaCO3 (s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2O (l)+ CO2(g) MgCO3 (s) + H2SO4(aq) -> MgSO4(aq) + H2O (l)+ CO2(g) Cu CO3 (s) + H2SO4(aq) -> CuSO4(aq) + H2O (l) + CO2(g) Ag2CO3 (s) + 2HNO3(aq) -> 2AgNO3(aq) + H2O (l) + CO2(g) Na2CO3 (s) + 2HNO3(aq) -> 2NaNO3(aq) + H2O (l) + CO2(g) K2CO3 (s) + 2HCl(aq) -> 2KCl(aq) + H2O (l) + CO2(g) NaHCO3 (s) + HNO3(aq) -> NaNO3(aq) + H2O (l) + CO2(g) KHCO3 (s) + HCl(aq) -> KCl(aq) + H2O (l) + CO2(g) (iv)neutralization/reaction of soluble base/alkali with dilute acid By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point. The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7324754374018538, "ocr_used": false, "chunk_length": 1268, "token_count": 503}}
{"text": "PbCO3 (s) + 2HNO3(aq) -> Pb(NO3) 2(aq) + H2O (l)+ CO2(g) ZnCO3 (s) + 2HNO3(aq) -> Zn(NO3) 2(aq) + H2O (l)+ CO2(g) CaCO3 (s) + 2HNO3(aq) -> Ca(NO3) 2(aq) + H2O (l)+ CO2(g) MgCO3 (s) + H2SO4(aq) -> MgSO4(aq) + H2O (l)+ CO2(g) Cu CO3 (s) + H2SO4(aq) -> CuSO4(aq) + H2O (l) + CO2(g) Ag2CO3 (s) + 2HNO3(aq) -> 2AgNO3(aq) + H2O (l) + CO2(g) Na2CO3 (s) + 2HNO3(aq) -> 2NaNO3(aq) + H2O (l) + CO2(g) K2CO3 (s) + 2HCl(aq) -> 2KCl(aq) + H2O (l) + CO2(g) NaHCO3 (s) + HNO3(aq) -> NaNO3(aq) + H2O (l) + CO2(g) KHCO3 (s) + HCl(aq) -> KCl(aq) + H2O (l) + CO2(g) (iv)neutralization/reaction of soluble base/alkali with dilute acid By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point. The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g. NaOH (aq) + HNO3(aq) -> NaNO3(aq) + H2O (l) KOH (aq) + HNO3(aq) -> KNO3(aq) + H2O (l) KOH (aq) + HCl(aq) -> KCl(aq) + H2O (l) 2KOH (aq) + H2SO4(aq) -> K2SO4(aq) + 2H2O (l) 2 NH4OH (aq) + H2SO4(aq) -> (NH4)2SO4(aq) + 2H2O (l) NH4OH (aq) + HNO3(aq) -> NH4NO3(aq) + H2O (l) (iv)direct synthesis/combination.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6632284611440202, "ocr_used": false, "chunk_length": 1319, "token_count": 629}}
{"text": "The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g. NaOH (aq) + HNO3(aq) -> NaNO3(aq) + H2O (l) KOH (aq) + HNO3(aq) -> KNO3(aq) + H2O (l) KOH (aq) + HCl(aq) -> KCl(aq) + H2O (l) 2KOH (aq) + H2SO4(aq) -> K2SO4(aq) + 2H2O (l) 2 NH4OH (aq) + H2SO4(aq) -> (NH4)2SO4(aq) + 2H2O (l) NH4OH (aq) + HNO3(aq) -> NH4NO3(aq) + H2O (l) (iv)direct synthesis/combination. When a metal burn in a gas jar containing a non metal , the two directly combine to form a salt. e.g. 2Na(s) + Cl2(g) -> 2NaCl(s) 2K(s) + Cl2(g) -> 2KCl(s) Mg(s) + Cl2(g) -> Mg Cl2 (s) Ca(s) + Cl2(g) -> Ca Cl2 (s) Some salts once formed undergo sublimation and hydrolysis. Care should be taken to avoid water/moisture into the reaction flask during their preparation.Such salts include aluminium(III)chloride(AlCl3) and iron (III)chloride(FeCl3)\n1. Heated aluminium foil reacts with chlorine to form aluminium(III)chloride that sublimes away from the source of heating then deposited as solid again 2Al(s) + 3Cl2(g) -> 2AlCl3 (s/g) Once formed aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas. AlCl3(s)+ 3H2 O(g) -> Al(OH)3 (aq) + 3HCl(g) 2.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7742881730085682, "ocr_used": false, "chunk_length": 1359, "token_count": 486}}
{"text": "Care should be taken to avoid water/moisture into the reaction flask during their preparation.Such salts include aluminium(III)chloride(AlCl3) and iron (III)chloride(FeCl3)\n1. Heated aluminium foil reacts with chlorine to form aluminium(III)chloride that sublimes away from the source of heating then deposited as solid again 2Al(s) + 3Cl2(g) -> 2AlCl3 (s/g) Once formed aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas. AlCl3(s)+ 3H2 O(g) -> Al(OH)3 (aq) + 3HCl(g) 2. Heated iron filings reacts with chlorine to form iron(III)chloride that sublimes away from the source of heating then deposited as solid again 2Fe(s) + 3Cl2(g) -> 2FeCl3 (s/g) Once formed , aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas. FeCl3(s)+ 3H2 O(g) -> Fe(OH)3 (aq) + 3HCl(g) (b)Insoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble. This is called double decomposition or precipitation. The mixture is filtered and the residue is washed with distilled water then dried.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8597469426565828, "ocr_used": false, "chunk_length": 1233, "token_count": 333}}
{"text": "FeCl3(s)+ 3H2 O(g) -> Fe(OH)3 (aq) + 3HCl(g) (b)Insoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble. This is called double decomposition or precipitation. The mixture is filtered and the residue is washed with distilled water then dried. CuSO4(aq) + Na2CO3 (aq) -> CuCO3 (s) + Na2 SO4(aq) BaCl2(aq) + K2SO4 (aq) -> BaSO4 (s) + 2KCl (aq) Pb(NO3)2(aq) + K2SO4 (aq) -> PbSO4 (s) + 2KNO3 (aq) 2AgNO3(aq) + MgCl2 (aq) -> 2AgCl(s) + Mg(NO3)2 (aq) Pb(NO3)2(aq) + (NH4) 2SO4 (aq) -> PbSO4 (s) + 2NH4NO 3(aq) BaCl2(aq) + K2SO3 (aq) -> BaSO3 (s) + 2KCl (aq) 14. Salts may lose their water of crystallization , decompose ,melt or sublime on heating on a Bunsen burner flame. The following shows the behavior of some salts on heating gently /or strongly in a laboratory school burner: (a)effect of heat on chlorides All chlorides have very high melting and boiling points and therefore are not affected by laboratory heating except ammonium chloride. Ammonium chloride sublimes on gentle heating. It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating. NH4Cl(s) NH4Cl(g) NH3(g) + HCl(g) (sublimation) (dissociation) (b)effect of heat on nitrate(V) (i) Potassium nitrate(V)/KNO3 and sodium nitrate(V)/NaNO3 decompose on heating to form Potassium nitrate(III)/KNO2 and sodium nitrate(III)/NaNO2 and producing Oxygen gas in each case.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7909720483938256, "ocr_used": false, "chunk_length": 1410, "token_count": 468}}
{"text": "Ammonium chloride sublimes on gentle heating. It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating. NH4Cl(s) NH4Cl(g) NH3(g) + HCl(g) (sublimation) (dissociation) (b)effect of heat on nitrate(V) (i) Potassium nitrate(V)/KNO3 and sodium nitrate(V)/NaNO3 decompose on heating to form Potassium nitrate(III)/KNO2 and sodium nitrate(III)/NaNO2 and producing Oxygen gas in each case. 2KNO3 (s) -> 2KNO2(s) + O2(g) 2NaNO3 (s) -> 2NaNO2(s) + O2(g) (ii)Heavy metal nitrates(V) salts decompose on heating to form the oxide and a mixture of brown acidic nitrogen(IV)oxide and oxygen gases. e.g. 2Ca(NO3)2 (s) -> 2CaO(s) + 4NO2(g) + O2(g) 2Mg(NO3)2(s) -> 2MgO(s) + 4NO2(g) + O2(g) 2Zn(NO3)2(s) -> 2ZnO(s) + 4NO2(g) + O2(g) 2Pb(NO3)2(s) -> 2PbO(s) + 4NO2(g) + O2(g) 2Cu(NO3)2(s) -> 2CuO(s) + 4NO2(g) + O2(g) 2Fe(NO3)2(s) -> 2FeO(s) + 4NO2(g) + O2(g) (iii)Silver(I)nitrate(V) and mercury(II) nitrate(V) are lowest in the reactivity series. They decompose on heating to form the metal(silver and mercury)and the Nitrogen(IV)oxide and oxygen gas. i.e.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7319930519814719, "ocr_used": false, "chunk_length": 1072, "token_count": 432}}
{"text": "2Ca(NO3)2 (s) -> 2CaO(s) + 4NO2(g) + O2(g) 2Mg(NO3)2(s) -> 2MgO(s) + 4NO2(g) + O2(g) 2Zn(NO3)2(s) -> 2ZnO(s) + 4NO2(g) + O2(g) 2Pb(NO3)2(s) -> 2PbO(s) + 4NO2(g) + O2(g) 2Cu(NO3)2(s) -> 2CuO(s) + 4NO2(g) + O2(g) 2Fe(NO3)2(s) -> 2FeO(s) + 4NO2(g) + O2(g) (iii)Silver(I)nitrate(V) and mercury(II) nitrate(V) are lowest in the reactivity series. They decompose on heating to form the metal(silver and mercury)and the Nitrogen(IV)oxide and oxygen gas. i.e. 2AgNO3(s) -> 2Ag (s) + 2NO2(g) + O2(g) 2Hg(NO3)2 (s) -> 2Hg (s) + 4NO2(g) + O2(g) (iv)Ammonium nitrate(V) and Ammonium nitrate(III) decompose on heating to Nitrogen(I)oxide(relights/rekindles glowing splint) and nitrogen gas respectively.Water is also formed.i.e. NH4NO3(s) -> N2O (g) + H2O(l) NH4NO2(s) -> N2 (g) + H2O(l) (c) effect of heat on nitrate(V) Only Iron(II)sulphate(VI), Iron(III)sulphate(VI) and copper(II)sulphate(VI) decompose on heating. They form the oxide, and produce highly acidic fumes of acidic sulphur(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6991919191919193, "ocr_used": false, "chunk_length": 990, "token_count": 443}}
{"text": "2AgNO3(s) -> 2Ag (s) + 2NO2(g) + O2(g) 2Hg(NO3)2 (s) -> 2Hg (s) + 4NO2(g) + O2(g) (iv)Ammonium nitrate(V) and Ammonium nitrate(III) decompose on heating to Nitrogen(I)oxide(relights/rekindles glowing splint) and nitrogen gas respectively.Water is also formed.i.e. NH4NO3(s) -> N2O (g) + H2O(l) NH4NO2(s) -> N2 (g) + H2O(l) (c) effect of heat on nitrate(V) Only Iron(II)sulphate(VI), Iron(III)sulphate(VI) and copper(II)sulphate(VI) decompose on heating. They form the oxide, and produce highly acidic fumes of acidic sulphur(IV)oxide gas. 2FeSO4 (s) -> Fe2O3(s) + SO3(g) + SO2(g) Fe2(SO4) 3(s) -> Fe2O3(s) + SO3(g) CuSO4 (s) -> CuO(s) + SO3(g)\n(d) effect of heat on carbonates(IV) and hydrogen carbonate(IV). (i)Sodium carbonate(IV)and potassium carbonate(IV)do not decompose on heating. (ii)Heavy metal nitrate(IV)salts decompose on heating to form the oxide and produce carbon(IV)oxide gas. Carbon (IV)oxide gas forms a white precipitate when bubbled in lime water. The white precipitate dissolves if the gas is in excess. e.g. CuCO3 (s) -> CuO(s) + CO2(g) CaCO3 (s) -> CaO(s) + CO2(g) PbCO3 (s) -> PbO(s) + CO2(g) FeCO3 (s) -> FeO(s) + CO2(g) ZnCO3 (s) -> ZnO(s) + CO2(g) (iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7797259420741203, "ocr_used": false, "chunk_length": 1352, "token_count": 503}}
{"text": "e.g. CuCO3 (s) -> CuO(s) + CO2(g) CaCO3 (s) -> CaO(s) + CO2(g) PbCO3 (s) -> PbO(s) + CO2(g) FeCO3 (s) -> FeO(s) + CO2(g) ZnCO3 (s) -> ZnO(s) + CO2(g) (iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e. 2NaHCO 3(s) -> Na2CO3(s) + CO2(g) + H2O(l) 2KHCO 3(s) -> K2CO3(s) + CO2(g) + H2O(l) (iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e. Ca(HCO3) 2(aq) -> CaCO3(s) + CO2(g) + H2O(l) Mg(HCO3) 2(aq) -> MgCO3(s) + CO2(g) + H2O(l)\n1 INTRODUCTION TO ELECTROLYSIS (ELECTROLYTIC CELL) 1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity. A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong. 2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include: (i)all mineral acids (ii)all strong alkalis/sodium hydroxide/potassium hydroxide. (iii)all soluble salts 3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions. Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8109333333333334, "ocr_used": false, "chunk_length": 1500, "token_count": 488}}
{"text": "Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (ii)all hydrocarbons(alkanes /alkenes/alkynes) (iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar) 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state. 6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. 2 An electrode that does not influence/alter the products of electrolysis is called an inert electrode. Common inert electrodes include: (i)Platinum (ii)Carbon graphite Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells). 7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte 8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e. M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.90071807104654, "ocr_used": false, "chunk_length": 2123, "token_count": 503}}
{"text": "M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules. X+ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water) 2X+ (l) + 2e -> X (l) (for cations from molten electrolytes) The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode. 11. The below set up shows an electrolytic cell. 3 BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.858471803367213, "ocr_used": false, "chunk_length": 1643, "token_count": 461}}
{"text": "3 BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide\n4 (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8443163853825738, "ocr_used": false, "chunk_length": 1579, "token_count": 469}}
{"text": "II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide\n4 (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8433792032555151, "ocr_used": false, "chunk_length": 1334, "token_count": 398}}
{"text": "II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas. d)To determine the products of electrolysis of molten Aluminium (III)oxide (i)Decomposition of electrolyte into free ions; Al2O3 (l) -> 2Al 3+(l) + 3O2-(l) (Compound decomposed into free cation and anion in liquid state)\n5 (ii)At the cathode/negative electrode(-); 4Al 3+ (l) + 12e -> 4Al (l) (Cation / Al 3+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 6O2-(l) -> 3O2 (g) + 12e (Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid aluminium metal. II.At the anode colourless gas that relights/rekindles glowing splint. 13.In industries electrolysis has the following uses/applications: (a)Extraction of reactive metals from their ores. Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods. (b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8470857510162955, "ocr_used": false, "chunk_length": 1693, "token_count": 487}}
{"text": "Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods. (b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there: (i)At the cathode; Cu2+ (aq) + 2e -> Cu(s) (Pure copper deposits on the strip (ii)At the anode; Cu(s) ->Cu2+ (aq) + 2e (impure copper erodes/dissolves) (c)Electroplating The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating, the cathode is made of the metal to be coated/impure. Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves)\n6 (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon)\nCARBON AND ITS COMPOUNDS Carbon is an element in Group IV(Group 4)of the Periodic table .It has atomic number 6 and electronic configuration 2:4 and thus has four valence electrons (tetravalent).It does not easily ionize but forms strong covalent bonds with other elements including itself. (a)Occurrence Carbon mainly naturally occurs as: (i)allotropes of carbon i.e graphite, diamond and fullerenes. (ii)amorphous carbon in coal, peat ,charcoal and coke.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8909573431243789, "ocr_used": false, "chunk_length": 1962, "token_count": 511}}
{"text": "Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves)\n6 (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon)\nCARBON AND ITS COMPOUNDS Carbon is an element in Group IV(Group 4)of the Periodic table .It has atomic number 6 and electronic configuration 2:4 and thus has four valence electrons (tetravalent).It does not easily ionize but forms strong covalent bonds with other elements including itself. (a)Occurrence Carbon mainly naturally occurs as: (i)allotropes of carbon i.e graphite, diamond and fullerenes. (ii)amorphous carbon in coal, peat ,charcoal and coke. (iii)carbon(IV)oxide gas accounting 0.03% by volume of normal air in the atmosphere. (b)Allotropes of Carbon Carbon naturally occur in two main crystalline allotropic forms, carbon-graphite and carbon-diamond Carbon-diamond Carbon-graphite Shiny crystalline solid Black/dull crystalline solid Has a very high melting/boiling point because it has a very closely packed giant tetrahedral structure joined by strong covalent bonds Has a high melting/boiling point because it has a very closely packed giant hexagonal planar structure joined by strong covalent bonds Has very high density(Hardest known Soft\n2 natural substance) Abrassive Slippery Poor electrical conductor because it has no free delocalized electrons Good electrical conductor because it has free 4th valency delocalized electrons Is used in making Jewels, drilling and cutting metals Used in making Lead-pencils,electrodes in batteries and as a lubricant Has giant tetrahedral structure Has giant hexagonal planar structure c)Properties of Carbon (i)Physical properties of carbon Carbon occur widely and naturally as a black solid It is insoluble in water but soluble in carbon disulphide and organic solvents.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9044833745777493, "ocr_used": false, "chunk_length": 2143, "token_count": 512}}
{"text": "(ii)amorphous carbon in coal, peat ,charcoal and coke. (iii)carbon(IV)oxide gas accounting 0.03% by volume of normal air in the atmosphere. (b)Allotropes of Carbon Carbon naturally occur in two main crystalline allotropic forms, carbon-graphite and carbon-diamond Carbon-diamond Carbon-graphite Shiny crystalline solid Black/dull crystalline solid Has a very high melting/boiling point because it has a very closely packed giant tetrahedral structure joined by strong covalent bonds Has a high melting/boiling point because it has a very closely packed giant hexagonal planar structure joined by strong covalent bonds Has very high density(Hardest known Soft\n2 natural substance) Abrassive Slippery Poor electrical conductor because it has no free delocalized electrons Good electrical conductor because it has free 4th valency delocalized electrons Is used in making Jewels, drilling and cutting metals Used in making Lead-pencils,electrodes in batteries and as a lubricant Has giant tetrahedral structure Has giant hexagonal planar structure c)Properties of Carbon (i)Physical properties of carbon Carbon occur widely and naturally as a black solid It is insoluble in water but soluble in carbon disulphide and organic solvents. It is a poor electrical and thermal conductor. (ii)Chemical properties of carbon I. Burning Experiment Introduce a small piece of charcoal on a Bunsen flame then lower it into a gas jar containing Oxygen gas. Put three drops of water. Swirl. Test the solution with blue and red litmus papers. Observation -Carbon chars then burns with a blue flame -Colourless and odourless gas produced -Solution formed turn blue litmus paper faint red. Red litmus paper remains red. Explanation Carbon burns in air and faster in Oxygen with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon burns in limited supply of air with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon (IV) oxide gas dissolve in water to form weak acidic solution of Carbonic (IV)acid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9163310691580653, "ocr_used": false, "chunk_length": 2018, "token_count": 449}}
{"text": "Explanation Carbon burns in air and faster in Oxygen with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon burns in limited supply of air with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas. Carbon (IV) oxide gas dissolve in water to form weak acidic solution of Carbonic (IV)acid. Chemical Equation C(s) + O2(g) -> CO2(g) (in excess air) 2C(s) + O2(g) -> 2CO(g) (in limited air) CO2(g) + H2O (l) -> H2CO3 (aq) (very weak acid) II. Reducing agent Experiment Mix thoroughly equal amounts of powdered charcoal and copper (II)oxide into a crucible. Heat strongly. Observation Colour change from black to brown Explanation\n3 Carbon is a reducing agent. For ages it has been used to reducing metal oxide ores to metal, itself oxidized to carbon(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8586653038168761, "ocr_used": false, "chunk_length": 791, "token_count": 209}}
{"text": "Heat strongly. Observation Colour change from black to brown Explanation\n3 Carbon is a reducing agent. For ages it has been used to reducing metal oxide ores to metal, itself oxidized to carbon(IV)oxide gas. Carbon reduces black copper(II)oxide to brown copper metal Chemical Equation 2CuO(s) + C(s) -> 2Cu(s) + CO2(g) (black) (brown) 2PbO(s) + C(s) -> 2Pb(s) + CO2(g) (brown when hot/ (grey) yellow when cool) 2ZnO(s) + C(s) -> 2Zn(s) + CO2(g) (yellow when hot/ (grey) white when cool) Fe2O3(s) + 3C(s) -> 2Fe(s) + 3CO2(g) (brown when hot/cool (grey) Fe3O4 (s) + 4C(s) -> 3Fe(s) + 4CO2(g) (brown when hot/cool (grey)\n4 B: COMPOUNDS OF CARBON The following are the main compounds of Carbon (i)Carbon(IV)Oxide(CO2) (ii)Carbon(II)Oxide(CO) (iii)Carbonate(IV) (CO32-)and hydrogen carbonate(IV(HCO3-) (iv)Sodium carbonate(Na2CO3) (i) Carbon(IV)Oxide (CO2) (a)Occurrence Carbon(IV)oxide is found: -in the air /atmosphere as 0.03% by volume. -a solid carbon(IV)oxide mineral in Esageri near Eldame Ravine and Kerita near Limuru in Kenya. (b)School Laboratory preparation In the school laboratory carbon(IV)oxide can be prepared in the school laboratory from the reaction of marble chips(CaCO3)or sodium hydrogen carbonate(NaHCO3) with dilute hydrochloric acid. 5 (c)Properties of carbon(IV)oxide gas(Questions) 1.Write the equation for the reaction for the school laboratory preparation of carbon (IV)oxide gas. Any carbonate reacted with dilute hydrochloric acid should be able to generate carbon (IV)oxide gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8233369552094651, "ocr_used": false, "chunk_length": 1506, "token_count": 485}}
{"text": "(b)School Laboratory preparation In the school laboratory carbon(IV)oxide can be prepared in the school laboratory from the reaction of marble chips(CaCO3)or sodium hydrogen carbonate(NaHCO3) with dilute hydrochloric acid. 5 (c)Properties of carbon(IV)oxide gas(Questions) 1.Write the equation for the reaction for the school laboratory preparation of carbon (IV)oxide gas. Any carbonate reacted with dilute hydrochloric acid should be able to generate carbon (IV)oxide gas. Chemical equations CaCO3(s) + 2HCl(aq) -> CaCO3 (aq) + H2O(l) + CO2 (g) ZnCO3(s) + 2HCl(aq) -> ZnCO3 (aq) + H2O(l) + CO2 (g) MgCO3(s) + 2HCl(aq) -> MgCO3 (aq) + H2O(l) + CO2 (g) CuCO3(s) + 2HCl(aq) -> CuCO3 (aq) + H2O(l) + CO2 (g) NaHCO3(s) + HCl(aq) -> Na2CO3 (aq) + H2O(l) + CO2 (g) KHCO3(s) + HCl(aq) -> K2CO3 (aq) + H2O(l) + CO2 (g) 2.What method of gas collection is used in preparation of Carbon(IV)oxide gas. Explain. Downward delivery /upward displacement of air/over mercury Carbon(IV)oxide gas is about 1½ times denser than air. 3.What is the purpose of : (a)water? To absorb the more volatile hydrogen chloride fumes produced during the vigorous reaction. (b)sodium hydrogen carbonate? To absorb the more volatile hydrogen chloride fumes produced during the vigorous reaction and by reacting with the acid to produce more carbon (IV)oxide gas . 6 Chemical equation NaHCO3(s) + HCl(aq) -> Na2CO3 (aq) + H2O(l) + CO2 (g) (c)concentrated sulphuric(VI)acid? To dry the gas/as a drying agent 4.Describe the smell of carbon(IV)oxide gas Colourless and odourless 5. Effect on lime water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8014464677773473, "ocr_used": false, "chunk_length": 1566, "token_count": 505}}
{"text": "6 Chemical equation NaHCO3(s) + HCl(aq) -> Na2CO3 (aq) + H2O(l) + CO2 (g) (c)concentrated sulphuric(VI)acid? To dry the gas/as a drying agent 4.Describe the smell of carbon(IV)oxide gas Colourless and odourless 5. Effect on lime water. Experiment Bubbled carbon(IV)oxide gas into a test tube containing lime water for about three minutes Observation White precipitate is formed. White precipitate dissolved when excess carbon(IV)oxide gas is bubbled . Explanation Carbon(IV)oxide gas reacts with lime water(Ca(OH)2) to form an insoluble white precipitate of calcium carbonate. Calcium carbonate reacts with more Carbon(IV) oxide gas to form soluble Calcium hydrogen carbonate. Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) CaCO3 (aq) + H2O(l) + CO2 (g) -> Ca(HCO3) 2 (aq) 6. Effects on burning Magnesium ribbon Experiment Lower a piece of burning magnesium ribbon into a gas jar containing carbon (IV)oxide gas. Observation The ribbon continues to burn with difficulty White ash/solid is formed. Black speck/solid/particles formed on the side of gas jar. Explanation Carbon(IV)oxide gas does not support combustion/burning.Magnesium burn to produce/release enough heat energy to decompose Carbon(IV) oxide gas to carbon and oxygen.Magnesium continues to burn in Oxygen forming white Magnesium Oxide solid/ash.Black speck/particle of carbon/charcoal residue forms on the sides of reaction flask. During the reaction Carbon(IV) oxide is reduced(Oxidizing agent)to carbon while Magnesium is Oxidized to Magnesium Oxide. Chemical equation 2Mg(s) + CO2 (g) -> C (s) + 2MgO(l)\n7 7. Dry and wet litmus papers were separately put in a gas jar containing dry carbon (IV)oxide gas. State and explain the observations made.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8638777139208174, "ocr_used": false, "chunk_length": 1728, "token_count": 456}}
{"text": "Chemical equation 2Mg(s) + CO2 (g) -> C (s) + 2MgO(l)\n7 7. Dry and wet litmus papers were separately put in a gas jar containing dry carbon (IV)oxide gas. State and explain the observations made. Observation Blue dry litmus paper remain blue Red dry litmus paper remain Red Blue wet/damp/moist litmus paper turn red Red wet/damp/moist litmus paper remain red Explanation Dry Carbon (IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+ and thus has no effect on litmus papers. Wet/damp/moist litmus papers contains water that dissolves/react with dry carbon (IV) oxide gas to form the weak solution of carbonic (IV) acid(H2CO3). Carbonic (IV) acid dissociate/ionizes to a few /little free H+ and CO32-. The few H+ (aq) ions are responsible for turning blue litmus paper to faint red showing the gas is very weakly acidic. Chemical equation H2CO3(aq) -> 2H+ (aq) + CO32-(aq) 8. Explain why Carbon (IV)oxide cannot be prepared from the reaction of: (i) marble chips with dilute sulphuric(VI)acid. Explanation Reaction forms insoluble calcium sulphate(VI)that cover/coat unreacted marble chips stopping further reaction Chemical equation CaCO3(s) + H2SO4 (aq) -> CaSO4 (s) + H2O(l) + CO2 (g) PbCO3(s) + H2SO4 (aq) -> PbSO4 (s) + H2O(l) + CO2 (g) BaCO3(s) + H2SO4 (aq) -> BaSO4 (s) + H2O(l) + CO2 (g) (ii) Lead(II)carbonate with dilute Hydrochloric acid. Reaction forms insoluble Lead(II)Chloride that cover/coat unreacted Lead(II) carbonate stopping further reaction unless the reaction mixture is heated. Lead(II)Chloride is soluble in hot water. Chemical equation PbCO3(s) + 2HCl (aq) -> PbCl2 (s) + H2O(l) + CO2 (g) 9.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8282699471541195, "ocr_used": false, "chunk_length": 1649, "token_count": 501}}
{"text": "Reaction forms insoluble Lead(II)Chloride that cover/coat unreacted Lead(II) carbonate stopping further reaction unless the reaction mixture is heated. Lead(II)Chloride is soluble in hot water. Chemical equation PbCO3(s) + 2HCl (aq) -> PbCl2 (s) + H2O(l) + CO2 (g) 9. Describe the test for the presence of Carbon (IV)oxide. Using burning splint Lower a burning splint into a gas jar suspected to contain Carbon (IV)oxide gas.The burning splint is extinguished. Using Lime water. Bubble the gas suspected to be Carbon (IV)oxide gas.A white precipitate that dissolve in excess bubbling is formed. 8 Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) CaCO3 (aq) + H2O(l) + CO2 (g) -> Ca(HCO3) 2 (aq) 10.State three main uses of Carbon (IV)oxide gas (i)In the Solvay process for the manufacture of soda ash/sodium carbonate (ii)In preservation of aerated drinks (iii)As fire extinguisher because it does not support combustion and is denser than air. (iv)In manufacture of Baking powder. (ii) Carbon(II)Oxide (CO) (a)Occurrence Carbon(II)oxide is found is found from incomplete combustion of fuels like petrol charcoal, liquefied Petroleum Gas/LPG. (b)School Laboratory preparation In the school laboratory carbon(II)oxide can be prepared from dehydration of methanoic acid/Formic acid(HCOOH) or Ethan-1,2-dioic acid/Oxalic acid(HOOCCOOH) using concentrated sulphuric(VI) acid. Heating is necessary.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8579751637046774, "ocr_used": false, "chunk_length": 1406, "token_count": 395}}
{"text": "(ii) Carbon(II)Oxide (CO) (a)Occurrence Carbon(II)oxide is found is found from incomplete combustion of fuels like petrol charcoal, liquefied Petroleum Gas/LPG. (b)School Laboratory preparation In the school laboratory carbon(II)oxide can be prepared from dehydration of methanoic acid/Formic acid(HCOOH) or Ethan-1,2-dioic acid/Oxalic acid(HOOCCOOH) using concentrated sulphuric(VI) acid. Heating is necessary. METHOD 1:Preparation of Carbon (IV)Oxide from dehydration of Oxalic/ethan-1,2-dioic acid METHOD 2:Preparation of Carbon (IV)Oxide from dehydration of Formic/Methanoic acid\n9 (c)Properties of Carbon (II)Oxide(Questions) 1.Write the equation for the reaction for the preparation of carbon(II)oxide using; (i)Method 1; Chemical equation HOOCCOOH(s) –Conc.H2SO4--> CO(g) + CO2 (g) + H2O(l) H2C2O4(s) –Conc.H2SO4--> CO(g) + CO2 (g) + H2O(l) (ii)Method 2; Chemical equation HCOOH(s) –Conc.H2SO4--> CO(g) + H2O(l) H2CO2(s) –Conc.H2SO4--> CO(g) + H2O(l) 2.What method of gas collection is used during the preparation of carbon (II) oxide. Over water because the gas is insoluble in water. Downward delivery because the gas is 1 ½ times denser than air . 3.What is the purpose of : (i) Potassium hydroxide/sodium hydroxide in Method 1 To absorb/ remove carbon (II) oxide produced during the reaction. 2KOH (aq) + CO2 (g) -> K2CO3 (s) + H2O(l) 2NaOH (aq) + CO2 (g) -> Na2CO3 (s) + H2O(l) (ii) Concentrated sulphuric(VI)acid in Method 1 and 2.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8066287309904437, "ocr_used": false, "chunk_length": 1444, "token_count": 484}}
{"text": "Downward delivery because the gas is 1 ½ times denser than air . 3.What is the purpose of : (i) Potassium hydroxide/sodium hydroxide in Method 1 To absorb/ remove carbon (II) oxide produced during the reaction. 2KOH (aq) + CO2 (g) -> K2CO3 (s) + H2O(l) 2NaOH (aq) + CO2 (g) -> Na2CO3 (s) + H2O(l) (ii) Concentrated sulphuric(VI)acid in Method 1 and 2. Dehydrating agent –removes the element of water (Hydrogen and Oxygen in ratio 2:1) present in both methanoic and ethan-1,2-dioic acid. 10 4. Describe the smell of carbon(II)oxide. Colourless and odourless. 5. State and explain the observation made when carbon(IV)oxide is bubbled in lime water for a long time. No white precipitate is formed. 6. Dry and wet/moist/damp litmus papers were separately put in a gas jar containing dry carbon(IV)oxide gas. State and explain the observations made. Observation -blue dry litmus paper remains blue -red dry litmus paper remains red - wet/moist/damp blue litmus paper remains blue - wet/moist/damp red litmus paper remains red Explanation Carbon(II)oxide gas is a molecular compound that does not dissociate /ionize to release H+ ions and thus has no effect on litmus papers. Carbon(II)oxide gas is therefore a neutral gas. 7. Carbon (II)oxide gas was ignited at the end of a generator as below. (i)State the observations made in flame K. Gas burns with a blue flame (ii)Write the equation for the reaction taking place at flame K. 2CO(g) + O2 (g) -> 2CO2 (g) 8. Carbon(II)oxide is a reducing agent. Explain Experiment Pass carbon(II)oxide through glass tube containing copper (II)oxide. Ignite any excess poisonous carbon(II)oxide. Observation Colour change from black to brown. Excess carbon (II)oxide burn with a blue flame. Flame K Dry carbon(II)oxide\n11 Explanation Carbon is a reducing agent.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.853395945642408, "ocr_used": false, "chunk_length": 1792, "token_count": 504}}
{"text": "Observation Colour change from black to brown. Excess carbon (II)oxide burn with a blue flame. Flame K Dry carbon(II)oxide\n11 Explanation Carbon is a reducing agent. It is used to reduce metal oxide ores to metal, itself oxidized to carbon(IV)oxide gas. Carbon(II)Oxide reduces black copper(II)oxide to brown copper metal Chemical Equation CuO(s) + CO(g) -> Cu(s) + CO2(g) (black) (brown) PbO(s) + CO(g) -> Pb(s) + CO2(g) (brown when hot/ (grey) yellow when cool) ZnO(s) + CO(g) -> Zn(s) + CO2(g) (yellow when hot/ (grey) white when cool) Fe2O3(s) + 3CO(s) -> 2Fe(s) + 3CO2(g) (brown when hot/cool (grey) Fe3O4 (s) + 4CO(g) -> 3Fe(s) + 4CO2(g) (brown when hot/cool (grey) These reaction are used during the extraction of many metals from their ore. 9. Carbon (II) oxide is a pollutant. Explain. Carbon(II)oxide is highly poisonous/toxic.It preferentially combine with haemoglobin to form stable carboxyhaemoglobin in the blood instead of oxyhaemoglobin.This reduces the free haemoglobin in the blood causing nausea , coma then death. 10.The diagram below show a burning charcoal stove/burner/jiko. Use it to answer the questions that follow. 12 Explain the changes that take place in the burner Explanation Charcoal stove has air holes through which air enters. Air oxidizes carbon to carbon(IV)oxide gas at region I. This reaction is exothermic(-∆H) producing more heat. Chemical equation C(s) + O2(g) -> CO2(g) Carbon(IV)oxide gas formed rises up to meet more charcoal which reduces it to Carbon(II)oxide gas. Chemical equation 2CO2 (g) + O2(g) -> 2CO (g) At the top of burner in region II, Carbon (II)oxide gas is further oxidized to Carbon(IV)oxide gas if there is plenty of air but escape if the air is limited poisoning the living things around.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8516539156994727, "ocr_used": false, "chunk_length": 1751, "token_count": 498}}
{"text": "This reaction is exothermic(-∆H) producing more heat. Chemical equation C(s) + O2(g) -> CO2(g) Carbon(IV)oxide gas formed rises up to meet more charcoal which reduces it to Carbon(II)oxide gas. Chemical equation 2CO2 (g) + O2(g) -> 2CO (g) At the top of burner in region II, Carbon (II)oxide gas is further oxidized to Carbon(IV)oxide gas if there is plenty of air but escape if the air is limited poisoning the living things around. Chemical equation 2CO (g) + O2(g) -> 2CO2 (g) (excess air) 11.Describe the test for the presence of carbon(II)oxide gas. Experiment Burn/Ignite the pure sample of the gas. Pass/Bubble the products into lime water/Calcium hydroxide . Observation Colourless gas burns with a blue flame. A white precipitate is formed that dissolve on further bubbling of the products. Chemical equation 2CO (g) + O2(g) -> 2CO2 (g) (gas burns with blue flame) Chemical equation Ca(OH) 2 (aq) + CO2 (g) -> CaCO3 (s) + H2O(l) Chemical equation CO2 (g) + CaCO3 (s) + H2O(l) -> Ca(HCO3) 2 (aq) 12. State the main uses of carbon (II)oxide gas. (i) As a fuel /water gas (ii)As a reducing agent in the blast furnace for extracting iron from iron ore(Magnetite/Haematite) (iii)As a reducing agent in extraction of Zinc from Zinc ore/Zinc blende (iv) As a reducing agent in extraction of Lead from Lead ore/Galena (v) As a reducing agent in extraction of Copper from Copper iron sulphide/Copper pyrites. 13 (iii)Carbonate(IV) (CO32-)and hydrogen carbonate(IV(HCO3-) 1.Carbonate (IV) (CO32-) are normal salts derived from carbonic(IV)acid (H2CO3) and hydrogen carbonate (IV) (HCO3-) are acid salts derived from carbonic(IV)acid.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8327941176470589, "ocr_used": false, "chunk_length": 1632, "token_count": 484}}
{"text": "State the main uses of carbon (II)oxide gas. (i) As a fuel /water gas (ii)As a reducing agent in the blast furnace for extracting iron from iron ore(Magnetite/Haematite) (iii)As a reducing agent in extraction of Zinc from Zinc ore/Zinc blende (iv) As a reducing agent in extraction of Lead from Lead ore/Galena (v) As a reducing agent in extraction of Copper from Copper iron sulphide/Copper pyrites. 13 (iii)Carbonate(IV) (CO32-)and hydrogen carbonate(IV(HCO3-) 1.Carbonate (IV) (CO32-) are normal salts derived from carbonic(IV)acid (H2CO3) and hydrogen carbonate (IV) (HCO3-) are acid salts derived from carbonic(IV)acid. Carbonic(IV)acid(H2CO3) is formed when carbon(IV)oxide gas is bubbled in water. It is a dibasic acid with two ionizable hydrogens. H2CO3(aq) ->2H+(aq) + CO32-(aq) H2CO3(aq) -> H+(aq) + HCO3 - (aq) 2.Carbonate (IV) (CO32-) are insoluble in water except Na2CO3 , K2CO3 and (NH4)2CO3 3.Hydrogen carbonate (IV) (HCO3-) are soluble in water. Only five hydrogen carbonates exist. Na HCO3 , KHCO3 ,NH4HCO3 Ca(HCO3)2 and Mg(HCO3)2 Ca(HCO3)2 and Mg(HCO3)2 exist only in aqueous solutions. 3.The following experiments show the effect of heat on Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) salts: Experiment In a clean dry test tube place separately about 1.0 of the following: Zinc(II)carbonate(IV), sodium hydrogen carbonate(IV), sodium carbonate(IV), Potassium carbonate(IV) ammonium carbonate(IV), potassium hydrogen carbonate(IV), Lead(II)carbonate(IV), Iron(II)carbonate(IV), and copper(II)carbonate(IV). Heat each portion gently the strongly. Test any gases produced with lime water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8200176856661675, "ocr_used": false, "chunk_length": 1619, "token_count": 494}}
{"text": "3.The following experiments show the effect of heat on Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) salts: Experiment In a clean dry test tube place separately about 1.0 of the following: Zinc(II)carbonate(IV), sodium hydrogen carbonate(IV), sodium carbonate(IV), Potassium carbonate(IV) ammonium carbonate(IV), potassium hydrogen carbonate(IV), Lead(II)carbonate(IV), Iron(II)carbonate(IV), and copper(II)carbonate(IV). Heat each portion gently the strongly. Test any gases produced with lime water. Observation (i)Colorless droplets form on the cooler parts of test tube in case of sodium carbonate(IV) and Potassium carbonate(IV). (ii)White residue/solid left in case of sodium hydrogen carbonate(IV), sodium carbonate(IV), Potassium carbonate(IV) and potassium hydrogen carbonate(IV). (iii)Colour changes from blue/green to black in case of copper(II)carbonate(IV). (iv) Colour changes from green to brown/yellow in case of Iron (II)carbonate(IV). (v) Colour changes from white when cool to yellow when hot in case of Zinc (II) carbonate(IV). (vi) Colour changes from yellow when cool to brown when hot in case of Lead (II) carbonate(IV). (vii)Colourless gas produced that forms a white precipitate with lime water in all cases. Explanation 1. Sodium carbonate(IV) and Potassium carbonate(IV) exist as hydrated salts with 10 molecules of water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid. Chemical equation Na2CO3 .10H2O(s) -> Na2CO3 (s) + 10H2O(l)\n14 K2CO3 .10H2O(s) -> K2CO3 (s) + 10H2O(l) 2. Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) salts decompose on heating except Sodium carbonate(IV) and Potassium carbonate(IV). (a) Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV) decompose on heating to form sodium carbonate(IV) and Potassium carbonate(IV).Water and carbon(IV)oxide gas are also produced.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8635341854331507, "ocr_used": false, "chunk_length": 1908, "token_count": 505}}
{"text": "Chemical equation Na2CO3 .10H2O(s) -> Na2CO3 (s) + 10H2O(l)\n14 K2CO3 .10H2O(s) -> K2CO3 (s) + 10H2O(l) 2. Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) salts decompose on heating except Sodium carbonate(IV) and Potassium carbonate(IV). (a) Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV) decompose on heating to form sodium carbonate(IV) and Potassium carbonate(IV).Water and carbon(IV)oxide gas are also produced. Chemical equation 2NaHCO3 (s) -> Na2CO3 (s) + H2O(l) + CO2 (g) (white) (white) 2KHCO3 (s) -> K2CO3 (s) + H2O(l) + CO2 (g) (white) (white) (b) Calcium hydrogen carbonate(IV) and Magnesium hydrogen carbonate(IV) decompose on heating to form insoluble Calcium carbonate(IV) and Magnesium carbonate(IV).Water and carbon(IV)oxide gas are also produced. Chemical equation Ca(HCO3)2 (aq) -> CaCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) Mg(HCO3)2 (aq) -> MgCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) (c) Ammonium hydrogen carbonate(IV) decompose on heating to form ammonium carbonate(IV) .Water and carbon(IV)oxide gas are also produced. Chemical equation 2NH4HCO3 (s) -> (NH4)2CO3 (s) + H2O(l) + CO2 (g) (white) (white) (d)All other carbonates decompose on heating to form the metal oxide and produce carbon(IV)oxide gas e.g.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7801440166993295, "ocr_used": false, "chunk_length": 1293, "token_count": 445}}
{"text": "Chemical equation 2NaHCO3 (s) -> Na2CO3 (s) + H2O(l) + CO2 (g) (white) (white) 2KHCO3 (s) -> K2CO3 (s) + H2O(l) + CO2 (g) (white) (white) (b) Calcium hydrogen carbonate(IV) and Magnesium hydrogen carbonate(IV) decompose on heating to form insoluble Calcium carbonate(IV) and Magnesium carbonate(IV).Water and carbon(IV)oxide gas are also produced. Chemical equation Ca(HCO3)2 (aq) -> CaCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) Mg(HCO3)2 (aq) -> MgCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) (c) Ammonium hydrogen carbonate(IV) decompose on heating to form ammonium carbonate(IV) .Water and carbon(IV)oxide gas are also produced. Chemical equation 2NH4HCO3 (s) -> (NH4)2CO3 (s) + H2O(l) + CO2 (g) (white) (white) (d)All other carbonates decompose on heating to form the metal oxide and produce carbon(IV)oxide gas e.g. Chemical equation MgCO3 (s) -> MgO (s) + CO2 (g) (white solid) (white solid) Chemical equation BaCO3 (s) -> BaO (s) + CO2 (g) (white solid) (white solid) Chemical equation CaCO3 (s) -> CaO (s) + CO2 (g) (white solid) (white solid) Chemical equation CuCO3 (s) -> CuO (s) + CO2 (g) (blue/green solid) (black solid) Chemical equation ZnCO3 (s) -> ZnO (s) + CO2 (g) (white solid) (white solid when cool/\n15 Yellow solid when hot) Chemical equation PbCO3 (s) -> PbO (s) + CO2 (g) (white solid) (yellow solid when cool/ brown solid when hot) 4.The following experiments show the presence of Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) ions in sample of a salt: (a)Using Lead(II) nitrate(V) I.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7778885197489849, "ocr_used": false, "chunk_length": 1548, "token_count": 537}}
{"text": "Chemical equation Ca(HCO3)2 (aq) -> CaCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) Mg(HCO3)2 (aq) -> MgCO3 (s) + H2O(l) + CO2 (g) (Colourless solution) (white) (c) Ammonium hydrogen carbonate(IV) decompose on heating to form ammonium carbonate(IV) .Water and carbon(IV)oxide gas are also produced. Chemical equation 2NH4HCO3 (s) -> (NH4)2CO3 (s) + H2O(l) + CO2 (g) (white) (white) (d)All other carbonates decompose on heating to form the metal oxide and produce carbon(IV)oxide gas e.g. Chemical equation MgCO3 (s) -> MgO (s) + CO2 (g) (white solid) (white solid) Chemical equation BaCO3 (s) -> BaO (s) + CO2 (g) (white solid) (white solid) Chemical equation CaCO3 (s) -> CaO (s) + CO2 (g) (white solid) (white solid) Chemical equation CuCO3 (s) -> CuO (s) + CO2 (g) (blue/green solid) (black solid) Chemical equation ZnCO3 (s) -> ZnO (s) + CO2 (g) (white solid) (white solid when cool/\n15 Yellow solid when hot) Chemical equation PbCO3 (s) -> PbO (s) + CO2 (g) (white solid) (yellow solid when cool/ brown solid when hot) 4.The following experiments show the presence of Carbonate (IV) (CO32-) and Hydrogen carbonate (IV) (HCO3-) ions in sample of a salt: (a)Using Lead(II) nitrate(V) I. Using a portion of salt solution in a test tube .add four drops of Lead(II)nitrate(V)solution.Preserve. Observation inference White precipitate/ppt CO32- ,SO32- ,SO42- ,Cl- - II. To the preserved solution ,add six drops of dilutte nitric(V)acid. Preserve.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7888981350599609, "ocr_used": false, "chunk_length": 1456, "token_count": 488}}
{"text": "Observation inference White precipitate/ppt CO32- ,SO32- ,SO42- ,Cl- - II. To the preserved solution ,add six drops of dilutte nitric(V)acid. Preserve. Observation inference White precipitate/ppt persists White precipitate/ppt dissolves SO42- ,Cl- CO32- ,SO32- II. To the preserved sample( that forms a precipitate ),heat to boil. Observation inference White precipitate/ppt persists White precipitate/ppt dissolves SO42- Cl- II. To the preserved sample( that do not form a precipitate ),add three drops of acidified potassium manganate(VII)/lime water Observation inference Effervescence/bubbles/fizzing colourless gas produced Acidified KMnO4 decolorized/no white precipitate on lime water Effervescence/bubbles/fizzing colourless gas produced Acidified KMnO4 not decolorized/ white precipitate on lime water SO32- CO32-\n16 Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8516988896509537, "ocr_used": false, "chunk_length": 1478, "token_count": 397}}
{"text": "To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Explanations Using Lead(II)nitrate(V)\n17 (i)Lead(II)nitrate(V) solution reacts with chlorides(Cl-), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV). Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8320344556934358, "ocr_used": false, "chunk_length": 1604, "token_count": 492}}
{"text": "Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g) PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g) (iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed; - Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling) - Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling. (iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8152039210543571, "ocr_used": false, "chunk_length": 1208, "token_count": 412}}
{"text": "Chemical/ionic equation: PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g) PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g) (iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed; - Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling) - Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling. (iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8027412906910337, "ocr_used": false, "chunk_length": 1236, "token_count": 428}}
{"text": "Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 18 Using Barium(II)nitrate(V)/ Barium(II)Chloride (i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV). Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.801603481163567, "ocr_used": false, "chunk_length": 1440, "token_count": 505}}
{"text": "Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g) BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g) (iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7820002315082765, "ocr_used": false, "chunk_length": 1304, "token_count": 472}}
{"text": "Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l)\n19 These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. (iii) Sodium carbonate(IV) (Na2CO3) (a)Extraction of sodium carbonate from soda ash Sodium carbonate naturally occurs in Lake Magadi in Kenya as Trona.trona is the double salt ; sodium sesquicarbonate. NaHCO3 .Na2CO3 .H2O.It is formed from the volcanic activity that takes place in Lake Naivasha, Nakuru ,Bogoria and Elementeita .All these lakes drain into Lake Magadi through underground rivers. Lake Magadi has no outlet. Solubility of Trona decrease with increase in temperature.High temperature during the day causes trona to naturally crystallize .It is mechanically scooped/dredged/dug and put in a furnace. Inside the furnace, trona decompose into soda ash/sodium carbonate.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8397268032650342, "ocr_used": false, "chunk_length": 1305, "token_count": 385}}
{"text": "Lake Magadi has no outlet. Solubility of Trona decrease with increase in temperature.High temperature during the day causes trona to naturally crystallize .It is mechanically scooped/dredged/dug and put in a furnace. Inside the furnace, trona decompose into soda ash/sodium carbonate. Chemical equation 2NaHCO3 .Na2CO3 .H2O (s) -> 3Na2CO3 (s) + 5H2O(l) + CO2 (g) (trona) (soda ash) Soda ash is then bagged and sold as Magadi soda.It is mainly used: (i)in making glass to lower the melting point of raw materials (sand/SiO2 from 1650oC and CaO from 2500oC to around 1500oC) (ii)in softening hard water (iii)in the manufacture of soapless detergents. (iv)Swimming pool “pH increaser” Sodium chloride is also found dissolved in the lake. Solubility of sodium chloride decrease with decreases in temperature/ sodium chloride has lower solubility at lower temperatures. When temperatures decrease at night it crystallize out .The crystals are then mechanically dug/dredged /scooped then packed for sale as animal/cattle feeds and seasoning food. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide\n20 b)The Solvay process for industrial manufacture of sodium carbonate(IV) (i)Raw materials. -Brine /Concentrated Sodium chloride from salty seas/lakes. -Ammonia gas from Haber. -Limestone /Calcium carbonate from chalk /limestone rich rocks. -Water from rivers/lakes. (ii)Chemical processes Ammonia gas is passed up to meet a downward flow of sodium chloride solution / brine to form ammoniated brine/ammoniacal brine mixture in the ammoniated brine chamber The ammoniated brine mixture is then pumped up, atop the carbonator/ solvay tower. In the carbonator/ solvay tower, ammoniated brine/ammoniacal brine mixture slowly trickle down to meet an upward flow of carbon(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.894740230902196, "ocr_used": false, "chunk_length": 1973, "token_count": 505}}
{"text": "-Water from rivers/lakes. (ii)Chemical processes Ammonia gas is passed up to meet a downward flow of sodium chloride solution / brine to form ammoniated brine/ammoniacal brine mixture in the ammoniated brine chamber The ammoniated brine mixture is then pumped up, atop the carbonator/ solvay tower. In the carbonator/ solvay tower, ammoniated brine/ammoniacal brine mixture slowly trickle down to meet an upward flow of carbon(IV)oxide gas. The carbonator is shelved /packed with quartz/broken glass to (i) reduce the rate of flow of ammoniated brine/ammoniacal brine mixture. (ii)increase surface area of the liquid mixture to ensure a lot of ammoniated brine/ammoniacal brine mixture react with carbon(IV)oxide gas. Insoluble sodium hydrogen carbonate and soluble ammonium chloride are formed from the reaction. Chemical equation CO2(g) + H2O(l) + NaCl (aq) + NH3(g) -> NaHCO3(s) + NH4Cl(aq)\n21 The products are then filtered. Insoluble sodium hydrogen carbonate forms the residue while soluble ammonium chloride forms the filtrate. Sodium hydrogen carbonate itself can be used: (i) as baking powder and preservation of some soft drinks. (ii) as a buffer agent and antacid in animal feeds to improve fibre digestion. (iii) making dry chemical fire extinguishers. In the Solvay process Sodium hydrogen carbonate is then heated to form Sodium carbonate/soda ash, water and carbon (IV) oxide gas. Chemical equation 2NaHCO3 (s) -> Na2CO3(s) + CO2(g) + H2O(l) Sodium carbonate is stored ready for use in: (i) during making glass/lowering the melting point of mixture of sand/SiO2 from 1650oC and CaO from 2500oC to around 1500oC (ii) in softening hard water (iii) in the manufacture of soapless detergents. (iv) swimming pool “pH increaser”. Water and carbon(IV)oxide gas are recycled back to the ammoniated brine/ammoniacal brine chamber. More carbon(IV)oxide is produced in the kiln/furnace. Limestone is heated to decompose into Calcium oxide and carbon(IV)oxide.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8859512977464803, "ocr_used": false, "chunk_length": 1963, "token_count": 508}}
{"text": "Water and carbon(IV)oxide gas are recycled back to the ammoniated brine/ammoniacal brine chamber. More carbon(IV)oxide is produced in the kiln/furnace. Limestone is heated to decompose into Calcium oxide and carbon(IV)oxide. Chemical equation CaCO3 (s) -> CaO(s) + CO2(g) Carbon(IV)oxide is recycled to the carbonator/solvay tower. Carbon (IV)oxide is added water in the slaker to form Calcium hydroxide. This process is called slaking. Chemical equation CaO(s) + H2O (l) -> Ca(OH)2 (aq) Calcium hydroxide is mixed with ammonium chloride from the carbonator/solvay tower in the ammonia regeneration chamber to form Calcium chloride , water and more ammonia gas. Chemical equation Ca(OH)2 (aq) +2NH4Cl (aq) -> CaCl2(s) + 2NH3(g) + H2O(l) NH3(g) and H2O(l) are recycled. Calcium chloride may be used: (i)as drying agent in the school laboratory during gas preparation (except ammonia gas)\n22 (ii)to lower the melting point of solid sodium chloride / rock salt salts during the Downs process for industrial extraction of sodium metal. Detailed Summary flow diagram of Solvay Process\n23 AmmoniatedbrineBrineAmmonia regenerationchamberSolvay Tower/CarbonatorKiln/FurnaceAmmoniumchlorideCarbon(IV)OxideHaberprocessSlakerCalcium hydroxideCalcium oxideWaterSodium hydrogen CarbonateRoasterSodiumcarbonateCalcium chlorideBrine saturated with ammoniaCoke & Limestone\n24 Practice 1. The diagram below shows part of the Solvay process used in manufacturing sodium carbonate. Use it to answer the questions that follow. (a)Explain how Sodium Chloride required for this process is obtained from the sea. Sea water is pumped /scooped into shallow pods. Evaporation of most of the water takes place leaving a very concentrated solution. (b)(i) Name process: I. Filtration II.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8859429135336331, "ocr_used": false, "chunk_length": 1759, "token_count": 456}}
{"text": "Evaporation of most of the water takes place leaving a very concentrated solution. (b)(i) Name process: I. Filtration II. Decomposition (ii) Write the equation for the reaction in process: Process I Chemical equation CO2(g) + H2O(l) + NaCl (aq) + NH3(g) -> NaHCO3(s) + NH4Cl(aq) Process II Chemical equation 2NaHCO3 (s) -> Na2CO3(s) + CO2(g) + H2O(l) (c)(i) Name two substances recycled in the solvay process Ammonia gas , Carbon(IV)Oxide and Water. (ii)Which is the by-product of this process? Calcium(II)Chloride /CaCl2 (iii)State two uses that the by-product can be used for: 1. As a drying agent in the school laboratory preparation of gases. 2. In the Downs cell/process for extraction of Sodium to lower the melting point of rock salt. Carbon (IV)oxide Ammonium chloride Process I Saturated sodium chloride solution Ammonia Sodium hydrogen carbonate Process II Sodium carbonate\n25 (iv)Write the chemical equation for the formation of the byproducts in the Solvay process. Chemical equation Ca(OH)2 (aq) +2NH4Cl (aq) -> CaCl2(s) + 2NH3(g) + H2O(l) (d)In an experiment to determine the % purity of Sodium carbonate produced in the Solvay process ,2.15g of the sample reacted with exactly 40.0cm3 of 0.5M Sulphuric(VI)acid. (i)Calculate the number of moles of sodium carbonate that reacted. Chemical equation Na2CO3 (aq) +H2SO4 (aq) -> Na2SO4 (aq)+ CO2(g) + H2O(l) Mole ratio Na2CO3 :H2SO4 => 1:1 Moles H2SO4 = Molarity x Volume => 0.5 x 40.0 = 0.02 Moles 1000 1000 Moles of Na2CO3 = 0.02 Moles (ii)Determine the % of sodium carbonate in the sample.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8003436426116839, "ocr_used": false, "chunk_length": 1552, "token_count": 486}}
{"text": "Chemical equation Ca(OH)2 (aq) +2NH4Cl (aq) -> CaCl2(s) + 2NH3(g) + H2O(l) (d)In an experiment to determine the % purity of Sodium carbonate produced in the Solvay process ,2.15g of the sample reacted with exactly 40.0cm3 of 0.5M Sulphuric(VI)acid. (i)Calculate the number of moles of sodium carbonate that reacted. Chemical equation Na2CO3 (aq) +H2SO4 (aq) -> Na2SO4 (aq)+ CO2(g) + H2O(l) Mole ratio Na2CO3 :H2SO4 => 1:1 Moles H2SO4 = Molarity x Volume => 0.5 x 40.0 = 0.02 Moles 1000 1000 Moles of Na2CO3 = 0.02 Moles (ii)Determine the % of sodium carbonate in the sample. Molar mass of Na2CO3 = 106g Mass of Na2CO3 = moles x Molar mass => 0.02 x 106 = 2.12 g % of Na2CO3 = ( 2.12 g x 100) = 98.6047% 2.15 (e) State two uses of soda ash. (i) during making glass/lowering the melting point of mixture of sand/SiO2 from 1650oC and CaO from 2500oC to around 1500oC (ii) in softening hard water (iii) in the manufacture of soapless detergents. (iv) swimming pool “pH increaser”. (f)The diagram below shows a simple ammonia soda tower used in manufacturing sodium carbonate .Use it to answer the questions that follow: Raw materialExcess Carbon(IV)oxide Metal plates Substance\n26 (i)Name the raw materials needed in the above process -Ammonia -Water -Carbon(IV)oxide -Limestone -Brine/ Concentrated sodium chloride (ii)Identify substance A Ammonium chloride /NH4Cl (iii) Write the equation for the reaction taking place in: I.Tower.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7697090872738179, "ocr_used": false, "chunk_length": 1429, "token_count": 478}}
{"text": "(i) during making glass/lowering the melting point of mixture of sand/SiO2 from 1650oC and CaO from 2500oC to around 1500oC (ii) in softening hard water (iii) in the manufacture of soapless detergents. (iv) swimming pool “pH increaser”. (f)The diagram below shows a simple ammonia soda tower used in manufacturing sodium carbonate .Use it to answer the questions that follow: Raw materialExcess Carbon(IV)oxide Metal plates Substance\n26 (i)Name the raw materials needed in the above process -Ammonia -Water -Carbon(IV)oxide -Limestone -Brine/ Concentrated sodium chloride (ii)Identify substance A Ammonium chloride /NH4Cl (iii) Write the equation for the reaction taking place in: I.Tower. Chemical equation CO2(g) + NaCl (aq) + H2O(l) + NH3(g) -> NaHCO3(s) + NH4Cl(aq) II. Production of excess carbon (IV)oxide. Chemical equation CaCO3 (s) -> CaO(s) + CO2(g) III. The regeneration of ammonia Chemical equation Ca(OH)2 (aq) +2NH4Cl (aq) -> CaCl2(s) + 2NH3(g) + H2O(l) (iv)Give a reason for having the circular metal plates in the tower. -To slow the downward flow of brine. -To increase the rate of dissolving of ammonia. -To increase the surface area for dissolution (v)Name the gases recycled in the process illustrated above. Ammonia gas , Carbon(IV)Oxide and Water. 2. Describe how you would differentiate between carbon (IV)oxide and carbon(II)oxide using chemical method. Method I -Bubble both gases in lime water/Ca(OH)2 -white precipitate is formed if the gas is carbon (IV) oxide - No white precipitate is formed if the gas is carbon (II) oxide Method II\n27 -ignite both gases - Carbon (IV) oxide does not burn/ignite - Carbon (II) oxide burn with a blue non-sooty flame. Method III -Lower a burning splint into a gas containing each gas separately.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.857041403799083, "ocr_used": false, "chunk_length": 1758, "token_count": 483}}
{"text": "Describe how you would differentiate between carbon (IV)oxide and carbon(II)oxide using chemical method. Method I -Bubble both gases in lime water/Ca(OH)2 -white precipitate is formed if the gas is carbon (IV) oxide - No white precipitate is formed if the gas is carbon (II) oxide Method II\n27 -ignite both gases - Carbon (IV) oxide does not burn/ignite - Carbon (II) oxide burn with a blue non-sooty flame. Method III -Lower a burning splint into a gas containing each gas separately. -burning splint is extinguished if the gas is carbon (IV) oxide -burning splint is not extinguished if the gas is carbon (II) oxide. 3.Using Magnesium sulphate(VI)solution ,describe how you can differentiate between a solution of sodium carbonate from a solution of sodium hydrogen carbonate -Add Magnesium sulphate(VI) solution to separate portions of a solution of sodium carbonate and sodium hydrogen carbonate in separate test tubes -White precipitate is formed in test tube containing sodium carbonate -No white precipitate is formed in test tube containing sodium hydrogen carbonate. Chemical equation Na2CO3 (aq) +MgSO4 (aq) -> Na2SO4 (aq) + MgCO3(s) (white ppt) Ionic equation CO32- (aq) + Mg2+ (aq) -> MgCO3(s) (white ppt) Chemical equation 2NaHCO3 (aq) +MgSO4 (aq) -> Na2SO4 (aq) + Mg(HCO3)2 (aq) (colourless solution) 4. The diagram below shows a common charcoal burner .Assume the burning take place in a room with sufficient supply of air. (a)Explain what happens around:\n28 (i)Layer A Sufficient/excess air /oxygen enter through the air holes into the burner .It reacts with/oxidizes Carbon to carbon(IV)oxide Chemical equation C(s) + O2(g) -> CO2 (g) (ii)Layer B Hot carbon(IV)oxide rises up and is reduced by more carbon/charcoal to carbon (II)oxide. Chemical equation C(s) + CO2(g) -> 2CO (g) (ii)Layer C Hot carbon(II)oxide rises up and burns with a blue flame to be oxidized by the excess air to form carbon(IV)oxide.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8634641252760082, "ocr_used": false, "chunk_length": 1922, "token_count": 510}}
{"text": "The diagram below shows a common charcoal burner .Assume the burning take place in a room with sufficient supply of air. (a)Explain what happens around:\n28 (i)Layer A Sufficient/excess air /oxygen enter through the air holes into the burner .It reacts with/oxidizes Carbon to carbon(IV)oxide Chemical equation C(s) + O2(g) -> CO2 (g) (ii)Layer B Hot carbon(IV)oxide rises up and is reduced by more carbon/charcoal to carbon (II)oxide. Chemical equation C(s) + CO2(g) -> 2CO (g) (ii)Layer C Hot carbon(II)oxide rises up and burns with a blue flame to be oxidized by the excess air to form carbon(IV)oxide. 2CO (g) + O2(g) -> 2CO2(g) (b)State and explain what would happen if the burner is put in an enclosed room. The hot poisonous /toxic carbon(II)oxide rising up will not be oxidized to Carbon(IV)oxide. (c)Using a chemical test , describe how you would differentiate two unlabelled black solids suspected to be charcoal and copper(II)oxide. Method I -Burn/Ignite the two substances separately. -Charcoal burns with a blue flame - Copper(II)oxide does not burn Method II -Add dilute sulphuric(VI)acid/Nitric(V)acid/Hydrochloric acid separately. -Charcoal does not dissolve. - Copper(II)oxide dissolves to form a colourless solution. 5. Excess Carbon(II)oxide was passed over heated copper(II)oxide as in the set up shown below for five minutes. 29 Dry carbon(IV)oxideHEATCopper(II)oxideBlue flame A (a)State and explain the observations made in the combustion tube. Observation Colour change from black to brown Explanation Carbon (II)oxide reduces black copper(II)oxide to brown copper metal itself oxidized to Carbon(IV)oxide. Chemical equation CO(g) + CuO (s) -> Cu(s) + CO2(g) (black) (brown) (b) (i)Name the gas producing flame A Carbon(II)oxide (ii)Why should the gas be burnt?", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8772382866862534, "ocr_used": false, "chunk_length": 1784, "token_count": 485}}
{"text": "29 Dry carbon(IV)oxideHEATCopper(II)oxideBlue flame A (a)State and explain the observations made in the combustion tube. Observation Colour change from black to brown Explanation Carbon (II)oxide reduces black copper(II)oxide to brown copper metal itself oxidized to Carbon(IV)oxide. Chemical equation CO(g) + CuO (s) -> Cu(s) + CO2(g) (black) (brown) (b) (i)Name the gas producing flame A Carbon(II)oxide (ii)Why should the gas be burnt? It is toxic/poisonous (iii)Write the chemical equation for the production of flame A 2CO(g) + O2(g) -> 2CO2(g) (c)State and explain what happens when carbon(IV)oxide is prepared using Barium carbonate and dilute sulphuric(VI)acid. Reaction starts then stops after sometime producing small/little quantity of carbon(IV)oxide gas. Barium carbonate react with dilute sulphuric(VI)acid to form insoluble Barium sulphate(VI) that cover/coat unreacted Barium carbonate stopping further reaction to produce more Carbon(IV)oxide. 30 (d) Using dot () and cross(x) to represent electrons show the bonding in a molecule of : (i) Carbon(II)oxide jgthungu@gmail.com56x●xxOxxCx ●●●lone pairs of electrons in carbon and oxygen atoms6 bonded pairs of electrons2covalent bonds1 dative bond (ii) Carbon(IV)Oxide. jgthungu@gmail.com30●x O●●C●x●x●xlone pairs of electrons in oxygen atom4 bonded pairs of electronsO●●●●●●CO2 (triatomic molecule) (e) Carbon (IV)oxide is an environmental pollutant of global concern. Explain. -It is a green house gas thus causes global warming. -It dissolves in water to form acidic carbonic acid which causes “acid rain” (f)Explain using chemical equation why lime water is used to test for the presence of Carbon (IV) oxide instead of sodium hydroxide.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8781231671554252, "ocr_used": false, "chunk_length": 1705, "token_count": 462}}
{"text": "Explain. -It is a green house gas thus causes global warming. -It dissolves in water to form acidic carbonic acid which causes “acid rain” (f)Explain using chemical equation why lime water is used to test for the presence of Carbon (IV) oxide instead of sodium hydroxide. 31 Using lime water/calcium hydroxide: - a visible white precipitate of calcium carbonate is formed that dissolves on bubbling excess Carbon (IV) oxide gas Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) (white precipitate) CaCO3 (aq) + H2O(l) + CO2 (g) -> Ca(HCO3) 2 (aq) Using sodium hydroxide: - No precipitate of sodium carbonate is formed Both sodium carbonate and sodium hydrogen carbonate are soluble salts/dissolves. Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (s) + H2O(l) (No white precipitate) Na2CO3 (s) + H2O(l) + CO2 (g) -> 2NaHCO3 (s) (g)Ethan-1,2-dioic acid and methanoic acid may be used to prepare small amount of carbon(II)oxide in a school laboratory. (i) Explain the modification in the set up when using one over the other. Before carbon(II)oxide is collected: -when using methanoic acid, no concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide. -when using ethan-1,2-dioic acid, concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide. (ii)Write the equation for the reaction for the formation of carbon(II)oxide from: I.Methanoic acid. Chemical equation HCOOH(aq) -> CO(g) + H2O(l) II. Ethan-1,2-dioic acid Chemical equation HOOCCOOH(aq) -> CO2(g)+CO(g)+H2O(l) (h)Both carbon(II)oxide and carbon(IV)oxide affect the environment. Explain why carbon(II)oxide is more toxic/poisonous. -Both gases are colourless,denser than water and odourless.", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8428872897154096, "ocr_used": false, "chunk_length": 1699, "token_count": 505}}
{"text": "Ethan-1,2-dioic acid Chemical equation HOOCCOOH(aq) -> CO2(g)+CO(g)+H2O(l) (h)Both carbon(II)oxide and carbon(IV)oxide affect the environment.\n\nExplain why carbon(II)oxide is more toxic/poisonous.\n\n-Both gases are colourless,denser than water and odourless.\n\n-Carbon(II)oxide is preferentially absorbed by human/mammalian haemoglobin when inhaled forming stable carboxyhaemoglobin instead of oxyhaemoglobin.This reduces the free haemoglobin in the blood leading to suffocation and quick death.\n\n-Carbon(IV)oxide is a green house gas that increases global warming.\n\n-Carbon(II)oxide is readily oxidized to carbon(IV)oxide 6.Study the flow chart below and use it to answer the questions that follow.\n\n32 (a)Name: (i)the white precipitate A Calcium carbonate (ii) solution B Calcium hydrogen carbonate (iii) gas C Carbon(IV)oxide (iv) white residue B Calcium oxide (v) solution D Calcium hydroxide/lime water (b)Write a balanced chemical equation for the reaction for the formation of: (i) the white precipitate A from solution D Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) (ii) the white precipitate A from solution B Chemical equation Ca(HCO3)2(aq) -> CO2 (g) + CaCO3 (s) + H2O(l) (iii) solution B from the white precipitate A Chemical equation CO2 (g) + CaCO3 (s) + H2O(l) -> Ca(HCO3)2(aq)\n33 (iv) white residue B from the white precipitate A Chemical equation CaCO3(s) -> CO2 (g) + CaO (s) (iv) reaction of white residue B with water Chemical equation CaO (s) + H2O(l) -> Ca(OH)2(aq)", "metadata": {"source": "KCSE-FORM-2-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8394027601440974, "ocr_used": false, "chunk_length": 1502, "token_count": 442}}
{"text": "PRACTICAL CHEMISTRY\nIntroduction\nScientific subjects are, by their nature, experimental. It is accordingly important that an assessment of a student’s knowledge and understanding of Chemistry should contain a component relating to practical work and experimental skills. This booklet has been produced to help students preparing for and taking practical Examinations. The material contained in this booklet does not extend the curriculum specification content. Rather it seeks to help the candidate succeed in practical examination by explaining in more depth what is required of him or her in carrying out the exercises, making observations and measurements with appropriate precision and recording these methodically. This booklet advices candidate on how he or she should interpret, explain, evaluate and communicate the results of the exercises clearly and logically using relevant chemical knowledge and understanding and using appropriate specialist vocabulary. 1.1APPARATUS IN A CHEMISTRY LABORATORY\n1.2 SPECIAL LABORATORY APPARATUS AND TECHNIQUES\n(a) USING THE BURETTE\nA burette is used to deliver solution in precisely measured, variable volumes. Burettes are used primarily for titration, to deliver one reactant until the precise end point of the reaction is reached. To fill a burette, close the stopcock at the bottom and use a funnel. You may need to lift up on the funnel slightly, to allow the solution to flow in freely. You can also fill a burette using a disposable transfer pipette. This works better than a funnel for the small, 10 mL burettes. Be sure the transfer pipette is dry or conditioned with the titrant, so the concentration of solution will not be changed. Before titrating, condition the burette with titrant solution and check that the burette is flowing freely. To condition a piece of glassware, rinse it so that all surfaces are coated with solution, then drain. Conditioning two or three times will insure that a stray drop of water does not change the concentration of titrant.Rinse the tip of the burette with water from a wash bottle and dry it carefully. After a minute, check for solution on the tip to see if your burette is leaking. The tip should be clean and dry before you take an initial volume reading.When your burette is conditioned and filled, with no air bubbles or leaks, take an initial volume reading. A burette-reading card with a black rectangle can help you to take a more accurate reading. Read the bottom of the meniscus.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9194029075130241, "ocr_used": false, "chunk_length": 2483, "token_count": 505}}
{"text": "The tip should be clean and dry before you take an initial volume reading.When your burette is conditioned and filled, with no air bubbles or leaks, take an initial volume reading. A burette-reading card with a black rectangle can help you to take a more accurate reading. Read the bottom of the meniscus. Be sure your eye is at the level of meniscus, not above or below. Reading from an angle, rather than straight on, results in a parallax error. Deliver solution to the titration flask by turning the stopcock. The solution should be delivered quickly until a couple of mL from the endpoint. The endpoint should be approached slowly, a drop at a time. Use a wash bottle to rinse the tip of the burette and the sides of the flask. Your TA can show you how to deliver a partial drop of solution, when near the endpoint. (b) Volumetric (measuring) Flasks\nErlenmeyer flasks and beakers are used for mixing, transporting, and reacting, but not for accurate measurements. The volumes stamped on the sides are approximate and accurate to within about 5%. (c) Graduated Cylinders\nGraduated cylinders are useful for measuring liquid volumes to within about 1%. They are for general-purpose use, but not for quantitative analysis. If greater accuracy is needed, use a pipette or volumetric flask. (d) Pipette\nA pipette is used to measure small amounts of solution very accurately. A pipette bulb is used to draw solution into the pipette. Start by squeezing the bulb in your preferred hand. Then place the bulb on the flat end of the pipette.Place the tip of the pipette in the solution and release your grip on the bulb to pull solution into the pipette. Draw solution in above the mark on the neck of the pipette. If the volume of the pipette is larger than the volume of the pipette bulb, you may need to remove the bulb from the pipette and squeeze it and replace it on the pipette a second time, to fill the pipette volume completely. Quickly, remove the pipette bulb and put your index finger on the end of the pipette. Gently release the seal made by your finger until the level of the solution meniscus exactly lines up with the mark on the pipette. Practice this with water until you are able to use the pipette and bulb consistently and accurately.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9131538102766186, "ocr_used": false, "chunk_length": 2251, "token_count": 499}}
{"text": "Quickly, remove the pipette bulb and put your index finger on the end of the pipette. Gently release the seal made by your finger until the level of the solution meniscus exactly lines up with the mark on the pipette. Practice this with water until you are able to use the pipette and bulb consistently and accurately. (e) Volumetric Flask\nA volumetric flask is used to make up a solution of fixed volume very accurately. This volumetric flask measures 500 mL ± 0.2 mL. This is a relative uncertainty of 4 x 10-4 or 400 parts per million. To make up a solution, first dissolve the solid material completely, in less water than required to fill the flask to the mark. After the solid is completely dissolved, very carefully fill the flask to the 500 mL mark. Move your eye to the level of the mark on the neck of the flask and line it up so that the circle around the neck looks like a line, not an ellipse. Then add distilled water a drop at a time until the bottom of the meniscus lines up exactly with the mark on the neck of the flask. Take care that no drops of liquid are in the neck of the flask above the mark. After the final dilution, remember to mix your solution thoroughly, by inverting the flask and shaking. (f) Analytical Balance\nAn analytical balance measures masses to within 0.0001 g. Use these balances when you need this high degree of precision.Turn the balance on by pressing the control bar. The display lights up for several seconds, then resets to 0.0000.Place creased, small weighing paper on the balance pan.Close the sliding glass doors. Wait for the green dot on the left to go out. This is the stability indicator light, indicating that the weight is stable.Press the control bar to cancel out the weight of the container or paper. The display will again read 0.0000. Carefully add the substance to be weighed up to the desired mass. Do not attempt to reach a particular mass exactly. Before recording the mass, close the glass doors and wait until the stability detector lamp goes out. Record mass of solid. Don't pick up tare containers with bare hands since your fingerprints add mass. Use tongs to prevent this.Don't lean on the bench while weighing.Do record the mass of your container, if you will need it later.Do check the level indicator bubble before weighing.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8959228876127974, "ocr_used": false, "chunk_length": 2300, "token_count": 508}}
{"text": "Record mass of solid. Don't pick up tare containers with bare hands since your fingerprints add mass. Use tongs to prevent this.Don't lean on the bench while weighing.Do record the mass of your container, if you will need it later.Do check the level indicator bubble before weighing. The two rear balance feet serve as levelling screws. Use the brush provided to clean spills in the weighing chamber. Discard any disposable tare containers or weighing paper, in the nearest wastebasket.(g) Calorimetry\nCalorimetry is used to determine the heat released or absorbed in a chemical reaction. The calorimeters shown here can determine the heat of a solution reaction at constant (atmospheric) pressure. The calorimeter is a double Styrofoam cup fitted with a plastic top in which there is a hole for a thermometer. (It's crude, but very effective!) Key techniques for obtaining accurate results are starting with a dry calorimeter, measuring solution volumes precisely, and determining change in temperature accurately. (h) Using a Calorimeter\nSolutions volumes should be carefully measured with a graduated cylinder. Add solution completely, to a dry calorimeter. Don't forget to add the spin bar each time! Set up the calorimeter with the thermometer (0° to 50°C, graduated every 0.1°C) supported from a stand so that the bulb does not touch the bottom of the cup. Note that the thermometer used for calorimetry differs from the less accurate one in your glassware drawer. Clamp the calorimeter so that it rests on the stirrer. Be careful not to turn on the heat or you will melt the Styrofoam. The change in temperature is determined by measuring the initial temperature, T1, of the reactants, and the maximum temperature, T2, of the contents of the calorimeter during the exothermic reaction. Use a magnifying glass to measure temperature values precisely.Interpolate between the divisions of the thermometer and record temperatures to +/- 0.01 °C. See your lab manual for a discussion of how to determine accurately the change in temperature from your graph of temperature vs. time. (i) Top-loading Balance\nUse a top loading balance to weigh solid material when a precision of 0.1 g is adequate. For more accurate mass measurements or small amounts, use an analytical balance. Using a Top-loading Balance\nCheck if the balance is turned on.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9056672568693029, "ocr_used": false, "chunk_length": 2340, "token_count": 496}}
{"text": "(i) Top-loading Balance\nUse a top loading balance to weigh solid material when a precision of 0.1 g is adequate. For more accurate mass measurements or small amounts, use an analytical balance. Using a Top-loading Balance\nCheck if the balance is turned on. If not, press the on/off button and wait until the display reads 0.0 g. Place a container or large, creased weighing paper on the balance pan. Push tare button to zero the balance. Carefully add substance to the container or paper. Record the mass. Use the brush provided to clean any spills. Discard any disposable tare containers or weighing paper in the nearest wastebasket. GENERAL EXPERIMENTAL PROCESSES\nQuantitative Transfer\nQuantitative Transfer simply means that all the material to be transferred from one place to another must make the trip. For example, every particle of solid must be transferred from the weighing paper to the (clean) beaker.This can be done be carefully tipping the creased weighing paper to pour the solid into the beaker. Tapping the paper with a spatula will knock particles into the beaker.Finally, the paper should be rinsed into the beaker, to remove all traces of the solid. Transferring a Solution or Mixture\nIf you are transferring a solution or heterogeneous mixture to another vessel, rinse the container with solvent to be sure the transfer is quantitative. The rinsing should be transferred to the second vessel along with the rest of the mixture or solution. Titration\nA titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A burette is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction. COMMON LABORATORY EXPERIMENTAL PROCEDURES\n1. BUNSEN BURNER (i) Close the air hole and turn the gas tap full on. Light the gas and hold a piece of wire in different parts of the flame, moving it from the bottom to the top. Note the hottest place in the flame. Open the air hole. Again hold the wire in the flame, moving from the bottom to the top. Note the hottest place in the flame. Compare the two flames and note which has the hottest point. (ii) Close the air hole. Hold a test-tube with its bottom end just above the flame. Carbon deposits on the glass.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.911918998545554, "ocr_used": false, "chunk_length": 2343, "token_count": 510}}
{"text": "(ii) Close the air hole. Hold a test-tube with its bottom end just above the flame. Carbon deposits on the glass. To test whether unburned carbon gives the yellow colour to the flame, sprinkle powdered charcoal into the flame to see if this gives the same effect. (iii) Open the air hole again. Note whether carbon deposits on a test-tube held in this flame. Air mixing with the gas helps it to burn more rapidly and efficiently. To test what is in the cooler inner cone hold a splint of wood in the flame so that it passes through the inner cone. Note which part of the splint burns. Hold a piece of glass tubing with one end in the inner cone then ignite the gas that comes out of the other end. (iv) Investigate a candle flame and the flame of a spirit lamp in a similar way. Find the hottest part of the flame. Test for unburned carbon particles in the flame. Look for an inner cone of unburned gases. IDENTIFICATION OF PURE SUBSTANCES\n2. Melting points, mp, of naphthalene (i) Put a very small amount of naphthalene in a capillary tube sealed at one end. You can pull out the capillary tube from heated glass tubing. Use a rubber band to attach the capillary tube, sealed end down, to a thermometer. Heat a thermometer and capillary tube in a beaker of water on a tripod. Use the thermometer to stir the water but do not let water enter the capillary tube. Slowly raise the temperature of the water. Note the temperature of the naphthalene when it melts. Let it cool and note the temperature when the naphthalene solidifies Calculate the average of these two values. (ii) Use a clean test-tube and thermometer to repeat the experiment using stearic acid, mp 69o, or use any other substance, mp < 100oC. 3. Impurities affect the melting point of a substance Mix stearic acid with the naphthalene, thus making the naphthalene impure. Look for changes in the melting point. Impurities lower the melting point. 4. Boiling point of water We can identify a pure substance from its melting point or boiling point. Put water in a test-tube and hold a thermometer with the bulb just under the water. Add boiling chips to prevent bumping. Bring the water to the boil with a very small flame and read the thermometer.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9061058344640435, "ocr_used": false, "chunk_length": 2211, "token_count": 514}}
{"text": "Put water in a test-tube and hold a thermometer with the bulb just under the water. Add boiling chips to prevent bumping. Bring the water to the boil with a very small flame and read the thermometer. Note any change in the reading if the thermometer touches the bottom of the test-tube. 5. Boiling point of inflammable liquids (i) Use a different method of heating inflammable liquids, e.g. ethanol, bp 78.4oC and acetone, bp 56oC. Put 2 cm of the inflammable liquid in a test-tube. Put the test-tube in an empty beaker. Put a thermometer into the test-tube with the bulb in the liquid. Boil water in an electric jug or on an electrical hot plate. Pour the hot water into the beaker so that the level is higher than the inflammable liquid in the test-tube. Stir the inflammable liquid gently with the thermometer and read the thermometer when the inflammable liquid boils. (ii) Use a very small test-tube or seal one end of a piece of glass tubing, 8 cm length and 3 cm external diameter. Put the inflammable liquid into this tube. Put a capillary tube, sealed at one end, into the inflammable liquid with the sealed end up and the open end down in the inflammable liquid. Use a rubber band to attach the tube containing inflammable and capillary tube to the bulb of a thermometer. Hold the apparatus in a beaker of water and heat gently with a Bunsen burner flame. When the temperature rises, bubbles slowly come out of the capillary tube. At the boiling point the bubbles suddenly come out as a steady stream. Read the temperature. Let the water cool and read the temperature again when the steady stream of bubbles ceases. Calculate the boiling point as the average of the two readings. 6. Pressure affects the boiling point Put water in a sidearm test-tube or in a round-bottom flask with a 2-hole stopper. Insert a thermometer through a hole in the stopper so that the bulb of the thermometer reaches, but does not touch, the bottom of the test-tube or flask. Add boiling chips to prevent bumping. Boil the water and read the temperature on the thermometer. Stop heating. Connect a water pump to the sidearm or to the second hole of the 2-hole stopper. When the water stops boiling, turn on the water pump to reduce the pressure.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9013688037343686, "ocr_used": false, "chunk_length": 2234, "token_count": 509}}
{"text": "Stop heating. Connect a water pump to the sidearm or to the second hole of the 2-hole stopper. When the water stops boiling, turn on the water pump to reduce the pressure. Read the temperature, heat to boiling and read the temperature again. 7(a). Solubility in water Test different salts taken to show that each has a different solubility in water. Take 5 g samples and try to dissolve each in 15 mL of water in a test-tube. Attach the stopper then shake each test-tube vigorously for the same time to show that solubility is a characteristic of a particular substance, e.g. sugar, common salt, potassium nitrate, calcium sulphate. The solubility of a salt is usually expressed as the number of grams able to dissolve in 100 g water at 20oC, e.g. ammonium chloride 37.2 g, barium chloride 35.7 g, calcium chloride 42.7 g, copper sulphate 20.7 g, lead nitrate 54.4 g, magnesium sulphate 25.2 g, potassium chloride 34.0 g, potassium iodide 144.0 g, sodium bicarbonate 9.6 g, sodium chloride 36.0 g, sodium hydroxide 109.0 g, sodium nitrate 87.5 g. (b) Solubility of a substance in water at a given temperature Put about 50 cm3 of water in a beaker and add baking powder, sodium bicarbonate, gradually while stirring. Potassium sulphate is an alternative substance. Make a saturated solution by stirring until no more solute will dissolve. Read the temperature of the saturated solution. Weigh a clean evaporating dish, w1. Add some clear saturated solution and weigh again, w2. Carefully evaporate the solution in the evaporating dish to dryness and weigh again, w3. The mass of the sodium bicarbonate dissolved = w3 - w1. The mass of water = w2 - w1 - w3. Calculate the solubility of the sodium bicarbonate as g per 100 g water at room temperature.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8411965112113525, "ocr_used": false, "chunk_length": 1748, "token_count": 449}}
{"text": "The mass of the sodium bicarbonate dissolved = w3 - w1. The mass of water = w2 - w1 - w3. Calculate the solubility of the sodium bicarbonate as g per 100 g water at room temperature. 8(a) Solubility and particle size To show that small particles dissolve faster than the large particles, add 4 g of coarse able salt to one test-tube half filled with water, and 4 g of fine table salt to a second test-tube containing the same amount of water or use whole and crushed copper sulphate crystals. Stir or shake both tubes equally and simultaneously. Pause after every few seconds to observe the amount of undissolved salt left in each tube. (b) Solubility and solvents (i) Fill two test-tubes one third full with (a) water and (b) methylated spirits. To each test-tube add 1 g sodium chloride, attach a stopper and shake. Sodium chloride dissolves readily in water but not so readily in methylated spirits. (ii) Fill two test-tubes one third full with (a) 88g water and (b) a solution of 1 g potassium iodide in 5 mL water. To each test-tube add iodine crystals, attach a stopper and shake. Sodium chloride dissolves readily in water but not so readily in alcohol\nDETERMINATION OF DENSITY\n9(a) Density of a solid The density of a solid is the ratio of mass to volume. We can find the density of a regular solid with a balance and ruler. We can find the volume of an insoluble irregular solid with a measuring cylinder. Half fill a graduated cylinder with water. Read the volume, immerse the solid in the water and read the volume again. The difference in the two readings is the volume of the solid. (b) Density of a liquid Weigh a small container, fill with liquid and weigh again. Transfer the liquid to a measuring cylinder. Find the density by dividing the mass of the liquid by the volume. Heat of fusion and vaporization\n10(a) Separate by sublimation Separate iodine from a mixture of crystals of iodine and sodium chloride. Heat the mixture in an evaporating dish with a funnel placed over it. The iodine sublimes on to the cool sides of the funnel. SEPARATION OF MIXTURES\n(b) Separate by distillation Put 10 mL ink in a flat bottom conical flask.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8900591857475938, "ocr_used": false, "chunk_length": 2150, "token_count": 511}}
{"text": "Heat the mixture in an evaporating dish with a funnel placed over it. The iodine sublimes on to the cool sides of the funnel. SEPARATION OF MIXTURES\n(b) Separate by distillation Put 10 mL ink in a flat bottom conical flask. Add boiling chips to prevent bumping. Fit a stopper with a delivery tube reaching half way down a collecting test-tube or an U-tube, in a beaker of water. Heat the ink with a Bunsen burner flame. Drops of a colourless liquid appear in the collecting tube. Identify the liquid as water by its action of turning white anhydrous copper sulphate to blue hydrated copper sulphate. Do not allow ink to froth up or splash into the delivery tube. (c) Separate salt and sand Prepare a mixture of salt and sand. Put about 2 mL of the mixture in a test-tube. Add 5 mL water and shake until all the salt has dissolved. Pour the contents of the tube into a filter paper in a funnel over an evaporating basin. Wash the test-tube with water and add this to the filter paper. The sand will remain on the filter paper and may be dried and collected. The salt can be recovered from the filtrate by warming the evaporation basin to drive off the water. (d) Solvent extraction of oil from nuts Put groundnuts (peanuts) or pieces of chopped coconut into a mortar. Add 20 mL of acetone or methylated spirits. Grind the nuts in the solvent as finely as possible. Pour off the liquid into a test-tube and filter into an evaporating basin. Warm the evaporating basin for 10 minutes. The solvent evaporates leaving the oil extracted from the nuts. (e) Separate two immiscible liquids of different density Use a separating funnel or make a separating funnel with a piece of wide tubing fitted with a stopper, tube and rubber tube with a clip. Shake the mixture thoroughly in a closed container then run it into the separating funnel. Wait until a clear boundary appears between the two liquids and then run off the denser layer into a beaker below. 11(b). HEATING AND BURNING DIFFERENT SUBSTANCES Substances that gain mass when heated. Clean 25 cm of magnesium ribbon and cut into 1 cm pieces. Weigh a crucible plus lid, put the pieces of magnesium ribbon in the crucible and weigh again.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9047084253180318, "ocr_used": false, "chunk_length": 2185, "token_count": 502}}
{"text": "HEATING AND BURNING DIFFERENT SUBSTANCES Substances that gain mass when heated. Clean 25 cm of magnesium ribbon and cut into 1 cm pieces. Weigh a crucible plus lid, put the pieces of magnesium ribbon in the crucible and weigh again. Put the crucible on a pipe clay triangle supported on a tripod. Heat very gently then strongly. Hold the crucible lid in a pair of tongs close to the crucible. The magnesium ribbon darkens just before it begins to melt. At the first sign of burning, place the lid on the crucible and remove the Bunsen burner. Every 4 seconds, raise the lid to allow more air to enter but do not allow any white magnesium oxide smoke to escape. When the magnesium stops burning on raising the lid, remove the lid. Heat the crucible again strongly but still hold the crucible lid in a pair of tongs close to the crucible in case the magnesium starts to burn again. Leave the crucible to cool. When cool, weigh the crucible plus lid plus contents. Calculate the increase in mass of the magnesium. (b) Substances that lose mass when heated (i) Weigh a test-tube containing 1 cm potassium permanganate crystals and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the potassium permanganate crystals. Note any loss in mass due to the loss of water of crystallization. (ii) Weigh a test-tube containing 1 cm copper carbonate and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the copper carbonate. Note any loss in mass. (b) Substances that neither gain nor lose mass when heated Weigh a test-tube containing 1 cm dry zinc oxide and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the zinc oxide. Note any loss in mass. (c) Effect of heat on copper sulphate crystals Crush blue copper sulphate crystals and put them into a dry test-tube to a depth of 4 cm. Heat the tube gently.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.900549319776464, "ocr_used": false, "chunk_length": 2122, "token_count": 493}}
{"text": "Note any loss in mass. (c) Effect of heat on copper sulphate crystals Crush blue copper sulphate crystals and put them into a dry test-tube to a depth of 4 cm. Heat the tube gently. Note whether vapour collects on the cooler parts, change of colour from blue to white, and any liquid collecting in the receiving tube. Identity of the liquid by measuring the boiling point. When all the copper sulphate crystals have changed to white and the tube has cooled, hold the tube in your hand and pour the liquid back on to the white crystals. Note whether the blue colour restored and if any heat is given back. Blue copper sulphate crystals + heat -> white, anhydrous copper sulphate + water. This is a reversible change. 12.PREPARING, COLLECTING AND TESTING GASES (b) Hydrogen Be careful! A dangerous explosion may occur if you use any vessel bigger than a small test-tube when igniting the gas, particularly if it is mixed with air. Never dry hydrogen gas with concentrated sulphuric acid! (i) Place a few pieces of granulated zinc or zinc foil from the casing of an old dry cell in a boiling tube; add 2 drops of copper sulphate solution. An alternative to the thistle funnel at A is a syringe as shown at B. Pour enough 1 M sulphuric acid down the thistle funnel on to the zinc to cover the bottom of the funnel tube. Discard the first two or three test tubes of hydrogen, as they will contain displaced air. Collect test tubes of the gas and stopper them. Test the third test-tube of gas by holding a lighted taper or splint over the mouth as soon as you take out the stopper. Pure hydrogen burns with a quiet \"pop\" sound. (ii) Alternatively, add sulphuric acid from a syringe. Gas cannot escape through the syringe so you do not need to cover the tube of the syringe with acid. (iii) Hydrogen burns in air to form water vapour. When hydrogen ignites in a dry test-tube, note any vapour or mist on the sides of the test-tube. (iv) Investigate whether hydrogen is lighter than air by \"pouring\" the gas into a test-tube held either above the first tube or below it. Use a lighted taper to investigate where the hydrogen has gone. (v) Blow soap bubbles by holding the delivery tube of the apparatus in detergent or soap solution.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.905372166855313, "ocr_used": false, "chunk_length": 2225, "token_count": 518}}
{"text": "(iv) Investigate whether hydrogen is lighter than air by \"pouring\" the gas into a test-tube held either above the first tube or below it. Use a lighted taper to investigate where the hydrogen has gone. (v) Blow soap bubbles by holding the delivery tube of the apparatus in detergent or soap solution. The hydrogen bubbles will rise into the air, showing the low density of hydrogen gas. (b) Oxygen (i) Prepare oxygen safely by decomposition of hydrogen peroxide solution. Put 20 mL hydrogen peroxide into a 100 mL bottle. Fix a delivery tube to the bottle. Add two spatulas of manganese dioxide and oxygen bubbles off for collection. (ii) Oxygen is colourless and has no smell. To test whether the test-tube contains oxygen, light a splint of dry wood, blow out the flame leaving a glowing splint then put the glowing splint in a test-tube of oxygen. The glowing splint bursts into flame. This experiment is called the glowing splint test. (iii) Use an L-shape piece of nichrome wire with a shield to fit on the top to protect your hand. Fix steel wool into a loop in the lower end of the Nichrome wire. Heat the steel wool to red hot in a Bunsen burner flame then inserts it quickly into a test-tube of oxygen. (iv) Fix a small piece of charcoal into the loop in the lower end of the Nichrome wire. Ignite the charcoal in the Bunsen burner flame and then insert it quickly into another test-tube of oxygen. (v) Dip the loop in the lower end of the Nichrome wire into sulphur powder. Ignite the sulphur powder in a Bunsen burner flame and then insert it quickly into another test-tube of oxygen. (c) Hydrogen chloride Put rock salt, sodium chloride, into a 100 mL filtering flask. Coarse rock salt causes less frothing than the fine salt. Carefully add concentrated sulphuric acid down the thistle funnel. You can collect hydrogen chloride gas by upward displacement of air, as in the diagram. (i) Collect four test tubes of the gas and cork them. Remove the cork from one of these test tubes under water. Note the solubility of hydrogen chloride. (ii) Hold a piece of cotton wool soaked in ammonium hydroxide at the mouth of a test-tube of hydrogen chloride. (iii) Shake a test-tube of the gas with water to obtain a solution of hydrogen chloride.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9065711700053338, "ocr_used": false, "chunk_length": 2248, "token_count": 516}}
{"text": "Note the solubility of hydrogen chloride. (ii) Hold a piece of cotton wool soaked in ammonium hydroxide at the mouth of a test-tube of hydrogen chloride. (iii) Shake a test-tube of the gas with water to obtain a solution of hydrogen chloride. Test the solution with an acid/base indicator. Put a piece of magnesium ribbon in the solution. Collect any gas formed and test for hydrogen with the glowing splint test. (d) Ammonia (i) Put a mixture of calcium hydroxide and ammonium chloride into a test-tube to a depth of 4 cm. Fill a U-tube with lumps of calcium oxide mixed with cotton wool. The cotton wool is to prevent blocking of the tube. Gently heat the test-tube. The calcium oxide dries the ammonia gas. Test whether the receiver test-tube is full by holding a piece of red litmus paper at the opening. Collect test tubes of ammonia and cork them. The method of collection illustrates that ammonia gas is lighter than air. (ii) Fill a flask with ammonia. Fit a cork and tube into the flask as shown. The tube should have been drawn out into a jet. Warm the flask gently to expand the gas and then hold the flask upside down with the tube in the water. Water will spray into the flask from the jet. (e) Carbon dioxide Many reactions can be used to produce carbon dioxide gas. Marble chips or other carbonate rock treated with dilute acid provides a good source. The gas is not too soluble to be collected by water displacement, as shown above for the preparation of hydrogen. Alternatively, carbon dioxide can be collected by displacing air from dry bottles. To test if the bottle is full, lower a lighted splint or taper into the top of the jar. If the flame is extinguished at the entrance as at (ii), then the jar is full. Put a cardboard cover over the top to prevent diffusion of the gas. Check the density of the carbon dioxide by pouring\" the gas into another bottle either above or below the first bottle. Find where the gas is by testing with a lighted splint. Note: The presence of carbon dioxide can be confirmed by the fact that limewater becomes \"milky\" when the gas is Passed through it. (f) Cooking and carbon dioxide The purpose of baking powder or soda in cooking is to produce tiny bubbles of carbon dioxide. This expands the pastry, cake or dough, making it light and pleasant to eat.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9101484581929957, "ocr_used": false, "chunk_length": 2308, "token_count": 511}}
{"text": "Note: The presence of carbon dioxide can be confirmed by the fact that limewater becomes \"milky\" when the gas is Passed through it. (f) Cooking and carbon dioxide The purpose of baking powder or soda in cooking is to produce tiny bubbles of carbon dioxide. This expands the pastry, cake or dough, making it light and pleasant to eat. Yeast cells do the same thing in bread making, though this takes longer. Baking powder, or sodium bicarbonate, NHCO3, reacts with an acid such as lactic acid from sour milk to produce carbon dioxide. Commercial \"baking powders\" often contain a solid acid, which only reacts with the sodium bicarbonate when moist. (i) Put baking powder into water. Note whether carbon dioxide gas forms. Note whether carbon dioxide forms when you put sodium bicarbonate into water. Add baking powder in a test-tube with vinegar or lemon juice (acetic acid). Note whether carbon dioxide forms. (ii) Make a sugar solution and half fill a jar with this solution. Add a spoonful of yeast and leave to stand for 2 days. Construct a bubbler to fit on the top of the jar. Note whether the yeast forms a gas. Note whether carbon dioxide gas collects in the upper part of the jar. 12(a) RUSTING Take 7 test tubes and 11 clean nails. Prepare the tubes as shown below: Tube 1: Put 2 clean nails in the test-tube and half cover them with distilled water. These nails are in contact with air and water and form the control experiment. Tube 2: Put a few pieces of anhydrous calcium chloride or silica gel in the bottom of a dry test-tube, and also two nails. Put a plug of cotton wool in the top of the tube. These nails are in contact with air, but not moisture. Tube 3: Boil water for several minutes to expel dissolved air and pour into the test-tube whilst hot. Put 2 nails in the water. Put a little Vaseline or a few drops of olive oil on the surface of the hot water. The Vaseline will melt and form an airtight layer, solidifying as the water-cools. These nails are in contact with water but not air. Tube 4: Half cover 2 nails with water containing a little common salt dissolved in it. These nails are in contact with air, water and salt. Tube 5: Wrap a piece of zinc foil round part of a nail.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8933433531888558, "ocr_used": false, "chunk_length": 2207, "token_count": 514}}
{"text": "Tube 4: Half cover 2 nails with water containing a little common salt dissolved in it. These nails are in contact with air, water and salt. Tube 5: Wrap a piece of zinc foil round part of a nail. Put the nail in the test-tube and almost cover with tap water. Tube 6: Wrap a piece of tin foil round part of the nail. Put the nail in the test-tube and add tap water as you did for tube 5. Tube 7: Wrap a piece of copper wire round a nail and put it in the test-tube exactly like tubes 5 and 6. Stand these 7 test tubes in a rack and leave for several days. Note the conditions for rusting and which metal, (zinc, copper or tin), is best at preventing rusting. (b) INCREASE IN MASS OF IRON DURING RUSTING Counterbalance a piece of iron on a knife-edge, using a brass weight or stone. Leave in moist air or on a window ledge for a few days and note the effect of the rust on the longer arm of the lever. During rusting, metallic ion changes to Fe (OH) 3.xH2O. (c) WHAT COMBINES WITH IRON DURING RUSTING Moisten the inside of a test-tube with water, sprinkle into it a spatula measure of iron filings and rotate it horizontally so that the filings spread and adhere to the walls. Alternatively, push a small plug of moistened iron wool to the bottom of the tube. Invert the test-tube in a beaker about one third full of water. Use the beaker lip to support the tube. The water levels inside and outside the tube should be the same and the level should be marked on the tube. Leave the tube in this position for a few days. The iron will rust and the water level will rise up inside the tube, finally becoming steady. Again add water to the beaker until the levels inside and outside the tube are the same and mark the new level. It will be seen that one fifth of the air volume has been used up, suggesting that oxygen has been used up in the rusting of iron. The residual gas does not support combustion of a lighted splint.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8928696679859279, "ocr_used": false, "chunk_length": 1920, "token_count": 466}}
{"text": "Again add water to the beaker until the levels inside and outside the tube are the same and mark the new level. It will be seen that one fifth of the air volume has been used up, suggesting that oxygen has been used up in the rusting of iron. The residual gas does not support combustion of a lighted splint. 13 COLOURED EXTRACTS FROM FLOWERS AS INDICATORS OF ACIDS AND BASES (a) Extract coloured substances from plants Select brightly coloured flowers, such as the purple and red bougainvillaea, or coloured leaves. Squeeze or grind one of the coloured flowers or leaves in a mortar with a mixture made of 2 mL acetone and 2 mL ethanol. By this means the colouring matter will be extracted into the solvent. Filter and collect the filtrate. Repeat this experiment with one or two other flower colours. Keep these coloured solutions for use as \"indicators\" in the next experiment. (b) Plant extracts to indicate whether a substance is acidic or basic (i) Put a spot of the coloured flower extract on to a filter paper and leave to dry. Put one drop of lemon juice on to the spot then note any change of colour. Repeat the experiment with other fruit juices and vinegar. They are acidic substances. Note any colour change with dilute hydrochloric acid. Different colours suggest that some substances are more acidic than others. (ii) Put some of the original filtrate on to another piece of filter paper. When dry, note the colours given by sodium bicarbonate solution, washing soda, limewater and a dilute solution of sodium hydroxide. These are alkaline, or basic, substances. Note whether they all give the same colour: Plant extracts can act as indicators to test whether a substance is acidic or basic. (iii) Add a few drops of sodium bicarbonate solution to 1 mL of flower extract indicator in a test-tube. Then add lemon juice and note any colour change. (iv) Repeat the experiment with limewater and indicator followed by dilute hydrochloric acid. Note any colour change. Note whether you can get back the original colour by adding more limewater. Note how many times you can change the indicator colour before the test-tube is full. (v) Litmus, an extract of lichens, is another plant indicator. An acidic solution turns blue litmus red. An alkaline solution turns red litmus blue.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.911633071763138, "ocr_used": false, "chunk_length": 2289, "token_count": 501}}
{"text": "(v) Litmus, an extract of lichens, is another plant indicator. An acidic solution turns blue litmus red. An alkaline solution turns red litmus blue. (v) Universal Indicator can be in the form of a solution or dried on filter paper. Universal indicator not only indicates whether a substance is acidic or basic but also how acidic it is. Investigate the effect of Universal Indicator on the solutions above. To avoid using the name of a colour to indicate acidity, we use a scale of numbers from 0 to 14 called the pH scale. We can use the pH scale to express the degree of acidity. Acid solutions have a pH value less than 7. Alkaline or basic solutions have a pH value greater than 7. Solutions with a pH value of 7 are neither acidic nor basic, they are neutral. On the bottle or packet of Universal indicator, a colour chart shows the colour and the pH value associated with this colour as follows: (a) Colour (b) pH (c) Acid/Base (a) Red (b) 1-3 (c) very acidic (a) Orange (b) 4-5 (c) weak acid (a) Yellow (b) 6 (c) very weak acid (a) Green (b) 7 (c) neutral (a) Blue (b) 8 (c) very weak base (a) Indigo (b) 9-10 (c) weak base (a) Violet (b) 11- 14 (c) very basic Use 2 drops of Universal Indicator to 10 mL of solution to be tested. Test the pH value of lemon juice, vinegar, sodium bicarbonate solution, washing soda, lime water, sodium hydroxide solution, tap water, distilled water. 14. CRYSTALS(a) Crystal growth Sodium thiosulphate crystals grow rapidly from a super-saturated aqueous solution. The formula for the crystals is Na2S2O3.10H2O. On heating, these crystals dissolve in some of their water of crystallization. Put 3 cm of sodium thiosulphate crystals in a test-tube. Add 2 drops of water. Heat gently until all the crystals have dissolved. They appear to \"melt\". Leave to cool. Crystals may not form unless you drop a tiny seed crystal of sodium thiosulphate into the solution. Then crystal growth commences and spreads rapidly through the whole solution.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8662625665286813, "ocr_used": false, "chunk_length": 1976, "token_count": 513}}
{"text": "Leave to cool. Crystals may not form unless you drop a tiny seed crystal of sodium thiosulphate into the solution. Then crystal growth commences and spreads rapidly through the whole solution. Watch the growth from one centre. Hold the tube in the hand while crystallization occurs. (b) Crystals of naphthalene grow from the melt Put a little naphthalene on a glass slide. Hold over a flame until the crystals melt. Put a cover slip over the liquid and allow to cool. Watch the crystals grow using a hand lens. Sometimes crystals will grow from several points simultaneously to make boundaries where they meet. Draw the shape of the boundary between the forming crystals and the melt. View the crystals through Polaroid filters. (c) Grow large crystals (i) Use a 0.5 - 0.8 cm long seed crystal to start growing large crystals. Make seed crystals by slow evaporation of 30 mL of saturated solution in a glass dish. Dry a selected seed crystal. Tie a piece of clean cotton around it without touching the seed crystal with your hands because impurities easily affect the size and shape of the crystal. Hang the seed crystal 5 cm from the base of the container with a bent wire. Fill a jar (with a screw-on lid) with a solution of the salt less than saturation strength before you put the seed crystal in position. If little crystals grow on the surface of the seed crystal, then screw on the lid of the jar to make the little crystals dissolve. Prevent crystals growing on the sides of a crystallizing dish by rubbing Vaseline round the upper inside rim. Evaporation may be increased by sitting the crystal growing jar on a tin with a 5 watt bulb mounted inside it. An air flow over the solution surface given by a fan will also hasten crystal growth. You can preserve large crystals by painting with a clear varnish. (ii) You can also support the seed crystal at the end of a glass tube. Heat a 3 mm bore piece of glass tubing in a flame until the end softens sufficiently to squeeze with pliers to make a smaller hole. When the glass tubing cools, drop seed crystals on its end until one catches in the smaller hole. Keep this crystal in place by dropping other crystals on it. Support the glass tubing vertically so that the seed crystal at the end is immersed in the solution of the salt. The seed crystal should then grow. Turn the crystal regularly so that growth on all faces is equal. 15.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9061865390684338, "ocr_used": false, "chunk_length": 2393, "token_count": 511}}
{"text": "The seed crystal should then grow. Turn the crystal regularly so that growth on all faces is equal. 15. DIFFUSION\n(a) Diffusion of heavy carbon dioxide gas upwards (i) Fill a jar with carbon dioxide and invert it over a similar jar full of air. After a few moments separate the jars, pour a little lime water in the lower one and shake it. The lime water will turn milky indicating that the carbon dioxide has fallen into the lower jar because it is the heavier gas. (ii) Repeat the experiment with the carbon dioxide in the lower jar and invert a jar of air on top of it. If the jars are left for about 5 minutes carbon dioxide will be carried into the upper jar by diffusion; in the same way air will be carried into the lower jar. The lime water test will show the presence of carbon dioxide in the upper jar. (b) Diffusion rates of ammonia and hydrogen chloride gases The long glass tube should be horizontal. Corks should fit at both ends. Using a pair of tongs or tweezers, dip a piece of cotton wool into concentrated hydrochloric acid and another piece into concentrated ammonium hydroxide. Drain off excess liquid. As nearly as possible at the same time, put the ammonia cotton wool at one end of the tube and the acid cotton wool at the other. Close the ends of the tube with corks. After a while, look carefully for a white ring which will form where the ammonia gas and the hydrogen chloride gas meet after diffusing through the air towards each other. Ammonia is the less dense gas and the white ring of ammonium chloride should form nearer to the hydrogen chloride end than from the ammonia end of the tube. (c) Diffusion of liquids (i) Place a crystal of potassium dichromate, potassium dichromate (VI), or ammonium dichromate at the bottom of a beaker of water. To do this, put a glass tube into the beaker of water so that it touches the bottom, then to drop the crystal down the tube. Close the top of the tube with your finger and remove the tube gently, leaving the crystal in the beaker. The colour of the dissolving crystal will spread throughout the water in quite a short time. (ii) Fill a very small open bottle with a strong solution of potassium permanganate, potassium manganate (VII). Place this in a larger jar.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9105878255569705, "ocr_used": false, "chunk_length": 2241, "token_count": 498}}
{"text": "The colour of the dissolving crystal will spread throughout the water in quite a short time. (ii) Fill a very small open bottle with a strong solution of potassium permanganate, potassium manganate (VII). Place this in a larger jar. Fill the larger jar very carefully by pouring water down the side until the water level is above the top of the small bottle. Leave this for a few days. The potassium permanganate solution diffuses evenly through the water. (d) Size of a molecule We select an oil molecule because it has a density less than the density of water. Oil floats on the surface and does not dissolve in the water. If the water has a large enough surface area, we assumed that thin oil will spread out in a layer one molecule thick called a monomolecular layer and not form little \"hills\" of molecules. If we know the original volume of oil and the surface area that it forms, then we can calculate the thickness of a monomolecular layer dividing the volume by the area. Use a tray with area > 30 cm2 so as not to restrict the oil film. Sprinkle the surface of the water with a very fine light powder, e.g. talc powder. When you put oil on the water, it pushes the powder aside so you can easily see the area covered by the oil. Pour thin petroleum distillate oil into a burette. Find the volume of fifty drops by running oil from the burette drop by drop and counting the drops. Allow one more drop to fall on a piece of plastic. Touch the oil drop with the point of a glass rod and then touch the prepared water surface. The oil spreads out. Measure the approximate area over which it spreads. Estimate what fraction of oil was removed by the glass point by using the glass point to remove successive fractions from the drop until it has been used up. The volume of oil put on the water can be calculated and an estimate made of the thickness of the oil layer, about 10-6 mm. This is an approximate dimension of a single molecule of the oil. 16. ELECTRICITY\n(a) Solids that conduct electricity The source of the DC supply can be dry cells in series giving 6 volts. The bulb, which should be low power, indicates when the current is flowing. The electrodes may be carbon or steel, perhaps mounted in a wooden support, cork or rubber stopper so as to keep the electrodes a constant distance apart.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.906126248491253, "ocr_used": false, "chunk_length": 2307, "token_count": 504}}
{"text": "ELECTRICITY\n(a) Solids that conduct electricity The source of the DC supply can be dry cells in series giving 6 volts. The bulb, which should be low power, indicates when the current is flowing. The electrodes may be carbon or steel, perhaps mounted in a wooden support, cork or rubber stopper so as to keep the electrodes a constant distance apart. (i) Test the conductivity of solids by making a good contact between the surface of the solid and the two electrodes. The surface of the solid must first be cleaned. All metals conduct electricity. Carbon conducts electricity. Note whether non-metallic solids, e.g. plastics, naphthalene, wax, sugar, sodium chloride and sulphur, conduct electricity. (ii) Glass can be a conductor. Heat a glass rod until it becomes very hot and begins to soften. Test the hot, soft part with the conductivity apparatus. When molten, glass is a good conductor of electricity. (b) LIQUIDS THAT CONDUCT ELECTRICITY (i) First test liquids obtained by melting substances. Melt the following substances, but heat very gently and cautiously because otherwise they may ignite and burn: sulphur, wax, naphthalene, polyethylene material, tin, lead and, if available, a low melting point salt such as lead bromide, m.p. 488oC, or potassium iodide, m.p. 682oC . Test the conductivity of the melt by dipping in the electrodes and waiting a few moments for the electrodes to reach the same temperature. This ensures that the electrodes are in contact with the liquid and not the solidified melt. Scrape and clean the electrodes between each test. (ii) Test ethanol, or methylated spirits, acetone, carbon (IV) chloride, vinegar, sugar solution, copper (II) sulphate solution, sodium chloride solution, and other substances dissolved in water. Clean and dry the electrodes between each test. (iii) Test pure distilled water for conductivity. Put the electrodes into a beaker of distilled water. The bulb does not light up so pure water does not conduct electricity. Very gradually stir small crystals of common salt into the water. Note any changes to the bulb as the salt dissolves.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9077186305440099, "ocr_used": false, "chunk_length": 2102, "token_count": 457}}
{"text": "The bulb does not light up so pure water does not conduct electricity. Very gradually stir small crystals of common salt into the water. Note any changes to the bulb as the salt dissolves. Classify substances into the following groups: (a) those which conduct electricity in the solid state and those which do not; (b) those which conduct in the liquid state and those which do not; (c) those which conduct when dissolved in water and those which do not\n(c) ELECTROLYSIS OF LEAD BROMIDE There are very few suitable low melting point salts. Lead bromide has a low melting point and makes an interesting electrolysis experiment. Potassium bromide may have too high a melting point, 682oC, to melt easily. The lead bromide is melted in a 100 mL hard glass beaker, or in a crucible. The carbon electrodes are supported by a strip of wood with two holes bored 2 cm apart for the electrodes. Connect crocodile clips to the rods and complete the circuit with a torch bulb, to indicate when a current is flowing, and a 12-volt torch battery or cells wired in series. The electrodes can be labelled positive and negative. The only ions present in this melt are the bromide and lead ions. Bromine is readily seen coming off at the positive electrode, which is the anode. The fact that bromine appears only at the positive electrode helps in the understanding of the existence of a negative bromide ion. Lead has both a lower melting point and a greater density than lead bromide and therefore appears as a melt at the bottom of the beaker. The small globule of lead, which accumulates at the negative electrode, the cathode, can be seen after about 10 minutes of electrolysis. Decant off the molten lead bromide carefully into another crucible. The electric current has split up crystalline lead bromide into bromine gas and lead metal. (d) ELECTROLYSIS In aqueous solutions there are usually four ions present, two from the water and two from the dissolved salt. The products will be gaseous, or metals which are deposited on the negative electrode. It is composed of an open cylinder of glass approximately 8 cm high and 2.5 cm in diameter. A small bottle of similar size with the bottom cut off would do just as well.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9067752017539767, "ocr_used": false, "chunk_length": 2210, "token_count": 486}}
{"text": "The products will be gaseous, or metals which are deposited on the negative electrode. It is composed of an open cylinder of glass approximately 8 cm high and 2.5 cm in diameter. A small bottle of similar size with the bottom cut off would do just as well. The cylinder has a 2-hole rubber stopper carrying two carbon electrodes with connecting leads to a battery, or DC supply of 4 to 6 volts. If cork is used, this must be made leak proof by covering the whole of the bottom surface round the electrodes and the glass edge with Faraday's wax or a similar soft wax. The electrodes may be carbon rods from a dry cell or pencil leads. The alloy supports for the coiled filament in electric light bulbs have also been found suitable for electrodes. The electrodes should project about 2 cm into the cylinder and also 2 cm below for attaching the leads to the battery. Pencil leads are brittle, and if they are used it is better to fix the electrodes in the following way. Solder a piece of stout copper wire to a 4 cm length of braided copper screening wire. Drill two holes in the rubber stopper with a 1 mm drill. Insert the copper wire into the hole from above and pull it right through the stopper until the screening wire is also pulled a little way into the hole. Into the core of this screening wire insert the pencil lead securely. Then pull the screening wire with the lead further into the stopper so that the lead electrode is firmly held in the stopper. The excess copper wire is cut off. This procedure is repeated with the other electrode. The solution is placed in the glass cylinder. The two small tubes are then filled with the solution, and carefully inverted over the electrodes. The electrodes are connected to a safe DC supply with a small bulb in series. Increase the voltage until the bulb lights, showing that a current is flowing. When this happens, cut out the bulb from the circuit by closing the switch, as shown. This will allow a larger current to flow. The tubes collect any gas given off and the properties of the gas should be tested. Using carbon electrodes, the following results will be found. (d) ELECTROLYSIS OF WATER Pure water does not conduct electricity. For this reason 2 or 3 mL of dilute sulphuric acid or dilute sodium sulphate solution is added to the water in the electrolysis cell.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9038526692802865, "ocr_used": false, "chunk_length": 2328, "token_count": 497}}
{"text": "Using carbon electrodes, the following results will be found. (d) ELECTROLYSIS OF WATER Pure water does not conduct electricity. For this reason 2 or 3 mL of dilute sulphuric acid or dilute sodium sulphate solution is added to the water in the electrolysis cell. Connect the cell to the DC supply and watch for bubbles of gas at both electrodes. If none appear, add a little more acid or sodium sulphate solution. After 5 to 10 minutes there should be enough hydrogen and oxygen gas for testing. Predict at which electrode each gas will appear. (e) Electrolysis of solutions of ionic salts Most ionic salts can be used satisfactorily in electrolysis. Concentrations of 1M or less are suitable. Potassium iodide gives iodine at the anode and hydrogen gas at the cathode. Zinc sulphate gives a spongy mass of zinc at the cathode, oxygen gas at the anode. Lead acetate deposits lead on the cathode and oxygen gas is produced at the anode. If the lead acetate solution is cloudy, add a few drops of acetic acid. Sodium chloride gives hydrogen gas at the cathode and chlorine gas at the anode. Copper sulphate deposits copper at the cathode and oxygen gas is produced at the anode. 17. COMMON REACTIONS\n(a) DISPLACEMENT OF COPPER FROM AQUEOUS SOLUTION OF COPPER IONS (i) A metal higher in the activity order can displace copper metal from a solution of copper ions. Put 10 mL of molar copper sulphate solution in a small beaker. Clean magnesium ribbon and cut into 0.5 cm pieces. Add these pieces to the copper sulphate solution one at a time. The reaction can be vigorous! Copper metal deposits and the blue colour gradually disappear as the magnesium displaces the copper ion. Note any heat given out by the reaction. When the solution is colourless, decant the solution from the red copper powder at the bottom of the beaker. Collect the copper and dry it. Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s) (ii) Repeat the experiment by attempting to displace copper metal using powdered zinc and iron metal. Note the comparative activity of the metals.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8989893433219533, "ocr_used": false, "chunk_length": 2038, "token_count": 477}}
{"text": "Collect the copper and dry it. Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s) (ii) Repeat the experiment by attempting to displace copper metal using powdered zinc and iron metal. Note the comparative activity of the metals. (b) REACTION OF SODIUM WITH WATER (i) A very safe way of demonstrating the reaction of sodium and water is to drop a very small piece of sodium into a swimming pool. (ii) Pour a 2 cm layer of kerosene on to the surface of water in a test-tube. Drop a 3 mm diameter piece of sodium into the kerosene. Be careful! Sodium sinks in the kerosene and float in the water. Adjust the layer of kerosene to be shallow enough to allow the top of the sodium to protrude above the surface. This reaction between sodium and the water is much slower than if the sodium had been dropped directly on to the water. You can watch the reaction through a hand lens held at the side, but never at the top. Sodium metal is lighter than water but heavier than kerosene. A small area of the sodium suddenly reacts causing a stream of bubbles to appear. The stream of bubbles at one side causes movement. The irregular shape of the sodium changes to a sphere. The sodium melts because the reaction gives off heat. Note any variations in light refraction and reflection below the sodium that indicates something dissolving in the water. Slight smoke where the hot sodium is above the kerosene level suggests a slight reaction with air. Test the gas bubbles for oxygen or hydrogen. (c) DISPLACEMENT OF HYDROGEN FROM ACIDS BY METALS (i) Pour 5 cm of the acids in the table below into test-tubes. Place a piece of metal foil in each test-tube. Note the formation of hydrogen and compare the different rates at which the bubbles are formed.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9024674989872206, "ocr_used": false, "chunk_length": 1724, "token_count": 395}}
{"text": "(c) DISPLACEMENT OF HYDROGEN FROM ACIDS BY METALS (i) Pour 5 cm of the acids in the table below into test-tubes. Place a piece of metal foil in each test-tube. Note the formation of hydrogen and compare the different rates at which the bubbles are formed. Rate of formation of hydrogen gas - very rapid, rapid, slight, very slight, none Metal (b) 3M hydrochloric acid (c) 3M sulphuric acid Magnesium (b) Very rapid (c) Rapid Aluminium (b) Slight (c) None Zinc (b) Moderate (c) Slight Iron (b) Very slight (c) Very slight Tin (b) None (c) None Lead (b) None (c) None Copper (b) None (c) None (ii) Recover the zinc after the reaction has ceased. Evaporate the solution to leave zinc sulphate crystals. Dissolve the colourless zinc sulphate crystals in water and put two carbon electrodes (central poles of dry cell batteries) in the solution. Connect the electrodes to a 6-volt or 12 volt DC supply. Zinc forms rapidly on the cathode. (d) PREPARE SULPHUR DIOXIDE Do the following preparations in a fume cupboard. (i) Prepare sulphur dioxide by burning sulphur in air. Put powdered sulphur in a porcelain jar; ignite it and collecting the gas formed in a funnel. Aspirate the gas into a bottle containing water. (ii) Prepare sulphur dioxide in a generator which allows dilute sulphuric or hydrochloric acid to drip slowly on to sodium sulphate. The acid is contained in a thistle funnel and a tap controls the flow on to sodium sulphite in a suitable flask. The sulphur dioxide produced can be collected in gas jars covered with cardboard discs, which have central holes for the delivery tube. (e) REDUCTION OF POTASSIUM PERMANGANATE WITH SULFUR DIOXIDE (i) Add 10 mL of 0.1M solution of potassium permanganate and 10 mL 3M solution of dilute sulphuric acid to 200 mL of water containing sulphur dioxide. The solution will gradually become colourless as the sulphur dioxide reacts with the permanganate.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.890713593800452, "ocr_used": false, "chunk_length": 1900, "token_count": 487}}
{"text": "The sulphur dioxide produced can be collected in gas jars covered with cardboard discs, which have central holes for the delivery tube. (e) REDUCTION OF POTASSIUM PERMANGANATE WITH SULFUR DIOXIDE (i) Add 10 mL of 0.1M solution of potassium permanganate and 10 mL 3M solution of dilute sulphuric acid to 200 mL of water containing sulphur dioxide. The solution will gradually become colourless as the sulphur dioxide reacts with the permanganate. The experiment can be continued further by stirring in a 0.25M solution of barium chloride when the solution will become \"milky\" due to the formation of barium sulphate. (ii) The generator used in experiment 2.75B is a convenient piece of apparatus for giving a continuous supply of sulphur dioxide for bleaching Bowers and other plants. The gas from the generator is passed through a jar containing the plant, and excess gas is absorbed in water. The colour of the bleached plant can easily be regenerated by placing the plant in a solution of hydrogen peroxide. This experiment could be used as an introduction to the processes of reduction and oxidation. (f) REACTION OF MAGNESIUM WITH CARBON DIOXIDE Fill a gas jar with carbon dioxide as described in experiment 2.38. Hold a piece of clean magnesium ribbon in a pair of tongs; ignite the magnesium with a Bunsen burner flame and plunge it into the carbon dioxide gas. The magnesium continues to burn. If the magnesium is taking oxygen from the carbon dioxide for burning then you would find carbon in the gas jar. Look for carbon specks in the gas jar. To make the carbon more visible, you can add drops of sulphuric acid to remove the magnesium oxide and any unburnt magnesium. (g) TITRATION OF ACIDS AND BASES Measure exactly 20 drops of a dilute acid such as vinegar and put these into a test-tube. Add one drop of indicator; either methyl orange or phenolphthalein is satisfactory. To this mixture of acid and indicator, add a dilute base drop by drop, and count the drops. Within experimental error, it will always take the same number of drops to neutralize the 20 drops of acid provided that the same dropper is used. A teat pipette makes a satisfactory dropper.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9006037805440295, "ocr_used": false, "chunk_length": 2169, "token_count": 500}}
{"text": "To this mixture of acid and indicator, add a dilute base drop by drop, and count the drops. Within experimental error, it will always take the same number of drops to neutralize the 20 drops of acid provided that the same dropper is used. A teat pipette makes a satisfactory dropper. If the concentration of the acid is known, the concentration of the base can be estimated by comparing the numbers of drops of acid and drops of base that just react. (h) MAKE SOAP FROM FATS Soap can be made from many oils and fats. The reaction is a double displacement involving a strong base such as sodium hydroxide and fats. (a) Obtain animal fat from a butcher. Boil this fat in water and the oil will separate on the surface. When cold, the fat will solidify and it can be separated from the water. Melt the fat again and strain through several layers of cloth. (b) Weigh this fat and then weigh out about one third as much sodium hydroxide pellets. Take care not to touch either the solid sodium hydroxide or the solution, because it is very caustic. Heat the fat in an iron saucepan or dish and, when it is molten, slowly add the sodium hydroxide solution with continuous stirring. Heat with a small flame to avoid boiling over. Allow the fat and the sodium hydroxide to boil for 30 minutes. Stir the mixture frequently. (c) The next stage is to weigh out common salt, sodium chloride; about twice the weight of sodium hydroxide used in (b) is needed. After the 30 minutes boiling, stir this salt well into the mixture. Then allow to cool. The soap separates as a layer at the top. Separate this soap from the liquid below, melt and pour into matchboxes where it will solidify again as small bars of soap. 17. ENERGY FROM CHEMICAL REACTIONS The following group of reactions involve ions in aqueous solution. When the water containing the reacting ions becomes hotter, then we have gained this heat and we can make it do work for us. During the reaction the ions have lost this heat, which we have gained. On the other hand, when the water containing the ions becomes colder, it is the ions which have gained the energy and the water has lost an equivalent amount. (a) REACTIONS THAT GIVE OUT HEAT ENERGY Be careful! The reaction is vigorous so do not do the experiment in a stoppered bottle!", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9076939347022344, "ocr_used": false, "chunk_length": 2284, "token_count": 513}}
{"text": "On the other hand, when the water containing the ions becomes colder, it is the ions which have gained the energy and the water has lost an equivalent amount. (a) REACTIONS THAT GIVE OUT HEAT ENERGY Be careful! The reaction is vigorous so do not do the experiment in a stoppered bottle! (i) Put white anhydrous copper sulphate powder to a depth of about 1 cm in a test-tube. Hold a thermometer with the bulb in the powder. Add water drop by drop. Record the changes of the thermometer reading. (ii) Put about 10 mL of strong aqueous copper sulphate solution into a wide test-tube or small beaker. Support a thermometer with the bulb in the solution. Add magnesium powder, or ribbon, a little at a time until the blue colour disappears. Note any changes in the thermometer reading. (iii) To a little water in a wide test-tube, add concentrated sulphuric acid, drop by drop, down the side of the tube. Stir gently with a thermometer after the addition of each drop. Note any changes in the thermometer reading. (b) REACTIONS THAT TAKE IN HEAT ENERGY Put 10 mL of water in a test-tube. Read the temperature of the water. Dissolve about 2 g of potassium nitrate in the water. The temperature should fall through 90oC. This means that, in the process of dissolving in the water, the particles have absorbed energy. This energy has been taken from the surrounding water in the form of heat. A similar result can be obtained by using potassium chloride instead. (c) HEAT OF A NEUTRALIZATION REACTION Dissolve 40 g of sodium hydroxide pellets in water and make up to 500 mL. This is a 2M solution Also prepare 500 mL of a 2M hydrochloric acid solution. Leave the solutions to cool to room temperature. Note the actual temperature of the solutions when cool. Then add the acid to the base quite rapidly and stir with a thermometer. Note the maximum temperature reached. The increase of temperature should be about 13oC. (iii) Since the volume of water has been doubled by adding one solution to the other, the final solution contains 1 mole of OH- (aq) ions which reacted with 1 mole of H+(aq) ions to form 1 mole of water molecules. We must assume that the specific heat of this moderately weak solution is the same as that of water.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8934853820783724, "ocr_used": false, "chunk_length": 2225, "token_count": 515}}
{"text": "The increase of temperature should be about 13oC. (iii) Since the volume of water has been doubled by adding one solution to the other, the final solution contains 1 mole of OH- (aq) ions which reacted with 1 mole of H+(aq) ions to form 1 mole of water molecules. We must assume that the specific heat of this moderately weak solution is the same as that of water. (d) HEAT OF A COPPER DISPLACEMENT REACTION (i) Put 25 mL 0.2 M copper sulphate solution in a 100 mL polythene fitted with a 1-hole stopper and thermometer. Replace the stopper, invert the bottle and shake it gently. Record the temperature of this solution. Turn the bottle the right way up, remove the stopper and add 0.5 g of zinc dust. The quantity of zinc powder is in excess to ensure that all the copper sulphate is used up in the reaction, so some zinc will remain when the reaction stops. Replace the stopper, invert the bottle, and shake gently. Record the highest temperature reached. Calculate the rise of temperature. This rise of temperature in not affected by the volume of 0.2 M copper sulphate used for the experiment. For a 1 M solution, multiply the rise in temperature by 5 (5 X 0.2M = 1.0 M). The reactants lost energy to the solution. The temperature change is usually between 9oC and 10oC. Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s) (ii) Repeat the experiment with 0.5 g of iron powder or iron filings. This amount is again an excess so that all the copper sulphate will be used up in the reaction. The temperature change is usually between 6oC and 7oC. (e) ELECTRICAL ENERGY FROM CHEMICAL REACTIONS - ELECTROCHEMICAL CELLS In the preceding experiment zinc metal became zinc ions and copper ions became copper metal due to transfer of electrons from zinc metal to the copper ion. To get electrical energy these electrons must flow in an external conductor from the zinc to copper. The potential or voltage will reflect the greater activity of zinc over copper. The current flowing will depend on the extent and rate of the reaction. 2.84 Electrical energy from the displacement of copper by zinc Put concentrated copper sulphate solution in a beaker.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8710493899577663, "ocr_used": false, "chunk_length": 2131, "token_count": 512}}
{"text": "The potential or voltage will reflect the greater activity of zinc over copper. The current flowing will depend on the extent and rate of the reaction. 2.84 Electrical energy from the displacement of copper by zinc Put concentrated copper sulphate solution in a beaker. Connect copper foil to the positive terminal of a 5 V voltmeter. Connect zinc foil to the negative terminal. Dip the two metals briefly into the copper sulphate solution. Note any changes in the voltmeter reading. Note the maximum reading. Note any changes at the copper foil and the zinc foil. The voltage falls to zero after a short time because copper deposited on the zinc and caused the reaction to stop. 18. RATE OF REACTION(a) SIZE OF PARTICLES AND RATE OF REACTION Marble chips can be broken up with a hammer and graded into 3 or 4 sizes: (a) coarse powder; (b) pieces about half the size of a rice grain; (c) pieces as large as rice grains; and (d) the original lumps of marble chips. Place four 100 X 16 mm test tubes in a stand. Weigh approximately 2 g of each grade, size, of marble chips and put the four grades separately into each of the four tubes. Obtain four balloons and blow them up several times to stretch them. Put 5 mL of bench hydrochloric acid into each of the four balloons and slip the mouth of the balloon over the top of the tube without letting any acid into the tube. When each balloon is in place, tip the acid into each test-tube at the same time and observe which balloon is the fastest and the slowest to be blown up. The smallest should give the carbon dioxide in the shortest time. Instead of using marble chips you can use granulated zinc. Do not use metals in powder form because the reaction may be too vigorous and even cause an explosion. Be careful! This reaction produces hydrogen gas! Instead of collecting the gas in a balloon or plastic bag, a more accurate method would be to collect the gas in a burette inverted over water and compare the volume of gas given off in unit time for each grade of marble chips. Another accurate method is to stand a conical flask containing the marble chips and acid on a balance and record the loss in mass every half minute. Carbon dioxide is a heavy gas and most balances will enable the loss in mass to be found as the gas escapes.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9027922041659686, "ocr_used": false, "chunk_length": 2286, "token_count": 494}}
{"text": "Instead of collecting the gas in a balloon or plastic bag, a more accurate method would be to collect the gas in a burette inverted over water and compare the volume of gas given off in unit time for each grade of marble chips. Another accurate method is to stand a conical flask containing the marble chips and acid on a balance and record the loss in mass every half minute. Carbon dioxide is a heavy gas and most balances will enable the loss in mass to be found as the gas escapes. (b) CONCENTRATION AND RATE OF REACTION The reaction between sodium thiosulphate and hydrochloric acid can take a noticeable time. Sulphur is produced during the reaction making the solution cloudy. The rate of reaction can be found by finding the time taken to reach a certain degree of cloudiness in the solution. The degree of cloudiness in this case may be defined as the point at which a black cross marked below the reaction vessel can no longer be seen by looking through the solution from above. In this experiment the concentration of sodium thiosulphate is made variable, whilst the concentration of acid is kept constant. Sodium thiosulphate may be bought as \"hypo\" which is used in photography. Make up 500 mL of aqueous solution containing 20 g sodium thiosulphate. 2 M hydrochloric acid is also needed. Bench dilute acid is usually of this strength. Using a measuring cylinder, put 50 mL of thiosulphate solution into a 100 mL beaker. Place the beaker on a black cross marked on a sheet of paper. Add 5 mL of the acid and note the time given by the second hand of a clock. Stir the acid into the solution. Note the time when the cross is no longer visible through the sulphur in the solution. Repeat the experiment with a smaller concentration of thiosulphate. Take 40 mL of thiosulphate solution and add 10 mL of distilled water. Stir and then add 5 mL of acid as before. The time for the cross to become invisible should be greater than for the last experiment. Repeat the experiment using 30 mL, 20 mL and 10 mL of this sulphate mixed with 20 mL, 30 mL and 40 mL of distilled water. Plot concentration of the thiosulphate solution against time taken for the reaction.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.898532094297819, "ocr_used": false, "chunk_length": 2169, "token_count": 490}}
{"text": "The time for the cross to become invisible should be greater than for the last experiment. Repeat the experiment using 30 mL, 20 mL and 10 mL of this sulphate mixed with 20 mL, 30 mL and 40 mL of distilled water. Plot concentration of the thiosulphate solution against time taken for the reaction. Concentration values may be taken as the volume of the original thiosulphate solution used. Since I/time, reciprocal of time, is the measure of the rate of the reaction, plot thiosulphate concentrations against I/time. The equation for the reaction can be written as: Na2S2O3 (aq) + 2HCl(aq) --> H2O (l) + SO2 (g) + S(s)\n(c) TEMPERATURE AND RATE OF A REACTION Use the reaction in experiment 2.92 to investigate the effect of temperature. Put 10 mL of sodium thiosulphate solution into the 100 mL beaker and stir in 40 mL of water. Use this concentration for the series of experiments with the temperature of the solution as the variable. Add 5 mL of acid as before and record the initial time and the temperature of the solution. Record the final time when the black cross below the beaker is no longer visible. Repeat the experiment, each time warming the thiosulphate solution to just over 30oC\n(c) CATAL YSTS AND RATE OF REACTION The variable in this reaction is the substance used as a catalyst in the decomposition of an aqueous solution of hydrogen peroxide. Set up the burette filled with water as in a standard water displacement experiment. 2 mL of 20-volume hydrogen peroxide will give enough oxygen almost to fill the burette. Weigh out 1 g each of copper (II) oxide, nickel oxide, manganese (lV) oxide and zinc oxide. Put 50 mL of water in the flask and add 2 mL of hydrogen peroxide solution. Add the 1 g of copper oxide. Immediately insert the bung with the delivery tube into the flask. Time the volume of oxygen given off at intervals of 15 seconds. Plot the volume of oxygen produced every 15 seconds against the time of the reaction. Repeat the experiment using the other oxides as catalysts. Plot a graph for each experiment. Manganese (IV) oxide is usually used as a catalyst in this reaction. The catalyst is not used up during the reaction.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8783347838700318, "ocr_used": false, "chunk_length": 2160, "token_count": 512}}
{"text": "Plot a graph for each experiment. Manganese (IV) oxide is usually used as a catalyst in this reaction. The catalyst is not used up during the reaction. A catalyst may slow down a reaction as well as speed it up. 19. BREAKDOWN OF STARCH TO SUGAR FEHLING'S TEST FOR REDUCING SUGARS Starch can be recognized by the deep blue colour which develops when it is in contact with iodine solution. This is a very sensitive test. Sugar does not react with iodine, but sugar will reduce copper (II) in Fehling's solution to red copper (I) oxide, and this is also a sensitive test. Starch does not react with Fehling's solution. Saliva contains enzyme catalysts, which convert starch to sugar. This experiment investigates the progress of this reaction. Put about 10 mL of dilute starch solution into a test-tube. Add to this 1 mL of saliva and stir this into the starch solution. Record the time of adding the saliva. At 5-minute intervals, remove 2 or 3 drops by means of a dropper and put them on a clean white tile, taking care to keep them from running into each other. The dropper must be washed well between each test. Put a little iodine solution on each drop. The decreasing intensity of the blue colour indicates that starch is being used up. Test for increasing amounts of sugar at the same time as testing for starch. To do this, put 2 or 3 drops of the reaction mixture into a small test-tube. Add 3 mL of Fehling's solution and warm this mixture almost to boiling point. The test should show that the amount of sugar is increasing. The enzyme in the saliva is therefore slowly breaking starch down into sugar, which is a smaller molecule. In a previous experiment yeast was used to break down sugar into ethanol, which is an even smaller molecule. Living yeast, which is a variety of fungus, produces enzymes which act as catalysts in the conversion: C6H12O6 -> 2C2H5OH + 2CO2\n20. BREAKDOWN OF ETHANOL TO ETHENE (ETHYLENE) Absorb ethanol on to cotton wool or asbestos wool and push this to the bottom of a hard glass test-tube. In the middle of the test-tube pack small pieces of unglazed porcelain. Fit a delivery tube to collect ethene gas over water.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8950573132244298, "ocr_used": false, "chunk_length": 2154, "token_count": 506}}
{"text": "BREAKDOWN OF ETHANOL TO ETHENE (ETHYLENE) Absorb ethanol on to cotton wool or asbestos wool and push this to the bottom of a hard glass test-tube. In the middle of the test-tube pack small pieces of unglazed porcelain. Fit a delivery tube to collect ethene gas over water. Have 3 test-tubes ready to collect ethene. First heat the porous pot strongly and then gently warm the cotton wool to produce ethanol vapour. This vapour will break down over the hot porous pot to produce ethene gas and water vapour. The ethene is insoluble in water, unlike ethanol, and will collect in the test-tubes. Test the 3 samples by: (a) burning ethene; (b) shaking with a few drops of dilute potassium permanganate solution made alkaline with sodium carbonate solution, the colour should disappear; (c) shaking with a little bromine water, the colour again disappears. Disconnect the delivery tube when you stop heating to avoid a suck back of water onto the hot porous pot. 20. BREAKDOWN OF POLYMERS Usually the smallest molecules are gaseous or liquid at room temperature and the large molecules are solids. Perspex and polystyrene are solid polymers, which can be broken down to smaller molecules by heat. Put pieces of Perspex or polystyrene in a hard glass test-tube. Connect a delivery tube as 2.97 From large molecules to small molecules A Perspex or polystyrene B receiving tube c cold water D a liquid collects. The collecting test-tube must be cooled thoroughly with cold water, as the fumes are harmful. Gently heat the test-tube containing the Perspex. The polymer will melt and give off vapours, which are collected in the receiving tube. Heating must be carefully controlled to enable all the fumes to be condensed in the receiving tube. A liquid is obtained. This suggests that the polymer has been broken down by heat to smaller molecules. The liquid does not return to the solid state unless a catalyst is used. The specific catalysts are not usually available in school laboratories. GENERAL EXAMINATION REQUIREMENTS\nApparatus List\nThis list given below has been drawn up in order to give guidance to schools concerning the apparatus that is expected to be generally available per student for examination purposes.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9138682099803084, "ocr_used": false, "chunk_length": 2215, "token_count": 487}}
{"text": "The liquid does not return to the solid state unless a catalyst is used. The specific catalysts are not usually available in school laboratories. GENERAL EXAMINATION REQUIREMENTS\nApparatus List\nThis list given below has been drawn up in order to give guidance to schools concerning the apparatus that is expected to be generally available per student for examination purposes. The list is not intended to be exhaustive: in particular, items (such as Bunsen burners, tripods) that are commonly regarded as standard equipment in a chemical laboratory are not included in this list. One burette, 50 cm3\nA measuring cylinder, 50 cm³ or 25 cm³\nA filter funnel\nOne pipette, 25cm3\nTwo conical flasks within the range 150cm3 0r 250cm3\nA beaker, squat form with lip: 250 cm³\nA thermometer, -10°C to + 110°C at 1 °C\nA polystyrene or other plastic beaker of approximate capacity 150 cm³\nClocks (or wall-clock) to measure to an accuracy of about 1s. (Where clocks are specified, candidates may use their own wristwatch if they prefer)\nWash bottle\nTest tubes (some of which should be Pyrex or hard glass). Approximately\n125 mm x 16 mm\nBoiling tubes, approximately 150 mm x 25 mm\nStirring rod\nACCURACY\nUnless a question instructs candidates differently they should assume that readings from equipment and apparatus ought to be made with the following precision:\nBurette readings should be to the nearest 0.05 cm3\nWeighings should be made to 0.01 g or 0.001 g depending on the precision of the balance\n0 to 1000C thermometers should be read to the nearest 0.50C and 0 to 500C thermometers to the nearest 0.20C\nTimers will normally be read to the nearest second. It is important that when candidates record reading they include the appropriate number of decimal places. For example a burette reading of exactly 24.7 cm3 should be recorded in a results table as 24.70cm3. A temperature of reading of exactly 350C should be recorded as 35.50C.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8469439813917427, "ocr_used": false, "chunk_length": 1925, "token_count": 481}}
{"text": "It is important that when candidates record reading they include the appropriate number of decimal places. For example a burette reading of exactly 24.7 cm3 should be recorded in a results table as 24.70cm3. A temperature of reading of exactly 350C should be recorded as 35.50C. When titres have to be averaged, it is important that the mean is expressed to either the nearest 0.05cm3 or to the second decimal point. For example, if a candidate records four different titres as listed below, the mean can be worked out. 26.50cm3 26.25cm3 26.60cm3 26.65cm3\nThe candidate is expected to ignore the second titre and average the remaining three\n26.50 + 26.60 + 26.65 = 26.583 this should be recorded as 26.60cm3\n3\n(to the nearest 0.05cm3)\nIn general, a final should always be given to the same number of significant figures as is suggested in the exercise. A significant proportion of marks for quantitative exercise will be awarded for accuracy. The marks will be awarded by comparing the candidate’s results with the teacher’s reports. CALCULATIONS\nUsually calculations will be structured. Candidates will be led through a series of steps leading to a final value. Since most of the marks for these steps will be for a correct method rather than the numerical answer, it is important that candidates include their working even if this seems to be trivial. No marks can be awarded for an incorrect answer without working but a correct method followed by an incorrect answer will receive credit. GRAPHS\nSome exercises in practical chemistry will require candidates to treat their readings graphically. The question will however instruct the candidate which axes to use for each quantity being plotted. Some useful points for candidates to keep in mind when constructing graphs are listed below. Candidates will normally be instructed to put the dependent variable, the quantity being measured e.g. temperature on the y- axis. The predetermined quantity e.g. volume will be on the x-axis. The scales should be chosen so that the results are spread out as far apart as the size of the grid allows but not at the expense of using a sensible scale. For example using 1cm to represent 3 units might spread the readings better than using 1 cm to represent 4 units but the scale may be difficult to read. It is always advisable ton use even scales.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8728284124946515, "ocr_used": false, "chunk_length": 2337, "token_count": 509}}
{"text": "The scales should be chosen so that the results are spread out as far apart as the size of the grid allows but not at the expense of using a sensible scale. For example using 1cm to represent 3 units might spread the readings better than using 1 cm to represent 4 units but the scale may be difficult to read. It is always advisable ton use even scales. The origin (0,0) need not necessarily be included on either scale if it is not relevant. For example if temperature readings between 210C and 280C are plotted, there is no need to begin the axes at zero. The axes must be clearly labelled with the quantity being plotted e.g. mass and its units e.g. kilograms\nThe points plotted may be joined with a straight line or a smooth curve. Since readings are all subject to experimental error, the line drawn may not necessarily pass through every point. Points should never be joined by a series of short straight lines. Doing a Titration\nBegin by preparing your burette, as described on the burette page. Your burette should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceeding with the titration.Take an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume. Prepare the solution to be analysed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely dissolved. Put a magnetic stirrer in the flask and add indicator. Use the burette to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change colour when the titrant hits the solution in the flask, but the colour change disappears upon stirring.Approach the endpoint more slowly and watch the colour of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the burette; to be sure all titrant is mixed in the flask. As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a Teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9086279176057116, "ocr_used": false, "chunk_length": 2372, "token_count": 505}}
{"text": "You can do this with a rapid spin of a Teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes.If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint. When you have reached the endpoint, read the final volume in the burette and record it in your notebook. Subtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution. Indicators\nAn indicator is a substance used in titrations, which has one colour in the presence of an excess of one reagent and a different colour in the presence of an excess of the other. Examples of indicators include;\nMethyl orange\nPhenolphthalein\nMethyl red\nBromothymol blue\nThe pH scale\nA much more useful measure of the strength of an acid solution was worked out by the Danish biochemist S. Sorensen. He worked in the laboratories of the Carlsberg breweries and was interested in checking the acidity of beer. The scale he introduced was the pH scale. The scale runs from 1 to 14, and the following general rules apply. Rules for the pH scale:\nAcid have a pH less than 7,\nThe more acidic a solution, the lower the pH. Neutral substances, such as pure water, have a pH of 7,\nAlkalis have a pH greater than 7,\nThe more alkaline a solution, the higher the pH. The pH of a solution can be measured in several ways. Universal indicator papers that are sensitive over the full range of values can be used. Some colour changes of common indicators are shown below. Indicator colour changes\nCandidates may be asked to carry out exercises involving:\nSimple quantitative experiments involving the measurement of volumes:\nSpeeds of reaction. Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9102424417031161, "ocr_used": false, "chunk_length": 2314, "token_count": 503}}
{"text": "Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration. Identification of ions and gasses as specified in the curriculum. HEALTH AND SAFETY\nCandidates must follow the health and Safety policy normally operates in their laboratories when carrying out the practical Examination. Eye protection must always be worn. Laboratory overalls are recommended. All substances should be regarded as being potentially toxic and hazardous. Hazard labels (e.g. flammable) should be read and appropriate precautions (e.g. keep liquid away from flame) taken. All substances spilled on the skin should be rinsed off immediately. Chemicals must never be tasted Gases and vapours should never be smelt unless the question instructs the candidates to do so and then only with great care. 2.0 QUANTITATIVE ANALYSIS\n2.1 Molar solutions and volumetric analysis. Amount of substance\nIn quantitative analysis, it is important to measure the amounts or moles of reacting substances accurately. To do that, we must use values of relative atomic masses expressed on a periodic table. If we are given the mass of a compound, we can determine the number of moles. The first step is to calculate the mass of one mole of the compound by summing up the relative atomic masses of the constituent atoms. E.g. 1 mole of calcium carbonate (CaCO3) will have a mass of 40 + 12 + 48 = 100g\n10g of CaCO3 will contain 10/100 = 0.1moles\nconcentration of a solution\nIt is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. So we need a standard way of comparing the concentrations of solutions. The concentration of a solution is determined from the number of moles of solute dissolved in one litre (1dm3) of solution. If one mole of a solute is dissolved in water and the volume of the solution made up to 1litre (1dm3), this solution in known as a molar solution or 1M solution. If two moles are dissolved in 1litre (1dm3), the resulting solution is 2M and so on.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8848039526862609, "ocr_used": false, "chunk_length": 2124, "token_count": 473}}
{"text": "The concentration of a solution is determined from the number of moles of solute dissolved in one litre (1dm3) of solution. If one mole of a solute is dissolved in water and the volume of the solution made up to 1litre (1dm3), this solution in known as a molar solution or 1M solution. If two moles are dissolved in 1litre (1dm3), the resulting solution is 2M and so on. You need to be able to calculate\n(i) The number of moles or mass of substance in an aqueous solution of given volume and concentration\n(ii) The concentration of an aqueous solution given the amount of substance and volume of water. (iii) Use the equation: molarity of Z = moles of Z / volume in dm3\nRemember: moles Z = mass Z / formula mass of Z\nConcentration can also be expressed directly in grams per litre (1dm3). 9.8g of H2SO4 can be expressed either as 0.1M or 9.8g/dm3\nMolarity (M) = g/dm3 / formula mass\nExample 1: 5.95g of potassium bromide were dissolved in 400cm3 of water. Calculate its molarity. [Ar values: K = 39, Br = 80]\nMoles = mass / formula mass, (KBr = 39 + 80 = 119)\nmol KBr = 5.95/119 = 0.05 mol\n400 cm3 = 400/1000 = 0.4 dm3\nmolarity = moles of solute / volume of solution\nmolarity of KBr solution = 0.05/0.4 = 0.125M\n2.3 Volumetric calculations (acid-alkali titrations)\nChemical Equations\nThese balancing numbers have an additional meaning where these amount calculations are concerned. The balancing numbers give the numbers of moles present for each chemical involved in the reaction. For example, sodium chloride may be prepared by the reaction of sodium hydroxide and hydrochloric acid, according to the following equation -\nNaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)\nThe equation specifies what amounts of sodium hydroxide and hydrochloric acid will react together and what amounts of products are produced.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8206263563703089, "ocr_used": false, "chunk_length": 1805, "token_count": 506}}
{"text": "[Ar values: K = 39, Br = 80]\nMoles = mass / formula mass, (KBr = 39 + 80 = 119)\nmol KBr = 5.95/119 = 0.05 mol\n400 cm3 = 400/1000 = 0.4 dm3\nmolarity = moles of solute / volume of solution\nmolarity of KBr solution = 0.05/0.4 = 0.125M\n2.3 Volumetric calculations (acid-alkali titrations)\nChemical Equations\nThese balancing numbers have an additional meaning where these amount calculations are concerned. The balancing numbers give the numbers of moles present for each chemical involved in the reaction. For example, sodium chloride may be prepared by the reaction of sodium hydroxide and hydrochloric acid, according to the following equation -\nNaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)\nThe equation specifies what amounts of sodium hydroxide and hydrochloric acid will react together and what amounts of products are produced. 1NaOH (aq) + 1HCl (aq) 1NaCl (aq) + 1H2O (l)\nThis equation states that 1 mol of sodium hydroxide and 1 mol of hydrochloric acid will react together to give of 1 mol sodium chloride and of 1 mol water. The ratio of NaOH: HCl: NaCl: H2O is 1: 1: 1: 1. What this means is that if the amount of any one of the components in the above reaction is known then the others may be worked out from the ratio above. Or, if the masses of chemicals reacting together are known then amounts can be calculated and the balancing numbers deduced from the amounts reacting together. 2.4 Titration Experiment\nTitrations can be used to find the concentration of an acid or alkali from the relative volumes used and the concentration of one of the two reactants. Titration involves the neutralisation of an acid with an alkali or a soluble carbonate. Since both reactants and products are colourless, an indicator is used to find the neutralisation point or end point i.e. the point at which the acid has been neutralised. This process is done in three stages. The acid is poured into the burette.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.8410515022488564, "ocr_used": false, "chunk_length": 1901, "token_count": 504}}
{"text": "the point at which the acid has been neutralised. This process is done in three stages. The acid is poured into the burette. A known volume of the alkali is transferred using a pipette into four separate conical flasks. Two or three drops of suitable indicator are added to each of the conical flasks. The acid is run into the flask until the indicator just changes colour. The difference between the initial and final burette readings gives the volume of the acid used commonly known as the titre. The first titration usually gives and approximate end point and is treated as the trial. The experiment is repeated for each of the other conical flask, to try to obtain the end point accurately. This is usually done by running out the acid to a point one unit away from the trial, the going drop-by-drop until an accurate end point is obtained. The volume of the acid required for each change is read off and recorded in a table similar to the one below. PROCEDURE DURING TITRATION\nTitrations require continuous shaking of the conical flask and its contents. Check the quantity of the pipette as indicated on the bulb of the pipette and remember to record it in the appropriate place. As you release the liquid from a pipette into a conical flask, one should not blow out the last drop remaining in the jet. Before you take any reading from the burette after filling it, first allow the solution to run out to fill the tap and jet of the burette, and then you begin taking your readings. Always write the reading immediately you take them. When completing the table of results, you will be expected to complete all columns as accurately as the as the limits of the apparatus can allow e.g. burette used is usually read to the nearest read to the nearest 0.05 cm 3, pipette is accurate to 0.05 cm 3 (1 drop). Use minimum amount of indicator possible (2 – 3 drops) and recognize the end point has been reached e.g. when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9020133806577253, "ocr_used": false, "chunk_length": 2212, "token_count": 475}}
{"text": "when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations. GENERAL NOTES FOR QUALITATIVE AND QUANTITIVE ANALYSIS\nIndicators\nAn indicator is a substance used in titrations, which has one colour in the presence of an excess of one reagent and a different colour in the presence of an excess of the other. Examples of indicators include;\nMethyl orange\nPhenolphthalein\nMethyl red\nBromothymol blue\nIndicator colour changes\nCandidates may be asked to carry out exercises involving:\nSimple quantitative experiments involving the measurement of volumes:\nSpeeds of reaction. Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration. Identification of ions and gasses as specified in the curriculum. QUALITATIVE ANALYSIS\nQualitative analysis is mainly about identification of substances. No emphasis is laid on amount or quantity. In qualitative analysis exercises, candidates should use approximately 1cm depth of a solution (1-2cm3) for each test and add reagents slowly, ensuring good mixing, until no further change is seen. Candidates should indicate at what stage a change occurs, writing any deductions alongside the observations on which they are based. Answers should include details of colour changes and precipitates formed and the names and chemical tests for any gases evolved. Marks for deductions or conclusions can only be gained if the appropriate observations are recorded. QUALITATIVE ANALYSIS NOTES\nTABLE 1:TESTING FOR GASES\nTABLE 2: TEST FOR ANIONS\nTABLE 3: TESTS FOR CATIONS\nTABLE 4: ORGANIC TESTS\nTABLE 5: MISCELLANEOUS TESTS\nTABLE 6. FLAME TESTS\nA tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9093969938107869, "ocr_used": false, "chunk_length": 2175, "token_count": 465}}
{"text": "FLAME TESTS\nA tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made. TABLE 7: HEAT ANALYSIS\nTABLE 8: SOLUBLE AND INSOLUBLE SALTS\nTABLE 9: GENERAL PRELIMINARY TESTS\nTABLE 10 : IGNITION ANALYSIS\nCandidates may be asked to heat an unknown alone in an ignition tube. The colour changes or the identity of any gases evolved may provide evidence as to the identity of the unknown. TABLE 10: RESIDUE WHEN HOT AND COLD\nResidue remaining after ignition may have a different colour when hot and cold\nTABLE 11: HEATING A SOLID\nGases or vapours may be evolved during heating of the solid\nTABLE 12: SOLUBILITY\nThe patterns of solubility for various types of salts\nKEY\nPpt = precipitate\nSol = solution\nInsol = insoluble\nXs = excess\nDil = dilute\n1.HYDROGEN CHLORIDE\nPhysical Properties\nChemical Properties\nFootnotes:\nDilute hydrochloric acid is one of the three common dilute acids used in the laboratory. Concentrated hydrochloric acid is used in the manufacture of many chemicals. 2.CARBON DIOXIDE\nPhysical Properties\nChemical Properties\nFootnotes:\nMakes up 0.04% of the gases in the air. Rainwater is slightly acidic because carbon dioxide dissolves in it. Needed by green plants during photosynthesis. Product of respiration in living things. Product of combustion\nIs one of the \"greenhouse\" gases present in the atmosphere. Used in fire extinguishers since it is heavier than air and forms a \"blanket\" around the fire. This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material. 3.CHLORINE GAS\nPhysical Properties\nChemical Properties\nFootnotes:\nOne of the \"family\" of halogen gases (iodine and bromine are in the same family)\nUsed in water purification. Used to make bleaching powder, disinfectants and antiseptics\nAlso used to make some explosives, poison gases and pesticides.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9049784444575681, "ocr_used": false, "chunk_length": 2065, "token_count": 497}}
{"text": "This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material.\n\n3.CHLORINE GAS\nPhysical Properties\nChemical Properties\nFootnotes:\nOne of the \"family\" of halogen gases (iodine and bromine are in the same family)\nUsed in water purification.\n\nUsed to make bleaching powder, disinfectants and antiseptics\nAlso used to make some explosives, poison gases and pesticides.\n\n4.HYDROGEN CHLORIDE GAS\nPhysical Properties\nChemical Properties\nFootnotes:\nDilute hydrochloric acid is one of the three common dilute acids used in the laboratory.\n\nConcentrated hydrochloric acid is used in the manufacture of many chemicals.\n\n5.HYDROGEN GAS\nPhysical Properties\nChemical Properties\nFootnotes:\nThe lightest gas known.\n\nOnce used in airships but replaced by helium which is not explosive.\n\nUsed to make ammonia which is needed in the manufacture of fertilizers and explosives.\n\n6.NITROGEN\nPhysical Properties\nChemical Properties\nFootnotes:\nMakes up around 79% of the gases in the air.\n\nUsed to make ammonia gas, which in turn is used to make explosives and fertilizers.\n\nUsed in light bulbs and thermometers because it is not reactive.\n\nAlso used as the atmosphere in rooms where explosives are stored.\n\n7.OXYGEN\nPhysical Properties\nChemical Properties\nFootnotes:\nMakes up 20% of the gases in the air.\n\nNeeded by the majority of living organisms for respiration\nIs produced by green plants as a by-product of photosynthesis.\n\n8.SULPHUR DIOXIDE GAS\nPhysical Properties\nChemical Properties\nFootnotes:\nIt is used as a bleaching agent.\n\nIt is one of the gases responsible for air pollution.", "metadata": {"source": "PRACTICAL-CHEMISTRY-4.docx", "file_type": "docx", "language": "en", "quality_score": 0.9075968523002421, "ocr_used": false, "chunk_length": 1652, "token_count": 373}}
{"text": "18.0.0 ACIDS, BASES AND SALTS (25 LESSONS) A.ACIDS AND BASES At a school laboratory: (i)An acid may be defined as a substance that turn litmus red. (ii)A base may be defined as a substance that turn litmus blue. Litmus is a lichen found mainly in West Africa. It changes its colour depending on whether the solution it is in, is basic/alkaline or acidic.It is thus able to identify/show whether 1. An acid is a substance that dissolves in water to form H+/H3O+ as the only positive ion/cation. This is called the Arrhenius definition of an acid. From this definition, an acid dissociate/ionize in water releasing H+ thus: HCl(aq) -> H+ (aq) + Cl- (aq) HNO3(aq) -> H+ (aq) + NO3- (aq) CH3COOH(aq) -> H+ (aq) + CH3COO-(aq) H2SO4(aq) -> 2H+ (aq) + SO42-(aq) H2CO3(aq) -> 2H+ (aq) + CO32-(aq) H3PO4(aq) -> 3H+ (aq) + PO43-(aq) 2.A base is a substance which dissolves in water to form OH- as the only negatively charged ion/anion. This is called Arrhenius definition of a base. From this definition, a base dissociate/ionize in water releasing OH- thus: KOH(aq) -> K+(aq) + OH-(aq) NaOH(aq) -> Na+(aq) + OH-(aq) NH4OH(aq) -> NH4+(aq) + OH-(aq) Ca(OH)2(aq) -> Ca2+(aq) + 2OH-(aq) Mg(OH)2(aq) -> Mg2+(aq) + 2OH-(aq) 3. An acid is a proton donor. A base is a proton acceptor. This is called Bronsted-Lowry definition of acids and bases. From this definition, an acid donates H+ . H+ has no electrons and neutrons .It contains only a proton. Examples I.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8052565137839746, "ocr_used": true, "chunk_length": 1444, "token_count": 485}}
{"text": "From this definition, an acid donates H+ . H+ has no electrons and neutrons .It contains only a proton. Examples I. From the equation: HCl(aq) + H2O(l) === H3O+(aq) + Cl- (aq) (a)(i)For the forward reaction from left to right, H2O gains a proton to form H3O+ and thus H2O is a proton acceptor .It is a Bronsted-Lowry base (ii) For the backward reaction from right to left, H3O+ donates a proton to form H2O and thus H3O+ is an ‘opposite’ proton donor. It is a BronstedLowry conjugate acid (b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl- and thus HCl is a proton donor . It is a Bronsted-Lowry acid (ii) For the backward reaction from right to left, Cl- gains a proton to form HCl and thus Cl- is an ‘opposite’ proton acceptor. It is a Bronsted-Lowry conjugate base. Every base /acid from Bronsted-Lowry definition thus must have a conjugate product/reactant. II. From the equation: HCl(aq) + NH3(aq) === NH4+(aq) + Cl- (aq) (a)(i)For the forward reaction from left to right, NH3 gains a proton to form NH4+ and thus NH3 is a proton acceptor . It is a Bronsted-Lowry base (ii) For the backward reaction from right to left, NH4+ donates a proton to form NH3 and thus NH4+ is an ‘opposite’ proton donor. It is a Bronsted-Lowry conjugate acid (b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl- and thus HCl is a proton donor . It is a Bronsted-Lowry acid (ii) For the backward reaction from right to left, Cl- gains a proton to form HCl and thus Cl- is an ‘opposite’ proton acceptor. It is a Bronsted-Lowry conjugate base. 4. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8621322665883735, "ocr_used": true, "chunk_length": 1703, "token_count": 500}}
{"text": "It is a Bronsted-Lowry conjugate base. 4. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g. (a)Hydrogen chloride gas dissolves in water to form hydrochloric acid Hydrochloric acid dissociates/ionizes in water to free H+(aq)/H3O+(aq) ions. The free H3O+(aq) / H+(aq) ions are responsible for: (i)turning blue litmus paper/solution red. (ii)show pH value 1/2/3/4/5/6 (iii)are good electrolytes/conductors of electricity/undergo electrolysis. (iv)react with metals to produce /evolve hydrogen gas and a salt. i.e. Ionically: -For a monovalent metal: 2M(s) + 2H+(aq) -> 2M+(aq) + H2(g) -For a divalent metal: M(s) + 2H+(aq) -> M2+(aq) + H2(g) -For a trivalent metal: 2M(s) + 6H+(aq) -> 2M3+(aq) + 3H2(g) Examples: -For a monovalent metal: 2Na(s) + 2H+(aq) -> 2Na+(aq) + H2(g) -For a divalent metal: Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g) -For a trivalent metal: 2Al(s) + 6H+(aq) -> 2Al3+(aq) + 3H2(g) (v)react with metal carbonates and hhydrogen carbonates to produce /evolve carbon(IV)oxide gas ,water and a salt. i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.777106786900807, "ocr_used": true, "chunk_length": 1075, "token_count": 396}}
{"text": "i.e. Ionically: -For a monovalent metal: M2O(s) + 2H+(aq) -> 2M+(aq) + H2O (l) MOH(aq) + H+(aq) -> M+(aq) + H2O (l)\n-For a divalent metal: MO(s) + 2H+(aq) -> M2+(aq) + H2O (l) M(OH) 2(s) + 2H+(aq) -> M2+(aq) + 2H2O(l) -For a trivalent metal: M2O3(s) + 6H+(aq) -> 2M3+(aq) + 3H2O (l) M(OH) 3(s) + 3H+(aq) -> M3+(aq) + 3H2O(l) Examples: -For a monovalent metal: K2O(s) + 2H+(aq) -> 2K+(aq) + H2O (l) NH4OH(aq) + H+(aq) -> NH4+(aq) + H2O (l) -For a divalent metal: ZnO (s) + 2H+(aq) -> Zn2+(aq) + H2O (l) Pb(OH) 2(s) + 2H+(aq) -> Pb2+(aq) + 2H2O(l) (b)Hydrogen chloride gas dissolves in methylbenzene /benzene but does not dissociate /ionize into free ions. It exists in molecular state showing none of the above properties. (c)Ammonia gas dissolves in water to form aqueous ammonia which dissociate/ionize to free NH4+ (aq) and OH-(aq) ions. This dissociation/ionization makes aqueous ammonia to: (i)turn litmus paper/solution blue. (ii)have pH 8/9/10/11 (iii)be a good electrical conductor (iv)react with acids to form ammonium salt and water only. NH4OH(aq) + HCl(aq) -> NH4Cl(aq) + H2O(l) (d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules 6.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7382419911340562, "ocr_used": true, "chunk_length": 1223, "token_count": 508}}
{"text": "This dissociation/ionization makes aqueous ammonia to: (i)turn litmus paper/solution blue. (ii)have pH 8/9/10/11 (iii)be a good electrical conductor (iv)react with acids to form ammonium salt and water only. NH4OH(aq) + HCl(aq) -> NH4Cl(aq) + H2O(l) (d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules 6. Solvents are either polar or non-polar. A polar solvent is one which dissolves ionic compounds and other polar solvents. Water is polar solvent that dissolves ionic and polar substance by surrounding the free ions as below: H ð+ H ð+ O ð- H ð+ H ð+ H ð+ O ð- H ð+ H ð+ H ð+\nO ð- H+ O ð- O ð- Cl - O ð- H ð+ Hð+ H ð+ O ð- H ð+ H+ H ð+ H ð+ H ð+ O ð- Beaker Cl- Cl- H+ water H+ Cl- H+ Free ions Note:Water is polar .It is made up of : Oxygen atom is partially negative and two hydrogen atoms which are partially positive. They surround the free H+ and Cl- ions. A non polar solvent is one which dissolved non-polar substances and covalent compounds. If a polar ionic compound is dissolved in non-polar solvent ,it does not ionize/dissociate into free ions as below: H-Cl H-Cl methyl benzene H-Cl H-Cl Covalent bond 7. Some acids and bases are strong while others are weak. (a)A strong acid/base is one which is fully/wholly/completely dissociated / ionized into many free H+ /OH- ions i.e. I. Strong acids exists more as free H+ ions than molecules. e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8688947508292568, "ocr_used": true, "chunk_length": 1436, "token_count": 448}}
{"text": "I. Strong acids exists more as free H+ ions than molecules. e.g. HCl(aq) H+(aq) + Cl- (aq) (molecules) (cation) (anion) HNO3(aq) H+(aq) + NO3-(aq) (molecules) (cation) (anion) H2SO4(aq) 2H+(aq) + SO42-(aq) (molecules) (cation) (anion) II. Strong bases/alkalis exists more as free OH- ions than molecules. e.g. KOH(aq) K+(aq) + OH- (aq) (molecules) (cation) (anion) NaOH(aq) Na+(aq) + OH-(aq) (molecules) (cation) (anion) (b) A weak base/acid is one which is partially /partly dissociated /ionized in water into free OH- (aq) and H+(aq) ions. I. Weak acids exists more as molecules than as free H+ ions. e.g. CH3COOH(aq) H+(aq) + CH3COO- (aq) (molecules) (cation) (anion) H3PO4(aq) 3H+(aq) + PO43-(aq) (molecules) (cation) (anion) H2CO3(aq) 2H+(aq) + CO32-(aq) (molecules) (cation) (anion) II. Weak bases/alkalis exists more as molecules than free OH- ions. e.g. NH4OH(aq) NH4+(aq) + OH- (aq) (molecules) (cation) (anion) Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq) (molecules) (cation) (anion) Mg(OH)2(aq) Mg2+(aq) + 2OH-(aq) (molecules) (cation) (anion)\n8. The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution. An acid/base/alkali with more acid/base/alkali in a decimeter(litre) of solution is said to be concentrated while that with less is said to be dilute. 9.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7956641932198458, "ocr_used": true, "chunk_length": 1334, "token_count": 503}}
{"text": "The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution. An acid/base/alkali with more acid/base/alkali in a decimeter(litre) of solution is said to be concentrated while that with less is said to be dilute. 9. (a) (i)strong acids have pH 1/2/3 while weak acids have high pH 4/5/6. (ii)a neutral solution have pH 7 (iii)strong alkalis/bases have pH 12/13/14 while weak bases/alkalis have pH 11/10 /9 / 8. (b) pH is a measure of H+(aq) concentration in a solution. The higher the H+(aq)ions concentration ; -the higher the acidity -the lower the pH -the lower the concentration of OH-(aq) -the lower the alkalinity At pH 7 , a solution has equal concentration of H+(aq) and OH-(aq). Beyond pH 7,the concentration of the OH-(aq) increases as the H+(aq) ions decreases. 10.(a) When acids /bases dissolve in water, the ions present in the solution conduct electricity. The more the dissociation the higher the yield of ions and the greater the electrical conductivity of the solution. A compound that conducts electricity in an electrolyte and thus a compound showing high electrical conductivity is a strong electrolyte while a compound showing low electrical conductivity is a weak electrolyte. (b) Practically, a bright light on a bulb ,a high voltage reading from a voltmeter high ammeter reading from an ammeter, a big deflection on a galvanometer is an indicator of strong electrolyte(acid/base) and the opposite for weak electrolytes(acids/base) 11. Some compounds exhibit/show both properties of acids and bases/alkalis. A substance that reacts with both acids and bases is said to be amphotellic. The examples below show the amphotellic properties of: (a) Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2) (i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8710280069915858, "ocr_used": true, "chunk_length": 2022, "token_count": 518}}
{"text": "Some compounds exhibit/show both properties of acids and bases/alkalis. A substance that reacts with both acids and bases is said to be amphotellic. The examples below show the amphotellic properties of: (a) Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2) (i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. Basic oxide + Acid -> salt + water Examples: Chemical equation ZnO(s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + H2O(l) ZnO(s) + 2HCl(aq) -> ZnCl2 (aq) + H2O(l) ZnO(s) + H2SO4(aq) -> ZnSO4 (aq) + H2O(l) Ionic equation ZnO(s) + 2H+ (aq) -> Zn 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Basic oxide + Base/alkali + Water -> Complex salt Examples: Chemical equation 1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. ZnO(s) + 2NaOH(aq) + H2O(l) -> Na2Zn(OH) 4(aq) 2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8339259053183105, "ocr_used": true, "chunk_length": 1365, "token_count": 430}}
{"text": "Basic oxide + Acid -> salt + water Examples: Chemical equation ZnO(s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + H2O(l) ZnO(s) + 2HCl(aq) -> ZnCl2 (aq) + H2O(l) ZnO(s) + H2SO4(aq) -> ZnSO4 (aq) + H2O(l) Ionic equation ZnO(s) + 2H+ (aq) -> Zn 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Basic oxide + Base/alkali + Water -> Complex salt Examples: Chemical equation 1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. ZnO(s) + 2NaOH(aq) + H2O(l) -> Na2Zn(OH) 4(aq) 2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. ZnO(s) + 2KOH(aq) + H2O(l) -> K2Zn(OH) 4(aq) Ionic equation ZnO(s) + 2OH-(aq) + H2O(l) -> 2[Zn(OH) 4]2- (aq) (ii)When Zinc(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7959231104063677, "ocr_used": true, "chunk_length": 1208, "token_count": 431}}
{"text": "i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only. Basic hydroxide + Acid -> salt + water Examples: Chemical equation Zn(OH) 2 (s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + 2H2O(l)\nZn(OH) 2 (s) + 2HCl(aq) -> ZnCl2 (aq) + 2H2O(l) Zn(OH) 2 (s) + H2SO4(aq) -> ZnSO4 (aq) + 2H2O(l) Ionic equation Zn(OH) 2 (s) + 2H+ (aq) -> Zn 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties by reacting with a base to form a complex salt. Basic hydroxide + Base/alkali -> Complex salt Examples: Chemical equation 1.When Zinc hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2NaOH(aq) -> Na2Zn(OH) 4(aq) 2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2KOH(aq) -> K2Zn(OH) 4(aq) Ionic equation Zn(OH) 2 (s) + 2OH-(aq) -> 2[Zn(OH) 4]2- (aq) (b) Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2) (i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7869779875420488, "ocr_used": true, "chunk_length": 1314, "token_count": 499}}
{"text": "Zn(OH) 2 (s) + 2NaOH(aq) -> Na2Zn(OH) 4(aq) 2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2KOH(aq) -> K2Zn(OH) 4(aq) Ionic equation Zn(OH) 2 (s) + 2OH-(aq) -> 2[Zn(OH) 4]2- (aq) (b) Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2) (i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. All other Lead salts are insoluble. Chemical equation PbO(s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + H2O(l) Ionic equation PbO(s) + 2H+ (aq) -> Pb 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2NaOH(aq) + H2O(l) -> Na2Pb(OH) 4(aq) 2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.809679948818058, "ocr_used": true, "chunk_length": 1297, "token_count": 452}}
{"text": "Chemical equation PbO(s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + H2O(l) Ionic equation PbO(s) + 2H+ (aq) -> Pb 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2NaOH(aq) + H2O(l) -> Na2Pb(OH) 4(aq) 2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2KOH(aq) + H2O(l) -> K2Pb(OH) 4(aq) Ionic equation PbO(s) + 2OH-(aq) + H2O(l) -> 2[Pb(OH) 4]2- (aq) (ii)When Lead(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only. Chemical equation Pb(OH) 2 (s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + 2H2O(l) Ionic equation Pb(OH) 2 (s) + 2H+ (aq) -> Pb 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8067634228648248, "ocr_used": true, "chunk_length": 1435, "token_count": 493}}
{"text": "Chemical equation Pb(OH) 2 (s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + 2H2O(l) Ionic equation Pb(OH) 2 (s) + 2H+ (aq) -> Pb 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. Pb(OH) 2 (s) + 2NaOH(aq) -> Na2Pb(OH) 4(aq) 2.When Lead(II) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt. Pb(OH) 2 (s) + 2KOH(aq) -> K2Pb(OH) 4(aq) Ionic equation Pb(OH) 2 (s) + 2OH-(aq) -> 2[Pb(OH) 4]2- (aq)\n(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3) (i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8031108902763076, "ocr_used": true, "chunk_length": 1129, "token_count": 401}}
{"text": "Pb(OH) 2 (s) + 2KOH(aq) -> K2Pb(OH) 4(aq) Ionic equation Pb(OH) 2 (s) + 2OH-(aq) -> 2[Pb(OH) 4]2- (aq)\n(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3) (i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. Chemical equation Al2O3 (s) + 6HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l) Al2O3 (s) + 6HCl(aq) -> AlCl3 (aq) + 3H2O(l) Al2O3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al2O3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al2O3 (s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) 2.When Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7752362056986599, "ocr_used": true, "chunk_length": 1229, "token_count": 461}}
{"text": "Chemical equation Al2O3 (s) + 6HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l) Al2O3 (s) + 6HCl(aq) -> AlCl3 (aq) + 3H2O(l) Al2O3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al2O3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al2O3 (s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) 2.When Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt. Al2O3 (s) + 2KOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) Ionic equation Al2O3 (s) + 2OH-(aq) + 3H2O(l) -> 2[Al(OH) 4]- (aq) (ii)When Aluminium(III)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7709299464650836, "ocr_used": true, "chunk_length": 1167, "token_count": 441}}
{"text": "i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only. Chemical equation Al(OH) 3 (s) + 3HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l)\nAl(OH)3 (s) + 3HCl(aq) -> AlCl3 (aq) + 3H2O(l) 2Al(OH)3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al(OH)3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + NaOH(aq) -> NaAl(OH) 4(aq) 2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7852933979347134, "ocr_used": true, "chunk_length": 834, "token_count": 306}}
{"text": "It reacts with a base to form a complex salt. Chemical equation 1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + NaOH(aq) -> NaAl(OH) 4(aq) 2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + KOH(aq) -> KAl(OH) 4(aq) Ionic equation Al(OH) 3 (s) + OH-(aq) -> [Al(OH) 4]- (aq) Summary of amphotellic oxides/hydroxides Oxide Hydroxide Formula of simple salt from nitric (V)acid Formula of complex salt from sodium hydroxide ZnO Zn(OH)2 Zn(NO3)2 Na2Zn(OH)4 [Zn(OH)4]2-(aq) Sodium tetrahydroxozincate(II) PbO Pb(OH)2 Pb(NO3)2 Na2Pb(OH)4 [Pb(OH)4]2-(aq) Sodium tetrahydroxoplumbate(II) Al2O3 Al(OH) 3 Al(NO3)3 NaAl(OH)4 [Al(OH)4]-(aq) Sodium tetrahydroxoaluminate(II) 12.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid. A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical. (b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid. Some acids are therefore: (i)monobasic acids generally denoted HX e.g. HCl, HNO3,HCOOH,CH3COOH. (ii)dibasic acids ; generally denoted H2X e.g. H2SO4, H2SO3, H2CO3,HOOCOOH. (iii)tribasic acids ; generally denoted H3X e.g. H3PO4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8177295633654501, "ocr_used": true, "chunk_length": 1459, "token_count": 509}}
{"text": "H2SO4, H2SO3, H2CO3,HOOCOOH. (iii)tribasic acids ; generally denoted H3X e.g. H3PO4. (c) Some salts are normal salts while other are acid salts. (i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen\nacid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8349312471311394, "ocr_used": true, "chunk_length": 1126, "token_count": 359}}
{"text": "(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen\nacid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7721563967238708, "ocr_used": true, "chunk_length": 1734, "token_count": 627}}
{"text": "(ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen\nacid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.769527400764635, "ocr_used": true, "chunk_length": 1706, "token_count": 619}}
{"text": "Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen\nacid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution. Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7681698631816649, "ocr_used": true, "chunk_length": 1705, "token_count": 628}}
{"text": "e.g. CuCO3 (s) -> CuO(s) + CO2(g) CaCO3 (s) -> CaO(s) + CO2(g) PbCO3 (s) -> PbO(s) + CO2(g) FeCO3 (s) -> FeO(s) + CO2(g) ZnCO3 (s) -> ZnO(s) + CO2(g) (iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e. 2NaHCO 3(s) -> Na2CO3(s) + CO2(g) + H2O(l) 2KHCO 3(s) -> K2CO3(s) + CO2(g) + H2O(l) (iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e. Ca(HCO3) 2(aq) -> CaCO3(s) + CO2(g) + H2O(l) Mg(HCO3) 2(aq) -> MgCO3(s) + CO2(g) + H2O(l) 15. Salts contain cation(positively charged ion) and anions(negatively charged ion).When dissolved in polar solvents/water. The cation and anion in a salt is determined/known usually by precipitation of the salt using a precipitating reagent. The colour of the precipitate is a basis of qualitative analysis of a compound. 16.Qualitative analysis is the process of identifying an unknown compound /salt by identifying the unique qualities of the salt/compound. It involves some of the following processes. (a)Reaction of cation with sodium/potassium hydroxide solution. Both sodium/potassium hydroxide solutions are precipitating reagents. The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution. NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8252584080266292, "ocr_used": true, "chunk_length": 1582, "token_count": 477}}
{"text": "Both sodium/potassium hydroxide solutions are precipitating reagents. The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution. NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide. The table below shows the observations, inferences / deductions and explanations from the following test tube experiments: Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three\ndrops of 2M sodium hydroxide solution then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8040622490515881, "ocr_used": true, "chunk_length": 1617, "token_count": 499}}
{"text": "Add three\ndrops of 2M sodium hydroxide solution then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)\nThe hydroxides formed react with more OH- ions to form complex salts/ions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7652735454330151, "ocr_used": true, "chunk_length": 1195, "token_count": 419}}
{"text": "Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)\nThe hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7489506154960174, "ocr_used": true, "chunk_length": 1381, "token_count": 519}}
{"text": "Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)\nThe hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) Green precipitate insoluble in excess On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7238768249294565, "ocr_used": true, "chunk_length": 1045, "token_count": 419}}
{"text": "Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)\nThe hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) Green precipitate insoluble in excess On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide. Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) (b)Reaction of cation with aqueous ammonia Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt. Like NaOH/KOH the OH- ion in NH4OH react with the cation to form a characteristic hydroxide .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7697075185563396, "ocr_used": true, "chunk_length": 1339, "token_count": 488}}
{"text": "Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) (b)Reaction of cation with aqueous ammonia Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt. Like NaOH/KOH the OH- ion in NH4OH react with the cation to form a characteristic hydroxide . Below are the observations ,inferences and explanations of the reactions of aqueous ammonia with salts from the following test tube reactions. Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M aqueous ammonia then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7887681648445584, "ocr_used": true, "chunk_length": 1434, "token_count": 469}}
{"text": "Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M aqueous ammonia then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides. Pb2+ (aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+ (aq) + 3OH-(aq) -> Al(OH) 3(s) Ba2+ (aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+ (aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+ (aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ions Zn2+ ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7207368136692105, "ocr_used": true, "chunk_length": 1139, "token_count": 447}}
{"text": "Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides. Pb2+ (aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+ (aq) + 3OH-(aq) -> Al(OH) 3(s) Ba2+ (aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+ (aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+ (aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ions Zn2+ ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions. Zn(OH) 2(s) + 4NH3(aq) ->[ Zn(NH3) 4]2+(aq)+ 2OH-(aq) Blue precipitate Cu2+ Cu2+ ions react with OH- from 2M aqueous\nthat dissolves in excess ammonia solution to form a deep/royal blue solution ammonia to form blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7399489046828205, "ocr_used": true, "chunk_length": 1259, "token_count": 487}}
{"text": "Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions. Zn(OH) 2(s) + 4NH3(aq) ->[ Zn(NH3) 4]2+(aq)+ 2OH-(aq) Blue precipitate Cu2+ Cu2+ ions react with OH- from 2M aqueous\nthat dissolves in excess ammonia solution to form a deep/royal blue solution ammonia to form blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions. Cu(OH) 2 (s) + 4NH3(aq) ->[ Cu(NH3) 4]2+(aq)+ 2OH-(aq) Green precipitate insoluble in excess. On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M aqueous ammonia to form insoluble green precipitate of Iron(II) hydroxide. Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) Note (i) Only Zn2+ ions/salts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia. (ii) Pb2+ and Al3+ ions/salts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7747445255474453, "ocr_used": true, "chunk_length": 1370, "token_count": 490}}
{"text": "Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) Note (i) Only Zn2+ ions/salts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia. (ii) Pb2+ and Al3+ ions/salts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia. (iii) Cu2+ ions/salts form a blue precipitate that dissolve to form a deep/royal blue solution in excess of 2M aqueous ammonia but only blue insoluble precipitate in 2M sodium hydroxide (c)Reaction of cation with Chloride (Cl-)ions All chlorides are soluble in water except Silver chloride and Lead (II)chloride (That dissolve in hot water).When a soluble chloride like NaCl, KCl, NH4Cl is\nadded to about 2cm3 of a salt containing Ag+ or Pb2+ions a white precipitate of AgCl or PbCl2 is formed. The following test tube reactions illustrate the above. Experiment Put about 2cm3 of silver nitrate(V) andLead(II)nitrate(V)solution into separate test tubes. Add five drops of NaCl /KCl / NH4Cl/HCl. Heat to boil. Observation Inference Explanation (i)White precipitate does not dissolve on heating Ag+ ions Ag+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating. Note Both Pb2+ and Al3+ ions forms an insoluble white precipitate in excess aqueous ammonia.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8449801796559625, "ocr_used": true, "chunk_length": 1774, "token_count": 511}}
{"text": "Observation Inference Explanation (i)White precipitate does not dissolve on heating Ag+ ions Ag+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating. Note Both Pb2+ and Al3+ ions forms an insoluble white precipitate in excess aqueous ammonia. A white precipitate on adding Cl- ions/salts shows Pb2+. No white precipitate on adding Cl- ions/salts shows Al3+. Adding a chloride/ Cl- ions/salts can thus be used to separate the identity of Al3+ and Pb2+. (d)Reaction of cation with sulphate(VI)/SO42- and sulphate(IV)/SO32- ions All sulphate(VI) and sulphate(IV)/SO32- ions/salts are soluble/dissolve in water except Calcium sulphate(VI)/CaSO4, Calcium sulphate(IV)/CaSO3, Barium sulphate(VI)/BaSO4, Barium sulphate(IV)/BaSO3, Lead(II) sulphate(VI)/PbSO4 and Lead(II) sulphate(IV)/PbSO3.When a soluble sulphate(VI)/SO42- salt like Na2SO4, H2SO4, (NH4)2SO4 or Na2SO3 is added to a salt containing Ca2+, Pb2+, Ba2+ ions, a white precipitate is formed. The following test tube experiments illustrate the above. Procedure Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, BaCl2 and Pb(NO3)2, in separate boiling tubes. Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV). Observation Inference Explanation White Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as\nprecipitate Pb2+ ions they are sparingly soluble.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8269444054390638, "ocr_used": true, "chunk_length": 1603, "token_count": 473}}
{"text": "Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV). Observation Inference Explanation White Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as\nprecipitate Pb2+ ions they are sparingly soluble. Ca2+(aq)+ SO32-(aq) -> CaSO3(s) Ca2+(aq)+ SO42-(aq) -> CaSO4(s) Ba2+(aq)+ SO32-(aq) -> BaSO3(s) Ba2+(aq)+ SO42-(aq) -> BaSO4(s) Pb2+(aq)+ SO32-(aq) -> PbSO3(s) Pb2+(aq)+ SO42-(aq) -> PbSO4(s) (e)Reaction of cation with carbonate(IV)/CO32- ions All carbonate salts are insoluble except sodium/potassium carbonate(IV) and ammonium carbonate(IV). They dissociate /ionize to release CO32- ions. CO32- ions produce a white precipitate when the soluble carbonate salts is added to any metallic cation. Procedure Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, MgCl2 ,Pb(NO3)2 andZnSO4 in separate boiling tubes. Add six drops of Potassium /sodium carbonate(IV)/ ammonium carbonate (IV)solution. Observation Inference Explanation Green precipitate Cu2+ ,Fe2+,ions CO32-(aq) Copper(II)carbonate(IV) and Iron(II) carbonate (IV) are precipitated as insoluble green precipitates. Cu2+(aq)+ CO32-(aq) -> CuCO3(s) Fe2+(aq)+ CO32-(aq) -> FeCO3(s) When sodium carbonate(IV)is added to CuCO3(s) the CO32-(aq) ions are first hydrolysed to produce CO2(g) and OH(aq)ions. CO32-(aq) + H2O (l) -> CO2 (g) + 2OH- (aq) The OH-(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.771639976414479, "ocr_used": true, "chunk_length": 1509, "token_count": 491}}
{"text": "Cu2+(aq)+ CO32-(aq) -> CuCO3(s) Fe2+(aq)+ CO32-(aq) -> FeCO3(s) When sodium carbonate(IV)is added to CuCO3(s) the CO32-(aq) ions are first hydrolysed to produce CO2(g) and OH(aq)ions. CO32-(aq) + H2O (l) -> CO2 (g) + 2OH- (aq) The OH-(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper. Cu2+(aq)+ CO32-(aq)+2OH- (aq)\n->CuCO3.Cu(OH)2 (s) White precipitate CO32- White ppt of the carbonate(IV)salt is precipitated Ca2+(aq) + CO32- (aq) -> CaCO3(s) Mg2+(aq) + CO32- (aq) -> MgCO3(s) Pb2+(aq) + CO32- (aq) -> PbCO3(s) Zn2+(aq) + CO32- (aq) -> ZnCO3(s) Note (i)Iron(III)carbonate(IV) does not exist. (ii)Copper(II)Carbonate(IV) exist only as the basic CuCO3.Cu(OH) 2 (iii)Both BaCO3 and BaSO3 are insoluble white precipitate. If hydrochloric acid is added to the white precipitate; I. BaCO3 produces CO2 gas. When bubbled/directed into lime water solution,a white precipitate is formed. II. I. BaSO3 produces SO2 gas. When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII). (f) Reaction of cation with sulphide / S2- ions All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed. Procedure Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7748539337420784, "ocr_used": true, "chunk_length": 1493, "token_count": 509}}
{"text": "When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII). (f) Reaction of cation with sulphide / S2- ions All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed. Procedure Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes. Add six drops of Potassium /sodium sulphide solution. Observation Inference Explanation Black ppt S2- ions CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates Cu2+(aq) + S2-(aq) -> CuS(s) Pb2+(aq) + S2-(aq) -> PbS(s) Fe2+(aq) + S2-(aq) -> FeS(s) Zn2+(aq) + S2-(aq) -> ZnS(s) Sample qualitative analysis guide\nYou are provided with solid Y(aluminium (III)sulphate(VI)hexahydrate).Carry out the following tests and record your observations and inferences in the space provided. 1(a) Appearance Observations inference (1mark) White crystalline solid Coloured ions Cu2+ , Fe2+ ,Fe3+ absent (b)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly. Observations inference (1mark) Colourless droplets formed on the cooler Hydrated compound/compound part of the test tube containing water of crystallization Solid remains a white residue (c)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.847747908714367, "ocr_used": true, "chunk_length": 1759, "token_count": 480}}
{"text": "Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali. Observation Inference (1mark) White ppt, soluble in excess Zn2+ , Pb2+ , Al3+ (ii)To the second portion, add three drops of aqueous ammonia then add excess of the alkali. Observation Inference (1mark) White ppt, insoluble in excess Pb2+ , Al3+ (iii)To the third portion, add three drops of sodium sulphate(VI)solution. Observation Inference (1mark) No white ppt Al3+\n(iv)I.To the fourth portion, add three drops of Lead(II)nitrate(IV)solution. Preserve Observation Inference (1mark) White ppt CO32-, SO42-, SO32-, Cl-, II.To the portion in (iv) I above , add five drops of dilute hydrochloric acid. Observation Inference (1mark) White ppt persist/remains SO42-, Cl-, III.To the portion in (iv) II above, heat to boil. Observation Inference (1mark) White ppt persist/remains SO42-, Note that: (i)From test above, it can be deduced that solid Y is hydrated aluminium(III)sulphate(VI) solid (ii)Any ion inferred from an observation below must be derived from previous correct observation and inferences above. e.g. Al3+ in c(iii) must be correctly inferred in either/or in c(ii) or c(i)above SO42- in c(iv)III must be correctly inferred in either/or in c(iv)II or c(iv)I above (iii)Contradiction in observations and inferences should be avoided.e.g. “White ppt soluble in excess” to infer presence of Al3+ ,Ba2+ ,Pb3+ (iv)Symbols of elements/ions should be correctly capitalized. e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.843181996568853, "ocr_used": true, "chunk_length": 1766, "token_count": 484}}
{"text": "e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV. The actual solutions, not in that order are lead nitrate, zinc sulphate potassium chloride and calcium chloride. a)When aqueous sodium carbonate was added to each sample separately, a white precipitate was formed in I, III and IV only. Identify solution II. b)When excess sodium hydroxide was added to each sample separately, a white precipitate was formed in solutions III and I only. Identify solution I 17.When solids/salts /solutes are added to a solvent ,some dissolve to form a solution. Solute + Solvent -> Solvent If a solution has a lot of solute dissolved in a solvent ,it is said to be concentrated. If a solution has little solute dissolved in a solvent ,it is said to be dilute. There is a limit to how much solute can dissolve in a given /specified amount of solvent/water at a given /specified temperature. The maximum mass of salt/solid/solute that dissolve in 100g of solvent/water at a specified temperature is called solubility of a salt. When no more solute can dissolve in a given amount of solvent at a specified temperature, a saturated solution is formed. For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8635973250279767, "ocr_used": true, "chunk_length": 1809, "token_count": 469}}
{"text": "For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 50.0 x 100 ) = 20.0 g /100g H2O\nMass/volume of water/solvent 250.0 (c)If the solubility of potassium chlorate(V) is 5g/100g H2O at 80oC,how much can dissolve in 5cm3 of water at 80oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 5 x 5 = 0.25g of KClO3 dissolve 100 (d)If the solubility of potassium chlorate(V) is 72g/100g H2O at 20oC,how much can saturate 25g of water at 20oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7524959202378558, "ocr_used": true, "chunk_length": 1443, "token_count": 509}}
{"text": "Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0. (f)What volume of water should be added to 22.0g of water at 10oC if the solubility of KNO3 at 10oC is 5.0g/100g H2O? Solubility is mass/100g H2O => 22.0g + x = 100cm3/100g H2O X= 100 – 22 = 78 cm3 of H2O 18. A graph of solubility against temperature is called solubility curve. It shows the influence of temperature on solubility of different substances/solids/salts. Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature KNO3 KClO3 Saturated solution of KClO3 Solubility\nNote: (i)solubility of KNO3 and KClO3 increase with increase in temperature. (ii)solubility of KNO3 is always higher than that of KClO3 at any specified temperature. (iii)solubility of NaCl decrease with increase in temperature. (iv)NaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature. (v)At point A both NaCl and KNO3 are equally soluble. (vi)At point B both NaCl and KClO3 are equally soluble. (vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.810722649715624, "ocr_used": true, "chunk_length": 1708, "token_count": 514}}
{"text": "(vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature. (b) For salts whose solubility decreases with increase in temperature, crystals form when the salt solution at lower temperatures is heated to a higher temperature. The examples below shows determination of the mass of crystals deposited with changes in temperature. 1.The solubility of KClO3 at 100oC is 60g/100g water .What mass of KClO3 will be deposited at: (i)75 oC if the solubility is now 39g/100g water. At 100oC = 60.0g Less at 75oC = - 39.0g Mass of crystallized out 21.0g (i)35 oC if the solubility is now 28 g/100g water. At 100oC = 60.0g Less at 35oC = - 28.0.0g Mass of crystallized out 32.0g 2. KNO3 has a solubility of 42 g/100g water at 20oC.The salt was heated and added 38g more of the solute which dissolved at100oC. Calculate the solubility of KNO3 at 100oC. Solubility of KNO3 at 100oC = solubility at 20oC + mass of KNO3 added => 42g + 38g = 80g KNO3 /100g H2O 3. A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7898333540604402, "ocr_used": true, "chunk_length": 1462, "token_count": 455}}
{"text": "A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved. Mass of solvent at 20oC = mass of solution – mass of solute => 65 - 5 = 60g Solubility before adding salt = mass of solute x 100 Volume of solvent => 5 x 100 = 8.3333g/100g water 60 Mass of solute to equalize with solubility = 25 – 8.3333g = 16.6667g Mass of solute undissolved = 60.0 - 16.6667g = 43.3333 g 4. Study the table below Salt Solubility in gram at\n50oC 20oC KNO3 90 30 KClO3 20 6 (i)What happens when the two salts are dissolved in water then cooled from 50oC to 20oC. (90 – 30) = 60.0 g of KNO3 crystals precipitate (20 – 6) = 14.0 g of KClO3 crystals precipitate (ii)State the assumption made in (i) above. Solubility of one salt has no effect on the solubility of the other. 5. 10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7476392961876833, "ocr_used": true, "chunk_length": 1320, "token_count": 451}}
{"text": "10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate? (K= 39.0,C=12.0.O= 16.0) Molar mass K2CO3= 138 Moles K2CO3 = mass of K2CO3 => 7.93 = 0.0515 moles Molar mass K2CO3 138 (d)How many moles of water are present in the hydrate for every one mole of K2CO3 ? (H=1.0.O= 16.0) Molar mass H2O = 18 Moles H2O = mass of H2O => 2.07 = 0.115 moles Molar mass H2O 18 Mole ratio H2O : K2CO3 = 0.115 moles 2 = 2 0.0515 moles 1 (e)What is the formula of the hydrated salt? K2CO3 .2 H2O 6. The table below shows the solubility of Potassium nitrate(V) at different temperatures. Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC\n(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6828707375099128, "ocr_used": true, "chunk_length": 1261, "token_count": 488}}
{"text": "Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC\n(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III. 55oC From a correctly plotted graph = 104g (ii)the temperature at which the following mass of KNO3 dissolved: I. 22g From a correctly plotted graph =13.0oC II. 30g From a correctly plotted graph =17.5oC III.100g From a correctly plotted graph =54.5oC (c)Explain the shape of your graph. Solubility of KNO3 increase with increase in temperature/More KNO3 dissolve as temperature rises. (d)Show on the graph the supersaturated and unsaturated solutions. Above the solubility curve write; “supersaturated” Below the solubility curve write; “unsaturated” (e)From your graph, calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to: I. 20oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 32 g Mass of KNO3crystals = 148 g II. 35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC\nSolubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7664094045489395, "ocr_used": true, "chunk_length": 1505, "token_count": 471}}
{"text": "35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC\nSolubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures. Temperature(oC) 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 Solubility of A 28.0 31.0 34.0 37.0 40.0 43.0 45.0 48.0 51.0 Solubility of B 13.0 21.0 32.0 46.0 64.0 85.0 110.0 138.0 169.0 (a)On the same axis plot a graph of solubility (y-axis) against temperature for each salt. (b)At what temperature are the two salts equally soluble. The point of intersection of the two curves = 24oC (c)What happens when a mixture of 100g of salt B with 100g if water is heated to 80oC From the graph, the solubility of B at 80oC is 169g /100g water. All the 100g crystals of B dissolve. (d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6990350877192982, "ocr_used": true, "chunk_length": 1140, "token_count": 411}}
{"text": "(d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g\n(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.765179095713447, "ocr_used": true, "chunk_length": 786, "token_count": 257}}
{"text": "Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g\n(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g\n(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7860447761194032, "ocr_used": true, "chunk_length": 2144, "token_count": 663}}
{"text": "Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g\n(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g\n(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7867514060500479, "ocr_used": true, "chunk_length": 2230, "token_count": 690}}
{"text": "Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g\n(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g\n(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.794531446084696, "ocr_used": true, "chunk_length": 2166, "token_count": 657}}
{"text": "I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g\n(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7692875754782266, "ocr_used": true, "chunk_length": 2058, "token_count": 639}}
{"text": "When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures. (b)Plot a graph of mass of KClO3 per 100g water against temperature at which crystals form. (c)From the graph, show and determine ; (i)the solubility of KClO3 at I. 50oC From a well plotted graph = 14.5 g KClO3/100g water II. 35oC From a well plotted graph = 9.0 g KclO3/100g water (ii)the temperature at which the solubility is: I.10g/100g water From a well plotted graph = 38.0 oC II.45g/100g water From a well plotted graph = 77.5 oC (d)Explain the shape of the graph. Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0\n(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.750891514403811, "ocr_used": true, "chunk_length": 1516, "token_count": 490}}
{"text": "Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0\n(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below. Mass of ammonium chloride(grams) 4.0 4.5 5.5 6.5 9.0 Temperature at which solid dissolved(oC) 30.0 50.0 70.0 90.0 120.0 Solubility of NH4Cl 16.0 18.0 22.0 26.0 36.0 (a)Complete the table (b)Plot a solubility curve (c)What happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC. (From the graph )Solubility at 80oC = 24.0 g Less (From the graph )Solubility at 40oC = 16.8 g Mass of crystallized/precipitated = 7.2 g 20. Solubility and solubility curves are therefore used (i) to know the effect of temperature on the solubility of a salt (ii)to fractional crystallize two soluble salts by applying their differences in solubility at different temperatures. (iii)determine the mass of crystal that is obtained from crystallization. 21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8164103890319638, "ocr_used": true, "chunk_length": 1802, "token_count": 499}}
{"text": "21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya. Temperatures around the lake are very high (30-40oC) during the day. The solubility of trona decrease with increase in temperature therefore solid crystals of trona grows on top of the lake (upto or more than 30metres thick) A bucket dredger mines the trona which is then crushed ,mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS. The CRS is then heated to chemically decompose trona to soda ash(Sodium carbonate) Chemical equation 2Na2CO3.NaHCO3.2H2O(s) -> 3Na2CO3 (s) + CO2(g) + 5H2O(l) Soda ash(Sodium carbonate) is then stored .It is called Magadi Soda. Magadi Soda is used : (i) make glass (ii) for making soapless detergents (iii) softening hard water. (iv) Common salt is colledcted at night because its solubility decreases with decrease in temperature. It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash\n23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8969850342510581, "ocr_used": true, "chunk_length": 2132, "token_count": 502}}
{"text": "It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash\n23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation. At the final pod ,the crystals are scapped together,piled in a heap and washed with brine (concentrated sodium chloride). It contains MgCl2 and CaCl2 . MgCl2 and CaCl2are hygroscopic. They absorb water from the atmosphere and form a solution. This makes table salt damp/wet on exposure to the atmosphere. 24.Some water form lather easily with soap while others do not. Water which form lather easily with soap is said to be “soft” Water which do not form lather easily with soap is said to be “hard” Hardness of water is caused by the presence of Ca2+ and Mg2+ ions. Ca2+ and Mg2+ ions react with soap to form an insoluble grey /white suspension/precipitate called Scum/ curd. Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water\nTemporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8807869549805034, "ocr_used": true, "chunk_length": 2002, "token_count": 481}}
{"text": "Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water\nTemporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e. In limestone areas; H2CO3(aq) + CaCO3(s) -> Ca(HCO3)2 (aq) In dolomite areas; H2CO3(aq) + MgCO3(s) -> Mg(HCO3)2 (aq) (b)permanent hardness of water Permanent hardness of water is caused by the presence of dissolved calcium sulphate(VI)/CaSO4 and magnesium sulphate(VI)/Mg SO4 Permanent hardness of water is caused by water dissolving CaSO4 and MgSO4 from ground rocks. Hardness of water can be removed by the following methods: (a)Removing temporary hardness of water (i)Boiling/heating. Boiling decomposes insoluble calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 to insoluble CaCO3 and MgCO3 that precipitate away. i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8298115046403969, "ocr_used": true, "chunk_length": 1788, "token_count": 497}}
{"text": "i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)\nCO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7551240122560878, "ocr_used": true, "chunk_length": 1378, "token_count": 482}}
{"text": "i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)\nCO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again. (iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7630206023464904, "ocr_used": true, "chunk_length": 1459, "token_count": 511}}
{"text": "(iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e. Na2X(aq) + Ca2+ (aq) -> Na+ (aq) + CaX(s) Na2X(aq) + Mg2+ (aq) -> Na+ (aq) + MgX(s) Hard water containing Mg2+ and Ca2+\nIon exchange resin as Sodium permutit ------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. When all the Na+ ions in the resin is fully exchanged with Ca2+ and Ng2+ ions in the permutit column ,it is said to be exhausted. Brine /concentrated sodium chloride solution is passed through the permutit column to regenerated /recharge the column again. Hard water containing Mg2+ and Ca2+ Ion exchange resin as Sodium permutit\n------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. (vi)Deionization /demineralization This is an advanced ion exchange method of producing deionized water .Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances. Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8430658539650622, "ocr_used": true, "chunk_length": 1744, "token_count": 492}}
{"text": "Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water. Chemical equation H+(aq) + OH- (aq) -> H2O(l) When exhausted the cation exchanger is regenerated by adding H+(aq) from sulphuric(VI)acid/hydrochloric acid. When exhausted the anion exchanger is regenerated by adding OH-(aq) from sodium hydroxide. Advantages of hard water Hard water has the following advantages: (i)Ca2+(aq) in hard water are useful in bone and teeth formation (ii) is good for brewing beer (iii)contains minerals that cause it to have better /sweet taste (iv)animals like snails and coral polyps use calcium to make their shells and coral reefs respectively. (v)processing mineral water Disadvantages of hard water\nHardness of water: (i)waste a lot of soap during washing before lather is formed. (ii)causes stains/blemishes/marks on clothes/garments (iii)causes fur on electric appliances like kettle ,boilers and pipes form decomposition of carbonates on heating .This reduces their efficiency hence more/higher cost of power/electricity. Sample revision questions In an experiment, soap solution was added to three separate samples of water. The table below shows the volumes of soap solution required to form lather with 1000cm3 of each sample of water before and after boiling. Sample I Sample II Sample III Volume of soap before water is boiled (cm3) 27.0 3.0 10.0 Volume of soap after water is boiled(cm3) 27.0 3.0 3.0 a) Which water sample is likely to be soft? Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8784533292088779, "ocr_used": true, "chunk_length": 1930, "token_count": 482}}
{"text": "Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard. 2.Study the scheme below and use it to aanswer the questions that follow:\n(a)Write the formula of: (i)Cation in solution K Al3+ (ii)white ppt L Al(OH)3 (iii) colourless solution M [Al(OH)4]- (iv) colourless solution N AlCl3 (v)white ppt P Al(OH)3 (b)Write the ionic equation for the reaction for the formation of: (i)white ppt L Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (v)white ppt P Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (c)What property is illustrated in the formation of colourless solution M and N. Amphotellic\n19.0.0 ENERGY CHANGES IN CHEMICAL AND PHYSICAL PROCESSES (25 LESSONS) 1.Introduction to Energy changes Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy: (i)Kinetic Energy(KE) ;the energy in motion. (ii)Potential Energy(PE); the stored/internal energy. Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb. Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8718519519663539, "ocr_used": true, "chunk_length": 1983, "token_count": 509}}
{"text": "The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt. All thermochemical reactions should be carried out at standard conditions of: (i) 298K /25oC temperature (ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure. 2.Exothermic and endothermic processes/reactions Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic. Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise) Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease) To demonstrate/illustrate exothermic and endothermic processes/reactions a) Dissolving Potassium nitrate(V)/ammonium chloride crystals Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8505925494524843, "ocr_used": true, "chunk_length": 1924, "token_count": 479}}
{"text": "Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”. (iii)Dissolution of both potassium nitrate(V) and ammonium chloride crystals is an endothermic process because initial(T1) temperature is higher than the final temperature(T2) thus causes a fall/drop in temperature. b) Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals\n3 Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals. CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8255759665074246, "ocr_used": true, "chunk_length": 1927, "token_count": 508}}
{"text": "CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2). (ii)Dissolution of both Sodium hydroxide pellets and concentrated sulphuric (VI) acid is an exothermic process because final (T2) temperature is higher than the initial temperature (T1) thus causes a rise in temperature. The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperature/thermometer readings. Dissolving both potassium nitrate(V) and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer reading. Dissolving both Sodium hydroxide pellets and concentrated sulphuric (VI) acid causes heat loss to the surrounding that causes rise in thermometer reading. At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8745945302502753, "ocr_used": true, "chunk_length": 1461, "token_count": 348}}
{"text": "At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e. Enthalpy/energy/ change in heat content ∆H = Hfinal – Hinitial For chemical reactions: ∆H = Hproducts – Hreactants For exothermic reactions, the heat contents of the reactants is more than/higher than the heat contents of products, therefore the ∆H is negative (-∆H)\n4 For endothermic reactions, the heat contents of the reactants is less than/lower than the heat contents of products, therefore the ∆H is negative (+∆H) Graphically, in a sketch energy level diagram: (i)For endothermic reactions the heat content of the reactants should be relatively/slightly lower than the heat content of the products (ii)For exothermic reactions the heat content of the reactants should be relatively/slightly higher than the heat content of the products Sketch energy level diagrams for endothermic dissolution Energy (kJ) H2 KNO3(aq) +∆H = H2 – H1 H1 KNO3(s) Reaction path/coordinate/progress Energy (kJ) H2 NH4Cl (aq) +∆H = H2 – H1 H1 NH4Cl (s) Reaction path/coordinate/progress Sketch energy level diagrams for exothermic dissolution\n5 H2 NaOH (s) Energy(kJ) -∆H = H2 – H1 H1 NaOH (aq) Reaction path/coordinate/progress H2 H2SO4 (l) Energy (kJ) -∆H = H2 – H1 H1 H2SO4 (aq) Reaction path/coordinate/progress 3.Energy changes in physical processes Melting/freezing/fusion/solidification and boiling/vaporization/evaporation are the two physical processes. Melting /freezing point of pure substances is fixed /constant. The boiling point of pure substance depend on external atmospheric pressure. Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8689167802161532, "ocr_used": true, "chunk_length": 1937, "token_count": 493}}
{"text": "Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e A (s) ========A(l)\n6 Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour. Condensation/liquidification is the physical change of gas/vapour to liquid. Boiling/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e B (l) ========B(g) Practically (i) Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together. Solids are made up of very strong bonds holding the particles very close to each other (Kinetic Theory of matter).On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom. Melting/freezing/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding. (ii)Freezing/fusion/solidification involves cooling a a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter). Freezing /fusion / solidification is an exothermic (-∆H)process that require particles holding the liquid together to lose energy to the surrounding. (iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together. Gaseous particles have high degree of freedom (Kinetic Theory of matter). Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding. (iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process. The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9136551945619013, "ocr_used": true, "chunk_length": 2198, "token_count": 485}}
{"text": "(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process. The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g. H2O(s) -> H2O(l) ∆H = +6.0kJ mole-1 (endothermic process) H2O(l) -> H2O(s) ∆H = -6.0kJ mole-1 (exothermic process) The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. e.g. H2O(l) -> H2O(g) ∆H = +44.0kJ mole-1 (endothermic process) H2O(g) -> H2O(l) ∆H = -44.0kJ mole-1 (exothermic process)\n7 The following experiments illustrate/demonstrate practical determination of melting and boiling a) To determine the boiling point of water Procedure: Measure 20cm3 of tap water into a 50cm3 glass beaker. Determine and record its temperature.Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minutes. Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 2.From the graph show and determine the boiling point of water Note: Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8054535377280634, "ocr_used": true, "chunk_length": 1793, "token_count": 510}}
{"text": "Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 2.From the graph show and determine the boiling point of water Note: Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC. 8 3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O) Working: Mass of water = density x volume => (20 x 1) /1000 = 0.02kg Quantity of heat produced = mass of water x specific heat capacity of water x temperature change =>0.02kg x 4.2 x ( 96 – 25 ) = 5.964kJ Heat of vaporization of one mole H2O = Quantity of heat Molar mass of H2O =>5.964kJ = 0.3313 kJ mole -1 18 To determine the melting point of candle wax Procedure Weigh exactly 5.0 g of candle wax into a boiling tube. Heat it on a strongly Bunsen burner flame until it completely melts. Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7595258358662614, "ocr_used": true, "chunk_length": 1316, "token_count": 383}}
{"text": "Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes. Sample results Time(seconds) 0 30 60 90 120 150 180 210 240 Temperature(oC) 93.0 85.0 78.0 70.0 69.0 69.0 69.0 67.0 65.0 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time 93 oC Temperature(0C) melting point 69oC time(seconds) 2.From the graph show and determine the melting point of the candle wax\n9 4.Energy changes in chemical processes Thermochemical reactions measured at standard conditions of 298K(25oC) and 101300Pa/101300Nm2/ 1 atmospheres/760mmHg/76cmHg produce standard enthalpies denoted ∆Hᶿ. Thermochemical reactions are named from the type of reaction producing the energy change. Below are some thermochemical reactions: (a) Standard enthalpy/heat of reaction ∆Hᶿr (b) Standard enthalpy/heat of combustion ∆Hᶿc (c) Standard enthalpy/heat of displacement ∆Hᶿd (d) Standard enthalpy/heat of neutralization ∆Hᶿn (e) Standard enthalpy/heat of solution/dissolution ∆Hᶿs (f) Standard enthalpy/heat of formation ∆Hᶿf (a)Standard enthalpy/heat of reaction ∆Hᶿr The molar standard enthalpy/heat of reaction may be defined as the energy/heat change when one mole of products is formed at standard conditions A chemical reaction involves the reactants forming products. For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8338963285071899, "ocr_used": true, "chunk_length": 1715, "token_count": 486}}
{"text": "For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy. The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494\n10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7971568227454181, "ocr_used": true, "chunk_length": 1634, "token_count": 509}}
{"text": "The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494\n10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7405569921995552, "ocr_used": true, "chunk_length": 1773, "token_count": 618}}
{"text": "(ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)\n11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6891285497767391, "ocr_used": true, "chunk_length": 1583, "token_count": 597}}
{"text": "Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)\n11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6836973990177859, "ocr_used": true, "chunk_length": 1626, "token_count": 620}}
{"text": "The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)\n11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7008144114844016, "ocr_used": true, "chunk_length": 1322, "token_count": 490}}
{"text": "The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)\n11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products. 12 (b)Standard enthalpy/heat of combustion ∆Hᶿc The molar standard enthalpy/heat of combustion(∆Hᶿc) is defined as the energy/heat change when one mole of a substance is burnt in oxygen/excess air at standard conditions. Burning is the reaction of a substance with oxygen/air. It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7968701766420855, "ocr_used": true, "chunk_length": 1372, "token_count": 427}}
{"text": "It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use. A fuel may be solid (e.g coal, wood, charcoal) liquid (e.g petrol, paraffin, ethanol, kerosene) or gas (e.g liquefied petroleum gas/LPG, Water gasCO2/H2, biogas-methane, Natural gas-mixture of hydrocarbons) To determine the molar standard enthalpy/heat of combustion(∆Hᶿc) of ethanol Procedure Put 20cm3 of distilled water into a 50cm3 beaker. Clamp the beaker. Determine the temperature of the water T1.Weigh an empty burner(empty tin with wick). 13 Record its mass M1.Put some ethanol into the burner. Weigh again the burner with the ethanol and record its mass M2. Ignite the burner and place it below the clamped 50cm3 beaker. Heat the water in the beaker for about one minute. Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8083027816771837, "ocr_used": true, "chunk_length": 1673, "token_count": 491}}
{"text": "Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2. Given that the specific heat capacity of water is 4.2 kJ-1kg-1K-1,determine the heat produced during the burning. Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 10 = 840 Joules = 0.84 kJ 1000 3.Calculate the molar heat of combustion of ethanol Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 0.84 kJ = 96.5517 kJmole-1 0.0087 /8.7 x 10-3 moles 4.List two sources of error in the above experiment. (i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7593360059230367, "ocr_used": true, "chunk_length": 1512, "token_count": 482}}
{"text": "(i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel. Heating value = molar heat of combustion => 96.5517 kJmole-1 = 2.0989 kJg-1 Molar mass of fuel 46 g 6.Explain other factors used to determine the choice of fuel for domestic and industrial use. (i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost. (ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky. (iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic. (iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point. 7.Explain the methods used to reduce pollution from common fuels. (i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8782052883388274, "ocr_used": true, "chunk_length": 2069, "token_count": 498}}
{"text": "(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000\n15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.780100057206779, "ocr_used": true, "chunk_length": 1209, "token_count": 374}}
{"text": "Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000\n15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.734552441689726, "ocr_used": true, "chunk_length": 1438, "token_count": 507}}
{"text": "Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000\n15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7704582406576398, "ocr_used": true, "chunk_length": 1569, "token_count": 518}}
{"text": "1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7848854034643009, "ocr_used": true, "chunk_length": 1440, "token_count": 488}}
{"text": "Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)\n16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7760502724991214, "ocr_used": true, "chunk_length": 1502, "token_count": 525}}
{"text": "A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)\n16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal. To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper Procedure Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter. Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8149356811662185, "ocr_used": true, "chunk_length": 1289, "token_count": 413}}
{"text": "Determine and record the temperature of the solution T1.Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer. Determine and record the highest temperature change to the nearest 0.5oC- T2 . Repeat the experiment to complete table 1 below Table 1 Experiment I II Final temperature of solution(T2) 30.0oC 31.0oC Final temperature of solution(T1) 25.0oC 24.0oC Change in temperature(∆T) 5.0 6.0 Questions 1.(a) Calculate: (i)average ∆T Average∆T = change in temperature in experiment I and II =>5.0 + 6.0 = 5.5oC 2 (ii)the number of moles of solution used Moles used = molarity x volume of solution = 0.2 x 20 = 0.004 moles 1000 1000 (iii)the enthalpy change ∆H for the reaction Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 5.5 = 462 Joules = 0.462 kJ 1000 (iv)State two assumptions made in the above calculations. Density of solution = density of water = 1gcm-3 Specific heat capacity of solution=Specific heat capacity of solution=4.2 kJ-1kg-1K This is because the solution is assumed to be infinite dilute. 2. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions. Molar heat of displacement ∆Hd = Heat produced ∆H\n17 Number of moles of fuel => 0.462 kJ = 115.5 kJmole-1 0.004 3.Write an ionic equation for the reaction taking place. Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq) 4.State the observation made during the reaction. Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7828083028083028, "ocr_used": true, "chunk_length": 1638, "token_count": 497}}
{"text": "Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless. Brown solid deposits are formed at the bottom of reaction vessel/ beaker. 5.Illustrate the above reaction using an energy level diagram. Zn(s) + Cu2+(aq) Energy ∆H = -115.5 kJmole-1 (kJ) Cu(s) + Zn2+(aq) Reaction progress/path/coordinates 6. Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution. No effect. Cu2+ (aq) are displaced from their solution.The element used to displace it does not matter.The reaction however faster if a more reactive metal is used. 7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes\n18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8104416856995507, "ocr_used": true, "chunk_length": 1568, "token_count": 479}}
{"text": "7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes\n18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7678189300411523, "ocr_used": true, "chunk_length": 1330, "token_count": 438}}
{"text": "The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings. Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>2.204 kJ = 0.0206moles 126 moles Molarity of the solution = moles x 1000 Volume of solution used = 0.0206moles x 1000 = 0.5167 M 40 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature. Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>25.08kJmole-1x 0.5 moles = 1.254 kJ x 1000 =1254J Mass of solution (m) = Heat produced ∆H specific heat capacity (c)x ∆T => 1254J = 99.5238 g 4.2 x 3 Volume = mass x density = 99.5238 g x 1 = 99.5238cm3 Note: The solution assumes to be too dilute /infinite dilute such that the density and specific heat capacity is assumed to be that of water. Graphical determination of the molar enthalpy of displacement of copper Procedure:\n19 Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper. Record its temperature at time T= 0. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds . Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7986372661891101, "ocr_used": true, "chunk_length": 1669, "token_count": 483}}
{"text": "Place all the (1.5g) Zinc powder provided. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC. Sample results Time oC 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Temperature 25.0 25.0 25.0 25.0 25.0 xxx 36.0 35.5 35.0 34.5 Sketch graph of temperature against time 36.5 Extrapolation Temperature point ∆T oC 130 Time(seconds) Questions 1. Show and determine the change in temperature ∆T From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation ∆T = 36.5 – 25.0 = 11.5oC 2.Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1 Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 11.5 = 966 Joules = 0.966 kJ 1000 Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>.966 kJ = 0.007728moles 125 moles 7.728 x 10-3moles 2. What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.699671757236261, "ocr_used": true, "chunk_length": 1317, "token_count": 451}}
{"text": "What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 20 4.The actual concentration of copper(II)sulphate(VI) solution was 0.4M.Explain the differences between the two. Practical value is lower than theoretical. Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases ∆T hence lowering the practical number of moles and molarity against the theoretical value 5.a) In an experiment to determine the molar heat of reaction when magnesium displaces copper ,0.15g of magnesium powder were added to 25.0cm3 of 2.0M copper (II) chloride solution. The temperature of copper (II) chloride solution was 25oC.While that of the mixture was 43oC. i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8231368705209372, "ocr_used": true, "chunk_length": 1682, "token_count": 480}}
{"text": "i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water. Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH- ions to form salt and water only. 21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8485383504467474, "ocr_used": true, "chunk_length": 1703, "token_count": 485}}
{"text": "21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH- ions) and exist more as molecules. Neutralization is an exothermic(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations. (ii) (i)for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH- ions.The overall energy evolved is comparatively higher/more than weak acid-base/ alkali neutralizations. For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used. This is because ionically: OH-(aq)+ H+(aq) -> H2O(l) for any wholly dissociated acid/base/alkali Practically ∆Hᶿn can be determined as in the examples below: To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid Procedure Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper. Record its temperature T1.Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide. Rinse the bulb of the thermometer in distilled water. Determine the temperature of the sodium hydroxide T2.Average T2 andT1 to get the initial temperature of the mixture T3. Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8764674919341354, "ocr_used": true, "chunk_length": 2006, "token_count": 507}}
{"text": "Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture. Repeat the experiment to complete table 1. Table I . Sample results Experiment I II Temperature of acid T1 (oC) 22.5 22.5 Temperature of base T2 (oC) 22.0 23.0 Final temperature of solution T4(oC) 35.5 36.0 Initial temperature of solution T3(oC) 22.25 22.75\n22 (a)Calculate T6 the average temperature change T6 = 13.25 +13.75 = 13.5 oC 2 (b)Why should the apparatus be very clean? Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ∆Hᶿn. (c)Calculate the: (i)number of moles of the acid used number of moles = molarity x volume => 2 x 50 = 0.1moles 1000 1000 (ii)enthalpy change ∆H of neutralization. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7642897371355012, "ocr_used": true, "chunk_length": 1381, "token_count": 435}}
{"text": "∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above. The theoretical value is higher Heat/energy loss to the surrounding/environment lowers ∆T/T6 and thus ∆Hn Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn (e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of: (i) potassium hydroxide with nitric(V) acid The results would be the same/similar. Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH- ions. (ii) ammonia with ethanoic acid Temperature change( T5) 13.25 13.75\n23 The results would be lower/∆Hn would be less. Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8192202903746192, "ocr_used": true, "chunk_length": 1594, "token_count": 479}}
{"text": "Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7631828978622328, "ocr_used": true, "chunk_length": 1263, "token_count": 419}}
{"text": "If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T\n24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7180613002439283, "ocr_used": true, "chunk_length": 1796, "token_count": 665}}
{"text": "Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T\n24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7080896913220734, "ocr_used": true, "chunk_length": 1717, "token_count": 648}}
{"text": "∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T\n24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7328952074391989, "ocr_used": true, "chunk_length": 1398, "token_count": 504}}
{"text": "∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T\n24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0. Put one portion of the base into the beaker containing the acid. Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC. Repeat the procedure above with other portions of the base to complete table 1 below Table 1:Sample results.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7534898785425101, "ocr_used": true, "chunk_length": 1425, "token_count": 482}}
{"text": "(b)Plot a graph of volume of sodium hydroxide against temperature change. 28.7=T2 temperature(oC) ∆T From the graph show and determine : (i)the highest temperature change ∆T ∆T =T2-T1 => highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0 :T1 =>∆T = 6.7 – 0.0 = 6.70C (ii)the volume of sodium hydroxide used for complete neutralization From a correctly plotted graph – 16.75cm3 (c)Calculate the number of moles of the alkali used Moles NaOH = molarity x volume =>2M x 16.75cm3 = 0.0335 moles 1000 1000 (d)Calculate ∆H for the reaction ∆H = mass of solution(acid+base) x c x ∆T =>(25.0 + 16.75) x 4.2 x 6.7 = 1174.845 J = 1.174845kJ 1000 (e)Calculate the molar enthalpy of neutralization of the alkali. ∆Hn = ∆Hn = 1.174845kJ = 35.0701kJ Number of moles 0.0335 Volume of sodium hydroxide(cm3) 22. oC T1\n26 (d)Standard enthalpy/heat of solution ∆Hᶿs The standard enthalpy of solution ∆Hᶿsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution. An infinite dilute solution is one which is too dilute to be diluted further. Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8030781253591377, "ocr_used": true, "chunk_length": 1565, "token_count": 482}}
{"text": "Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl). The table below shows some ∆Hl in kJ for the process MX(s) -> M+ (g) + X- (g) Li Na K Ca Mg F +1022 +900 +800 +760 +631 Cl +846 +771 +690 +2237 +2493 Br +800 +733 +670 +2173 +2226 (ii)surrounding the free ions by polar water molecules. This process is called hydration. The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8202182627811919, "ocr_used": true, "chunk_length": 1406, "token_count": 445}}
{"text": "The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ\n27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8050316122233931, "ocr_used": true, "chunk_length": 1168, "token_count": 423}}
{"text": "The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ\n27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7560774801690011, "ocr_used": true, "chunk_length": 1486, "token_count": 593}}
{"text": "(ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ\n27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7495472370766488, "ocr_used": true, "chunk_length": 1122, "token_count": 441}}
{"text": "(a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ\n27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7056038143991302, "ocr_used": true, "chunk_length": 1396, "token_count": 603}}
{"text": "Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.696976254199647, "ocr_used": true, "chunk_length": 1033, "token_count": 449}}
{"text": "Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature. (d)Dissolving magnesium bromide crystal/s// MgBr2 (s) ->MgBr2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Br-(g) ∆Hl =+2226 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Br-(g) + aq -> 2Br-(aq) ∆Hh = (- 335x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 335 x 2 kJ)) + +2226 kJ = -364.0 kJmole-1 MgBr2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.733331892888524, "ocr_used": true, "chunk_length": 1268, "token_count": 500}}
{"text": "Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined. To determine the ∆Hs ammonium nitrate Place 100cm3 of distilled water into a plastic beaker/calorimeter. Determine its temperature and record it at time =0 in table I below. Put all the 5.0g of ammonium nitrate (potassium nitrate/ammonium chloride can also be used)provided into the plastic beaker/calorimeter, stir using a thermometer and record the highest temperature change to the nearest 0.5oCafter every ½ minute to complete table I. Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1\n29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7739621151817552, "ocr_used": true, "chunk_length": 1593, "token_count": 482}}
{"text": "Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1\n29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7009419516935695, "ocr_used": true, "chunk_length": 1029, "token_count": 367}}
{"text": "oC T1\n29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working\n30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.711867596116837, "ocr_used": true, "chunk_length": 1745, "token_count": 661}}
{"text": "∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working\n30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7302600036879956, "ocr_used": true, "chunk_length": 1479, "token_count": 541}}
{"text": "The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working\n30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7395358551230437, "ocr_used": true, "chunk_length": 1405, "token_count": 504}}
{"text": "Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working\n30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7563769256515329, "ocr_used": true, "chunk_length": 1557, "token_count": 533}}
{"text": "∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”. Hess’ law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess(1802-1850). He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between. i.e. A(s) --∆H1-->C(s) = A(s) --∆H2-->B(s)--∆H3-->C(s) Applying Hess’ law of constant heat summation then:\n31 A(s) ∆H2 B(s) ∆H1 ∆H3 C(s) The above is called an energy cycle diagram. It can be used to calculate any of the missing energy changes since: (i) ∆H1 =∆H2 + ∆H3 (ii) ∆H2 =∆H1 + -∆H3 (iii) ∆H3 = - ∆H1 + ∆H2 Examples of applying Hess’ law of constant heat summation 1.Calculate the molar enthalpy of formation of methane (CH4) given that ∆Hᶿc of carbon-graphite is -393.5kJmole-1,Hydrogen is -285.7 kJmole-1 and that of methane is 890 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form methane. Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8361214183190687, "ocr_used": true, "chunk_length": 1447, "token_count": 418}}
{"text": "Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. C(s)+ 2H2 (g)+2O2 (g) --∆H1--> CH4(g) +2O2(g) --∆H2--> CO2(g)+2H2O(l) C(s)+ 2H2 (g)+2O2 (g) --∆H3--> CO2(g)+2H2O(l) Energy cycle diagram C(s) + 2H2 (g) + 2O2(g) ∆H1=∆Hᶿc =-890.4kJ CH4(g)+2O2(g) ∆H3=∆Hᶿc =-393.5kJ ∆H3=∆Hᶿc =-285.7kJ x 2 ∆H2= ∆Hᶿf= x CO2(g) + 2H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 -393.5 + (-285.7 x 2) = -890.4kJ + x x = -74.5 kJ\n32 Heat of formation ∆Hᶿf CH4 = -74.5 kJmole-1 2. Calculate the molar enthalpy of formation of ethyne (C2H2) given : ∆Hᶿc of carbon-graphite = -394kJmole-1,Hydrogen = -286 kJmole-1 , (C2H2) = -1300 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form ethyne. Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6950072875307837, "ocr_used": true, "chunk_length": 1010, "token_count": 447}}
{"text": "Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. 2C(s)+ H2 (g)+2 ½ O2 (g) --∆H1--> C2 H2 (g) +2 ½ O2(g) --∆H2--> CO2(g)+H2O(l) 2C(s)+ H2 (g)+ 2 ½ O2 (g) --∆H3--> 2CO2(g)+H2O(l) Energy cycle diagram 2C(s) + H2 (g) +2½O2(g) ∆H1=∆Hᶿf =x C2 H2+2½O2(g) ∆H3=∆Hᶿc =-394kJx 2 ∆H3=∆Hᶿc =-286kJ ∆H2= ∆Hᶿc= -1300kJ 2CO2(g) + H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 ( -394 x 2) + -286 = -1300kJ + x x = +244 kJ Heat of formation ∆Hᶿf CH4 = +244 kJmole-1 3. Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿc of carbon-graphite = -393.5kJmole-1, ∆Hᶿc of carbon(II)oxide (CO)= -283 kJmole-1 Working Carbon-graphite reacts with oxygen first to form carbon (II)oxide (CO). Carbon(II)oxide (CO) then burn in the excess oxygen to form carbon(IV)oxide. Carbon-graphite can burn in excess oxygen to form carbon (IV) oxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6904088734232274, "ocr_used": true, "chunk_length": 968, "token_count": 434}}
{"text": "Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1. Table 1:Sample results\n35 Experiment I II Highest /lowest temperature T2 27.0 29.0 Initial temperature T1 24.0 25.0 Change in temperature ∆T 3.0 4.0 Experiment II Weigh accurately 8.0g of anhydrous copper(II)sulphate(VI). Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.827422185277547, "ocr_used": true, "chunk_length": 802, "token_count": 211}}
{"text": "Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T\n36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7108569872711185, "ocr_used": true, "chunk_length": 1239, "token_count": 468}}
{"text": "Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T\n36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6854885938032006, "ocr_used": true, "chunk_length": 1320, "token_count": 532}}
{"text": "Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T\n36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6501005747126437, "ocr_used": true, "chunk_length": 1440, "token_count": 625}}
{"text": "∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information. Ammonium chloride dissolves in water to form ammonium chloride solution. Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution. The heat change taking place can be calculated from the heat of reactions:\n37 (i) NH3(aq) + HCl(aq) -> NH4Cl(s) (ii) NH4Cl(s) + (aq) -> NH4Cl(aq) (iii) NH3(aq) + HCl(aq) -> NH4Cl(aq) Experiment procedure I Measure 50cm3 of water into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table I. Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7362162162162161, "ocr_used": true, "chunk_length": 1295, "token_count": 462}}
{"text": "Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I. Sample results TableI Experiment I II final temperature(oC) 19.0 20.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 3.0 2.0 Experiment procedure II Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table II. Measure 25cm3 of 2M hydrochloric acid solution. Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8253415393648365, "ocr_used": true, "chunk_length": 751, "token_count": 187}}
{"text": "Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:\n38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6582804053251078, "ocr_used": true, "chunk_length": 1367, "token_count": 595}}
{"text": "Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:\n38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:\n39 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.650154596702481, "ocr_used": true, "chunk_length": 1347, "token_count": 592}}
{"text": "Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:\n38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:\n39 1. Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6527639112011989, "ocr_used": true, "chunk_length": 1354, "token_count": 595}}
{"text": "Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that. ∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6\n40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1\n2\n20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.657868050165448, "ocr_used": true, "chunk_length": 1883, "token_count": 809}}
{"text": "∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6\n40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1\n2\n20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6624398139289781, "ocr_used": true, "chunk_length": 1909, "token_count": 817}}
{"text": "Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6\n40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1\n2\n20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7307258938244854, "ocr_used": true, "chunk_length": 1562, "token_count": 580}}
{"text": "41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1\n2\n20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering. Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation. Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids. The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries. Theories of rates of reaction. The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e. Rate of reaction = Change in concentration/amount of reactants Time taken for the change to occur Rate of reaction = Change in concentration/amount of products formed Time taken for the products to form For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product. The SI unit of time is second(s) but minutes and hours are also used. (a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8790676935660133, "ocr_used": true, "chunk_length": 2083, "token_count": 489}}
{"text": "(a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions (iii)the speed at which particles collide is called collision frequency. The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products. (iv)the higher the chances of successful collisions, the faster the reaction. (v)the average distance between solid particles from one another is too big for them to meet and collide successfully. (vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place. The solute particle distance is reduced as the particle ions are free to move in the solvent medium. (vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products. (b)The Activation Energy(Ea) theory\nThe Enthalpy of activation(∆Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles. Bond breaking is an endothermic process that require an energy input. The higher the bond energy the slower the reaction to start of. Activation energy does not influence whether a reaction is exothermic or endothermic. The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr\nThe activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9231133939154486, "ocr_used": true, "chunk_length": 2531, "token_count": 498}}
{"text": "Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr\nThe activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy. Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion. 3. Measuring the rate of a chemical reaction. The rate of a chemical reaction can be measure as: (i)Volume of a gas in unit time; - if reaction is producing a gas as one of the products. - if reaction is using a gas as one reactants (ii)Change in mass of reactants/products for solid products/reactants in unit time. (iii)formation of a given mass of precipitate in unit time (iv)a certain mass of reactants to completely form products/diminish. Reactants may be homogenous or heterogenous. -Homogenous reactions involve reactants in the same phase/state e.g. solid-solid,gas-gas,liquid-liquid. -Heterogenous reactions involve reactants in the different phase/state e.g. solid-liquid,gas-liquid,solid-gas. 4. Factors influencing/altering/affecting/determining rate of reaction The following factors alter/influence/affect/determine the rate of a chemical reaction: (a)Concentration (b)Pressure (c) Temperature (d)Surface area\n(e)Catalyst a) Influence of concentration on rate of reaction The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9103601783396353, "ocr_used": true, "chunk_length": 2350, "token_count": 509}}
{"text": "An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7624523362698574, "ocr_used": true, "chunk_length": 1614, "token_count": 478}}
{"text": "Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock. (ii)Consistent use of a minimum of four decimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator. (c) accuracy against a school value based on candidate’s teachers-results submitted. (d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results. Sample questions 1. On separate graph papers plot a graph of: (i)volume of acid used(x-axis) against time. Label this graph I (ii) volume of acid used(x-axis) against 1/t. Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7569472938017285, "ocr_used": true, "chunk_length": 1454, "token_count": 437}}
{"text": "Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)\nSketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)\n1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6847556258790436, "ocr_used": true, "chunk_length": 1440, "token_count": 503}}
{"text": "Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)\nSketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)\n1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)\nReaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6996065808297567, "ocr_used": true, "chunk_length": 1864, "token_count": 659}}
{"text": "Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)\nSketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)\n1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)\nReaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6978404613234251, "ocr_used": true, "chunk_length": 1849, "token_count": 660}}
{"text": "Sketch sample Graph I Time (seconds)\nSketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)\n1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)\nReaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6916010670927429, "ocr_used": true, "chunk_length": 1835, "token_count": 669}}
{"text": "Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)\nReaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8299342922150705, "ocr_used": true, "chunk_length": 949, "token_count": 281}}
{"text": "Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water (cm3) Volume of sodium thiosulphate (cm3) Concentation of sodium thisulphate in molesdm-3 Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds) T-1 5.0 0.0 25.0 0.4 20.0 5.0 x 10-2 5.0 5.0 20.0 0.32 23.0 4.35 x 10-2 5.0 10.0 15.0 0.24 27.0 3.7 x 10-2 5.0 15.0 10.0 0.16 32.0 3.13 x 10-2 Note concentration of diluted solution is got: C1V1=C2V2 => 0.4 x 25 = C2x 25 =0.4M C1V1=C2V2 => 0.4 x 20 = C2x 25 =0.32M C1V1=C2V2 => 0.4 x 15 = C2x 25 =0.24M C1V1=C2V2 => 0.4 x 10 = C2x 25 =0.16M Sample questions 1. On separate graph papers plot a graph of: (i)Concentration of sodium thiosulphate against time. Label this graph I\n(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II 2. Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.693093991523311, "ocr_used": true, "chunk_length": 1337, "token_count": 476}}
{"text": "Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M From correct graph at concentration 0.192M => 2.4 x10-2 I/t = 2.4 x10-2 t = 41.6667seconds (ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water. At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6860092540586804, "ocr_used": true, "chunk_length": 1089, "token_count": 386}}
{"text": "At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M Mole ratio Na2S2 O3 :HCl =1:2 Moles of Na2S2 O3 = 0.192M x 12 => 2.304 x 10-3 moles 1000 Mole ratio HCl =2.304 x 10-1 moles = 1.152 x 10-3 moles 2 Molarity o f HCl = 1.152 x 10-3 moles x 1000 = 0.2304M 5.0\nMethod 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below: Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6868485980062016, "ocr_used": true, "chunk_length": 1348, "token_count": 491}}
{"text": "cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below. Sample results: Table II Time(seconds) 0 30 60 90 120 150 180 210 240 Volume of gas produced(cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 Sample practice questions 1.Plot a graph of volume of gas produced (y-axis) against time Magnesium ribbon Hydrochloric acid Graduated gas jar Hydrogen gas\n2.Explain the shape of the graph. The rate of reaction is faster when the concentration of the acid is high . As time goes on, the concentration of the acid decreases and therefore less gas is produced. When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens. 3.Calculate the rate of reaction at 120 seconds From a tangent at 120 seconds rate of reaction = Change in volume of gas Change in time => From the tangent at 120seconds V2 - V1 = 96-84 = 12 = 0.2cm3sec-1 T2 - T1 150-90 60 4. Write an ionic equation for the reaction taking place. Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g) 5. On the same axis sketch then explain the curve that would be obtained if: (i) 0.1 M hydrochloric acid is used –Label this curve I (ii)1.0 M hydrochloric acid is used –Label this curve II\nObservation: Curve I is to the right Curve II is to the left Explanation A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction. An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction. Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8179694256796243, "ocr_used": true, "chunk_length": 1912, "token_count": 495}}
{"text": "Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained. Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products. An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react. At industrial level, the following are some reactions that are affected by pressure: (a)Haber process for manufacture of ammonia N2(g) + 3H2(g) -> 2NH3(g) (b)Contact process for manufacture of sulphuric(VI)acid 2SO2(g) + O2(g) -> 2SO3(g) (c)Ostwalds process for the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l) The influence of pressure on reaction rate is not felt in solids and liquids. This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter) c)Influence of temperature on rate of reaction\nAn increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency. Increase in temperature increases the particles which can overcome the activation energy (Ea). A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half. Practical determination of effect of Temperature on reaction rate Method 1 Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Determine and record its temperature as room temperature in table 2 below. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8859239905985059, "ocr_used": true, "chunk_length": 2166, "token_count": 501}}
{"text": "Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC. Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate. Sample results:Table 2. Temperature of Na2S2O3 Room temperature 30 40 50 60 Time taken for mark X to be obscured /invisible (seconds) 50.0 40.0 20.0 15.0 10.0 Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1 Sample practice questions 1. Plot a graph of temperature(x-axis) against 1/t\n2(a)From your graph determine the temperature at which: (i)1/t is ; I. 0.03 Reading directly from a correctly plotted graph = 32.25 oC II. 0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.742122345905665, "ocr_used": true, "chunk_length": 1163, "token_count": 353}}
{"text": "0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III. 25 seconds 25 seconds => 1/t =1/25 =0.04 Reading directly from a correctly plotted graph 0.04 => 36.0 oC (b) From your graph determine the time taken for the cross to become invisible at: (i) 57.5 oC Reading directly from a correctly plotted graph at 57.5 oC= 0.094 =>1/t = 0.094 t= 1/0.094 => 10.6383 seconds\n(ii) 45 oC Reading directly from a correctly plotted graph at 45 oC = 0.062 =>1/t = 0.062 t= 1/0.094 => 16.1290 seconds (iii) 35 oC Reading directly from a correctly plotted graph at 35 oC = 0.047 =>1/t = 0.047 t= 1/0.047 => 21.2766 seconds Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 5centimeter length of magnesium ribbon with sand paper/steel wool. Cut the piece into five equal one centimeter smaller pieces. Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker . Put one piece of the magnesium ribbon into the acid, swirl. Immediately start off the stop watch/clock. Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above. Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7374353958410766, "ocr_used": true, "chunk_length": 1649, "token_count": 506}}
{"text": "Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above . Record each time to complete table 2 below using different temperatures of the acid. Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions\n1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7316266947347319, "ocr_used": true, "chunk_length": 1410, "token_count": 475}}
{"text": "Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions\n1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)\nMole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6632147513503446, "ocr_used": true, "chunk_length": 2065, "token_count": 771}}
{"text": "Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions\n1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)\nMole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6653719120799779, "ocr_used": true, "chunk_length": 2073, "token_count": 771}}
{"text": "Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)\nMole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6933469292953327, "ocr_used": true, "chunk_length": 2029, "token_count": 710}}
{"text": "1/t Temperature(oC)\nMole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products. d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting\nsolid lumps into smaller pieces/chips then crushing the chips into powder.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7170520089940366, "ocr_used": true, "chunk_length": 1544, "token_count": 495}}
{"text": "d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting\nsolid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area. An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions. Reaction of chalk/calcium carbonate on dilute hydrochloric acid Procedure Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 . Using a watch glass weigh three separate 2.5g a piece of white chalk. Place the conical flask C1 on an electronic balance. Reset the balance scale to 0.0. Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0. Simultaneously start of the stop watch. Determine and record the scale reading after every 30 seconds to complete Table I . Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieces/chips for C2 and crushing the chalk to powder for C3 Sample results:Table 1. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 2.0 1.8 1.4 1.2 1.0 0.8 0.5 0.5 Loss in mass 0.0 0.5 0.7 1.1 1.3 1.5 1.7 2.0 2.0 Sample results:Table 1I.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7920976592947179, "ocr_used": true, "chunk_length": 1618, "token_count": 439}}
{"text": "Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5 Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0 Sample questions: 1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III 2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III. 3.Explain why there is a loss in mass in all experiments. Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere. 4.Write an ionic equation for the reaction that take place CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g) 5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7483385238080258, "ocr_used": true, "chunk_length": 1149, "token_count": 380}}
{"text": "Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion. 6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0) Method I Mole ratio CaCO3(s) : CO2(g) = 1:1 Moles CaCO3(s) used = Mass CaCO3(s) = 0.025 moles Molar mass CaCO3(s) Moles CO2(g) = 0.025 moles Volume of CO2(g) = moles x molar gas volume =>0.025 moles x 24 dm3 = 0.600 dm3/600cm3\nMethod II Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g) Mass of CaCO3(s) =2.5 g produce 2.5 x 24 = 0.600dm3 100 7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve. From tangent at 180 seconds on curve I Rate = M2-M1 => 2.08 – 1.375 = 0.625 = 0.006944g sec-1 T2- T1 222-132 90 8.What is the effect of particle size on the rate of reaction? A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency. Theoretical examples 1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7200267649764842, "ocr_used": true, "chunk_length": 1509, "token_count": 505}}
{"text": "Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass? Carbon (IV) oxide gas was produced that escape to the surrounding (b)Calculate the average rate of loss in mass between: (i) 0 to 2 minutes Average rate =M2-M1 => 1.80 – 0.0 = 1.8 = 9.00g min-1 T2- T1 2.0 – 0.0 2 (i) 6 to 8 minutes Average rate =M2-M1 => 3.20 – 2.95 = 0.25 = 0.125g min-1 T2- T1 8.0 – 6.0 2 (iii) Explain the difference between the average rates of reaction in (i) and(ii) above. Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products. Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products. (c)Write the equation for the reaction that takes place. CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.772822299651568, "ocr_used": true, "chunk_length": 1722, "token_count": 489}}
{"text": "CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster. (iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles. (e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen. It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent. (f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed . (i)Name the white precipitate. Calcium(II)sulphate(VI) (ii)Write an ionic equation for the formation of the white precipitate Ca2+(aq) + SO42-(aq)->CaSO4(s) (iii)State one use of the white precipitate -Making plaster for building -Manufacture of plaster of Paris\n-Making sulphuric(VI)acid (g)(i) Plot a graph of total loss in mass(y-axes) against time (ii)From the graph, determine the rate of reaction at time 2 minutes. From a tangent/slope at 2 minutes; Rate of reaction = Average rate =M2-M1 => 2.25 – 1.30 = 0.95 = 0.3958g min-1 T2- T1 3.20 – 0.8 2.4 (iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II e) Influence of catalyst on rate of reaction Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8601477136471875, "ocr_used": true, "chunk_length": 2036, "token_count": 506}}
{"text": "Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(∆Ha)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(∆Ha)/activation energy (Ea) by: (i) forming short lived intermediate compounds called activated complex that break up to form the final product/s (ii) being absorbed by the reactants thus providing the surface area on which reaction occurs. A catalyst has no effect on the enthalpy of reaction ∆Hr but only lowers the Enthalpy of activation(∆Ha)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Ea Catalysed A A B B Energy kJ Ea uncatalysed A-A B-B A-B A-B\nEnergy level diagram showing the activation energy for endothermic processes /reactions. Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9032731888657177, "ocr_used": true, "chunk_length": 1681, "token_count": 393}}
{"text": "Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities. (b)Ostwalds process A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr\nPlatinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) ----Pt/Rh--> 4NO (g) + 6H2O(l) (c)Haber process Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas N2(g) + 3H2(g) ---Pt or Fe---> 2NH3(g) (d)Hydrogenation/Hardening of oil to fat Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing - C=C- or –C=C- to saturated compounds without double or triple bond This process is used is used in hardening oil to fat. (e)Decomposition of hydrogen peroxide Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas. This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8643993010048056, "ocr_used": true, "chunk_length": 1635, "token_count": 458}}
{"text": "This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen. H2 SO4 (aq) + Zn(s) ----CuSO4--> ZnSO4 (aq) + H2(g) (g) Substitution reactions When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow. e.g. CH4(g) + Cl2(g) ---u.v. light--> CH3Cl(g) + HCl(g) (h)Photosynthesis Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet light. 6CO2(g) + 6H2O(l) ---u.v. light--> C6H12O6(g) + O2(g) (i)Photography\nPhotographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight. 2AgBr(s) ---u.v/sun light--> 2Ag(s) + Br2(l) When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred. Practical determination of effect of catalyst on decomposition of hydrogen peroxide Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water. Divide it into two equal portions. (i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8573737373737375, "ocr_used": true, "chunk_length": 1815, "token_count": 486}}
{"text": "Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask. (ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask. Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below. Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1. (iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide. Mass of MnO2 before reaction(g) Mass of MnO2 after reaction(g) 2.0 2.0 Plot a graph of volume of gas produced against time(x-axes) b) On the same axes, plot a graph of the uncatalysed reaction. (c) Explain the changes in mass of manganese(IV)oxide before and after the reaction. The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Volume of gas (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 96.0 Catalysed reaction Uncatalysed reaction\nB.EQUILIBRIA (CHEMICAL CYBERNETICS) Equilibrium is a state of balance. Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8264064769195272, "ocr_used": true, "chunk_length": 1797, "token_count": 492}}
{"text": "Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e. A + B -> C + D Reactions in which the reactants form products and the products can reform the reactants are said to be reversible. A + B C + D Reversible reactions may be: (a)Reversible physical changes (b)Reversible chemical changes (c)Dynamic equilibrium (a)Reversible physical changes Reversible physical change is one which involves: (i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original. (ii) colour changes. Some substances/compounds change their colours without change in chemical substance. Examples of reversible physical changes (i) colour change on heating and cooling: I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back. ZnO(s) ZnO(s) (white when cold) (yellow when hot) II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back. PbO(s) PbO(s)\n(brown when hot) (yellow when cold) (ii)Sublimation I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid. I2(s) I2(g) (grey crystalline solid (purple vapour undergo sublimation) undergo deposition) II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced CO2(s) CO2(g) (white powdery solid (colourless/odourless gas undergo sublimation) undergo deposition) (iii)Melting/ freezing and boiling/condensation Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9045560587347332, "ocr_used": true, "chunk_length": 2082, "token_count": 502}}
{"text": "Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid. Melting boiling Freezing condensing (iv)Dissolving/ crystallization/distillation Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g. NaCl(s) + aq NaCl(aq) (b)Reversible chemical changes These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products. Examples of Reversible chemical changes (i)Heating Hydrated salts/adding water to anhydrous salts. H2O(s) H2O(l) H2O(s)\nWhen hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated. When anhydrous salts are added (back) some water they form hydrated compound/salts. Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate (i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes. (ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9103919141612041, "ocr_used": true, "chunk_length": 2075, "token_count": 508}}
{"text": "Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white. (ii)colourless liquid forms on the cooler parts of boiling / test tube (i)colour changes from white to blue (ii)boiling tube becomes warm /hot. Cobalt(II)chloride hexahydrate Pink crystalline solid/solution (i)colour changes from pink to blue. (ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used) (i)colour changes from blue to pink (ii)boiling tube becomes warm/hot. When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube. This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is\nformed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively. CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq) (white/anhydrous) (blue/hydrated) CoCl2(s) + 6H2 O(l) CoCl2.6H2 O(s/aq) (blue/anhydrous) (pink/hydrated) (ii)Chemical sublimation Some compounds sublime from solid to gas by dissociating into new different compounds. e.g. Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8987672693900837, "ocr_used": true, "chunk_length": 1869, "token_count": 487}}
{"text": "Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made. When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes. This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl- ions) and ammonia gas (and hence NH4+ ions) (ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made. When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change. NH3(g) + HCl(g) NH4Cl(s) (Turns moist (Turns moist (forms white fumes) litmus paper blue) litmus paper red)\n(c)Dynamic equilibria For reversible reactions in a closed system: (i) at the beginning; -the reactants are decreasing in concentration with time -the products are increasing in concentration with time (ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium. Sketch showing the changes in concentration of reactants and products in a closed system For a system in equilibrium: (i) a reaction from left to right (reactants to products) is called forward reaction. (ii) a reaction from right to left (products to reactants) is called backward reaction. (iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium. A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9099573134363268, "ocr_used": true, "chunk_length": 2311, "token_count": 479}}
{"text": "A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants. Concentration Mole dm-3 Reaction progress/path/coordinate\nThe influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that: “if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change” Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium: (a)Concentration (b)Pressure (c)Temperature (d)Catalyst (a)Influence of concentration on dynamic equilibrium An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration. Examples of influence of concentration on dynamic equilibrium (i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation: 2H+ (aq) + 2CrO42- (aq) Cr2O72- (aq) + H2O(l) (Yellow) (Orange) I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More Cr2O72- ions formed in the solution mixture make it to be more orange in colour. II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8943837222284248, "ocr_used": true, "chunk_length": 2120, "token_count": 483}}
{"text": "II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH- (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour. 2OH- (aq) + 2Cr2O72- (aq) CrO42- (aq) + H2O(l) (Orange) (Yellow) I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH- (aq). H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the 2OH- (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour. II. If a base /OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH- (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH- (aq) ions added. More of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour. (i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube. Note that the solution mixture is orange. Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully. Note that the solution mixture is remains orange. Add about six drops of 2M sodium hydroxide solution. Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8608989361608457, "ocr_used": true, "chunk_length": 1769, "token_count": 490}}
{"text": "Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:\nOH- H+ When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist. When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist. (ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture Measure 2cm3 of bromine water into a boiling tube. Note its colour. Bromine water is yellow Add three drops of 2M sulphuric(VI)acid. Note any colour change Colour becomes more yellow Add seven drops of 2M sodium hydroxide solution. Note any colour change. Solution mixture becomes colourless/Bromine water is decolourized. Explanation When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br-), hydrobromite ion(OBr-) and hydrogen ion(H+) as in the equation: H2O(l) + Br2(aq) OBr- (aq) + H+ (aq) + Br- (aq) If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow. If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-\nPlace 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8906384957810113, "ocr_used": true, "chunk_length": 2076, "token_count": 505}}
{"text": "If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-\nPlace 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes. To each test tube add two drops of water. Record your observations in Table 1 below. To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below. To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below. To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1 Indicator Colour of indicator in Water Acid(2M sulphuric (VI) acid) Base(2M sodium hydroxide) Phenolphthalein Colourless Colourless Pink Methyl orange Yellow Red Orange Litmus solution Colourless Red Blue Explanation An indicator is a substance which shows whether another substance is an acid , base or neutral. Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8909461109913723, "ocr_used": true, "chunk_length": 1932, "token_count": 473}}
{"text": "i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In-) molecule shows/appears. The following examples illustrate the above. (i)Phenolphthalein indicator exist as: HPh H+ (aq) + Ph-(aq) (colourless molecule) (Pink anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph-) molecule shows/appears. (ii)Methyl Orange indicator exists as: HMe H+ (aq) + Me-(aq) (Red molecule) (Yellow/Orange anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8873629686896018, "ocr_used": true, "chunk_length": 1990, "token_count": 514}}
{"text": "The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed. If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium The following examples show the influence of pressure on dynamic equilibrium: (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture\nNitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) Gay Lussacs law 2Volume 1Volume Avogadros law 2molecule 1molecule 2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown. (ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture. Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas. Chemical equation : I2(g) + H2(g) ===== 2HI (g) Gay Lussacs law 1Volume 1Volume 2Volume Avogadros law 1molecule 1molecule 2molecule (1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas. Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8848099467763291, "ocr_used": true, "chunk_length": 1933, "token_count": 469}}
{"text": "Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process. The yield of ammonia is thus favoured by high pressures Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g) Gay Lussacs law 1Volume 3Volume 2Volume Avogadros law 1molecule 3molecule 2molecule (1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of ammonia increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of ammonia decrease. (iv)Contact process. Increase in pressure of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures. Chemical equation : 2SO2(g) + O2 (g) -> 2SO3 (g) Gay Lussacs law 2Volume 1Volume 2Volume Avogadros law 2molecule 1molecule 2molecule (2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease. (v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8722084492790285, "ocr_used": true, "chunk_length": 1809, "token_count": 470}}
{"text": "(v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures. Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (g) Gay Lussacs law 4Volume 5Volume 4Volume 6Volume Avogadros law 4molecule 5molecule 4molecule 6Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour. Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease. Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase. Note If the water vapour is condensed on cooling, then: Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (l) Gay Lussacs law 4Volume 5Volume 4Volume 0Volume Avogadros law 4molecule 5molecule 4molecule 0Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour. Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase. Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8597004733809528, "ocr_used": true, "chunk_length": 1804, "token_count": 488}}
{"text": "The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH). Endothermic reaction are thus favoured by high temperature/heating Exothermic reaction are favoured by low temperature/cooling. If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position. (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow. This show that (i)the forward reaction to the right is exothermic(-ΔH). On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat. (ii)the backward reaction to the right is endothermic(+ΔH). On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat. (c)Influence of Catalyst on dynamic equilibrium A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g. Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase. CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9049321987276655, "ocr_used": true, "chunk_length": 2116, "token_count": 494}}
{"text": "CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount. The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions Optimum condition thus require understanding the effect of various factors on: (i)rate of reaction(Chemical kinetics) (ii)dynamic equilibrium(Chemical cybernetics)\n1.Optimum condition in Haber process Chemical equation N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ Equilibrium/Reaction rate considerations (i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8963243945700087, "ocr_used": true, "chunk_length": 2109, "token_count": 465}}
{"text": "(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation. 2.Optimum condition in Contact process Chemical equation 2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΔH = -197kJ Equilibrium/Reaction rate considerations\n(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8917480998914225, "ocr_used": true, "chunk_length": 1842, "token_count": 444}}
{"text": "Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation. 3.Optimum condition in Ostwalds process Chemical equation 4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΔH = -950kJ Equilibrium/Reaction rate considerations (i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained. (ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules. An optimum pressure of about 9 atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 900oC is normally used. (iv)Platinum can be used as catalyst. Platinum is very expensive.It is: -promoted with Rhodium to increase the surface area/area of contact. -added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used. The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation. C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8925655567232528, "ocr_used": true, "chunk_length": 1962, "token_count": 474}}
{"text": "C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2. Explain the effect of increase in pressure on the following: (i) N2(g) + O2(g) ===== 2NO(g) Gay Lussacs law 1Volume 1Volume 2 Volume Avogadros law 1 molecule 1 molecule 2 molecule 2 volume on reactant side produce 2 volume on product side. Increase in pressure thus have no effect on position of equilibrium. (ii) 2H2(g) + CO(g) ===== CH3OH (g)\nGay Lussacs law 2Volume 1Volume 1 Volume Avogadros law 2 molecule 1 molecule 1 molecule 3 volume on reactant side produce 1 volume on product side. Increase in pressure shift the equilibrium forward to the left. More yield of CH3OH is formed. 4. Explain the effect of increasing temperature on the following: 2SO2(g) + O2 (g) ===== 2SO3 (g) ΔH = -189kJ Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat. 5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below Time(Seconds) Mass of flask(grams) Loss in mass(grams) 0 600 20 599.50 40 599.12 60 598.84 80 598.66 100 598.54 120 598.50 140 598.50 160 598.50 (a)Complete the table by calculating the loss in mass (b)What does the “600” gram reading on the balance represent The initial mass of brass and the acid before any reaction take place. (c)Plot a graph of Time (x-axes) against loss in mass. (d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the\nreact is complete. No more hydrogen is evolved.The mass of flask remain constant.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8105358100531423, "ocr_used": true, "chunk_length": 1908, "token_count": 507}}
{"text": "(d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the\nreact is complete. No more hydrogen is evolved.The mass of flask remain constant. (d)At what time was the loss in mass equal to: (i)1.20g Reading from a correctly plotted graph = (ii)1.30g Reading from a correctly plotted graph = (iii)1.40g Reading from a correctly plotted graph = (e)What was the loss in mass at: (i)50oC Reading from a correctly plotted graph = (ii) 70oC Reading from a correctly plotted graph = (iii) 90oC g Reading from a correctly plotted graph =\n1 21.0.0 ELECTROCHEMISTRY (25 LESSONS) Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with: i) Reduction and oxidation ii) Electrochemical (voltaic) cell iii) Electrolysis (electrolytic) cell (i)REDUCTION AND OXIDATION (REDOX) 1. In teams of oxygen transfer: i) Reduction is removal of oxygen. ii) Oxidation is addition of oxygen. iii) Redox is simultaneous addition and removal of oxygen. iv) Reducing agent is the species that undergoes oxidation, therefore gains oxygen. v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates oxygen. e.g. When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below: CuO (s) + H2 (g) -> Cu (s) + H2O (l)\n2 (Oxidising agent) (Reducing agent) 2. In terms of hydrogen transfer: i) Oxidation is the removal of hydrogen. ii) Reduction is the addition of hydrogen. iii) Redox is simultaneous addition and removal of hydrogen. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates hydrogen. v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8745873394399128, "ocr_used": true, "chunk_length": 2054, "token_count": 511}}
{"text": "v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas. Cl2 (g) + H2S (g) -> S(S) + 2HCl (g) (Oxidizing agent) (Reducing agent) 3. In terms of electron transfer: i) Oxidation is donation/ loss/ removal of electrons. ii) Reduction is gain/ accept/ addition of electrons. iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates electrons. v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts electrons. Example a) Displacement of metals from their solutions: Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below. Metal added to Iron (II) sulphate (VI) solution Colour changes Copper Solution remains green Zinc Green colour fades Magnesium Green colour fades Explanation -When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution. 3 -When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution. -Copper is less reactive than iron therefore cannot displace iron its solution. -Zinc is more reactive than iron therefore can displace iron from its solution. -Magnesium is more reactive than iron therefore can displace iron from its solution. In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8995999026578083, "ocr_used": true, "chunk_length": 2209, "token_count": 495}}
{"text": "In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Zn(s) -> Zn2+(aq) + Fe(s) (redox/both donation and gain of electrons) 2. Mg(s) -> Mg2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Mg(s) -> Mg2+(aq) + Fe(s) (redox/both donation and gain of electrons) 3. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7677850825390301, "ocr_used": true, "chunk_length": 1399, "token_count": 483}}
{"text": "Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6. 2Mg(s) -> 2Mg2+(aq) + 4e (oxidation/donation of electrons) O2(g) + 4e -> 2O2-(aq) (reduction/gain of electrons) O2(g) + 2Mg(s) -> 2Mg2+(aq) + 2O2-(aq) (redox/both donation and gain of electrons)\n4 Note (i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired. (i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions from their solution. b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7751261204919743, "ocr_used": true, "chunk_length": 1476, "token_count": 507}}
{"text": "b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions from their solution to form Fe3+(aq). (iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement. (iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent. 4. (a)In terms of oxidation number: i) Oxidation is increase in oxidation numbers. ii) Reduction is decrease in oxidation numbers. iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance. iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1\n5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8502520706721031, "ocr_used": true, "chunk_length": 1857, "token_count": 470}}
{"text": "iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1\n5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7363969247840216, "ocr_used": true, "chunk_length": 1147, "token_count": 365}}
{"text": "v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1\n5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0\n6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7281426326423708, "ocr_used": true, "chunk_length": 2324, "token_count": 765}}
{"text": "(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1\n5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0\n6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7070452388028661, "ocr_used": true, "chunk_length": 2693, "token_count": 937}}
{"text": "Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0\n6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6\n7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6721632505860432, "ocr_used": true, "chunk_length": 3144, "token_count": 1162}}
{"text": "Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0\n6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II. Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6\n7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7080101338772604, "ocr_used": true, "chunk_length": 3540, "token_count": 1229}}
{"text": "Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound is thus Sulphuric(VI)acid -H2SO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Sulphuric(IV)acid III. Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6\n7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6962426349850286, "ocr_used": true, "chunk_length": 2436, "token_count": 870}}
{"text": "Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Carbonic(IV)acid IV.Manganese in; -MnO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Manganese(IV)oxide -KMnO4 = ((+1 + x + (-2 x 4)) thus x= 0-( +1 +-8) =+7 The chemical name of this compound is thus Potassium manganate(VII) V.Chromium in; - Cr2O3 => 2x + (-2 x 3)= 0 thus 2x = 0 – (-6) = +6 / 2= +3 The chemical name of this compound is thus Chromium(III)oxide -K2Cr2O7 => (+1 x 2) + 2x + (-2 x7)= 0 thus 2x = 0 – +2 +-14 = +12 / 2= +6 The chemical name of this compound is thus Potassium dichromate(VI) -K2CrO4 => (+1 x 2) + x + (-2 x4)= 0 thus 2x = 0 – +2 +-8 = +12 / 2= +6\n7 The chemical name of this compound is thus Potassium chromate(VI) 6.The sum of the oxidation numbers of atoms of elements making a charged radical/complex ion is equal to its charge. Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7076713595725003, "ocr_used": true, "chunk_length": 2104, "token_count": 729}}
{"text": "Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in SO42- = ( +6 + (-2 x 4)) = -2 The chemical name of this radical is thus sulphate(VI) ion b) NO3- has- -one atom of N with oxidation number +4( refer to Rule 4) -three atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in NO3- = ( +4 + (-2 x 3)) = -1 The chemical name of this radical is thus nitrate(IV) ion. Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)\n8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents\n9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL\n10 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6862545798065257, "ocr_used": true, "chunk_length": 4509, "token_count": 1719}}
{"text": "Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)\n8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents\n9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL\n10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6743052410282401, "ocr_used": true, "chunk_length": 3962, "token_count": 1560}}
{"text": "Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)\n8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents\n9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL\n10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6770902426938838, "ocr_used": true, "chunk_length": 3994, "token_count": 1571}}
{"text": "Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 thus 2x = -2 – +2 +-14 = +12 / 2= +6 The chemical name of this compound/ion//radical is thus dichromate(VI) ion -CrO42- => x + (-2 x4)= -2 thus x = -2 + (-2 x 4) = +6 The chemical name of this compound/ion//radical is thus chromate(VI) ion (c)Using the concept/idea of oxidation numbers as increase and decrease in oxidation numbers , the oxidizing and reducing species/agents can be determined as in the following examples; (i) Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Oxidation numbers -> +2 0 +2 0 Oxidizing species/agents =>Cu2+ ;its oxidation number decrease from+2 to 0 in Cu(s)\n8 Reducing species/agents => Zn2+ ;its oxidation number increase from 0 to +2 in Zn(s) (ii) 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2 (l) Oxidation numbers -> -1 0 -1 0 Oxidizing agent =>Cl2(g) ;its oxidation number decrease from 0 to-1 in 2Cl- (aq) Reducing agents => Zn2+ ;its oxidation number increase from -1 to 0 in Zn(s) (iii) Br2 (l) + Zn(s) -> Zn2+ (aq) + 2Br- (aq) Oxidation numbers -> 0 0 +2 -1 Oxidizing agent => Br2 (l) ;its oxidation number decrease from 0 to-1 in 2Br-(aq) Reducing agents => Zn(s) ;its oxidation number increase from 0 to +2 in Zn2+ (iv) 2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g) Oxidation numbers -> 2 (+1 -1) 0 +2 2(-1) 0 Oxidizing agent => H+ in HCl;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Mg(s) ;its oxidation number increase from 0 to +2 in Mg2+ (v) 2H2O (l) + 2Na(s) -> 2NaOH (aq) + H2 (g) Oxidation numbers -> +1 -2 0 +1 -2 +1 0 Oxidizing agent => H+ in H2O;its oxidation number decrease from +1to 0 in H2 (g) Reducing agents => Na(s) ;its oxidation number increase from 0 to +1 in Na+ (vi) 5Fe2+ (aq) + 8H+ (aq) + MnO4- -> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) +2 +1 +7 -2 +3 +2 +1 -2 Oxidizing agent => Mn in MnO4- ;its oxidation number decrease from +7to+2 in Mn2+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (vii) 6Fe2+ (aq) + 14H+ (aq) + Cr2O72-(aq) -> 6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l) +2 +1 +6 -2 +3 +3 +1 -2 Oxidizing agent: Cr in Cr2O72- ;its oxidation number decrease from +6 to+3 in Cr3+ Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (viii) 2Fe2+ (aq) + 2H+ (aq) + H2O2(aq) -> 2Fe3+ (aq) + 2H2O (l) +2 +1 +1 -1 +3 +1 -2 Oxidizing agent: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Reducing agents => Fe2+ ;its oxidation number increase from +2 to +3 in Fe3+ (ix) Cr2O72-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Cr3+ (aq) + 2H2O (l) + 5O2(g) +6 -2 +1 +1 -1 +3 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Cr in Cr2O72- its oxidation number decrease from +6 to +3 in Cr3+ Reducing agents\n9 O in H2O2;its oxidation number increase from -1 to O in O2(g) O in Cr2O72- its oxidation number increase from -2 to O in O2(g) (x) 2MnO4-(aq) + 6H+ (aq) + 5H2O2(aq) -> 2Mn2+ (aq) + 8H2O (l) + 5O2(g) +7 -2 +1 +1 -1 +2 +1 -2 0 Oxidizing agents: O in H2O2;its oxidation number decrease from -1 to -2 in H2O Mn in MnO4- its oxidation number decrease from +7 to +2 in Mn2+ Reducing agents O in H2O2;its oxidation number increase from -1 to O in O2(g) O in MnO4- its oxidation number increase from -2 to O in O2(g) (ii)ELECTROCHEMICAL (VOLTAIC) CELL\n10 1. When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6816794293930435, "ocr_used": true, "chunk_length": 3951, "token_count": 1533}}
{"text": "When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal rod/plate. 2.The metal rod becomes therefore negatively charged while its own solution positively charged. As the positive charges of the solution increase, some of them recombine with the electrons to form back the metal atoms M+(aq) + e -> M(s) ( monovalent metal) M2+(aq) + 2e -> M(s) (divalent metal) M3+(aq) + 3e -> M(s) (Trivalent metal) 3. When a metal rod/plate is put in a solution of its own salt, it constitutes/forms a half-cell. The tendency of metals to ionize differ from one metal to the other. The difference can be measured by connecting two half cells to form an electrochemical/voltaic cell as in the below procedure: To set up an electrochemical /voltaic cell To compare the relative tendency of metals to ionize Place 50cm3 of 1M Zinc(II) sulphate(VI) in 100cm3 beaker. Put a clean zinc rod/plate into the solution. Place 50cm3 of 1M Copper(II) sulphate(VI) in another 100cm3 beaker. Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8582855436081244, "ocr_used": true, "chunk_length": 1488, "token_count": 406}}
{"text": "Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V\n11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8801372829153193, "ocr_used": true, "chunk_length": 1171, "token_count": 320}}
{"text": "Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V\n11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+\n12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8638132358518734, "ocr_used": true, "chunk_length": 2529, "token_count": 705}}
{"text": "Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V\n11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+\n12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8644867670322248, "ocr_used": true, "chunk_length": 2540, "token_count": 704}}
{"text": "Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V\n11 (ii)Zinc half cell with Silver rod/plate/coin dipped in 50cm3 of IM silver(I) nitrate(V) solution (iii)Copper half cell with Iron rod/plate/spoon dipped in 50cm3 of IM Iron (II) sulphate(VI) solution Record the observations in the table below Changes on the 1st metal rod (A) Changes on the 2nd metal rod (B) Changes on the 1st solution (A(aq)) Changes on the 2nd solution (B(aq)) Voltage/voltmeter reading(Volts) Using Zn/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Zinc(II)sulphate (VI)colour remain colourless Blue Copper (II)sulphate (VI)colour fades. Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+\n12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8662375415282392, "ocr_used": true, "chunk_length": 2580, "token_count": 703}}
{"text": "Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoretical value=2.04V) Using Ag/Cu half cell -The rod increase in size /mass /deposited -silver coin/ rod /plate increase in size /mass/ deposited Blue Copper (II)sulphate (VI)colour remains Silver(I)nitrate (V)colour remain colourless 0.20 (Theoretical value=0.46V) Using Fe/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Iron(II)sulphate (VI)colour becomes more green Blue Copper (II)sulphate (VI)colour fades.Brown solid/residue/ deposit 0.60 (Theoretical value=0.78V) From the above observations ,it can be deduced that: (i)in the Zn/Cu half-cell the; -Zinc rod/plate ionizes /dissolves faster than the copper rod/plate to form Zn2+\n12 Ionic equation Zn(s) -> Zn2+(aq) + 2e -blue copper ions in the Copper (II)sulphate solution gains the donated electrons to form brown copper metal/atoms Ionic equation Cu2+(aq) + 2e -> Cu(s) This reaction shows /imply the Zinc rod has a higher tendency to ionize than copper.The Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons. The copper rod is therefore relatively more positive with respect to Zinc rod. When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8664302846056311, "ocr_used": true, "chunk_length": 1954, "token_count": 512}}
{"text": "When the two half cells are connected , electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions. This means a net accumulation/increase of Zn2+ positive ions on the negative half cell and a net decrease in Cu2+ positive ions on the positive half cell. The purpose of the salt bridge therefore is: (i)complete the circuit (ii)maintain balance of charges /ions on both half cells. For the negative half cell the NO3- /Cl- from salt bridge decrease/neutralise the increased positive(Zn2+) ion. For the positive half cell the Na+ / K+ from salt bridge increase the decreased positive(Cu2+) ion. The voltmeter should theoretically register/read a 1.10Volts as a measure of the electromotive force (e.m.f) of the cell .Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed. A combination of two half cells that can generate an electric current from a redox reaction is called a voltaic/electrochemical cell. By convention a voltaic/electrochemical cell is represented; M(s) / M2+(aq) // N2+ (aq) / N(s) (metal rod of M)(solution ofM)(solution ofN)(metal rod ofN) Note; a)(i)Metal M must be the one higher in the reactivity series. (ii)It forms the negative terminal of the cell. (iii)It must diagrammatically be drawn first on the left hand side when illustrating the voltaic/electrochemical cell. b)(i)Metal N must be the one lower in the reactivity series. (ii)It forms the positive terminal of the cell. (iii)It must diagrammatically be drawn second/after/ right hand side when illustrating the voltaic/electrochemical cell. Illustration of the voltaic/electrochemical cell. (i)Zn/Cu cell\n13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8689777962726274, "ocr_used": true, "chunk_length": 1979, "token_count": 513}}
{"text": "(i)Zn/Cu cell\n13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation. Zn(s) / 1M, Zn2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +1.10 V 5.cell diagram (ii)Mg/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Mg(s) -> Mg2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Mg(s) -> Mg2+(aq) + Cu(s) 4.cell representation. Mg(s) / 1M, Mg2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +2.04 V 5.cell diagram. Voltmeter\n14 (iii)Fe/Cu cell 1. Magnesium rod ionizes /dissolves to form Mg2+ ions at the negative terminal Fe(s) -> Fe2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Fe(s) -> Fe2+(aq) + Cu(s) 4.cell representation. Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V\n15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7096889123893695, "ocr_used": true, "chunk_length": 1276, "token_count": 491}}
{"text": "Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V\n15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2. Silver ions in solution gain the donated electrons to form silver atoms/metal 2Ag+(aq) + 2e -> 2Ag(s) 3.Overall redox equation 2Ag+(aq) + Cu(s) -> Cu2+(aq) + 2Ag(s) 4.cell representation. Cu(s) / 1M, Cu2+(aq) // 1M,2Ag+(aq) / 2Ag(s) E0 = +0.46 V 5.cell diagram. Standard electrode potential (Eᶿ) The standard electrode potential (Eᶿ) is obtained if the hydrogen half cell is used as reference. The standard electrode potential (Eᶿ) consist of inert platinum electrode immersed/dipped in 1M solution of (sulphuric(VI) acid) H+ ions. Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7667841719921807, "ocr_used": true, "chunk_length": 1171, "token_count": 404}}
{"text": "Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equilibrium/balance exist between the adsorbed layer of molecular hydrogen and H+ ions in solution to form a half cell. ½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)\n16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8450319561345301, "ocr_used": true, "chunk_length": 1233, "token_count": 333}}
{"text": "½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)\n16 The standard electrode potential (Eᶿ) is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode. If the other electrode has a higher/greater tendency to lose electrons than the hydrogen electrode, the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative (Eᶿ) values. If the other electrode has a lower/lesser tendency to lose electrons than the hydrogen electrode, the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive (Eᶿ) values. Table showing the standard electrode potential (Eᶿ) of some reactions Reaction (Eᶿ) values in volts F2 (g)+ 2e -> 2F- (aq) +2.87 H2 O2 (aq)+ H+ (aq) +2e -> H2 O (l) +1.77 Mn O4- (aq)+ 4H+ (aq) +3e -> MnO2 (s) +H2 O (l) +1.70 2HClO (aq)+ 2H+ (aq) +2e -> Cl2 (aq) +2H2 O (l) +1.59 Mn O4- (aq)+ 4H+ (aq) +5e -> Mn2+ (aq) +H2 O (l) +1.51 Cl2 (g)+ 2e -> 2Cl- (aq) +1.36 Mn O2 (s)+ 4H+ (aq) +2e -> Mn2+ (aq) +2H2 O (l) +1.23 Br2 (aq)+ 2e -> 2Br- (aq) +1.09 NO3- (aq)+ 2H+ (aq) + e -> NO2 (g) + H2 O (l) +0.80 Ag+ (aq) + e -> Ag(s) +0.80 Fe3+ (aq) + e -> Fe2+ (aq) +0.77 2H+ (aq)+ O2 (g) -> H2 O2 (aq) +0.68 I2 (aq)+ 2e -> 2I- (aq) +0.54 Cu2+ (aq) + 2e -> Cu(s) +0.34 2H+ (aq) + 2e -> H2(g) +0.00 Pb2+ (aq) + 2e -> Pb(s) -0.13 Fe2+ (aq) + 2e -> Fe(s) -0.44 Zn2+ (aq) + 2e -> Zn(s) -0.77 Al3+ (aq) + 3e -> Al(s) -1.66 Mg2+ (aq) + 2e -> Mg(s) -2.37 Na+ (aq) + e -> Na(s) -2.71 K+ (aq) + e -> K(s) -2.92 Note: (i)Eᶿ values generally show the possibility/feasibility of a reduction process/oxidizing strength.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6536812449904372, "ocr_used": true, "chunk_length": 1795, "token_count": 741}}
{"text": "(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons. It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons. 17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+\n18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8003719561587674, "ocr_used": true, "chunk_length": 1446, "token_count": 502}}
{"text": "17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+\n18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:\n19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7158068492972918, "ocr_used": true, "chunk_length": 2930, "token_count": 1278}}
{"text": "(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+\n18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:\n19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7218678986295343, "ocr_used": true, "chunk_length": 3005, "token_count": 1291}}
{"text": "If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left Hand Side diagram) Zn(s) ->Zn2+ (aq) + 2e Eᶿ = +0.77(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿoxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 0.77) = +1.10V Overall redox equation: Cu2+ (aq) + Zn(s) -> Zn2+ (aq) + Cu(s) Eᶿ = +1.10V Overall conventional cell representation: Zn(s) / Zn2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +1.10V Overall conventional cell diagram: Voltmeter(1.10V) 1M Cu2+ (aq) 1M Zn2+ (aq) Zn2+\n18 Zinc and copper reaction has a positive(+) overall Eᶿ therefore is possible/feasible and thus Zinc can displace/reduce Copper solution. b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:\n19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7236173380239894, "ocr_used": true, "chunk_length": 3019, "token_count": 1287}}
{"text": "b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.34 – (- 2.37) = +2.71V Overall redox equation: Cu2+ (aq) + Mg(s) -> Mg2+ (aq) + Cu(s) Eᶿ = +2.71V Overall conventional cell representation: Mg(s) / Mg2+ (aq) 1M, // 1M,Cu2+ (aq) / Cu(s) Eᶿ = +2.71V c)Ag and Pb From the table above: 2Ag+ (aq) + 2e -> 2Ag(s) Eᶿ = +0.80V(higher Eᶿ /Right Hand Side diagram) Pb2+ (aq) + 2e ->Pb(s) Eᶿ = -0.13V(lower Eᶿ/ Left Hand Side diagram) Pb(s) ->Pb2+ (aq) + 2e Eᶿ = +0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +0.80 – (- 0.13) = +0.93V Overall redox equation: 2Ag+ (aq) + Pb(s) -> Pb2+ (aq) + 2Ag(s) Eᶿ = +0.93V Overall conventional cell representation: Pb(s) / Pb2+ (aq) 1M, // 1M,2Ag+ (aq) / Ag(s) Eᶿ = +0.93V d)Chlorine and Bromine From the table above: 2e + Cl2(g) ->2Cl- (aq) Eᶿ = +1.36V(higher Eᶿ /Right Hand Side diagram) 2e + Br2(aq) ->2Br- (aq) Eᶿ = +0.13V(lower Eᶿ/ Left Hand Side diagram) 2Br- (aq) -> Br2(aq) + 2e Eᶿ = -0.13(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = - 0.13 – (- 1.36) = +1.23V Overall redox equation:\n19 2Br- (aq) + Cl2(g) -> 2Cl- (aq) + Br2(aq) Eᶿ = +1.23V Overall conventional cell representation: Cl2(g) / 2Cl- (aq) 1M, // 1M, 2Br- (aq) / Br2(aq) Eᶿ = +1.23V Chlorine displaces bromine from bromine water. When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7185240755188018, "ocr_used": true, "chunk_length": 2125, "token_count": 911}}
{"text": "When chlorine gas is thus bubbled in bromine water, the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed. This reaction is feasible /possible because the overall redox reaction has a positive Eᶿ value. e)Strongest oxidizing agent and the strongest reducing agent. From the table above: 2e + F2(g) ->2F- (aq) Eᶿ = +2.87V(highest Eᶿ /strongest oxidizing agent) 2e + 2K+ (aq) ->2K (aq) Eᶿ = -2.92V(lowest Eᶿ/ strongest reducing agent) 2K (aq) -> 2K+ (aq) + 2e Eᶿ = +2.92V (reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ Eᶿ oxidized- Eᶿ reduced Substituting: Overall Eᶿ = +2.87 – (-2.92) = +5.79V Overall redox equation: F2(g) + 2K(s) -> 2F- (aq) + 2K+ (aq) Eᶿ = +5.79V Overall conventional cell representation: 2K(s) / 2K+ (aq),1M, // 1M, 2F- (aq) / F2(g) Eᶿ = +5.79V The redox reactions in an electrochemical/voltaic is commercially applied to make the: (a)Dry /primary/Laclanche cell. (b)Wet /secondary /accumulators. (a)Dry/primary/Laclanche cell Examine a used dry cell. Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8073396707409946, "ocr_used": true, "chunk_length": 1423, "token_count": 475}}
{"text": "Note the positive and the negative terminal of the cell. Carefully using a knife cut a cross section from one terminal to the other. The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of; -Ammonium chloride -Zinc chloride -powdered manganese (IV) oxide mixed with Carbon. Zinc acts/serve as the negative terminal where it ionizes/dissociates:\n20 Zn(s) -> Zn2+(aq) + 2e Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas. 2NH4+(aq) + 2e -> 2NH3(g) + H2(g) Ammonia forms a complex salt / compound /(Zn(NH3) 4)2+ (aq) / tetramminezinc(II) complex with the Zinc chloride in the paste. Manganese (IV) oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell. Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8807133675019617, "ocr_used": true, "chunk_length": 1204, "token_count": 297}}
{"text": "Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulators 1. Wet/Secondary/Accumulators are rechargeable unlike dry /primary /Laclanche cells.Wet/Secondary/Accumulators are made up of: (i)Lead plate that forms the negative terminal (ii)Lead(IV) oxide that forms the positive terminal 2.The two electrodes are dipped in concentrated sulphuric(VI) acid of a relative density 1.2/1.3 3.At the negative terminal,lead ionizes /dissolves; Pb(s) -> Pb2+ + 2e 4.At the positive terminal,\n21 (i) Lead(IV) oxide reacts with the hydrogen ions in sulphuric(VI)acid to form Pb2+ (aq) ions; PbO2(s) + 4H+(aq) + 2e -> Pb2+ (aq) + H2O(l) (ii) Pb2+ (aq) ions formed instantly react with sulphate (VI) ions/ SO42- (aq) from sulphuric (VI)acid to form insoluble Lead(II) sulphate (VI). Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8264819739540685, "ocr_used": true, "chunk_length": 1523, "token_count": 462}}
{"text": "Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive form making the cell less efficient. As the battery discharges ,lead and lead(IV)oxide are depleted/finished/reduced and the concentration of sulphuric(VI)acid decreases. 7. During recharging, the electrode reaction is reversed as below: 2PbSO4(s) + 2H2O(l) ->PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq) 8. A car battery has six Lead-acid cells making a total of 12 volts. (iii)ELECTROLYSIS (ELECTROLYTIC CELL) 1.Electrolysis is defined simply as the decomposition of a compound by an electric current/electricity. A compound that is decomposed by an electric current is called an electrolyte. Some electrolytes are weak while others are strong. 2.Strong electrolytes are those that are fully ionized/dissociated into (many) ions. Common strong electrolytes include: (i)all mineral acids (ii)all strong alkalis/sodium hydroxide/potassium hydroxide. (iii)all soluble salts\n22 3.Weak electrolytes are those that are partially/partly ionized/dissociated into (few) ions. Common weak electrolytes include: (i)all organic acids (ii)all bases except sodium hydroxide/potassium hydroxide. (iii)Water 4. A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8508899823959527, "ocr_used": true, "chunk_length": 1679, "token_count": 500}}
{"text": "A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (ii)all hydrocarbons(alkanes /alkenes/alkynes) (iii)Chemicals of life(e.g. proteins, carbohydrates, lipids, starch, sugar) 5. An electrolytes in solid state have fused /joined ions and therefore do not conduct electricity but the ions (cations and anions) are free and mobile in molten and aqueous (solution, dissolved in water) state. 6.During electrolysis, the free ions are attracted to the electrodes. An electrode is a rod through which current enter and leave the electrolyte during electrolysis. An electrode that does not influence/alter the products of electrolysis is called an inert electrode. Common inert electrodes include: (i)Platinum (ii)Carbon graphite Platinum is not usually used in a school laboratory because it is very expensive. Carbon graphite is easily/readily and cheaply available (from used dry cells). 7.The positive electrode is called Anode.The anode is the electrode through which current enter the electrolyte/electrons leave the electrolyte 8.The negative electrode is called Cathode. The cathode is the electrode through which current leave the electrolyte / electrons enter the electrolyte 9. During the electrolysis, free anions are attracted to the anode where they lose /donate electrons to form neutral atoms/molecules. i.e. M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode\n23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9017307341956642, "ocr_used": true, "chunk_length": 1996, "token_count": 467}}
{"text": "M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + e (for cations from electrolytes in aqueous state / solution / dissolved in water) The neutral atoms /molecules form the products of electrolysis at the anode. This is called discharge at anode\n23 10. During electrolysis, free cations are attracted to the cathode where they gain /accept/acquire electrons to form neutral atoms/molecules. X+ (aq) + 2e -> X(s) (for cations from electrolytes in aqueous state / solution / dissolved in water) 2X+ (l) + 2e -> X (l) (for cations from molten electrolytes) The neutral atoms /molecules form the products of electrolysis at the cathode. This is called discharge at cathode. 11. The below set up shows an electrolytic cell. BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode\n24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)\n25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8544676616915423, "ocr_used": true, "chunk_length": 1650, "token_count": 464}}
{"text": "BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode\n24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electrolysis of molten Lead(II)chloride (i)Decomposition of electrolyte into free ions; PbCl2 (l) -> Pb 2+(l) + 2Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Pb 2+(l) + 2e -> Pb (l) (Cation / Pb 2+ gains / accepts / acquires electrons to form free atom)\n25 (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid lead metal. II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8439667617847434, "ocr_used": true, "chunk_length": 1581, "token_count": 469}}
{"text": "II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (Cation / Zn2+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Br-(l) -> Br2 (g) + 2e (Anion / Br- donate/lose electrons to form free atom then a liquid molecule which change to gas on heating) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid Zinc metal. II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8454584629472522, "ocr_used": true, "chunk_length": 1332, "token_count": 396}}
{"text": "II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l) + 2e -> Na (l) (Cation / Na+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 2Cl-(l) -> Cl2 (g) + 2e (Anion / Cl- donate/lose electrons to form free atom then a gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid sodium metal. II.At the anode pale green chlorine gas. 26 d)To determine the products of electrolysis of molten Aluminium (III)oxide (i)Decomposition of electrolyte into free ions; Al2O3 (l) -> 2Al 3+(l) + 3O2-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 4Al 3+ (l) + 12e -> 4Al (l) (Cation / Al 3+ gains / accepts / acquires electrons to form free atom) (iii)At the anode/positive electrode(+); 6O2-(l) -> 3O2 (g) + 12e (Anion /6O2- donate/lose 12 electrons to form free atom then three gas molecule) (iv)Products of electrolysis therefore are; I.At the cathode grey beads /solid aluminium metal. II.At the anode colourless gas that relights/rekindles glowing splint. 13. For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8509854170480201, "ocr_used": true, "chunk_length": 1821, "token_count": 514}}
{"text": "For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) Position of cations and anions in the electrochemical series 1. Most electropositive cations require more energy to reduce (gain electrons) and thus not readily discharged. The higher elements /metals in the electrochemical series the less easily/readily it is discharged at the cathode in the electrolytic cell. Table I showing the relative ease of discharge of cations in an electrolytic cell K+(aq) + e -> K(s) (least readily/easily discharged) Na+(aq) + e -> Na(s) Ca2+(aq) + 2e -> Ca(s) Mg2+(aq) + 2e -> Mg(s) Al3+(aq) + 3e -> Al(s) Zn2+(aq) + 2e -> Zn(s) Fe2+(aq) + 2e -> Fe(s) Pb2+(aq) + 2e -> Pb(s) 2H+(aq) + 2e -> H2(g) (hydrogen is usually “metallic”) Cu2+(aq) + 2e -> Cu(s) Hg2+(aq) + 2e -> Hg(s) Ag+(aq) + e -> Ag(s) (most readily/easily discharged) 2.The OH- ion is the most readily/easily discharged anion . All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell\n27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7517610863004147, "ocr_used": true, "chunk_length": 1462, "token_count": 510}}
{"text": "All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never discharged. The ease of discharge of halogen ions increase down the group. Table II showing the relative ease of discharge of anions in an electrolytic cell\n27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) 2 I-(aq) -> I2(aq) + 2e 2 Br-(aq) -> Br2(aq) + 2e 2 Cl-(aq) -> Cl2(aq) + 2e 2 F-(aq) -> F2(aq) + 2e SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43- not/never/rarely discharged. 3.(a)When two or more cations are attracted to the cathode, the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above. This is called selective/preferential discharge at cathode. (b)When two or more anions are attracted to the anode, the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above. This is called selective/preferential discharge at anode. 4.The following experiments shows the influence /effect of selective/preferential discharge on the products of electrolysis: (i)Electrolysis of acidified water/dilute sulphuric(VI) acid Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)\n28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.821908091908092, "ocr_used": true, "chunk_length": 1716, "token_count": 514}}
{"text": "Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq)\n28 H2 SO4(aq) -> SO42-(aq) + 2H+(aq) II. Name the ions in acidified water that are attracted/move to: Cathode- H+(aq) from either sulphuric(VI) acid (H2 SO4) or water (H2O) Anode- SO42-(aq) from sulphuric (VI) acid (H2 SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of acidified water. Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8404643905826265, "ocr_used": true, "chunk_length": 1681, "token_count": 504}}
{"text": "The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Why is electrolysis of dilute sulphuric(VI) acid called “electrolysis of (acidified) water”? The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. This implies/means that water in the electrolyte is being decomposed into hydrogen and Oxygen gases. The electrolysis of dilute sulphuric acid is therefore called “electrolysis of acidified water.” VI. Explain the changes in concentration of the electrolyte during electrolysis of acidified water” The concentration of dilute sulphuric (VI) acid increases. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape. The concentration /mole of acid present in a given volume of solution thus continue increasing/rising. (ii)Electrolysis of Magnesium sulphate(VI) solution Fill the Hoffmann voltameter with dilute sulphuric(VI) acid. Connect the Hoffmann voltameter to a d.c. electric supply. Note the observations at each electrode. Answer the following questions:\n29 I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8735458357777632, "ocr_used": true, "chunk_length": 1587, "token_count": 423}}
{"text": "Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from water (H2O) Anode- SO42-(aq) from Magnesium sulphate(VI) solution (Mg SO4) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Mg2+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of SO42- ions at the anode) IV. Name the products of electrolysis of Magnesium sulphate(VI) solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. The four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by the four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8384741338395825, "ocr_used": true, "chunk_length": 1720, "token_count": 501}}
{"text": "VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /mole of acid present in a given volume of Magnesium sulphate(VI) solution thus continue increasing/rising. 30 The set – up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes. Name a suitable pair of electrodes for this experiment Identify the ions and cations in the solution On the diagram label the cathode Write ionic equations for the reactions that took place at the anode. Explain the change that occurred to the concentration of magnesium sulphate solution during the experience. During the electrolysis a current of 2 amperes was passed through the solution for 4 hours. Calculate the volume of the gas produced at the anode.(1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3) One of the uses of electrolysis is electroplating What is meant by electroplating? Give tow reasons why electroplating is necessary. b) Concentration of the electrolytes 1.High concentrations of cations and/or anions at the electrodes block the ion/s that is likely to be discharged at the electrode. This is called over voltage. A concentrated solution therefore produces different products of electrolysis from a dilute one. 2. The following experiments show the influence/effect of concentration of electrolyte on the products of electrolysis. (i)Electrolysis of dilute and concentrated(brine)sodium chloride solution\n31 I. Dissolve about 0.5 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Transfer the set up into a fume chamber/open and continue to make observations for a further 10 minute. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.890062729305133, "ocr_used": true, "chunk_length": 2214, "token_count": 498}}
{"text": "Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodiumchloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute sodium chloride solution Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution The concentration of dilute sodium chloride solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8604965870815822, "ocr_used": true, "chunk_length": 1899, "token_count": 508}}
{"text": "Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of salt present in a given volume of sodium chloride solution continue increasing/rising. 32 II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an electrolytic cell. Note the observations continuously at each electrode for 30 minutes in a fume chamber/open. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from sodium chloride solution (NaCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2H+(aq) + 2e -> H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2(g) + 4e (Cl- ions with a higher concentration block the discharge of OH- ions at the anode) IV. Name the products of electrolysis of concentrated sodium chloride solution/brine Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas(pale green gas that bleaches damp/moist/wet litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8665011091063126, "ocr_used": true, "chunk_length": 1909, "token_count": 504}}
{"text": "Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode. VI. Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solution/brine The concentration of concentrated sodium chloride solution/brine increases. The ratio of Cl2 (g): H2 (g) is 1:1 as they are combined in water. Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode. 33 The concentration /moles of OH- (aq) and Na+ ion (as NaOH) present in a given volume of electrolyte continue increasing/rising. This makes the electrolyte strongly alkaline with high pH. As the electrolysis of brine continues the concentration of Cl- ions decrease and oxygen gas start being liberated at anode. The electrolyte pH is thus lowered and the concentration of brine starts again increasing. (ii)Electrolysis of dilute and concentrated Hydrochloric acid solution I. Prepare about 50cm3 of 0.05 M of dilute Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 10 minutes. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8705969283441913, "ocr_used": true, "chunk_length": 1887, "token_count": 493}}
{"text": "Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 4OH- (aq) -> H2O(l) +O2+ 4e (4OH- ions selectively discharged instead of Cl- ions at the anode) IV. Name the products of electrolysis of dilute Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Oxygen gas (colourless gas that relights /rekindles glowing splint) V. Explain the difference in volume of products at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of oxygen (O2)gas at the anode are gained/acquired/accepted by four H+(aq) ions to form 2 molecule/2volume/2mole of Hydrogen (H2)gas at the cathode. The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode. 34 VI. Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8560962846677134, "ocr_used": true, "chunk_length": 1729, "token_count": 487}}
{"text": "Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid The concentration of dilute Hydrochloric acid increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products. The concentration /moles of HCl present in a given volume of dilute Hydrochloric acid continue increasing/rising. II. Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution. Place the solution in an electrolytic cell. Note the observations at each electrode for 30 minutes CautionThis experiment should be done in the open/fume chamber. Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in 2M Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- Cl-(aq) from dilute Hydrochloric acid (HCl) and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8688387984662627, "ocr_used": true, "chunk_length": 1752, "token_count": 467}}
{"text": "Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of OH- ions) IV. Name the products of electrolysis of 2M Hydrochloric acid Cathode-Hydrogen gas (colourless gas that extinguishes burning splint with explosion/ “pop” sound Anode-Chlorine gas (Pale green gas that bleaches blue/red moist/wet/damp litmus papers) V. Explain the difference in volume of products at the cathode and anode. Two(2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode/ The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode. 35 VI. Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid The concentration of Hydrochloric acid decreases. The ratio of H2 (g): Cl2 (g) is 1:1 as they are combined in Hydrochloric acid. Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode. There is a net accumulation of excess OH- (aq) ions in solution. This makes the electrolyte strongly alkaline with high pH. c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8801588470205028, "ocr_used": true, "chunk_length": 2007, "token_count": 504}}
{"text": "c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The examples below illustrate the influence of the nature of electrode on the products of electrolysis: (i)Electrolysis of copper(II) sulphate(VI) solution I. Using carbon-graphite electrodes Weigh Carbon -graphite electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 22.4 g Brown solid deposit at the cathode after electrolysis - Bubbles of colourless gas that relight splint - Blue colour of electrolyte fades/become less blue - Blue colour of electrolyte fades /become less blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)\n36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8714617206123435, "ocr_used": true, "chunk_length": 1766, "token_count": 449}}
{"text": "Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq)\n36 II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode 2Cu2+ (aq) + 4e -> 2Cu(g) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode 4OH- (aq) -> H2O(l) + O2+ 4e (OH- ions ions are higher than SO42- ions in the electrochemical series therefore selectively discharged at the cathode.)) IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution Cathode-2 moles of copper metal as brown solid coat Anode-Oxygen gas (Colourless gas that relights /rekindles glowing splint) V. Explain the changes that take place at the cathode and anode. Four(4) electrons donated/lost by OH- ions to form 1 molecule/1volume/1mole of Oxygen (O2)gas at the anode are gained/acquired/accepted by two Cu2+(aq) ions to form 2 moles of brown copper solid that deposit itself at the cathode. The moles of oxygen gas at the anode is equal to the moles of copper produced at the cathode VI. Explain the changes in electrolyte during electrolysis of 1M copper (II) sulphate(VI) solution. (i)The pH of copper(II) sulphate(VI) solution lowers/decreases. The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8456776203545731, "ocr_used": true, "chunk_length": 1768, "token_count": 506}}
{"text": "The salt becomes more acidic. Water in the electrolyte is decomposed only into Oxygen gas (from the OH- ions) that escapes as products at the anode. There is a net accumulation of excess H+ (aq) ions in solution. This makes the electrolyte strongly acidic with low pH. (ii) Cu2+ (aq) ions are responsible for the blue colour of the electrolyte/ copper(II) sulphate (VI) solution. As electrolysis continues, blue Cu2+ (aq) ions gain electrons to form brown Copper. The blue colour of electrolyte therefore fades/become less blue. (iii)Copper is deposited at the cathode. This increases the mass of the cathode.OH- ions that produce Oxygen gas at anode come from water. Oxygen escapes out/away without increasing the mass of anode. 37 II. Using copper electrodes Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass current for about 20 minutes. Observe each electrode and any changes in electrolyte. Remove the electrodes from the electrolyte. Wash with acetone/propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8773094872778417, "ocr_used": true, "chunk_length": 1738, "token_count": 426}}
{"text": "Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Blue colour of electrolyte remain blue - Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate because less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. 38 IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8500228445439266, "ocr_used": true, "chunk_length": 1698, "token_count": 467}}
{"text": "Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode. Cu2+ ions have greater tendency to accept/gain/acquire electrons to form brown copper atoms/solid that deposit itself and increase the mass/size of the cathode.The copper deposited at the cathode is pure -H+ ions accumulate around the cathode. Electrolyte thus becomes strongly acidic around the cathode. -Cu2+ ions in solution are responsible for the blue colour of electrolyte. Blue colour of electrolyte fade around the cathode. (ii)Anode Copper atom at the anode easily ionizes to release electrons. The anode therefore keeps decreasing in mass/eroding. The amount of copper that dissolve/erode is equal to the mass of copper deposited. This is called electrode ionization. Electrode ionization is where the anode erodes/decrease and the cathode deposits/increase during electrolysis. The overall concentration of the electrolyte remains constant 14.In industries electrolysis has the following uses/applications: (a)Extraction of reactive metals from their ores. Potassium, sodium ,magnesium, and aluminium are extracted from their ores using electrolytic methods. (b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.91197192513369, "ocr_used": true, "chunk_length": 1760, "token_count": 394}}
{"text": "(b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there: (i)At the cathode; Cu2+ (aq) + 2e -> Cu(s) (Pure copper deposits on the strip (ii)At the anode; Cu(s) ->Cu2+ (aq) + 2e (impure copper erodes/dissolves) (c)Electroplating The label EPNS(Electro Plated Nickel Silver) on some steel/metallic utensils mean they are plated/coated with silver and/or Nickel to improve their appearance(add their aesthetic value)and prevent/slow corrosion(rusting of iron). Electroplating is the process of coating a metal with another metal using an electric current. During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (a)The SI unit of quantity of electricity is the coulomb(C).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.89112451873873, "ocr_used": true, "chunk_length": 1775, "token_count": 452}}
{"text": "During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes/dissociates to release electrons: Ag(s) ->Ag+ (aq) + e (impure silver erodes/dissolves) (iv) silver (Ag+)ions from electrolyte gain electrons to form pure silver deposits / coat /cover the spoon/impure Ag+ (aq) + e ->Ag(s) (pure silver deposits /coat/cover on spoon) 15.The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis. Faradays 1st law of electrolysis states that “the mass/amount of substance liberated/produced/used during electrolysis is directly proportional to the quantity of of electricity passed/used.” (a)The SI unit of quantity of electricity is the coulomb(C). The coulomb may be defined as the quantity of electricity passed/used when a current of one ampere flow for one second.i.e; 1Coulomb = 1 Ampere x 1Second The Ampere is the SI unit of current(I) The Second is the SI unit of time(t) therefore;\n40 Quantity of electricity(in Coulombs) = Current(I) x time(t) Practice examples 1. A current of 2 amperes was passed through an electrolytic cell for 20 minutes. Calculate the quantity of electric charge produced. Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x (20 x 60) = 2400 C 2. A current of 2 amperes was passed through an electrolytic.96500 coulombs of charge were produced. Calculate the time taken. Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8634435968145151, "ocr_used": true, "chunk_length": 1928, "token_count": 496}}
{"text": "Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used. Working: Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 96500 10 x 60 = 160.8333 Amperes (b)The quantity of electricity required for one mole of electrons at the anode/cathode is called the Faraday constant(F). It is about 96500 Coulombs.i.e The number of Faradays used /required is equal to the number of electrons used at cathode/anode during the electrolytic process. e.g. Cu2+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode. Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas 2O2- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas. 4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8252368287362162, "ocr_used": true, "chunk_length": 1842, "token_count": 511}}
{"text": "Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas 2O2- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas. 4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e\n42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7222998110808891, "ocr_used": true, "chunk_length": 3491, "token_count": 1211}}
{"text": "4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e\n42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.719135066582118, "ocr_used": true, "chunk_length": 3154, "token_count": 1098}}
{"text": "41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed through copper(II)sulphate(VI) for 30 minutes in an electrolytic cell. (Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e\n42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;\n43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7045483607301788, "ocr_used": true, "chunk_length": 3388, "token_count": 1208}}
{"text": "(Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 72000C -> 7200 x 63.5 = 2.3689 g of copper 2 x 96500 2.a)If 3.2 g of Lead were deposited when a current of 2.5 amperes was passed through an electrolytic cell of molten Lead(II)bromide for 20 minutes, determine the Faraday constant.(Pb = 207) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2.5 x (20 x 60) = 3000 C If 3.2g of Lead -> 3000C Then 207 g of Lead -> 207 x 3000 = 194062.5 C 3.2 Equation at the cathode: Pb2+ (l) + 2e -> Pb(l) From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C 1mole of electrons = 1 Faraday => 194062.5 = 97031.25 C 2 b)What is the volume of bromine vapour produced at the anode at room temperature(1mole of gas at room temperature and pressure = 24000cm3) Method 1 Equation at the anode: Br- (l) -> Br2(g) + 2e\n42 From the equation: 2 moles of electrons = 2 Faradays = 194062.5 C -> 24000cm3 3000 C -> 3000 x 24000 194062.5 =371.0145cm3 Method 2 Equation at the anode: Br- (l) -> Br2(g) + 2e Mole ratio of products at Cathode: anode = 1:1 Moles of Lead at cathode = 3.2 = 0.0155moles = moles of Bromine 207 1 moles of bromine vapour -> 24000cm3 0.0155moles of Bromine -> 0.0155 x 24000 = 372 cm3 Method 3 Equation at the anode: Br- (l) -> Br2(g) + 2e Ratio of Faradays used to form products at Cathode: anode = 2:2 => 2 x 97031.25 C produce 24000cm3 of bromine vapour Then: 3000 C -> 3000 x 24000cm3 = 371.0145cm3 2 x 97031.25 3.What mass of copper remain from 2.0 at the anode if a solution of copper(II)sulphate(VI) is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutes.(Cu= 63.5, 1Faraday = 96487 coulombs) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1 x (20 x 60) = 1200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 1200C -> 1200 x 63.5 = 0.3948g of copper deposited 2 x 96500 Mass of copper remaining = Original mass – mass dissolved/eroded => 2.0 -0.3948 = 1.6052 g of copper remain 4. Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;\n43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6896993419799843, "ocr_used": true, "chunk_length": 3130, "token_count": 1149}}
{"text": "Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;\n43 63.5 g -> 2 x 96500C 0.234 g -> 0.234 x 2 x 96500 = 711.2126 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 711.2126 C 4x 60 = 2.9634 Amperes 5. (a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7220017696195319, "ocr_used": true, "chunk_length": 1434, "token_count": 499}}
{"text": "(a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g\n44 of “P” was deposited at the cathode.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7398260917270134, "ocr_used": true, "chunk_length": 1488, "token_count": 504}}
{"text": "(Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solution of Zinc (II)sulphate(VI) above if a current of 4.0 Amperes is used. Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g\n44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7184309876170341, "ocr_used": true, "chunk_length": 1806, "token_count": 628}}
{"text": "Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C a) Calculate the quantity of electricity used. Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g\n44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7368487928843711, "ocr_used": true, "chunk_length": 1574, "token_count": 516}}
{"text": "Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produce a mass =molar mass of M thus; RAM of M = 0.26 g x 3 x 96500 = 55.7556(No units) 1350 7.An element “P” has a relative atomic mass 88.When a current of 0.5 amperes was passed through fused chloride of “P” for 32 minutes and 10seconds ,0.44 g\n44 of “P” was deposited at the cathode. Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours. Calculate the amount of current that was passed.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7379125055699721, "ocr_used": true, "chunk_length": 1256, "token_count": 415}}
{"text": "Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C = 1 mole of electrons = 1 Faradays = single charge 193000 C -> 193000 = 2 moles/Faradays/charges => symbol of ion = P2+ 96500 8. During purification of copper by electrolysis 1.48 g of copper was deposited when a current was passed through aqueous copper (II)sulphate(VI) for 2 ½ hours. Calculate the amount of current that was passed. (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu2+ (aq) + 2e-> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 63.5 g -> 2 x 96500C 1.48 g -> 1.48 x 2 x 96500 = 4255.1181 C 63.5 Current(I) in amperes = Quantity of electricity(in Coulombs) Time(t) in seconds Substituting/converting time to second= 4255.1181C (( 2 x 60) + 30) x60 = 0.4728 Amperes 17. Practically Faraday 1st law of electrolysis can be verified as below. Verifying Faraday 1st law of electrolysis Procedure. Weigh clean copper plates electrodes. Record the masses of the electrodes in table I below. Place the electrodes in 1M copper(II) sulphate(VI) solution in a beaker. Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7551882103998339, "ocr_used": true, "chunk_length": 1583, "token_count": 505}}
{"text": "Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 7.00 g Mass of anode before electrolysis 7.75 g Mass of cathode after 8.25 g Mass of anode after 6.50 g\n45 electrolysis electrolysis Change in mass at cathode after electrolysis 1.25 g Change in mass at anode after electrolysis 1.25 g Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate as less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8405214635551715, "ocr_used": true, "chunk_length": 1780, "token_count": 504}}
{"text": "Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode= 2 moles (ii)How many Faradays are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode : 2 moles = 2 Faradays (iii)Calculate the quantity of electric charge used Working:\n46 Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x 20 x 60 = 2400C VI. (i) Calculate the quantity of electricity required to deposit/erode one mole of copper at the cathode/anode(Cu=63.5) Since 1.25 g of copper -> 2400C Then 63.5 g (1mole of copper) -> 63.5 x 2400 = 121920 C 1.25 (ii)Determine the Faraday constant from the results in V(i) above From the equation at; Cathode Cu2+ (aq) + 2e -> Cu(s) Anode Cu (s) -> Cu2+(aq) + 2e 2 moles = 2 Faradays -> 121920 C 1 moles = 1 Faradays -> 121920 = 60960 C 2 (iii) The faraday constant obtained above is far lower than theoretical.Explain -high resistance of the wires used. -temperatures at 25oC were not kept constant -plates/electrodes used were not made of pure copper -plates/electrodes used were not thoroughly clean copper Further practice 1.An element P has a relative atomic mass of 88. When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.805529368310052, "ocr_used": true, "chunk_length": 1691, "token_count": 503}}
{"text": "When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs). 2.During electrolysis of aqueous copper (II) sulphate, 144750 coulombs of electricity were used. Calculate the mass of copper metal that was obtained (Cu = 64 ;1 Faraday = 96500 coulombs) ( 3 mks) 3.A nitrate of a metal M was electrolysed .1.18 g of metal was deposited when a current of 4 ampheres flow for 16 minutes.Determine the formula of the sulphate(VI)salt of the metal. (Faraday constant = 96500 , RAM of X = 59.0) Working Q = It =>( 4 x 16 x 60) = 3840 C 1.18 g of X => 3840 C 59.0 g => 59.0 x 3840 = 192000 C 1.18 96500 C = 1Faraday 192000 C= 192000 C x1 = 2F thus charge of M = M2+ 96500 C Valency of M is 2 thus formula of sulphate(VI)salt MSO4\n47 4. Below is the results obtained when a current of 2.0ampheres is passed through copper(II)sulphate(VI)solution for 15 minutes during electrolysis using copper electrode. Initial mass of cathode = 1.0 g Final mass of cathode = 1.6 g Change in mass of cathode = 0.60 g (i)Determine the change in mass at the anode. Explain your answer. Mass decrease = 0.6g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7462685962959216, "ocr_used": true, "chunk_length": 1230, "token_count": 405}}
{"text": "Initial mass of cathode = 1.0 g Final mass of cathode = 1.6 g Change in mass of cathode = 0.60 g (i)Determine the change in mass at the anode. Explain your answer. Mass decrease = 0.6g. Electrode ionization take place where the cathode increase in mass form the erosion of the anode (ii)Calculate the quantity of electricity required to deposit one mole of copper.(Cu =63.5) Q =It => 2 x 15 x 60 = 1800 coulombs Method 1 0.60 g of copper ->1800 coulombs 63.5 g -> 63.5 x 1800 = 190500 Coulombs 0.60 Method 2 Moles of Copper = Mass => 0.60 = 9.4488 x10 -3 moles Molar mass 63.5 9.4488 x10 -3 moles -> 1800 coulombs 1 Mole -> 1 x 1800 coulombs = 190500.381 coulombs 9.4488 x10 -3 moles (iii)Determine the oxidation number of copper produced at the cathode and hence the formula of its nitrate (V)salt (1 Faraday = 96500 Coulombs) 96500 Coulombs -> 1 Faraday 190500.381 coulombs -> 190500.381 coulombs x 1 96500 Coulombs = 1.9741 Faradays => 2F(whole number) Charge of copper = 2+ = Oxidation number => Valency of copper = 2 hence chemical formula of nitrate (V)salt = Cu (NO3)2\n1 22.0.0 METALS (20 LESSONS) a)Introduction to metals The rationale of studying metals cannot be emphasized.Since ages, the world over, metals like gold and silver have been used for commercial purposes. The periodicity of alkali and alkaline earth metals was discussed in year 2 of secondary school education. This topic generally deals with: (a)Natural occurrence of the chief ores of the most useful metals for industrial /commercial purposes. (b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7644287559190702, "ocr_used": true, "chunk_length": 1725, "token_count": 512}}
{"text": "(b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals. The metals given detailed emphasis here are; Sodium, Aluminium, Iron, Zinc, Lead and Copper. 2 The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. 1.SODIUM a) Natural occurrence Sodium naturally occurs as: (i)Brine-a concentrated solution of sodium chloride(NaCl(aq)) in salty seas and oceans. (ii)Rock salt-solid sodium chloride(NaCl(s) (iii)Trona-sodium sesquicarbonate(NaHCO3.Na2CO3.2H2O) especially in lake Magadi in Kenya. (iv)Chile saltpeter-sodium nitrate(NaNO3) Position on the earth’s crust If near the surface ,open cast mining / quarrying is used If deep on the earth’s crust deep mining is used If the ore is low grade oil, water, and air is blown forming a froth(froth flotation) to concentrate The ore first roasted if it is a carbonate or sulphide of Zinc, Iron, Tin, Lead, and Copper to form the oxide Electrolysis of the ore is used for reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium The oxide is reduced using carbon/ carbon(II) oxide in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron\n3 b)(i) Extraction of Sodium from brine/Manufacture of Sodium hydroxide/The flowing mercury cathode cell/ TheCaster-Keller process I.Raw materials (i) Brine-concentrated solution of sodium chloride (NaCl (aq)) from salty seas and oceans. (ii)Mercury (iii)Water from river/lakes II. Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9018406747891284, "ocr_used": true, "chunk_length": 2134, "token_count": 504}}
{"text": "Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. H2O(l) H+(aq) + OH-(aq) NaCl(aq) Na+(aq) + Cl-(aq) II. Name the ions present in brine that moves to the: (i)Mercury cathode; H+(aq) , Na+(aq) (ii)Titanium/graphite; OH-(aq), Cl-(aq) III. Write the equation for the reaction that take place during the electrolytic process at the; Cathode; 2Na+(aq) + 2e 2Na(s) Anode; 2Cl-(aq) Cl2(g) + 2e Note (i)Concentration of 2Cl-(aq) ions is higher than OH- ions causing overvoltage thus blocking OH- ions from being discharged at the anode. (ii)Concentration of Na+(aq) ions is higher than H+ ions causing overvoltage thus blocking H+ ions from being discharged at the cathode. 4 IV. Name the products of electrolysis in the flowing mercury-cathode cell. (i)Mercury cathode; Sodium metal as grey soft metal/solid (ii)Titanium/graphite; Chlorine gas as a pale green gas that turns moist blue/red litmus papers red then bleaches both. Chlorine gas is a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8873538531346895, "ocr_used": true, "chunk_length": 1889, "token_count": 498}}
{"text": "(iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas. 2Na Hg (l) + 2H2O(l) 2NaOH (aq) + 2Hg(l) + H2(g) Hydrogen gas is a very useful by-product in; (i)making ammonia gas in the Haber process (ii)manufacture of hydrochloric acid (iii)in weather balloons to forecast weather (iv)as rocket fuel As the electrolysis of brine continues, the concentration of Cl-ions decreases and oxygen gas start being liberated. Continuous feeding of the electrolyte is therefore very necessary. III.Uses of sodium hydroxide The sodium hydroxide produced is very pure and is used mainly in: (i)Making soapy and soapless detergents. (ii)making cellulose acetate/rayon IV. Diagram showing the Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 5 3 V. Environmental effects of Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 1.Most of the Mercury used at the cathode is recycled ; (i)to reduce the cost because mercury is expensive (ii)to reduce pollution because mercury kills marine life. (iii)because it causes chromosomal/genetic mutation to human beings. 2.Chlorine produced at the anode; (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. b)(ii) Extraction of sodium from rock salt/The Downs cell/process\n6 I. Raw materials (i)Rock salt/solid sodium chloride (ii)calcium(II)chloride II. Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8938557251010082, "ocr_used": true, "chunk_length": 1961, "token_count": 494}}
{"text": "Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC. A little calcium (II) chloride is added to lower the melting point of the electrolyte to about 600oC. The molten electrolyte is the electrolyzed in a carbon graphite anode suspended at the centre and surrounded by steel cathode. Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. NaCl(l) Na+(l) + Cl-(l) Note: In absence of water, the ions are in liquid state. II. Name the ions present in molten rock salt that move to the; (i)Steel cathode -Na+(l) (ii)Carbon graphite anode- Cl-(l) III. Write the equation for the reaction that take place during the electrolytic process at the; (i)Steel cathode 2Na+(l) + 2e 2Na(l) (ii)Carbon graphite anode 2Cl-(l) Cl2(g) + 2e IV. Name the products of electrolysis in the Downs cell at; (i)Cathode: Grey solid Sodium metal is less dense than the molten electrolyte and therefore float on top of the cathode to be periodically tapped off. (ii)Anode: Pale green chlorine gas that turns moist/damp/wet blue/red litmus papers red then bleaches/decolorizes both. Chlorine gas is again a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. 7 A steel diaphragm/gauze is suspended between the electrodes to prevent recombination of sodium at the cathode and chlorine gas at the anode back to sodium chloride. III. Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.890384068900693, "ocr_used": true, "chunk_length": 1794, "token_count": 459}}
{"text": "Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting. 2.Sodium is used in making very useful sodium compounds like; (i)Sodium hydroxide(NaOH) (ii)Sodium cyanide(NaCN) (iii)Sodium peroxide(Na2O2) (iv)Sodamide(NaNH2) 3.An alloy of Potassium and Sodium is used as coolant in nuclear reactors. V. Environmental effects of Downs cell. 1.Chlorine produced at the anode;\n8 (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. 2.Sodium metal rapidly react with traces of water to form alkaline Sodium hydroxide(NaOH(aq))solution. This raises the pH of rivers/lakes killing aquatic lifein case of leakages. VI. Test for presence of Na. If a compound has Na+ ions in solid/molten/aqueous state then it changes a non-luminous clear/colourless flame to a yellow coloration but does not burn Experiment Scoop a portion of sodium chloride crystals/solution in a clean metallic spatula. Introduce it to a clear /colourless Bunsen flame. Observation Inference Yellow coloration Na+ Practice (i)Calculate the time taken in hours for 230kg of sodium to be produced in the Downs cell when a current of 120kA is used. (ii)Determine the volume of chlorine released to the atmosphere.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8926671730385124, "ocr_used": true, "chunk_length": 1445, "token_count": 365}}
{"text": "10 (a)Calculate the concentration of the resulting solution in moles per litre. (b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7146546883773162, "ocr_used": true, "chunk_length": 822, "token_count": 301}}
{"text": "(b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25\n11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6585041276620084, "ocr_used": true, "chunk_length": 1239, "token_count": 512}}
{"text": "25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25\n11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium. Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7507563025210084, "ocr_used": true, "chunk_length": 1330, "token_count": 445}}
{"text": "Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C. (i) Name the salt that is added (1mk) Calcium chloride (ii) State why it is necessary to lower the temperature(1mk) To reduce the cost of production\n12 (c) Explain why aqueous sodium chloride is not suitable as an electrolyte for the manufacture of sodium in the Downs process( 2mk) The sodium produced react explosively/vigorously with water in the aqueous sodium chloride (d) Sodium metal reacts with air to form two oxide. Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8622911782138232, "ocr_used": true, "chunk_length": 1851, "token_count": 509}}
{"text": "Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities. The mixture is first dissolved in hot concentrated sodium/potassium hydroxide solution. The alkalis dissolve both bauxite and silicon(IV)oxide. This is because bauxite is amphotellic while silicon(IV)oxide is acidic. Iron(III)oxide (Fe2O3) is filtered of /removed as a residue. Carbon(IV)oxide is bubbled into the filtrate to precipitate aluminium (III) hydroxide (Al(OH)3) as residue. The aluminium (III) hydroxide (Al(OH)3) residue is filtered off. Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)\n13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8486068222990778, "ocr_used": true, "chunk_length": 1536, "token_count": 472}}
{"text": "Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)\n13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide. It is therefore dissolved first in molten cryolite /sodium hexafluoroaluminate (III)/Na3AlF6 to lower the melting point to about 800oC. The molten electrolyte is put in the Hall cell made up of a steel tank lined with carbon graphite and an anode suspended into the electrolyte. During the electrolysis: (i)At the cathode; 4Al3+(l) + 12e 4Al(l) (ii) At the anode; 6O2-(l) 3O2(g) + 12e Aluminium is denser than the electrolyte therefore sink to the bottom of the Hall cell. At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere. C(s) + O2(g) CO2(g) The anode thus should be continuously replaced from time to time. Flow chart summary of extraction of aluminium from Bauxite Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2 Powdered mixture Crush (increase surface area) Hot concentrated sodium hydroxide\n14 c) Diagram showing the Hall cell / process for extraction of Bauxite Iron(III)oxide- Fe2O3 as residue Sodium aluminate (NaAl(OH)4) and sodium silicate (Na2SiO3) as filtrate Carbon(IV)oxide Aluminium hydroxide (Al(OH)3) as residue Sodium silicate (Na2SiO3) Aluminium (III) Oxide Roast at 1000oC Cryolite Electrolysis Oxygen gas at anode Pure aluminium sinks in Hall cell\n15 d)Uses of aluminium (i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8466529156531458, "ocr_used": true, "chunk_length": 1696, "token_count": 504}}
{"text": "At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere. C(s) + O2(g) CO2(g) The anode thus should be continuously replaced from time to time. Flow chart summary of extraction of aluminium from Bauxite Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2 Powdered mixture Crush (increase surface area) Hot concentrated sodium hydroxide\n14 c) Diagram showing the Hall cell / process for extraction of Bauxite Iron(III)oxide- Fe2O3 as residue Sodium aluminate (NaAl(OH)4) and sodium silicate (Na2SiO3) as filtrate Carbon(IV)oxide Aluminium hydroxide (Al(OH)3) as residue Sodium silicate (Na2SiO3) Aluminium (III) Oxide Roast at 1000oC Cryolite Electrolysis Oxygen gas at anode Pure aluminium sinks in Hall cell\n15 d)Uses of aluminium (i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light. (ii)Making duralumin-an alloy which is harder and has a higher tensile strength (iii)Making utensils,sauce pans,spoons because it is light and good conductor of electricity. (iv)Making overhead electric cables because it is light,ductile and good conductor of electricity. (iv)Used in the thermite process for production of Manganese, Chromium amd Titanium. e) Environmental effects of extracting aluminium from Bauxite. Carbon(IV)oxide gas that escape to the atmosphere is a green house gas that causes global warming. Bauxite is extracted by open caste mining that causes soil/environmental degradation. 16 f) Test for presence of Al3+ If an ore is suspected to contain Al3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8802350445924098, "ocr_used": true, "chunk_length": 1779, "token_count": 457}}
{"text": "Bauxite is extracted by open caste mining that causes soil/environmental degradation. 16 f) Test for presence of Al3+ If an ore is suspected to contain Al3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia. Observation Inference White precipitate in excess 2M NaOH(aq) Pb2+ , Al3+, Zn2+ White precipitate in excess 2M NH3(aq) Pb2+ , Al3+ No black precipitate on adding Na2S(aq) Al3+ No white precipitate on adding either NaCl(aq),HCl(aq),H2SO4(aq),Na2SO4(aq) Al3+ Practice 1.An unknown rock X was discovered in Ukraine. Test with dilute sulphuric (VI)acid shows rapid effervescence with production of a colourless gas A that forms a white precipitate with lime water and colourless solution B. On adding 3cm3 of 2M sodium hydroxide, a white precipitate C is formed that dissolves to form a colourless solution D on adding more sodium hydroxide. On adding 2M aqueous ammonia, a white precipitate E is formed which persist in excess aqueous ammonia.On which on adding 5cm3 of 1M Lead(II)nitrate(V) to F a white precipitate G is formed which remains on heating. Identify: A Hydrogen/H2 B Aluminium sulphate(VI)/Al2(SO4) 3 C Aluminium hydroxide/ Al(OH4) 3 D Tetrahydroxoaluminate(III)/ [Al(OH4) 3] - E Aluminium hydroxide/ Al(OH) 3 F Aluminium chloride/ AlCl3\n17 2.Aluminium is obtained from the ore with the formula Al2O3. 2H2O. The ore is first heated and refined to obtain pure aluminium oxide (Al2O3). The oxide is then electrolysed to get Aluminium and oxygen gas using carbon anodes and carbon as cathode.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8396165276579427, "ocr_used": true, "chunk_length": 1647, "token_count": 477}}
{"text": "2H2O. The ore is first heated and refined to obtain pure aluminium oxide (Al2O3). The oxide is then electrolysed to get Aluminium and oxygen gas using carbon anodes and carbon as cathode. Give the common name of the ore from where aluminium is extracted from ½ mark What would be the importance of heating the ore first before refining it?1 mark To remove the water of crystallization The refined ore has to be dissolved in cryolite first before electrolysis. Why is this necessary? 1½ mark To lower the melting point of aluminium oxide from about 2015oC to 900oC so as to lower /reduce cost of production Why are the carbon anodes replaced every now and then in the cell for electrolysing aluminium oxide? 1 mark Oxygen produced at anode react with carbon to form carbon(IV)oxide gas that escape State two uses of aluminium In making aeroplane parts, buses, tankers, utensils, sauce pans,spoons Making overhead electric cables Making duralumin\n18 3. IRON a)Natural occurrence Iron is the second most common naturally occurring metal. It makes 4% of the earths crust as: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) b)The blast furnace for extraction of iron from Haematite and Magnetite a)Raw materials: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) (iv)Coke/charcoal/ carbon (v)Limestone b)Chemical processes: Iron is usually extracted from Haematite (Fe2O3), Magnetite(Fe3O4) Siderite (FeCO3).These ores contain silicon(IV)oxide(SiO2) and aluminium(III)oxide (Al2O3) as impurities. When extracted from siderite, the ore must first be roasted in air to decompose the iron(II)Carbonate to Iron(II)oxide with production of carbon(IV)oxide gas: FeCO3(s) FeO(s) + CO2(g) Iron(II)oxide is then rapidly oxidized by air to iron(III)oxide(Haematite).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8688378823542611, "ocr_used": true, "chunk_length": 1777, "token_count": 493}}
{"text": "IRON a)Natural occurrence Iron is the second most common naturally occurring metal. It makes 4% of the earths crust as: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) b)The blast furnace for extraction of iron from Haematite and Magnetite a)Raw materials: (i)Haematite(Fe2O3) (ii)Magnetite(Fe3O4) (iii)Siderite(FeCO3) (iv)Coke/charcoal/ carbon (v)Limestone b)Chemical processes: Iron is usually extracted from Haematite (Fe2O3), Magnetite(Fe3O4) Siderite (FeCO3).These ores contain silicon(IV)oxide(SiO2) and aluminium(III)oxide (Al2O3) as impurities. When extracted from siderite, the ore must first be roasted in air to decompose the iron(II)Carbonate to Iron(II)oxide with production of carbon(IV)oxide gas: FeCO3(s) FeO(s) + CO2(g) Iron(II)oxide is then rapidly oxidized by air to iron(III)oxide(Haematite). 4FeO(s) + O2(g) 2Fe2O3(s) Haematite (Fe2O3), Magnetite(Fe3O4), coke and limestone are all then fed from top into a tall (about 30metres in height) tapered steel chamber lined with refractory bricks called a blast furnace. The furnace is covered with inverted double cap to prevent/reduce amount of any gases escaping . Near the base/bottom, blast of hot air at about 1000K (827oC) is driven/forced into the furnace through small holes called Tuyeres. As the air enters ,it reacts with coke/charcoal/carbon to form carbon(IV)oxide gas. This reaction is highly exothermic. C(s)+ O2(g) CO2 (g) ∆H = -394kJ\n19 This raises the temperature at the bottom of the furnace to about 2000K(1650oC).As Carbon(IV)oxide gas rises up the furnace it reacts with more coke to form carbon(II)oxide gas.This reaction is endothermic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8454382608959982, "ocr_used": true, "chunk_length": 1639, "token_count": 503}}
{"text": "As the air enters ,it reacts with coke/charcoal/carbon to form carbon(IV)oxide gas. This reaction is highly exothermic. C(s)+ O2(g) CO2 (g) ∆H = -394kJ\n19 This raises the temperature at the bottom of the furnace to about 2000K(1650oC).As Carbon(IV)oxide gas rises up the furnace it reacts with more coke to form carbon(II)oxide gas.This reaction is endothermic. CO2 (g) + C(s) 2CO (g) ∆H = +173kJ Carbon(II)oxide gas is a strong reducing agent that reduces the ores at the upper parts of the furnace where temperatures are about 750K(500oC) i.e. For Haematite; Fe2O3 (s) + 3CO(g) 2Fe(s) + CO2(g) For Magnetite; Fe3O4 (s) + 4CO(g) 3Fe(s) + 4CO2(g) Iron is denser than iron ore. As it falls to the hotter base of the furnace it melts and can easily be tapped off. Limestone fed into the furnace decomposes to quicklime/calcium oxide and produce more carbon(IV)oxide gas. CaCO3(s) CaO(s) + CO2(g) Quicklime/calcium oxide reacts with the impurities silicon(IV)oxide(SiO2) and aluminium(III)oxide(Al2O3)in the ore to form calcium silicate and calcium aluminate. CaO(s) + SiO2(s) CaSiO3 (l) CaO(s) + Al2O3(s) Ca Al2O4 (l) Calcium silicate and calcium aluminate mixture is called slag.Slag is denser than iron ore but less dense than iron therefore float on the pure iron. It is tapped at different levels to be tapped off for use in: (i)tarmacing roads (ii) cement manufacture (iii)as building construction material (c)Uses of Iron Iron obtained from the blast furnace is hard and brittle. It is called Pig iron. It is remelted, added scrap steel then cooled. This iron is called cast iron. Iron is mainly used to make: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes,lamp posts because it is cheap.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8454173021141712, "ocr_used": true, "chunk_length": 1704, "token_count": 505}}
{"text": "It is remelted, added scrap steel then cooled. This iron is called cast iron. Iron is mainly used to make: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes,lamp posts because it is cheap. 20 (ii)nails, cutlery, scissors, sinks, vats, spanners,steel rods, and railway points from steel. Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium. It does not rust/corrode like iron. e) Environmental effects of extracting Iron from Blast furnace (i)Carbon(IV)oxide(CO2) gas is a green house gas that causes/increases global warming if allowed to escape/leak from the furnace. (ii)Carbon(II)oxide(CO)gas is a highly poisonous/toxic odourless gas that can kill on leakage. It is preferentially absorbed by the haemoglobin in mammals instead of Oxygen to form a stable compound that reduce free hemoglobin in the blood. (iii) Haematite (Fe2O3), Magnetite(Fe3O4) and Siderite (FeCO3) are extracted through quarrying /open cast mining that cause soil / environmental degradation . 21 f) Test for the presence of Iron Iron naturally exist in its compound as Fe2+ /Fe3+ If an ore is suspected to contain Fe2+ /Fe3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. I) an insoluble brown precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8718557000480935, "ocr_used": true, "chunk_length": 1579, "token_count": 423}}
{"text": "21 f) Test for the presence of Iron Iron naturally exist in its compound as Fe2+ /Fe3+ If an ore is suspected to contain Fe2+ /Fe3+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent like 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble green precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. I) an insoluble brown precipitate in excess of 2M sodium hydroxide /2M aqueous ammonia if Fe2+ ions are present. Observation Inference green precipitate in excess 2M NaOH(aq) Fe2+ green precipitate in excess 2M NH3(aq) Fe2+ brown precipitate in excess 2M NaOH(aq) Fe3+ brown precipitate in excess 2M NH3(aq) Fe3+ Practice questions\n22 4.COPPER a)Natural occurrence Copper is found as uncombined element/metal on the earths crust in Zambia, Tanzania, USA and Canada .The chief ores of copper are: (i)Copper pyrites(CuFeS2) (ii)Malachite(CuCO3.Cu(OH)2) (iii)Cuprite(Cu2O) b)Extraction of copper from copper pyrites. Copper pyrites are first crushed into fine powder. The powdered ore is the added water and oil. The purpose of water is to dissolve hydrophilic substances/particle. The purpose of oil is to make cover copper ore particle so as to make it hydrophobic Air is blown through the mixture. Air creates bubbles that stick around hydrophobic copper ore. The air bubbles raise through buoyancy small hydrophobic copper ore particles to the surface. A concentrated ore floats at the top as froth. This is called froth flotation. The concentrated ore is then skimmed off.The ore is then roasted in air to form copper(I)sulphide ,sulphur(IV)oxide and iron (II) oxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8628075628075628, "ocr_used": true, "chunk_length": 1716, "token_count": 467}}
{"text": "A concentrated ore floats at the top as froth. This is called froth flotation. The concentrated ore is then skimmed off.The ore is then roasted in air to form copper(I)sulphide ,sulphur(IV)oxide and iron (II) oxide. 2CuFeS2(s) + 4O2(g) Cu2S(s) + 3SO2(g) + 2FeO(s) Limestone (CaCO3) and silicon(IV)oxide (SiO2) are added and the mixture heated in absence of air.Silicon(IV)oxide (SiO2) reacts with iron (II) oxide to form Iron silicate which constitutes the slag and is removed. FeO(s) + SiO2(s) FeSiO3(s) The slag separates off from the copper(I)sulphide. Copper(I)sulphide is then heated in a regulated supply of air where some of it is converted to copper (I) oxide. 2Cu2S (s) + 3O2(g) 2Cu2S(s) + 2SO2(g) The mixture then undergo self reduction in which copper(I)oxide is reduced by copper(I)sulphide to copper metal. Cu2S (s) + 2Cu2O (s) 6Cu (s) + SO2(g) The copper obtained has Iron, sulphur and traces of silver and gold as impurities.It is therefore about 97.5% pure. It is refined by electrolysis/electrolytic method. During the electrolysis of refining copper, the impure copper is made the anode and a small pure strip is made the cathode. Electrode ionization takes place where: At the anode;\n23 Cu(s) Cu2+ (aq) + 2e Note: Impure copper anode dissolves/erodes into solution and decreases in size. At the Cathode; Cu2+ (aq) + 2e Cu(s) Note: The copper ions in the electrolyte(CuSO4) are reduced and deposited as copper metal at the cathode. The copper obtained is 99.98% pure. Valuable traces of silver and gold collect at the bottom of the electrolytic cell as sludge. It is used to finance the extraction of copper pyrites.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8452652324544117, "ocr_used": true, "chunk_length": 1634, "token_count": 490}}
{"text": "The copper obtained is 99.98% pure. Valuable traces of silver and gold collect at the bottom of the electrolytic cell as sludge. It is used to finance the extraction of copper pyrites. (c)Flow chart summary of extraction of copper from Copper pyrites Copper pyrites(CuFeS2) ore with impurities Fe2O3 and SiO2 Froth Crush (increase surface area) Oil Concentration chamber 1st roasting chamber Silicon(IV) oxide Smelting furnace 2nd roasting furnace Calcium aluminate (CaAl2O4)slag Limestone Sulphur(IV)Oxide Iron Silicate (FeSiO3)Slag Water Excess air Limited air Sulphur(IV)Oxide Self reduction Impure copper Rocky impurities Cu2S Cu2S, Cu2O Electrolysis using Copper electrodes\n24 Electrolytic purification of impure copper d) Uses of copper Copper is mainly used in: (i)making low voltage electric cables,contact switches, cockets and plugs because it is a good conductor of electricity. (ii)Making solder because it is a good thermal conductor. (iii)Making useful alloys e.g. -Brass is an alloy of copper and Zinc(Cu/Zn) -Bronze is an alloy of copper and Tin(Cu/Sn) -German silver is an alloy of copper ,Zinc and Nickel(Cu/Zn/Ni) (iv)Making coins and ornaments. e) Environmental effects of extracting copper from Copper pyrites (i)Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes head ache to human in high concentration. (ii)Sulphur(IV)oxide gas if allowed to escape dissolves in water /rivers/rain to form weak sulphuric(IV)acid lowering the pH of the water leading to marine pollution, accelerated corrosion/rusting of metals/roofs and breathing problems to human beings. Anode; Impure Copper eroded. Cathode; Pure Copper deposited. 25 (iii)Copper is extracted by open caste mining leading to land /environmental /soil degradation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8913553757007362, "ocr_used": true, "chunk_length": 1761, "token_count": 455}}
{"text": "Anode; Impure Copper eroded. Cathode; Pure Copper deposited. 25 (iii)Copper is extracted by open caste mining leading to land /environmental /soil degradation. f) Test for the presence of copper in an ore Copper naturally exist in its compound as Cu2+ /Cu+ Copper (I) / Cu+ is readily oxidized to copper(II)/ Cu2+ If an ore is suspected to contain Cu2+ /Cu+ it is; (i)added hot concentrated sulphuric(VI)/Nitric(V)acid to free the ions present. (ii)the free ions are then added a precipitating reagent; 2M sodium hydroxide /2M aqueous ammonia which forms; I) an insoluble blue precipitate in excess of 2M sodium hydroxide if Cu2+ ions are present. I) an insoluble blue precipitate in 2M aqueous ammonia that dissolve to royal/deep blue solution in excess if Cu2+ ions are present. Observation Inference blue precipitate in excess 2M NaOH(aq) Cu2+ blue precipitate,dissolve to royal/deep blue solution in excess 2M NH3(aq) Cu2+ g)Sample questions Copper is extracted from copper pyrites as in the flow chart outlined below. Study it and answer the questions that follow\n26 5.ZINC and LEAD a)Natural occurrence Zinc occurs mainly as: (i)Calamine-Zinc carbonate(ZnCO3) (ii)Zinc blende-Zinc sulphide(ZnS) Lead occurs mainly as Galena-Lead(II)Sulphide mixed with Zinc blende: b)Extraction of Zinc/Lead from Calamine ,Zinc blende and Galena. During extraction of Zinc , the ore is first roasted in air: For Calamine Zinc carbonate decompose to Zinc oxide and carbon(IV) oxide gas. ZnCO3(s) ZnO(s) + CO2(g) Zinc blende does not decompose but reacts with air to form Zinc oxide and sulphur(IV) oxide gas. Galena as a useful impurity also reacts with air to form Lead(II) oxide and sulphur(IV) oxide gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.871642417879106, "ocr_used": true, "chunk_length": 1695, "token_count": 468}}
{"text": "During extraction of Zinc , the ore is first roasted in air: For Calamine Zinc carbonate decompose to Zinc oxide and carbon(IV) oxide gas. ZnCO3(s) ZnO(s) + CO2(g) Zinc blende does not decompose but reacts with air to form Zinc oxide and sulphur(IV) oxide gas. Galena as a useful impurity also reacts with air to form Lead(II) oxide and sulphur(IV) oxide gas. 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) (Zinc blende) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) (Galena) The oxides are mixed with coke and limestone/Iron(II)oxide/ Aluminium (III) oxide and heated in a blast furnace. At the furnace temperatures limestone decomposes to quicklime/CaO and produce Carbon(IV)oxide gas. CaCO3(s) CaO(s) + CO2 (g) Carbon(IV)oxide gas reacts with more coke to form the Carbon(II)oxide gas. C(s) + CO2 (g) 2CO (g) Both Carbon(II)oxide and carbon/coke/carbon are reducing agents. 27 The oxides are reduced to the metals by either coke or carbon (II)oxide. ZnO(s) + C(s) Zn(g) + CO (g) PbO(s) + C(s) Pb(l) + CO (g) PbO(s) + CO(s) Pb(l) + CO2 (g) PbO(s) + CO(s) Pb(g) + CO2 (g) At the furnace temperature: (i)Zinc is a gas/vapour and is collected at the top of the furnace. It is condensed in a spray of molten lead to prevent reoxidation to Zinc oxide. On further cooling , Zinc collects on the surface from where it can be tapped off (ii)Lead is a liquid and is ale to trickle to the bottom of the furnace from where it is tapped off. Quicklime/CaO, Iron(II)Oxide, Aluminium(III)oxide are used to remove silica/silicon(IV)oxide as silicates which float above Lead preventing its reoxidation back to Lead(II)Oxide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.85069200281604, "ocr_used": true, "chunk_length": 1587, "token_count": 507}}
{"text": "It is condensed in a spray of molten lead to prevent reoxidation to Zinc oxide. On further cooling , Zinc collects on the surface from where it can be tapped off (ii)Lead is a liquid and is ale to trickle to the bottom of the furnace from where it is tapped off. Quicklime/CaO, Iron(II)Oxide, Aluminium(III)oxide are used to remove silica/silicon(IV)oxide as silicates which float above Lead preventing its reoxidation back to Lead(II)Oxide. CaO(s) + SiO2(s) CaSiO3(s/l) (Slag-Calcium silicate) FeO(s) + SiO2(s) FeSiO3(s/l) (Slag-Iron silicate) Al2O3(s) + SiO2(s) Al2SiO4(s/l) (Slag-Aluminium silicate) c)Flow chart on extraction of Zinc from Calamine ,Zinc blende. Zinc ore (calamine /Zinc blende Powdered ore Froth flotation Roasting chamber Oil Water CO2 from calamine SO2 from Zinc blende Reduction chamber Iron/aluminium/ Limestone Coke Slag (Iron silicate/ aluminium silicate/calcium Condenser Filtration Gl ikWater\n28 d) Flow chart on extraction of Lead from Galena e) Uses of Lead Lead is used in: (i)making gun-burettes. (ii)making protective clothes against nuclear (alpha rays/particle) radiation in a nuclear reactor. (iii)Mixed with tin(Sn) to make solder alloy f) Uses of Zinc Zinc is used in: Lead ore/Galena Powdered ore Froth flotation Roasting chamber oil Water Reduction chamber Iron/Limestone coke SO2(g) Slag(Iron silicate) Condenser Filtration LEAD VAPOUR Zinc residue\n29 (i)Galvanization-when iron sheet is dipped in molten Zinc ,a thin layer of Zinc is formed on the surface.Since Zinc is more reactive than iron ,it reacts with elements of air(CO2/ O2 / H2O) to form basic Zinc carbonate(ZnCO3.Zn(OH)2).This sacrificial method protects iron from corrosion/rusting. (ii)As negative terminal and casing in dry/Laclanche cells. (iii)Making brass alloy with copper(Cu/Zn) g) Environmental effects of extracting Zinc and Lead.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8784667008562104, "ocr_used": true, "chunk_length": 1846, "token_count": 511}}
{"text": "(iii)Mixed with tin(Sn) to make solder alloy f) Uses of Zinc Zinc is used in: Lead ore/Galena Powdered ore Froth flotation Roasting chamber oil Water Reduction chamber Iron/Limestone coke SO2(g) Slag(Iron silicate) Condenser Filtration LEAD VAPOUR Zinc residue\n29 (i)Galvanization-when iron sheet is dipped in molten Zinc ,a thin layer of Zinc is formed on the surface.Since Zinc is more reactive than iron ,it reacts with elements of air(CO2/ O2 / H2O) to form basic Zinc carbonate(ZnCO3.Zn(OH)2).This sacrificial method protects iron from corrosion/rusting. (ii)As negative terminal and casing in dry/Laclanche cells. (iii)Making brass alloy with copper(Cu/Zn) g) Environmental effects of extracting Zinc and Lead. (i) Lead and Lead salts are carcinogenic/causes cancer (ii)Carbon(IV)oxide is a green house gas that causes/accelerate global warming. (iii)Carbon(II)oxide is a colourless odourless poisonous /toxic gas that combines with haemoglobin in the blood to form stable carboxyhaemoglobin reducing free haemoglobin leading to death. (iv) Sulphur(IV)oxide is a gas that has a pungent poisonous smell that causes headache to human if in high concentration. (v)Any leakages in Sulphur(IV)oxide gas escapes to the water bodies to form weak sulphuric(VI)acid lowering the pH of the water. This causes marine pollution /death of aquatic life, accelerated rusting/corrosion of metals/roofs and breathing problems to human beings. h) Test for presence of Zinc/ Lead. If an ore is suspected to contain Zinc/Lead it is: I.added hot concentrated Nitric(V)acid to free the ions present. Note: Concentrated Sulphuric(VI)acid forms insoluble PbSO4 thus cannot be used to free the ions in Lead salts. II.the free ions are then added a precipitating reagent mostly 2M sodium hydroxide or 2M aqueous ammonia with the formation of; (i)a soluble precipitate in excess of 2M sodium hydroxide if Zn2+, Pb2+, Al3+ions are present.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8908028643239911, "ocr_used": true, "chunk_length": 1917, "token_count": 497}}
{"text": "If an ore is suspected to contain Zinc/Lead it is: I.added hot concentrated Nitric(V)acid to free the ions present. Note: Concentrated Sulphuric(VI)acid forms insoluble PbSO4 thus cannot be used to free the ions in Lead salts. II.the free ions are then added a precipitating reagent mostly 2M sodium hydroxide or 2M aqueous ammonia with the formation of; (i)a soluble precipitate in excess of 2M sodium hydroxide if Zn2+, Pb2+, Al3+ions are present. (ii)a white precipitate that dissolves to form a colorless solution in excess 2M aqueous ammonia if Zn2+ions are present. (iii)an insoluble white precipitate in excess 2M aqueous ammonia if Pb2+, Al3+ions are present. (iv) Pb2+ ions form a white precipitate when any soluble SO42-, SO32-, CO32-, Cl-, is added while Al3+ ions do not form a white precipitate\n30 (v) Pb2+ ions form a yellow precipitate when any soluble I-(e.g. Potassium/sodium Iodide) is added while Al3+ ions do not form a yellow precipitate. (vi) Pb2+ ions form a black precipitate when any soluble S-(e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8449728260869567, "ocr_used": true, "chunk_length": 1024, "token_count": 287}}
{"text": "(iv) Pb2+ ions form a white precipitate when any soluble SO42-, SO32-, CO32-, Cl-, is added while Al3+ ions do not form a white precipitate\n30 (v) Pb2+ ions form a yellow precipitate when any soluble I-(e.g. Potassium/sodium Iodide) is added while Al3+ ions do not form a yellow precipitate. (vi) Pb2+ ions form a black precipitate when any soluble S-(e.g. Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions\n31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron\n32 b) Summary of extraction of common metal.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.841050055712237, "ocr_used": true, "chunk_length": 1753, "token_count": 493}}
{"text": "(vi) Pb2+ ions form a black precipitate when any soluble S-(e.g. Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions\n31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron\n32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8310327196179554, "ocr_used": true, "chunk_length": 1989, "token_count": 587}}
{"text": "Potassium/sodium sulphide) is added while Al3+ ions do not form a black precipitate.i.e; Observation Inference White precipitate in excess 2M NaOH (aq) Zn2+, Pb2+, Al3+ ions White precipitate that dissolves to form a colourless solution in excess 2M NH3(aq) Zn2+ ions White precipitate in excess 2M NH3(aq) Pb2+, Al3+ ions White precipitate on adding about 4 drops of either Na2CO3(aq), Na2SO3(aq), Na2SO4(aq), H2SO4(aq), HCl(aq), NaCl(aq) Pb2+ions Yellow precipitate on adding about 4 drops of of KI(aq).NaI (aq) Pb2+ ions Black precipitate on adding aout 4 drops of Na2S(aq)/K2S(aq) Pb2+ ions\n31 6.GENERAL SUMMARY OF METALS a) Summary methods of extracting metal from their ore The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron\n32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8084054229958854, "ocr_used": true, "chunk_length": 2255, "token_count": 718}}
{"text": "Position on the earth’s crust If near the surface use open cast mining / quarrying If deep on the earth’s crust use deep mining Add oil, water, and blow air to form froth to concentrate the ore if it is a low grade Roast the ore first if it is a carbonate / sulphide of Zinc, Iron, Tin, Lead, and copper to form the oxide Electrolyse the ore if it is made of reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium Reduce the oxide using carbon in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron\n32 b) Summary of extraction of common metal. Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8071441217208377, "ocr_used": true, "chunk_length": 1466, "token_count": 476}}
{"text": "Metal Chief ore/s Chemical formula of ore Method of extraction Main equation during extraction Sodium Rock salt NaCl(s) Downs process Through electrolysis of molten NaCl (CaCl2 lower m.pt from 800oC-> 600oC) Cathode: 2Na+(l) + 2e -> 2Na(l) Anode: 2Cl-(l) -> Cl2(g) + 2e Sodium/ sodium hydroxide Brine NaCl(aq) Flowing mercury cathode cell Through electrolysis of concentrated NaCl(aq) Cathode: 2Na+(aq)+2e ->2Na(aq) Anode: 2Cl-(aq) -> Cl2(g) + 2e Aluminium Bauxite Al2O3.2H2O Halls process Through electrolysis of molten Al2O3. (Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air. Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)\n33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7300766050583658, "ocr_used": true, "chunk_length": 1408, "token_count": 573}}
{"text": "(Cryolite lower m.pt from 2015oC -> 800oC) Cathode: 4Al3+(l) + 12e -> 4Al(l) Anode: 6O2-(l) -> 3O2(g) + 12e Iron Haematite Magnetite Fe2O3 Fe3O4 Blast furnace Reduction of the ore by carbon(II)oxide Fe2O3(s)+ 3CO(g) 2Fe(l) +3CO2(g) Fe3O4(s)+ 4CO(g) 3Fe(l) +4CO2(g) Copper Copper pyrites CuFeS2 Roasting the ore in air to get Cu2S. Heating Cu2S ore in regulated supply of air. Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)\n33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal. Alloy name Constituents of the alloy Uses of the alloy Brass Copper and Zinc Making scews and bulb caps Bronze Copper and Tin Making clock springs,electrical contacts and copper coins Soldier Lead and Tin Soldering, joining electrical contacts because of its low melting points and high thermal conductivity Duralumin Aluminium, Copper and Magnesium Making aircraft , utensils ,windows frames because of its light weight and corrosion resistant. Steel Iron, Carbon ,Manganese and other metals Railway lines , car bodies girders and utensils.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7812811660912927, "ocr_used": true, "chunk_length": 1422, "token_count": 499}}
{"text": "Reduction of Cu2O by Cu2S 2CuFeS2 (s)+ 4O2(g) -> Cu2S(s)+3SO2(g) +2FeO(s) 2Cu2S (s)+ 3O2(g) -> 2Cu2O(s)+2SO2(g) Cu2S (s)+ 2Cu2O(s) -> 6Cu(s)+ SO2(g) Zinc Calamine ZnCO3 Roasting the ore in air to get ZnO ZnCO3(s)-> ZnO(s) + CO2(g)\n33 Blast furnace /reduction of the oxide by Carbon(II)Oxide/Carbon 2ZnS(s) +3O2(g) -> 2ZnO(s) + 2SO2(g) ZnO(s) + CO(g)-> Zn(s) + CO2(g) Lead Galena PbS Blast furnaceReduction of the oxide by carbon(II)oxide /Carbon PbO(s) + CO(g)-> Pb(s) + CO2(g) c) Common alloys of metal. Alloy name Constituents of the alloy Uses of the alloy Brass Copper and Zinc Making scews and bulb caps Bronze Copper and Tin Making clock springs,electrical contacts and copper coins Soldier Lead and Tin Soldering, joining electrical contacts because of its low melting points and high thermal conductivity Duralumin Aluminium, Copper and Magnesium Making aircraft , utensils ,windows frames because of its light weight and corrosion resistant. Steel Iron, Carbon ,Manganese and other metals Railway lines , car bodies girders and utensils. Nichrome Nichrome and Chromium Provide resistance in electric heaters and ovens German silver Copper,Zinc and Nickel Making coins d) Physical properties of metal. Metals form giant metallic structure joined by metallic bond from electrostatic attraction between the metallic cation and free delocalized electrons. This makes metals to have the following physical properties: (i)High melting and boiling points The giant metallic structure has a very close packed metallic lattice joined by strong electrostatic attraction between the metallic cation and free delocalized electrons.The more delocalized electrons the higher the melting/boiling points e.g. Aluminium has a melting point of about 2015oC while that of sodium is about 98oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8588945254960341, "ocr_used": true, "chunk_length": 1873, "token_count": 498}}
{"text": "Metals form giant metallic structure joined by metallic bond from electrostatic attraction between the metallic cation and free delocalized electrons. This makes metals to have the following physical properties: (i)High melting and boiling points The giant metallic structure has a very close packed metallic lattice joined by strong electrostatic attraction between the metallic cation and free delocalized electrons.The more delocalized electrons the higher the melting/boiling points e.g. Aluminium has a melting point of about 2015oC while that of sodium is about 98oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one. Aluminium has a boiling point of about 2470oC while that of sodium is about 890oC.This is mainly because aluminium has more/three delocalized electrons than sodium/has one. 34 (ii)High thermal and electrical conductivity All metals are good thermal and electrical conductors as liquid or solids. The more delocalized electrons the higher the thermal and electrical conductivity. e.g. Aluminium has an electrical conductivity of about 3.82 x 19-9 ohms per metre. Sodium has an electrical conductivity of about 2.18 x 19-9 ohms per metre. (iii)Shiny/Lustrous The free delocalized electrons on the surface of the metal absorb, vibrate and then scatter/re-emit/lose light energy. All metals are therefore usually shades of grey in colour except copper which is shiny brown.e.g. Zinc is bluish grey while iron is silvery grey. (iv)High tensile strength The free delocalized electrons on the surface of the metal atoms binds the surface immediately when the metal is coiled/folded preventing it from breaking /being brittle. (v)Malleable. Metals can be made into thin sheet. The metallic crystal lattice on being beaten/pressed/hammered on two sides extend its length and width/bredth and is then immediately bound by the delocalized electrons preventing it from breaking/being brittle. (vi)Ductile. Metals can be made into thin wires. The metallic crystal lattice on being beaten/pressed/hammered on all sides extend its length is then immediately bound by the delocalized electrons preventing it from breaking/being brittle.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9051906426906426, "ocr_used": true, "chunk_length": 2184, "token_count": 469}}
{"text": "(vi)Ductile. Metals can be made into thin wires. The metallic crystal lattice on being beaten/pressed/hammered on all sides extend its length is then immediately bound by the delocalized electrons preventing it from breaking/being brittle. Revision questions 1.Given some soil , dilute sulphuric(VI)acid,mortar,pestle,filter paper,filter funnel and 2M aqueous ammonia,describe with explanation,how you would show that the soil contain Zinc. Place the soil sample in the pestle. Crush using the mortar to reduce the particle size/increase its surface area. Add dilute sulphuric(VI)acid to free the ions in soil sample. Filter to separate insoluble residue from soluble filtrate To filtrate,add three drops of aqueous ammonia as precipitating reagent. A white precipitate of Zn(OH)2, Pb(OH)2 or Al(OH)3 is formed Add excess aqueous ammonia to the white precipitate. If it dissolves the Zn2+ ions are present. Zn(OH)2 react with excess ammonia to form soluble [Zn(OH)4]2+ complex. 2.In the extraction of aluminium,the oxide is dissolved in cryolite. (i)What is the chemical name of cryolite? Sodium hexafloroaluminate/Na3AlF6\n35 (ii)What is the purpose of cryolite? To lower the melting point of the electrolyte/Aluminium oxide from about 2015oC to 900oC (iii)Name the substance used for similar purpose in the Down cell Calcium chloride/CaCl2 (iv)An alloy of sodium and potassium is used as coolant in nuclear reactors.Explain. Nuclear reactors generate a lot of heat energy. sodium and potassium alloy reduce/lower the high temperature in the reactors. (v)Aluminium metal is used to make cooking utensils in preference to other metals.Explain. Aluminium (i) is a very good conductor of electricity because it has three delocalized electrons in its metallic structure (ii)is cheap,malleable,ductile and has high tensile strength (iii)on exposure to fire/heat form an impervious layer that prevent it from rapid corrosion. 3.Study the scheme below and use it to answer the questions that follow.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8933140847413322, "ocr_used": true, "chunk_length": 1992, "token_count": 490}}
{"text": "(v)Aluminium metal is used to make cooking utensils in preference to other metals.Explain. Aluminium (i) is a very good conductor of electricity because it has three delocalized electrons in its metallic structure (ii)is cheap,malleable,ductile and has high tensile strength (iii)on exposure to fire/heat form an impervious layer that prevent it from rapid corrosion. 3.Study the scheme below and use it to answer the questions that follow. (a)Identify: (i)solid residue L Iron(III)Oxide/Fe2O3 (ii)Solid N Aluminium hydroxide /Al(OH)3\n36 (iii)Filtrate M Sodium tetrahydroxoaluminate/ NaAl(OH)4 and sodium silicate/ NaSiO3 (iv)Solid P Aluminium oxide/ Al2O3 (v)Gas Q Oxygen/O2 (vi)Process K1 Filtration (vii)Process K2 Electrolysis (b)Write the equation for the reaction taking place in the formation of solid P from solid N 2Al(OH)3 -> Al2O3 (s) + 3H2O(l) (c)Name a substance added to solid N before process Process K2 take place. Cryolite/Sodium tetrahydroxoaluminate/ NaAl(OH)4 (d)State the effect of evolution of gas Q on (i)process K2 Oxygen produced at the anode reacts with the carbon anode to form carbon(IV) oxide which escape. The electrolytic process needs continuous replacement of the carbon anode. (ii)the environment Oxygen produced at the anode reacts with the carbon anode to form carbon(IV) oxide which escape to the atmosphere.CO2 is a green house gas that cause global warming. (e)An aluminium manufacturing factory runs for 24 hours. If the total mass of aluminium produced is 27000kg, (i)Calculate the current used. (Faraday constant=96500Coulombs, Al=27.0).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8638350677363645, "ocr_used": true, "chunk_length": 1579, "token_count": 435}}
{"text": "(e)An aluminium manufacturing factory runs for 24 hours. If the total mass of aluminium produced is 27000kg, (i)Calculate the current used. (Faraday constant=96500Coulombs, Al=27.0). (ii)assuming all the gas produced react with 200kg of anode ,calculate the loss in mass of the electrode.(Molar gas volume at room temperature = 24dm3,C=12.0) Working Equation at Cathode Al3+(l) + 3e -> Al(l) 27g Al -> 3 Faradays = 3 x 96500C (27000kg x 1000) g -> (27000kg x 1000) g x 3 x 96500C 27g =289500000000 Coulombs\n37 Current = Quantity of electricity =>289500000000 Coulombs Time in seconds 24 x 60 x 60 3350690Ampheres Working Equation at Anode 2O2-(l) + 4e -> O2(g) 4 Faradays -> 4 x 96500C24dm3 O2(g) - 289500000000 Coulombs -> 289500000000 Coulombs x 24dm3 4 x 96500C 18,000,000dm3 Chemical equation at anode O2(g) + C (s) -> CO2(g) Method 1 24dm3 of O2(g) -> 12.0g Carbon 18,000,000dm3 of O2(g) -> 18,000,000dm3 x 12 = 9000000g = 9000kg 24dm3 1000g Loss in mass of the carbon graphite anode = 9000kg NB:Mass of the carbon graphite anode remaining =27000kg - 9000kg =18000kg The flow chart below shows the extraction of iron metal.Use it to answer the questions that follow.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6781495907187935, "ocr_used": true, "chunk_length": 1171, "token_count": 413}}
{"text": "If the total mass of aluminium produced is 27000kg, (i)Calculate the current used. (Faraday constant=96500Coulombs, Al=27.0). (ii)assuming all the gas produced react with 200kg of anode ,calculate the loss in mass of the electrode.(Molar gas volume at room temperature = 24dm3,C=12.0) Working Equation at Cathode Al3+(l) + 3e -> Al(l) 27g Al -> 3 Faradays = 3 x 96500C (27000kg x 1000) g -> (27000kg x 1000) g x 3 x 96500C 27g =289500000000 Coulombs\n37 Current = Quantity of electricity =>289500000000 Coulombs Time in seconds 24 x 60 x 60 3350690Ampheres Working Equation at Anode 2O2-(l) + 4e -> O2(g) 4 Faradays -> 4 x 96500C24dm3 O2(g) - 289500000000 Coulombs -> 289500000000 Coulombs x 24dm3 4 x 96500C 18,000,000dm3 Chemical equation at anode O2(g) + C (s) -> CO2(g) Method 1 24dm3 of O2(g) -> 12.0g Carbon 18,000,000dm3 of O2(g) -> 18,000,000dm3 x 12 = 9000000g = 9000kg 24dm3 1000g Loss in mass of the carbon graphite anode = 9000kg NB:Mass of the carbon graphite anode remaining =27000kg - 9000kg =18000kg The flow chart below shows the extraction of iron metal.Use it to answer the questions that follow. (a)Identify: (i)gas P Carbon(IV)oxide/CO2 (ii)Solid Q\n38 Carbon/coke/charcoal (iii)Solid R Carbon/coke/charcoal (iv)Solid V Limestone/calcium carbonate/CaCO3 (v)Solid S Iron/Fe (b)Write the chemical equation for the reaction for the formation of: (i)Solid S Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g) (ii)Carbon(II)oxide C(s) + CO2 (g) -> 2CO (g) (iii)Slag SiO2(s) + CaO(s) -> CaSiO3(s) Al2O3 (s) + CaO(s) -> Ca Al2O4(s) (iv)Gas P C(s) + O2 (g) -> CO2 (g) (c)State two uses of: (i)Solid S Iron is used in making: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes, lamp posts because it is cheap.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.709278213035506, "ocr_used": true, "chunk_length": 1714, "token_count": 641}}
{"text": "(Faraday constant=96500Coulombs, Al=27.0). (ii)assuming all the gas produced react with 200kg of anode ,calculate the loss in mass of the electrode.(Molar gas volume at room temperature = 24dm3,C=12.0) Working Equation at Cathode Al3+(l) + 3e -> Al(l) 27g Al -> 3 Faradays = 3 x 96500C (27000kg x 1000) g -> (27000kg x 1000) g x 3 x 96500C 27g =289500000000 Coulombs\n37 Current = Quantity of electricity =>289500000000 Coulombs Time in seconds 24 x 60 x 60 3350690Ampheres Working Equation at Anode 2O2-(l) + 4e -> O2(g) 4 Faradays -> 4 x 96500C24dm3 O2(g) - 289500000000 Coulombs -> 289500000000 Coulombs x 24dm3 4 x 96500C 18,000,000dm3 Chemical equation at anode O2(g) + C (s) -> CO2(g) Method 1 24dm3 of O2(g) -> 12.0g Carbon 18,000,000dm3 of O2(g) -> 18,000,000dm3 x 12 = 9000000g = 9000kg 24dm3 1000g Loss in mass of the carbon graphite anode = 9000kg NB:Mass of the carbon graphite anode remaining =27000kg - 9000kg =18000kg The flow chart below shows the extraction of iron metal.Use it to answer the questions that follow. (a)Identify: (i)gas P Carbon(IV)oxide/CO2 (ii)Solid Q\n38 Carbon/coke/charcoal (iii)Solid R Carbon/coke/charcoal (iv)Solid V Limestone/calcium carbonate/CaCO3 (v)Solid S Iron/Fe (b)Write the chemical equation for the reaction for the formation of: (i)Solid S Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g) (ii)Carbon(II)oxide C(s) + CO2 (g) -> 2CO (g) (iii)Slag SiO2(s) + CaO(s) -> CaSiO3(s) Al2O3 (s) + CaO(s) -> Ca Al2O4(s) (iv)Gas P C(s) + O2 (g) -> CO2 (g) (c)State two uses of: (i)Solid S Iron is used in making: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes, lamp posts because it is cheap. (ii)nails, cutlery, scissors, sinks, vats, spanners, steel rods, and railway points from steel.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7110357441567633, "ocr_used": true, "chunk_length": 1727, "token_count": 648}}
{"text": "(ii)assuming all the gas produced react with 200kg of anode ,calculate the loss in mass of the electrode.(Molar gas volume at room temperature = 24dm3,C=12.0) Working Equation at Cathode Al3+(l) + 3e -> Al(l) 27g Al -> 3 Faradays = 3 x 96500C (27000kg x 1000) g -> (27000kg x 1000) g x 3 x 96500C 27g =289500000000 Coulombs\n37 Current = Quantity of electricity =>289500000000 Coulombs Time in seconds 24 x 60 x 60 3350690Ampheres Working Equation at Anode 2O2-(l) + 4e -> O2(g) 4 Faradays -> 4 x 96500C24dm3 O2(g) - 289500000000 Coulombs -> 289500000000 Coulombs x 24dm3 4 x 96500C 18,000,000dm3 Chemical equation at anode O2(g) + C (s) -> CO2(g) Method 1 24dm3 of O2(g) -> 12.0g Carbon 18,000,000dm3 of O2(g) -> 18,000,000dm3 x 12 = 9000000g = 9000kg 24dm3 1000g Loss in mass of the carbon graphite anode = 9000kg NB:Mass of the carbon graphite anode remaining =27000kg - 9000kg =18000kg The flow chart below shows the extraction of iron metal.Use it to answer the questions that follow. (a)Identify: (i)gas P Carbon(IV)oxide/CO2 (ii)Solid Q\n38 Carbon/coke/charcoal (iii)Solid R Carbon/coke/charcoal (iv)Solid V Limestone/calcium carbonate/CaCO3 (v)Solid S Iron/Fe (b)Write the chemical equation for the reaction for the formation of: (i)Solid S Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g) (ii)Carbon(II)oxide C(s) + CO2 (g) -> 2CO (g) (iii)Slag SiO2(s) + CaO(s) -> CaSiO3(s) Al2O3 (s) + CaO(s) -> Ca Al2O4(s) (iv)Gas P C(s) + O2 (g) -> CO2 (g) (c)State two uses of: (i)Solid S Iron is used in making: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes, lamp posts because it is cheap. (ii)nails, cutlery, scissors, sinks, vats, spanners, steel rods, and railway points from steel. Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7223637915599692, "ocr_used": true, "chunk_length": 1779, "token_count": 656}}
{"text": "(a)Identify: (i)gas P Carbon(IV)oxide/CO2 (ii)Solid Q\n38 Carbon/coke/charcoal (iii)Solid R Carbon/coke/charcoal (iv)Solid V Limestone/calcium carbonate/CaCO3 (v)Solid S Iron/Fe (b)Write the chemical equation for the reaction for the formation of: (i)Solid S Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g) (ii)Carbon(II)oxide C(s) + CO2 (g) -> 2CO (g) (iii)Slag SiO2(s) + CaO(s) -> CaSiO3(s) Al2O3 (s) + CaO(s) -> Ca Al2O4(s) (iv)Gas P C(s) + O2 (g) -> CO2 (g) (c)State two uses of: (i)Solid S Iron is used in making: (i)gates ,pipes, engine blocks, rails, charcoal iron boxes, lamp posts because it is cheap. (ii)nails, cutlery, scissors, sinks, vats, spanners, steel rods, and railway points from steel. Steel is an alloy of iron with carbon, and/or Vanadium, Manganese, Tungsten, Nickel ,Chromium. It does not rust/corrode like iron. (ii)Slag (i) tarmacing roads (ii) cement manufacture (iii) as building construction material 3.You are provided with sulphuric(VI)acid ,2M aqueous ammonia and two ores suspected to contain copper and iron. Describe with explanation how you would differentiate the two ores. Crush the two ores separately in using a mortar and pestle to reduce the particle size/increase the surface area. Add sulphuric(VI)acid to separate portion of the ore. Filter. 39 To a portion of the filtrate,add three drops of 2M aqueous ammonia then axcess Results A green precipitate insoluble in excess 2M aqueous ammonia confirms the ore contain Fe2+ ion. A brown precipitate insoluble in excess 2M aqueous ammonia confirms the ore contain Fe3+ ion. A blue precipitate that dissolve in excess 2M aqueous ammonia to form a deep/royal blue solution confirms the ore contain Cu2+ ion. 4.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8440781527531083, "ocr_used": true, "chunk_length": 1689, "token_count": 512}}
{"text": "A brown precipitate insoluble in excess 2M aqueous ammonia confirms the ore contain Fe3+ ion. A blue precipitate that dissolve in excess 2M aqueous ammonia to form a deep/royal blue solution confirms the ore contain Cu2+ ion. 4. Use the flow chart below showing the extraction of Zinc metal to answer the questions that follow (a)Name: (i)two ores from which Zinc can be extracted Calamine(ZnCO3) Zinc blende(ZnS) (ii)two possible identity of gas P Sulphur(IV)oxide(SO2) from roasting Zinc blende Carbon(IV)oxide(CO2) from decomposition of Calamine. (b)Write a possible chemical equation taking place in the roasting chamber. 2ZnS(s) + 3O2 (g) -> 2ZnO(s) + 2SO2(g) ZnCO3(s) -> ZnO(s) + CO2(g) (c)Explain the effect of the by-product of the roating on the environment. Sulphur (IV)oxide from roasting Zinc blende is an acidic gas that causes “acid rain” on dissolving in rain water. Carbon(IV)oxide(CO2) from decomposition of Calamine is a green house gas that causes global warming. (d)(i)Name a suitable reducing agent used in the furnace during extraction of Zinc. Carbon(II)oxide\n40 (ii)Write a chemical equation for the reduction process ZnO(s) + CO(g) -> Zn(s) + CO2(g) (e)(i)Before electrolysis, the products from roasting is added dilute sulphuric (VI)acid. Write the equation for the reaction with dilute sulphuric(VI)acid. ZnO(s) + H2SO4 (aq) -> Zn SO4(aq) + H2(g) (ii)During the electrolysis for extraction of Zinc,state the I. Anode used Aluminium sheet II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8552320300783643, "ocr_used": true, "chunk_length": 1468, "token_count": 429}}
{"text": "Write the equation for the reaction with dilute sulphuric(VI)acid. ZnO(s) + H2SO4 (aq) -> Zn SO4(aq) + H2(g) (ii)During the electrolysis for extraction of Zinc,state the I. Anode used Aluminium sheet II. Cathode used Lead plate coated with silver (ii)Write the equation for the electrolysis for extraction of Zinc at the: I.Cathode; Zn2+(aq) + 2e -> Zn(s) II.Anode; 4OH-(aq) -> 2H2O(l) + O2(s) + 4e (f)(i)What is galvanization Dipping Iron in molten Zinc to form a thin layer of Zinc to prevent iron from rusting. (ii)Galvanized iron sheet rust after some time. Explain The thin layer of Zinc protect Iron from rusting through sacrificial protection. When all the Zinc has reacted with elements of air, Iron start rusting. (g)State two uses of Zinc other than galvanization. Making brass(Zinc/copper alloy) Making german silver(Zinc/copper/nickel alloy) As casing for dry cells/battery (h)Calculate the mass of Zinc that is produced from the reduction chamber if 6400kg of Calamine ore is fed into the roaster.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8664389809767494, "ocr_used": true, "chunk_length": 1010, "token_count": 284}}
{"text": "When all the Zinc has reacted with elements of air, Iron start rusting. (g)State two uses of Zinc other than galvanization. Making brass(Zinc/copper alloy) Making german silver(Zinc/copper/nickel alloy) As casing for dry cells/battery (h)Calculate the mass of Zinc that is produced from the reduction chamber if 6400kg of Calamine ore is fed into the roaster. Assume the process is 80% efficient in each stage(Zn=64.0,C=12.0,O=16.0) Molar mass ZnCO3(s) =124g Molar mass Zn = 64g Molar mass ZnO = 80g Chemical equation ZnCO3(s) -> ZnO(s) + CO2(g) Method 1 124g ZnCO3 => 80g ZnO\n41 (6400kg x1000)g ZnCO3 => (6400 x1000) x 80 = 512,000,000 g of ZnO 124 100% => 512,000,000 g of ZnO 80% => 80 x 512,000,000 g = 409600000g of ZnO 100 Chemical equation ZnO(s) + CO(g) -> Zn(s) + CO2(g) 80g ZnO(s) => 64g Zn(s) 409600000g of ZnO => 409600000g x 64 = 327680000 g Zn 80 100% => 327680000 g Zn 80% => 80 x 327680000 g Zn = 262144000g of Zn 100 Mass of Zinc produced = 262144000g of Zn 5.An ore is suspected to bauxite. Describe the process that can be used to confirm the presence of aluminium in the ore. Crush the ore to fine powder to increase surface area/reduce particle size. Add hot concentrated sulphuric(VI)/nitric(V) acid to free the ions. Filter. Retain the filtrate Add excess aqueous ammonia to a sample of filtrate. A white precipitate confirms presence of either Al3+ or Pb2+. Add sodium sulphate,dilute sulphuric(VI)to another portion of filtrate.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7469137526888001, "ocr_used": true, "chunk_length": 1453, "token_count": 494}}
{"text": "Retain the filtrate Add excess aqueous ammonia to a sample of filtrate. A white precipitate confirms presence of either Al3+ or Pb2+. Add sodium sulphate,dilute sulphuric(VI)to another portion of filtrate. No white precipitate confirms presence of Al3+ Or Add potassium iodide to another portion of filtrate. No yellow precipitate confirms presence of Al3+ 6.The flow chart below illustrate the industrial extraction of Lead metal\n42 (a)(i)Name the chief ore that is commonly used in this process Galena(PbS) (ii)Explain what take place in the roasting furnace\n23.0.0 ORGANIC CHEMISTRY II (ALKANOLS AND ALKANOIC ACIDS) (20 LESSONS) B.ALKANOLS(Alcohols) (A) INTRODUCTION.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8693080054274085, "ocr_used": true, "chunk_length": 670, "token_count": 180}}
{"text": "Add sodium sulphate,dilute sulphuric(VI)to another portion of filtrate. No white precipitate confirms presence of Al3+ Or Add potassium iodide to another portion of filtrate. No yellow precipitate confirms presence of Al3+ 6.The flow chart below illustrate the industrial extraction of Lead metal\n42 (a)(i)Name the chief ore that is commonly used in this process Galena(PbS) (ii)Explain what take place in the roasting furnace\n23.0.0 ORGANIC CHEMISTRY II (ALKANOLS AND ALKANOIC ACIDS) (20 LESSONS) B.ALKANOLS(Alcohols) (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH H H Ethanol\n2 C2H5 OH H C – C –O - H │ H H 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ H H H H H H Hexanol 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol\n3 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7877956439569973, "ocr_used": true, "chunk_length": 1855, "token_count": 710}}
{"text": "No white precipitate confirms presence of Al3+ Or Add potassium iodide to another portion of filtrate. No yellow precipitate confirms presence of Al3+ 6.The flow chart below illustrate the industrial extraction of Lead metal\n42 (a)(i)Name the chief ore that is commonly used in this process Galena(PbS) (ii)Explain what take place in the roasting furnace\n23.0.0 ORGANIC CHEMISTRY II (ALKANOLS AND ALKANOIC ACIDS) (20 LESSONS) B.ALKANOLS(Alcohols) (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH H H Ethanol\n2 C2H5 OH H C – C –O - H │ H H 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ H H H H H H Hexanol 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol\n3 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7850443883749183, "ocr_used": true, "chunk_length": 1845, "token_count": 707}}
{"text": "No yellow precipitate confirms presence of Al3+ 6.The flow chart below illustrate the industrial extraction of Lead metal\n42 (a)(i)Name the chief ore that is commonly used in this process Galena(PbS) (ii)Explain what take place in the roasting furnace\n23.0.0 ORGANIC CHEMISTRY II (ALKANOLS AND ALKANOIC ACIDS) (20 LESSONS) B.ALKANOLS(Alcohols) (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH H H Ethanol\n2 C2H5 OH H C – C –O - H │ H H 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ H H H H H H Hexanol 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol\n3 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7837803641796052, "ocr_used": true, "chunk_length": 1818, "token_count": 702}}
{"text": "Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH H H Ethanol\n2 C2H5 OH H C – C –O - H │ H H 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ H H H H H H Hexanol 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol\n3 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7706916327349888, "ocr_used": true, "chunk_length": 1484, "token_count": 601}}
{"text": "(iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points. (vi)they show similar and gradual change in their chemical properties. B. ISOMERS OF ALKANOLS. Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines: (i)Like alkanes , identify the longest carbon chain to be the parent name. (ii)Identify the position of the -OH functional group to give it the smallest /lowest position. (iii) Identify the type and position of the side branches. Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol\n4 Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.845599420499819, "ocr_used": true, "chunk_length": 1255, "token_count": 349}}
{"text": "Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol\n4 Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer) CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3\n5 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7357997557997559, "ocr_used": true, "chunk_length": 1323, "token_count": 508}}
{"text": "2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer) CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3\n5 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS. For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation. 6 In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination. Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7965567646581128, "ocr_used": true, "chunk_length": 1457, "token_count": 455}}
{"text": "Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase. (C6H10O5)n (s) + H2O(l) --diastase enzyme --> C12H22O11(aq) (Starch) (Maltose) (ii)Hydrolysis of Maltose to glucose using the enzyme maltase. C12H22O11(aq)+ H2O(l) -- maltase enzyme -->2 C6H12O6(aq) (Maltose) (glucose) (iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase. C6H12O6(aq) -- zymase enzyme --> 2 C2H5OH(aq) + 2CO2(g) (glucose) (Ethanol) At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction. To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol). Methanol is much more poisonous /toxic than ethanol. Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours. School laboratory preparation of ethanol from fermentation of glucose Measure 100cm3 of pure water into a conical flask. Add about five spatula end full of glucose. Stir the mixture to dissolve. Add about one spatula end full of yeast. Set up the apparatus as below. 7 Preserve the mixture for about three days. D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols (a) Role of yeast Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process. (b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8820282942928439, "ocr_used": true, "chunk_length": 1968, "token_count": 490}}
{"text": "(b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water. More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate. Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) H2O(l) + CO2 (g) + CaCO3(s) -> Ca(HCO3) 2 (aq) (c)Effects on litmus paper Experiment\n8 Take the prepared sample and test with both blue and red litmus papers. Repeat the same with pure ethanol and methylated spirit. Sample Observation table Substance/alkanol Effect on litmus paper Prepared sample Blue litmus paper remain blue Red litmus paper remain red Absolute ethanol Blue litmus paper remain blue Red litmus paper remain red Methylated spirit Blue litmus paper remain blue Red litmus paper remain red Explanation Alkanols are neutral compounds/solution that have characteristic sweet smell and taste. They have no effect on both blue and red litmus papers. (d)Solubility in water. Experiment Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water. Repeat the same with pure ethanol and methylated spirit. Observation No layers formed between the two liquids. Explanation Ethanol is miscible in water.Both ethanol and water are polar compounds . The solubility of alkanols decrease with increase in the alkyl chain/molecular mass. The alkyl group is insoluble in water while –OH functional group is soluble in water. As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases. e)Melting/boiling point. Experiment Place pure ethanol in a long boiling tube .Determine its boiling point. Observation Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure. Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . 9 This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.901311683047928, "ocr_used": true, "chunk_length": 2224, "token_count": 488}}
{"text": "Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . 9 This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling. f)Density Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other. This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density). Summary table showing the trend in physical properties of alkanols Alkanol Melting point (oC) Boiling point (oC) Density gcm-3 Solubility in water Methanol -98 65 0.791 soluble Ethanol -117 78 0.789 soluble Propanol -103 97 0.803 soluble Butanol -89 117 0.810 Slightly soluble Pentanol -78 138 0.814 Slightly soluble Hexanol -52 157 0.815 Slightly soluble Heptanol -34 176 0.822 Slightly soluble Octanol -15 195 0.824 Slightly soluble Nonanol -7 212 0.827 Slightly soluble Decanol 6 228 0.827 Slightly soluble g)Burning Experiment Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit. Observation/Explanation Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting. Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8299145916795766, "ocr_used": true, "chunk_length": 1634, "token_count": 409}}
{"text": "Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes. 10 Chemica equation C2 H5OH(l) + 3O2 (g) -> 3H2O(l) + 2CO2 (g) ( excess air) C2 H5OH(l) + 2O2 (g) -> 3H2O(l) + 2CO (g) ( limited air) 2CH3OH(l) + 3O2 (g) -> 4H2O(l) + 2CO2 (g) ( excess air) 2 CH3OH(l) + 2O2 (g) -> 4H2O(l) + 2CO (g) ( limited air) 2C3 H7OH(l) + 9O2 (g) -> 8H2O(l) + 6CO2 (g) ( excess air) C3 H7OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) 2C4 H9OH(l) + 13O2 (g) -> 20H2O(l) + 8CO2 (g) ( excess air) C4 H9OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) Due to its flammability, ethanol is used; (i) as a fuel in spirit lamps (ii) as gasohol when blended with gasoline (h)Formation of alkoxides Experiment Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7431806857815828, "ocr_used": true, "chunk_length": 1115, "token_count": 440}}
{"text": "Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations\n11 Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9159360580092288, "ocr_used": true, "chunk_length": 984, "token_count": 231}}
{"text": "Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations\n11 Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)\n12 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7988172124608012, "ocr_used": true, "chunk_length": 2494, "token_count": 825}}
{"text": "Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations\n11 Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)\n12 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8003428805293402, "ocr_used": true, "chunk_length": 2547, "token_count": 842}}
{"text": "If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)\n12 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.759567638806224, "ocr_used": true, "chunk_length": 1857, "token_count": 681}}
{"text": "Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)\n12 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7537759002811287, "ocr_used": true, "chunk_length": 1817, "token_count": 672}}
{"text": "Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water. Smell the products. Repeat with methanol Sample observations Substance/alkanol Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid Absolute ethanol Sweet fruity smell Methanol Sweet fruity smell Explanation Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory. Alkanol + Alkanoic acid –Conc. H2SO4-> Ester + water Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters . Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g. Ethanol + Ethanoic acid -> Ethylethanoate + Water Ethanol + Propanoic acid -> Ethylpropanoate + Water Ethanol + Methanoic acid -> Ethylmethanoate + Water Ethanol + butanoic acid -> Ethylbutanoate + Water Propanol + Ethanoic acid -> Propylethanoate + Water Methanol + Ethanoic acid -> Methyethanoate + Water Methanol + Decanoic acid -> Methyldecanoate + Water Decanol + Methanoic acid -> Decylmethanoate + Water\n13 During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol. R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O e.g. 1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8911813971819885, "ocr_used": true, "chunk_length": 1791, "token_count": 484}}
{"text": "1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3COOH(l) --Conc. H2SO4 --> CH3COOCH2CH3(aq) +H2O(l) 2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water. Ethanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2CH3(aq) +H2O(l) 3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water. Methanol + Ethanoic acid --Conc. H2SO4 -->Methylethanoate + Water CH3OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO CH3(aq) +H2O(l) 4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water. Methanol + propanoic acid --Conc. H2SO4 -->Methylpropanoate + Water CH3OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. Propanol + Propanoic acid --Conc.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7280866456185836, "ocr_used": true, "chunk_length": 1132, "token_count": 479}}
{"text": "H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. Propanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C3H7OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C3H7(aq) +H2O(l) CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2 CH2CH3(aq) +H2O(l) (j)Oxidation Experiment Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of pH= 4/5/6 Weakly acidic\n14 KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH = 4/5/6 Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7788510911424904, "ocr_used": true, "chunk_length": 984, "token_count": 339}}
{"text": "Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of pH= 4/5/6 Weakly acidic\n14 KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH = 4/5/6 Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8256261163920707, "ocr_used": true, "chunk_length": 1436, "token_count": 440}}
{"text": "pH = 4/5/6 Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH\n15 Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7972083748753739, "ocr_used": true, "chunk_length": 1534, "token_count": 510}}
{"text": "methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH\n15 Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”. (k)Hydrolysis /Hydration and Dehydration I. Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8121212121212122, "ocr_used": true, "chunk_length": 1001, "token_count": 324}}
{"text": "(k)Hydrolysis /Hydration and Dehydration I. Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e. Alkenes + Water - H3PO4 catalyst-> Alkanol Examples (i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol Ethene + water ---60 atm/300oC/ H3PO4 --> Ethanol H2C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2OH(l) This is the main method of producing large quantities of ethanol instead of fermentation (ii) Propene + water ---60 atm/300oC/ H3PO4 --> Propanol CH3C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2OH(l) (iii) Butene + water ---60 atm/300oC/ H3PO4 --> Butanol CH3 CH2 C=CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2 CH2OH(l) II. Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances. Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e Alkanol --Conc. H2 SO4/180oC--> Alkene + Water Examples 1. At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7551297591297591, "ocr_used": true, "chunk_length": 1430, "token_count": 473}}
{"text": "At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene. Propanol ---180oC/ H2SO4 --> Propene + Water CH3 CH2 CH2OH(l) --180oC/ H2SO4 --> CH3CH =CH2 (g) + H2O(l) 3. Butanol undergoes dehydration to form Butene. Butanol ---180oC/ H2SO4 --> Butene + Water CH3 CH2 CH2CH2OH(l) --180oC/ H2SO4 --> CH3 CH2C =CH2 (g) + H2O(l) 3. Pentanol undergoes dehydration to form Pentene. Pentanol ---180oC/ H2SO4 --> Pentene + Water\n16 CH3 CH2 CH2 CH2 CH2OH(l)--180oC/ H2SO4-->CH3 CH2 CH2C =CH2 (g)+H2O(l) (l)Similarities of alkanols with Hydrocarbons I. Similarity with alkanes Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7404651777454737, "ocr_used": true, "chunk_length": 1127, "token_count": 399}}
{"text": "This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. CH2 CH2OH(l) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) CH2 CH2OH(l) + 2O2(g) -Limited air-> 2CO (g) + 3H2 O(l) CH3 CH3(g) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) 2CH3 CH3(g) + 5O2(g) -Limited air-> 4CO (g) + 6H2 O(l) II. Similarity with alkenes/alkynes Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond: (i)decolorize acidified KMnO4 (ii)turns Orange acidified K2Cr2O7 to green. Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH). Alkenes are oxidized to alkanols with duo/double functional groups. Examples 1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid. Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol. Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7876220474345101, "ocr_used": true, "chunk_length": 1367, "token_count": 506}}
{"text": "Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water. Alkenes decolorizes bromine and chlorine water to form halogenoalkanols Example\n17 When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol. Ethene + Bromine water -> Bromoethanol. H2C=CH2 + HOBr -> BrCH2 -CH2OH IV. Differences in melting and boiling point with Hydrocarbons Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne) This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding. Two alkanol molecules form a dimer joined by hydrogen bonding. Example In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen. This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen. Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer. H H H H C C O H H H H H O C C H H H Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting. E.USES OF SOME ALKANOLS (a)Methanol is used as industrial alcohol and making methylated spirit (b)Ethanol is used: 1. as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. Hydrogen bonds Covalent bonds\n18 B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8881367388284837, "ocr_used": true, "chunk_length": 1853, "token_count": 484}}
{"text": "as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. Hydrogen bonds Covalent bonds\n18 B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION. Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:\n19 Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where: (i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows. (ii) the members have R-COOH/R C-O-H as the functional group.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8808725079084824, "ocr_used": true, "chunk_length": 788, "token_count": 219}}
{"text": "Hydrogen bonds Covalent bonds\n18 B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION. Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:\n19 Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where: (i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows. (ii) the members have R-COOH/R C-O-H as the functional group. n General /molecular formular Structural formula IUPAC name 0 HCOOH H – C –O - H │ O Methanoic acid 1 CH3 COOH H H – C – C – O - H │ H O Ethanoic acid 2 CH3 CH2 COOH C2 H5 COOH H H H-C – C – C – O – H H H O Propanoic acid 3 CH3 CH2 CH2 COOH C3 H7 COOH H H H H- C - C – C – C – O – H H H H O Butanoic acid 4 CH3CH2CH2CH2 COOH C4 H9 COOH H H H H H - C – C - C – C – C – O – H H H H H O Pentanoic acid 5 CH3CH2 CH2CH2CH2 COOH C5 H11 COOH H H H H H H C - C – C - C – C – C – O – H H H H H H O Hexanoic acid 6 CH3CH2 CH2 CH2CH2CH2 COOH C6 H13 COOH H H H H H H H C C - C – C - C – C – C – O – H H H H H H H O Pentanoic acid\n20 O (iii)they have the same general formula represented by R-COOH where R is an alkyl group. (iv)each member differ by –CH2- group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting point.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8115719272408668, "ocr_used": true, "chunk_length": 1421, "token_count": 506}}
{"text": "(iv)each member differ by –CH2- group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting point. (vi)they show similar and gradual change in their chemical properties. (vii) since they are acids they show similar properties with mineral acids. (B) ISOMERS OF ALKANOIC ACIDS. Alkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines (i)Like alkanes. identify the longest carbon chain to be the parent name. (ii)Identify the position of the -C-O-H functional group to give it the smallest O /lowest position. (iii)Identify the type and position of the side group branches. Practice examples on isomers of alkanoic acids 1.Isomers of butanoic acid C3H7COOH CH3 CH2 CH2 COOH Butan-1-oic acid CH3 H2C C COOH 2-methylpropan-1-oic acid 2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does. 2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n21 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8137824048538335, "ocr_used": true, "chunk_length": 1568, "token_count": 516}}
{"text": "(iii)Identify the type and position of the side group branches. Practice examples on isomers of alkanoic acids 1.Isomers of butanoic acid C3H7COOH CH3 CH2 CH2 COOH Butan-1-oic acid CH3 H2C C COOH 2-methylpropan-1-oic acid 2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does. 2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n21 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS. In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 22 The oxidation converts the alkanol first to an alkanal the alkanoic acid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7788775498508012, "ocr_used": true, "chunk_length": 1153, "token_count": 429}}
{"text": "2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n21 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS. In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 22 The oxidation converts the alkanol first to an alkanal the alkanoic acid. NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7 General equation: R- CH2 – OH + [O] --H+/KMnO4--> R- CH –O + H2O(l) (alkanol) (alkanal) R- CH – O + [O] --H+/KMnO4--> R- C –OOH (alkanal) (alkanoic acid) Examples 1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7574202496532594, "ocr_used": true, "chunk_length": 1064, "token_count": 432}}
{"text": "In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 22 The oxidation converts the alkanol first to an alkanal the alkanoic acid. NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7 General equation: R- CH2 – OH + [O] --H+/KMnO4--> R- CH –O + H2O(l) (alkanol) (alkanal) R- CH – O + [O] --H+/KMnO4--> R- C –OOH (alkanal) (alkanoic acid) Examples 1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid . CH3- CH2 – OH + [O] --H+/KMnO4--> CH3- CH –O + H2O(l) (ethanol) (ethanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (ethanal) (ethanoic acid) 2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid CH3- CH2 CH2 – OH + [O] --H+/KMnO4--> CH3- CH2 CH –O + H2O(l) (propanol) (propanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (propanal) (propanoic acid) Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from: (a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e. Alkenes + Steam/water -- H2PO4 Catalyst--> Alkanol The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7981476306439379, "ocr_used": true, "chunk_length": 1354, "token_count": 481}}
{"text": "CH3- CH2 – OH + [O] --H+/KMnO4--> CH3- CH –O + H2O(l) (ethanol) (ethanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (ethanal) (ethanoic acid) 2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid CH3- CH2 CH2 – OH + [O] --H+/KMnO4--> CH3- CH2 CH –O + H2O(l) (propanol) (propanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (propanal) (propanoic acid) Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from: (a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e. Alkenes + Steam/water -- H2PO4 Catalyst--> Alkanol The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid. Alkanol + Air -- MnSO4 Catalyst/5 atm pressure--> Alkanoic acid Example Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol. 23 CH2=CH2 + H2O -> CH3 CH2OH (Ethene) (Ethanol) This is the industrial large scale method of manufacturing ethanol Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2OH + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanol) (Ethanoic acid) (b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8040654513932025, "ocr_used": true, "chunk_length": 1503, "token_count": 493}}
{"text": "Alkanol + Air -- MnSO4 Catalyst/5 atm pressure--> Alkanoic acid Example Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol. 23 CH2=CH2 + H2O -> CH3 CH2OH (Ethene) (Ethanol) This is the industrial large scale method of manufacturing ethanol Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2OH + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanol) (Ethanoic acid) (b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals. Alkyne + Water -- Mercury(II)sulphate(VI)catalyst--> Alkanal The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid. Alkanal + air/oxygen -- Manganese(II)sulphate(VI)catalyst--> Alkanoic acid Example Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal. CH = CH + H2O --HgSO4--> CH3 CH2O (Ethyne) (Ethanal) This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas. Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2O + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanal) (Oxygen from air) (Ethanoic acid) (D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8429242989139246, "ocr_used": true, "chunk_length": 1592, "token_count": 464}}
{"text": "CH = CH + H2O --HgSO4--> CH3 CH2O (Ethyne) (Ethanal) This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas. Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2O + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanal) (Oxygen from air) (Ethanoic acid) (D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS. I.Physical properties of alkanoic acids\n24 The table below shows some physical properties of alkanoic acids Alkanol Melting point(oC) Boiling point(oC) Density(gcm-3) Solubility in water Methanoic acid 18.4 101 1.22 soluble Ethanoic acid 16.6 118 1.05 soluble Propanoic acid -2.8 141 0.992 soluble Butanoic acid -8.0 164 0.964 soluble Pentanoic acid -9.0 187 0.939 Slightly soluble Hexanoic acid -11 205 0.927 Slightly soluble Heptanoic acid -3 223 0.920 Slightly soluble Octanoic acid 11 239 0.910 Slightly soluble Nonanoic acid 16 253 0.907 Slightly soluble Decanoic acid 31 269 0.905 Slightly soluble From the table note the following: (i) Melting and boiling point decrease as the carbon chain increases due to increase in intermolecular forces of attraction between the molecules requiring more energy to separate the molecules. (ii) The density decreases as the carbon chain increases as the intermolecular forces of attraction increases between the molecules making the molecule very close reducing their volume in unit mass. (iii) Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain. (iv) Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e..", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8081076920924115, "ocr_used": true, "chunk_length": 1743, "token_count": 481}}
{"text": "(iii) Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain. (iv) Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e.. 25 jgthungu@gmail.com104Hydrogen bondscovalent bondsR1C O δ-…….….…H δ+O δO δH δ+……..….O δCR2R1 and 2 are extensions of the molecule.For ethanoic acid the extension is made up of CH3 – to make the structure; jgthungu@gmail.com105For ethanoic acid the extension is made up of CH3 – to make the structure;Hydrogen bondscovalent bonds CH3CO δ-…………… H δ+O δO δH δ+…………O δCCH3Ethanoic acid has a higher melting/boiling point than ethanol .This is because ethanoic acid has two/more hydrogen bond than ethanol. II Chemical properties of alkanoic acids The following experiments shows the main chemical properties of ethanoic (alkanoic) acid. (a)Effect on litmus papers Experiment Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid. Sample observations Solution/acid Observations/effect on litmus papers Inference Ethanoic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion\n26 Succinic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Citric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Oxalic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Tartaric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Nitric(V)acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Explanation All acidic solutions contains H+/H3O+(aq) ions.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8747236551215919, "ocr_used": true, "chunk_length": 1725, "token_count": 484}}
{"text": "(a)Effect on litmus papers Experiment Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid. Sample observations Solution/acid Observations/effect on litmus papers Inference Ethanoic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion\n26 Succinic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Citric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Oxalic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Tartaric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Nitric(V)acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Explanation All acidic solutions contains H+/H3O+(aq) ions. The H+ /H3O+ (aq) ions is responsible for turning blue litmus paper/solution to red (b)pH Experiment Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid. Sample observations Solution/acid pH Inference Ethanoic acid 4/5/6 Weakly acidic Succinic acid 4/5/6 Weakly acidic Citric acid 4/5/6 Weakly acidic Oxalic acid 4/5/6 Weakly acidic Tartaric acid 4/5/6 Weakly acidic Sulphuric(VI)acid 1/2/3 Strongly acidic Explanations Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8567421277306364, "ocr_used": true, "chunk_length": 1635, "token_count": 459}}
{"text": "Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid. Sample observations Solution/acid pH Inference Ethanoic acid 4/5/6 Weakly acidic Succinic acid 4/5/6 Weakly acidic Citric acid 4/5/6 Weakly acidic Oxalic acid 4/5/6 Weakly acidic Tartaric acid 4/5/6 Weakly acidic Sulphuric(VI)acid 1/2/3 Strongly acidic Explanations Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water. All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO- Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H+ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e Examples 1. CH3COOH(aq) CH3COO-(aq) + H+(aq)\n27 (ethanoic acid) (ethanoate ion) (few H+ ion) 2. CH3 CH2COOH(aq) CH3 CH2COO-(aq) + H+(aq) (propanoic acid) (propanoate ion) (few H+ ion) 3. CH3 CH2 CH2COOH(aq) CH3 CH2 CH2COO-(aq) + H+(aq) (Butanoic acid) (butanoate ion) (few H+ ion) 4. HOOH(aq) HOO-(aq) + H+(aq) (methanoic acid) (methanoate ion) (few H+ ion) 5. H2 SO4 (aq) SO42- (aq) + 2H+(aq) (sulphuric(VI) acid) (sulphate(VI) ion) (many H+ ion) 6.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7932554389815136, "ocr_used": true, "chunk_length": 1412, "token_count": 495}}
{"text": "CH3 CH2 CH2COOH(aq) CH3 CH2 CH2COO-(aq) + H+(aq) (Butanoic acid) (butanoate ion) (few H+ ion) 4. HOOH(aq) HOO-(aq) + H+(aq) (methanoic acid) (methanoate ion) (few H+ ion) 5. H2 SO4 (aq) SO42- (aq) + 2H+(aq) (sulphuric(VI) acid) (sulphate(VI) ion) (many H+ ion) 6. HNO3 (aq) NO3- (aq) + H+(aq) (nitric(V) acid) (nitrate(V) ion) (many H+ ion) (c)Reaction with metals Experiment Place about 4cm3 of ethanoic acid in a test tube. Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.781712772629085, "ocr_used": true, "chunk_length": 635, "token_count": 248}}
{"text": "Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Inference Ethanoic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Succinic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Citric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Oxalic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Tartaric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion\n28 Nitric(V)acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Explanation Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group -COOH is /are displaced and not in the alkyl hydrocarbon chain. Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e. Examples 1.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8850063999109579, "ocr_used": true, "chunk_length": 1510, "token_count": 424}}
{"text": "Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e. Examples 1. For a monovalent metal with monobasic acid 2R – COOH + 2M -> 2R- COOM + 2H2(g) 2.For a divalent metal with monobasic acid 2R – COOH + M -> (R- COO) 2M + H2(g) 3.For a divalent metal with dibasic acid HOOC-R-COOH+ M -> MOOC-R-COOM + H2(g) 4.For a monovalent metal with dibasic acid HOOC-R-COOH+ 2M -> MOOC-R-COOM + H2(g) 5 For mineral acids (i)Sulphuric(VI)acid is a dibasic acid H2 SO4 (aq) + 2M -> M2 SO4 (aq) + H2(g) H2 SO4 (aq) + M -> MSO4 (aq) + H2(g) (ii)Nitric(V) and hydrochloric acid are monobasic acid HNO3 (aq) + 2M -> 2MNO3 (aq) + H2(g) HNO3 (aq) + M -> M(NO3 ) 2 (aq) + H2(g) Examples 1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas. Caution: This reaction is explosive. CH3COOH (aq) + Na(s) -> CH3COONa (aq) + H2(g) (Ethanoic acid) (Sodium ethanoate) 2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas. 2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)\n29 (Ethanoic acid) (Calcium ethanoate) 3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7409386088237264, "ocr_used": true, "chunk_length": 1149, "token_count": 459}}
{"text": "hydrogen gas. 2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)\n29 (Ethanoic acid) (Calcium ethanoate) 3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas. HOOC-COOH+ 2Na -> NaOOC - COONa + H2(g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate) Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate. 4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas. HOOC-R-COOH+ Mg -> ( OOC - COO) Mg + H2(g) (ethan-1,2-dioic acid) (magnesium ethan-1,2-dioate) 5.Magnesium reacts with (i)Sulphuric(VI)acid to form Magnesium sulphate(VI) H2 SO4 (aq) + Mg -> MgSO4 (aq) + H2(g) (ii)Nitric(V) and hydrochloric acid are monobasic acid 2HNO3 (aq) + Mg -> M(NO3 ) 2 (aq) + H2(g) (d)Reaction with hydrogen carbonates and carbonates Experiment Place about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7763242610477027, "ocr_used": true, "chunk_length": 1139, "token_count": 420}}
{"text": "Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Inference Ethanoic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Succinic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Citric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Oxalic acid (i)effervescence, fizzing, bubbles H3O+/H+(aq)ion\n30 (ii)colourless gas produced that forms a white precipitate with lime water Tartaric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Nitric(V)acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water. Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint. Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water. Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide Examples 1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8895431612412745, "ocr_used": true, "chunk_length": 1749, "token_count": 468}}
{"text": "Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water. Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide Examples 1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + H2O(l) + CO2 (g) (Ethanoic acid) (Sodium ethanoate) 2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. 2CH3COOH (aq) + Na2CO3 (s) -> 2CH3COONa (aq) + H2O(l) + CO2 (g) (Ethanoic acid) (Sodium ethanoate) 3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. HOOC-COOH+ Na2CO3 (s) -> NaOOC - COONa + H2O(l) + CO2 (g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate) 4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. HOOC-COOH+ 2NaHCO3 (s) -> NaOOC - COONa + H2O(l) + 2CO2 (g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)\n31 (e)Esterification Experiment Place 4cm3 of ethanol acid in a boiling tube. Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame. Pour the mixture into a beaker containing 50cm3 of water. Smell the products.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8078353079010604, "ocr_used": true, "chunk_length": 1436, "token_count": 495}}
{"text": "Warm/heat gently on Bunsen flame. Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Ethanoic acid Sweet fruity smell Succinic acid Sweet fruity smell Citric acid Sweet fruity smell Oxalic acid Sweet fruity smell Tartaric acid Sweet fruity smell Dilute sulphuric(VI)acid No sweet fruity smell Explanation Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids. Alkanol + Alkanoic acids -> Ester + water Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g. Ethanol + Ethanoic acid -> Ethylethanoate + Water Ethanol + Propanoic acid -> Ethylpropanoate + Water Ethanol + Methanoic acid -> Ethylmethanoate + Water Ethanol + butanoic acid -> Ethylbutanoate + Water Propanol + Ethanoic acid -> Propylethanoate + Water Methanol + Ethanoic acid -> Methyethanoate + Water Methanol + Decanoic acid -> Methyldecanoate + Water Decanol + Methanoic acid -> Decylmethanoate + Water\n32 During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol. R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O Examples 1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.885859885128315, "ocr_used": true, "chunk_length": 1777, "token_count": 496}}
{"text": "Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3COOH(l) --Conc. H2SO4 --> CH3COOCH2CH3(aq) +H2O(l) 2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water. Ethanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2CH3(aq) +H2O(l) 3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water. Methanol + Ethanoic acid --Conc. H2SO4 -->Methylethanoate + Water CH3OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO CH3(aq) +H2O(l) 4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water. Methanol + propanoic acid --Conc. H2SO4 -->Methylpropanoate + Water CH3OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. Propanol + Propanoic acid --Conc.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.730610738538108, "ocr_used": true, "chunk_length": 1129, "token_count": 477}}
{"text": "H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. Propanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C3H7OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C3H7(aq) +H2O(l) CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)\n33 C. DETERGENTS Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to: (i)dissolve substances which water can not e.g grease ,oil, fat (ii)be washed away after cleaning. There are two types of detergents: (a)Soapy detergents (b)Soapless detergents (a) SOAPY DETERGENTS Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e. Sodium hydroxide + octadecanoic acid -> Sodium octadecanoate + water NaOH(aq) + CH3 (CH2) 16 COOH(aq) -> CH3 (CH2) 16 COO – Na+ (aq) +H2 O(l) Commonly ,soap can thus be represented ; R- COO – Na+ where; R is a long chain alkyl group and -COO – Na+ is the alkanoate ion. In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8232105053553838, "ocr_used": true, "chunk_length": 1539, "token_count": 499}}
{"text": "In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced. Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol Fats/Oils are esters with fatty acids and glycerol parts in their structure; C17H35COOCH2 C17H35COOCH C17H35COOCH2\n34 When boiled with concentrated sodium hydroxide solution NaOH; (i)NaOH ionizes/dissociates into Na+ and OH- ions (ii)fat/oil split into three C17H35COO- and one CH2 CH CH2 (iii) the three Na+ combine with the three C17H35COO- to form the salt C17H35COO- Na+ (iv)the three OH-ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol) C17H35COOCH2 CH2OH C17H35COOCH +NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol Generally: CnH2n+1COOCH2 CH2OH CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO- Na+ + CHOH CnH2n+1COOCH2 CH2OH Ester Alkali Soap glycerol R - COOCH2 CH2OH R - COOCH +NaOH -> 3R-COO- Na+ + CHOH R- COOCH2 CH2OH Ester Alkali Soap glycerol During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out. The soap is then added colouring agents ,perfumes and herbs of choice.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8167330278216012, "ocr_used": true, "chunk_length": 1479, "token_count": 500}}
{"text": "Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol Fats/Oils are esters with fatty acids and glycerol parts in their structure; C17H35COOCH2 C17H35COOCH C17H35COOCH2\n34 When boiled with concentrated sodium hydroxide solution NaOH; (i)NaOH ionizes/dissociates into Na+ and OH- ions (ii)fat/oil split into three C17H35COO- and one CH2 CH CH2 (iii) the three Na+ combine with the three C17H35COO- to form the salt C17H35COO- Na+ (iv)the three OH-ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol) C17H35COOCH2 CH2OH C17H35COOCH +NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol Generally: CnH2n+1COOCH2 CH2OH CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO- Na+ + CHOH CnH2n+1COOCH2 CH2OH Ester Alkali Soap glycerol R - COOCH2 CH2OH R - COOCH +NaOH -> 3R-COO- Na+ + CHOH R- COOCH2 CH2OH Ester Alkali Soap glycerol During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out. The soap is then added colouring agents ,perfumes and herbs of choice. School laboratory preparation of soap Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8005621908769046, "ocr_used": true, "chunk_length": 1417, "token_count": 498}}
{"text": "The soap is then added colouring agents ,perfumes and herbs of choice. School laboratory preparation of soap Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of\n35 /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve. The action of soap Soapy detergents: (i)act by reducing the surface tension of water by forming a thin layer on top of the water. (ii)is made of a non-polar alkyl /hydrocarbon tail and a polar -COO-Na+ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar -COO-Na+ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water. Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is removed from the garment by rinsing with fresh water.The repulsion of the soap head prevent /ensure the droplets do not mix.Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules. Advantages and disadvantages of using soapy detergents Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution. Soapy detergents have the diadvatage in that: (i)they are made from fat and oils which are better eaten as food than make soap.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9014974460544147, "ocr_used": true, "chunk_length": 2120, "token_count": 510}}
{"text": "Advantages and disadvantages of using soapy detergents Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution. Soapy detergents have the diadvatage in that: (i)they are made from fat and oils which are better eaten as food than make soap. (ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate formed when soap reacts with Ca2+ and Mg2+ present in hard water. Chemical equation 2C17H35COO- Na+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2Na+(aq) (insoluble Calcium octadecanote/scum) 2C17H35COO- Na+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2Na+(aq) (insoluble Magnesium octadecanote/scum) This causes wastage of soap. Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive. (b)SOAPLESS DETERGENTS\n36 Soapless detergent usually called detergent is a long chain salt fromed from byproducts of fractional distillation of crude oil.Commonly used soaps include: (i)washing agents (ii)toothpaste (iii)emulsifiers/wetting agents/shampoo Soapless detergents are derived from reacting: (i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8683746536297164, "ocr_used": true, "chunk_length": 1645, "token_count": 488}}
{"text": "Potassium is more expensive than sodium and thus its soap is also more expensive. (b)SOAPLESS DETERGENTS\n36 Soapless detergent usually called detergent is a long chain salt fromed from byproducts of fractional distillation of crude oil.Commonly used soaps include: (i)washing agents (ii)toothpaste (iii)emulsifiers/wetting agents/shampoo Soapless detergents are derived from reacting: (i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent. alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water sulphate(VI) hydroxide alkyl hydrogen sulphate(VI) R –O-SO3H + NaOH -> R –O-SO3- Na+ + H2O Example Step I : Reaction of Octadecanol with Conc.H2SO4 C17H35CH2OH (aq) + H2SO4 -> C17H35CH2-O- SO3- H+ (aq) + H2O (l) octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water Step II: Neutralization by an alkali C17H35CH2-O- SO3- H+ (aq) + NaOH -> C17H35CH2-O- SO3- Na+ (aq) + H2O (l) Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water sulphate(VI) hydroxide hydrogen sulphate(VI) School laboratory preparation of soapless detergent Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8484350028810299, "ocr_used": true, "chunk_length": 1559, "token_count": 488}}
{"text": "Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent. alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water sulphate(VI) hydroxide alkyl hydrogen sulphate(VI) R –O-SO3H + NaOH -> R –O-SO3- Na+ + H2O Example Step I : Reaction of Octadecanol with Conc.H2SO4 C17H35CH2OH (aq) + H2SO4 -> C17H35CH2-O- SO3- H+ (aq) + H2O (l) octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water Step II: Neutralization by an alkali C17H35CH2-O- SO3- H+ (aq) + NaOH -> C17H35CH2-O- SO3- Na+ (aq) + H2O (l) Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water sulphate(VI) hydroxide hydrogen sulphate(VI) School laboratory preparation of soapless detergent Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water. Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent. The action of soapless detergents\n37 The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8412001791312136, "ocr_used": true, "chunk_length": 1508, "token_count": 476}}
{"text": "Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent. The action of soapless detergents\n37 The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e. vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COO-Na+ vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO3- Na+ (long hydrophobic /non-polar alkyl tail) (hydrophilic/polar/ionic head) The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water. The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching. The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment. The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away. Advantages and disadvantages of using soapless detergents Soapless detergents are non-biodegradable unlike soapy detergents. They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death. Soapless detergents have the advantage in that they: (i)do not form scum with hard water. (ii)are cheap to manufacture/buying (iii)are made from petroleum products but soapis made from fats/oil for human consumption. Sample revision questions 1. Study the scheme below Fat/oil KOH Boiling Sodium Chloride Filtration Filtrate Y Residue X\n38 (a)Identify the process Saponification (b)Fats and oils are esters.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9121821790623668, "ocr_used": true, "chunk_length": 1793, "token_count": 468}}
{"text": "(ii)are cheap to manufacture/buying (iii)are made from petroleum products but soapis made from fats/oil for human consumption. Sample revision questions 1. Study the scheme below Fat/oil KOH Boiling Sodium Chloride Filtration Filtrate Y Residue X\n38 (a)Identify the process Saponification (b)Fats and oils are esters. Write the formula of the a common structure of ester C17H35COOCH2 C17H35COOCH C17H35COOCH2 (c)Write a balanced equation for the reaction taking place during boiling C17H35COOCH2 CH2OH C17H35COOCH +3NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol (d)Give the IUPAC name of: (i)Residue X Potassium octadecanoate (ii)Filtrate Y Propan-1,2,3-triol (e)Give one use of fitrate Y Making paint (f)What is the function of sodium chloride To reduce the solubility of the soap hence helping in precipitating it out (g)Explain how residue X helps in washing. Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat Has a polar /ionic hydrophilic head that dissolves in water. 39 From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment. (h)State one: (i)advantage of continued use of residue X on the environment Is biodegradable and thus do not pollute the environment (ii)disadvantage of using residue X Uses fat/oil during preparation/manufacture which are better used for human consumption. (i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below. Water sample A B C Drops of residue X 15 2 15 Drops of residue X in boiled water 2 2 15 (i)State and explain which sample of water is: I. Soft Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating. II. Permanent hard Sample C .", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8724771084171516, "ocr_used": true, "chunk_length": 1822, "token_count": 503}}
{"text": "Soft Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating. II. Permanent hard Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water. III. Temporary hard Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water. (ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n40 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8052749934629131, "ocr_used": true, "chunk_length": 1043, "token_count": 362}}
{"text": "Boiling remove temporary hardness of water. (ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n40 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7451266159531788, "ocr_used": true, "chunk_length": 1412, "token_count": 576}}
{"text": "(ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n40 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.766888657429326, "ocr_used": true, "chunk_length": 1611, "token_count": 623}}
{"text": "Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n40 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7624753848907985, "ocr_used": true, "chunk_length": 1568, "token_count": 615}}
{"text": "Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc. H2SO4 Ice cold water Brown solid A 6M sodium hydroxide Substance B\n41 (a)Identify : (i)brown solid A Alkyl hydrogen sulphate(VI) (ii)substance B Sodium alkyl hydrogen sulphate(VI) (b)Write a general formula of: (i)Substance A.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8058988547162529, "ocr_used": true, "chunk_length": 1227, "token_count": 432}}
{"text": "(v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc. H2SO4 Ice cold water Brown solid A 6M sodium hydroxide Substance B\n41 (a)Identify : (i)brown solid A Alkyl hydrogen sulphate(VI) (ii)substance B Sodium alkyl hydrogen sulphate(VI) (b)Write a general formula of: (i)Substance A. O R-O-S O3 H // R- O - S - O - H O (ii)Substance B O R-O-S O3 - Na+ R- O - S - O - Na+ O (c)State one (i) advantage of continued use of substance B -Does not form scum with hard water -Is cheap to make -Does not use food for human as a raw material. (ii)disadvantage of continued use of substance B. Is non-biodegradable therefore do not pollute the environment (d)Explain the action of B during washing. Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat. Has a polar/ionic hydrophilic head that dissolves in water Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment. (e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B. 42 Product A Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI) H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O (f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8599437405574167, "ocr_used": true, "chunk_length": 1686, "token_count": 480}}
{"text": "(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B. 42 Product A Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI) H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O (f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation. Product A Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O 3.Below is part of a detergent H3C – (CH2 )16 – O - SO3 - K + (a)Write the formular of the polar and non-polar end Polar end H3C – (CH2 )16 – Non-polar end – O - SO3 - K + (b)Is the molecule a soapy or saopless detergent? Soapless detergent (c)State one advantage of using the above detergent -does not form scum with hard water -is cheap to manufacture 4.The structure of a detergent is H H H H H H H H H H H H H\n43 H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO-Na+ H H H H H H H H H H H H H a) Write the molecular formula of the detergent. (1mk) CH3(CH2)12COO-Na+ b) What type of detergent is represented by the formula?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.787000309389436, "ocr_used": true, "chunk_length": 1399, "token_count": 489}}
{"text": "Product A Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O 3.Below is part of a detergent H3C – (CH2 )16 – O - SO3 - K + (a)Write the formular of the polar and non-polar end Polar end H3C – (CH2 )16 – Non-polar end – O - SO3 - K + (b)Is the molecule a soapy or saopless detergent? Soapless detergent (c)State one advantage of using the above detergent -does not form scum with hard water -is cheap to manufacture 4.The structure of a detergent is H H H H H H H H H H H H H\n43 H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO-Na+ H H H H H H H H H H H H H a) Write the molecular formula of the detergent. (1mk) CH3(CH2)12COO-Na+ b) What type of detergent is represented by the formula? (1mk) Soapy detergent c) When this type of detergent is used to wash linen in hard water, spots (marks) are left on the linen. Write the formula of the substance responsible for the spots (CH3(CH2)12COO-)2Ca2+ / CH3(CH2)12COO-)2Mg2+ D. POLYMERS AND FIBRES Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8069814366424537, "ocr_used": true, "chunk_length": 1416, "token_count": 459}}
{"text": "POLYMERS AND FIBRES Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization. Polymers and fibres are either: (a)Natural polymers and fibres (b)Synthetic polymers and fibres Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include: -proteins/polypeptides making amino acids in animals -cellulose that make cotton,wool,paper and silk -Starch that come from glucose -Fats and oils -Rubber from latex in rubber trees. Synthetic polymers and fibres are man-made. They include: -polyethene\n44 -polychloroethene -polyphenylethene(polystyrene) -Terylene(Dacron) -Nylon-6,6 -Perspex(artificial glass) Synthetic polymers and fibres have the following characteristic advantages over natural polymers 1. They are light and portable 2. They are easy to manufacture. 3. They can easily be molded into shape of choice. 4. They are resistant to corrosion, water, air , acids, bases and salts. 5. They are comparatively cheap, affordable, colourful and aesthetic Synthetic polymers and fibres however have the following disadvantages over natural polymers 1. They are non-biodegradable and hence cause environmental pollution during disposal 2. They give out highly poisonous gases when burnt like chlorine/carbon(II)oxide 3. Some on burning produce Carbon(IV)oxide. Carbon(IV)oxide is a green house gas that cause global warming. 4. Compared to some metals, they are poor conductors of heat,electricity and have lower tensile strength. 5. To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used: 1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment. 2.Production of biodegradable synthetic polymers and fibres that rot away.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8990613151055843, "ocr_used": true, "chunk_length": 2051, "token_count": 477}}
{"text": "To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used: 1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment. 2.Production of biodegradable synthetic polymers and fibres that rot away. There are two types of polymerization: (a)addition polymerization (b)condensation polymerization (a)addition polymerization Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. 45 Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n46 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8928912257944965, "ocr_used": true, "chunk_length": 2008, "token_count": 498}}
{"text": "The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n46 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8646422955974843, "ocr_used": true, "chunk_length": 1272, "token_count": 367}}
{"text": "Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n46 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8597396423501218, "ocr_used": true, "chunk_length": 1796, "token_count": 512}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n46 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8532926829268292, "ocr_used": true, "chunk_length": 1640, "token_count": 480}}
{"text": "Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n47 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.867839923999525, "ocr_used": true, "chunk_length": 1203, "token_count": 328}}
{"text": "The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n47 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8663864267508654, "ocr_used": true, "chunk_length": 1626, "token_count": 458}}
{"text": "It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n47 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n48 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8471976972881405, "ocr_used": true, "chunk_length": 1736, "token_count": 528}}
{"text": "Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n47 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n48 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8469853960708048, "ocr_used": true, "chunk_length": 1730, "token_count": 527}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n48 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8474452969802136, "ocr_used": true, "chunk_length": 1709, "token_count": 520}}
{"text": "Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n48 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n49 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8267433831990795, "ocr_used": true, "chunk_length": 1738, "token_count": 557}}
{"text": "It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n49 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n50 104 The commercial name of polyphenylethene is polystyrene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8230240746027215, "ocr_used": true, "chunk_length": 1695, "token_count": 547}}
{"text": "PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n49 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n50 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8224198077086702, "ocr_used": true, "chunk_length": 1679, "token_count": 545}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n49 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n50 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8140977819998789, "ocr_used": true, "chunk_length": 1523, "token_count": 503}}
{"text": "Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n50 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8350857606223888, "ocr_used": true, "chunk_length": 1007, "token_count": 300}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:\n51 H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8275350627351769, "ocr_used": true, "chunk_length": 1703, "token_count": 506}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:\n51 H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8280262292617854, "ocr_used": true, "chunk_length": 1695, "token_count": 512}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n52 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8809150488002537, "ocr_used": true, "chunk_length": 1124, "token_count": 332}}
{"text": "Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n52 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n53 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E).", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8540134936272288, "ocr_used": true, "chunk_length": 1774, "token_count": 551}}
{"text": "(ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n52 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n53 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8512520824559378, "ocr_used": true, "chunk_length": 1709, "token_count": 534}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n52 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n53 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8443717722546914, "ocr_used": true, "chunk_length": 1618, "token_count": 509}}
{"text": "polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n53 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees. During harvesting an incision is made on the rubber tree to produce a milky white substance called latex. Latex is a mixture of rubber and lots of water. The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-\n54 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8362184761012252, "ocr_used": true, "chunk_length": 1689, "token_count": 506}}
{"text": "The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-\n54 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher. During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer. H CH3 H H H CH3 H H - C - C - C - C - C - C - C - C - H S H H S H H CH3 S H H CH3 S H - C - C - C - C - C - C - C - C - H H H H H H Vulcanized rubber is used to make tyres, shoes and valves. 6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers\n55 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8341451363870304, "ocr_used": true, "chunk_length": 1675, "token_count": 494}}
{"text": "6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers\n55 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (b)Condensation polymerization Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water). Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds During condensation polymerization: (i)the two monomers are brought together by high pressure to reduce distance between them. (ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8692218543906787, "ocr_used": true, "chunk_length": 1821, "token_count": 449}}
{"text": "(ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H H- O - C – (CH2 ) 4 – C – O - H + H –N – (CH2) 6 – N – H\n56 (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H H H- O - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + H 2O . Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8296139911903458, "ocr_used": true, "chunk_length": 1566, "token_count": 451}}
{"text": "Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H Cl - C – (CH2 ) 4 – C – Cl + H –N – (CH2) 6 – N – H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O H H Cl - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + HCl . Polymer bond linkage The two monomers each has six carbon chain hence the name “nylon-6,6” The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets. 2.Formation of Terylene\n57 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8424450063898632, "ocr_used": true, "chunk_length": 1599, "token_count": 456}}
{"text": "2.Formation of Terylene\n57 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H- O - C – C6H5 – C – O - H + H –O – CH2 CH2 – O – H (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H- O - C – C6H5 – C – O – (CH2) 6 – N – H + H 2O . Polymer bond linkage of terylene\n58 Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol. Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring. The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen During the formation of Terylene (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene\n59 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.840142590030859, "ocr_used": true, "chunk_length": 1676, "token_count": 497}}
{"text": "O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene\n59 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits. 60 Practice questions Organic chemistry 1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture (i) Name and write the formula of the main products Name…………………………………. Formula…………………………………….. (ii) Which homologous series does the product named in (i) above belong? 2. The structure of the monomer phenyl ethene is given below:- a) Give the structure of the polymer formed when four of the monomers are added together b) Give the name of the polymer formed in (a) above 3. Explain the environmental effects of burning plastics in air as a disposal method 4. Write chemical equation to represent the effect of heat on ammonium carbonate 5. Sodium octadecanoate has a chemical formula CH3(CH2)6 COO-Na+, which is used as soap. HC = CH2 O\n61 Explain why a lot of soap is needed when washing with hard water 6. A natural polymer is made up of the monomer: (a) Write the structural formula of the repeat unit of the polymer (b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer were obtained. Determine the number of the monomer molecules in this polymer. (C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH\n62 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain\n63 8.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8391644349223354, "ocr_used": true, "chunk_length": 1867, "token_count": 491}}
{"text": "(C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH\n62 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain\n63 8. Study the polymer below and use it to answer the questions that follow: (a) Give the name of the monomer and draw its structures (b) Identify the type of polymerization that takes place (c) State one advantage of synthetic polymers 9. Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture of ethanol and pentane 10. (a) What is absolute ethanol? GLUCOSE SOLUTION CRUDE ETHANOL 95% ETHANOL ABSOLUTE ETHANOL G H H H H H C C C C\n64 (b) State two conditions required for process G to take place efficiently 11. (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time\n65 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode\n66 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8186715455823197, "ocr_used": true, "chunk_length": 1756, "token_count": 481}}
{"text": "(a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time\n65 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode\n66 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V\n67 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U………………………………………………………..", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7891921395251824, "ocr_used": true, "chunk_length": 1671, "token_count": 490}}
{"text": "(ii) Write the equation of a reaction that occurs at the cathode\n66 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V\n67 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U……………………………………………………….. (II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K\n68 i) Name the following compounds:- I. Product T ………………………… II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7943584363957598, "ocr_used": true, "chunk_length": 1415, "token_count": 430}}
{"text": "(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K\n68 i) Name the following compounds:- I. Product T ………………………… II. K ……… ii) State one common physical property of substance G iii) State the type of reaction that occurred in step J\n69 iv) Give one use of substance K v) Write an equation for the combustion of compound P vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12) 14. Study the scheme given below and answer the questions that follow:- H2 (g) Ni High temp Polymer Q Polymerization Compound P CH3CH2CH3 CH3CH2CH2ONa + H2 Na(s) Propan-l-ol Step I Propylethanoate CH3CH2COOH Solution T + CO2 (g) Step III Na2CO3(aq) Conc. H2SO4 180oC Step II\n70 (a) (i) Name compound P …………………………………………………………………… (ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3 (b) State one use of polymer Q (c) Name one oxidising agent that can be used in step II ………………………………….. (d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8084887198033599, "ocr_used": true, "chunk_length": 1692, "token_count": 493}}
{"text": "(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne. Explain your answer (h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour. 71 Calculate the minimum volume of air needed for the complete combustion of ethene (Air contains 20% by volume of oxygen) 15. (a) Study the schematic diagram below and answer the questions that follow:- (i) Identify the following: Substance Q .............................................................................................................. Substance R............................................................................................................... Gas P.......................................................................................................................... CH3CH2COOCH2CH2CH3 CH3CHCH2 CH3CH2CH2ONa + Gas P CH3CH2CH2OH X V HCl Step 5 Step 1 R Na H+ Step 3 Q + H2O MnO4 Step 4Ni H2\n72 (ii) Name: Step 1................................................................................................. Step 4................................................................................................. (iii) Draw the structural formula of the major product of step 5 (iv) State the condition and reagent in step 3 16. Study the flow chart below and answer the questions that follow (a) (i) Name the following organic compounds: M……………………………………………………………..…….. L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P\n73 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6593966593482286, "ocr_used": true, "chunk_length": 2088, "token_count": 492}}
{"text": "L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P\n73 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17. a) Give the names of the following compounds: i) CH3CH2CH2CH2OH …………………………………………………………………… ii) CH3CH2COOH ………………………………………………………………… iii) CH3C – O- CH2CH3 …………………………………………………………………… 18. Study the scheme given below and answer the questions that follow;\n74 n i) Name the reagents used in: Step I: ……………………………………………………………………… Step II …………………………………………………………………… Step III ……………………………………………………………………… ii) Write an equation to show products formed for the complete combustion of CH = CH iii) Explain one disadvantage of continued use of items made form the compound formed in step III 19. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3% i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11) ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH\n75 iii) Explain one disadvantage of continued use of items made form the compound formed in step III\n76 21.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7406074937238213, "ocr_used": true, "chunk_length": 1657, "token_count": 466}}
{"text": "Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH\n75 iii) Explain one disadvantage of continued use of items made form the compound formed in step III\n76 21. Give the IUPAC name for each of the following organic compounds; i) CH3 - CH - CH2 - CH3 OH ii)CH3 – CH – CH2 – CH2 - CH3 C2H5 iii)CH3COOCH2CH2CH3 22. The structure below represents a cleansing agent. O R – S – O-Na+ O a) State the type of cleansing agent represented above b) State one advantage and one disadvantage of using the above cleansing agent. 23. The structure below shows part of polymer .Use it to answer the questions that follow. CH3 CH3 CH3   \n77 ― CH - CH2 – CH- CH2 - CH – CH2 ― a) Derive the structure of the monomer b) Name the type of polymerization represented above 24. The flow chart below represents a series of reactions starting with ethanoic acid:- (a) Identify substances A and B (b) Name the process I 25. a) Write an equation showing how ammonium nitrate may be prepared starting with ammonia gas (b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of ammonia (H=1, N=14, O=16) 26. (a) What is meant by the term, esterification? Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O\n78 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid\n79 28. The scheme below shows some reactions starting with ethanol.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8360929065234275, "ocr_used": true, "chunk_length": 1731, "token_count": 463}}
{"text": "Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O\n78 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid\n79 28. The scheme below shows some reactions starting with ethanol. Study it and answer the questions that follow:- (i) Name and draw the structure of substance Q Q P CH3COONa Ethanol CH3CH2ONa C H2SO4(l) Cr2O7(aq) / H+(aq) Na(s) Step 2 Step 4 CH3CH2OH/H2SO4 Step 3 2-\n80 (ii) Give the names of the reactions that take place in steps 2 and 4 (iii) What reagent is necessary for reaction that takes place in step 3 29. Substances A and B are represented by the formulae ROH and RCOOH respectively. They belong to two different homologous series of organic compounds. If both A and B react with potassium metal: (a) Name the common product produced by both (b) State the observation made when each of the samples A and B are reacted with sodium hydrogen carbonate (i) A (ii) B 30. Below are structures of particles. Use it to answer questions that follow. In each case only electrons in the outermost energy level are shown key P = Proton N = Neutron X = Electron W U V 19P Z Y\n81 (a) Identify the particle which is an anion 31. Plastics and rubber are extensively used to cover electrical wires. (a) What term is used to describe plastic and rubbers used in this way? (b) Explain why plastics and rubbers are used this way\n82 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8565245200523146, "ocr_used": true, "chunk_length": 1854, "token_count": 484}}
{"text": "(b) Explain why plastics and rubbers are used this way\n82 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams. The halflife period of the isotope is 20days (a) Find the initial mass of the isotope (b) Give one application of radioactivity in agriculture 34. The structure below represents a polymer. Study and answer the questions that follow:- R Conc. R Cleaning agent X H H -C – C - n\n83 (i) Name the polymer above.................................................................................. (ii) Determine the value of n if giant molecule had relative molecular mass of 4956 35. RCOO-Na+ and RCH2OSO3-Na+ are two types of cleansing agents; i) Name the class of cleansing agents to which each belongs ii) Which one of these agents in (i) above would be more suitable when washing with water from the Indian ocean. Explain iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference in their bleaching properties 36. The formula given below represents a portion of a polymer H H H H C C C C O H O H n\n84 (a) Give the name of the polymer (b) Draw the structure of the monomer used to manufacture the polymer\n24.0.0 RADIOACTIVITY (10 LESSONS) Contents A INTRODUCTION/CAUSES OF RADIOCTIVITY Alpha (α) particle Beta (β) particle Gamma(y) particle B .NUCLEAR FISSION AND NUCLEAR FUSSION C. HALF-LIFE PERIOD AND DECAY CURVES D .CHEMICAL vs NUCLEAR REACTIONS E .APPLICATION OF RADIOACTIVITY AND RADIO ISOTOPES. F. DANGERS OF RADIOACTIVITY AND RADIO ISOTOPES. G. COMPREHENSIVE REVISION QUESTIONS A: INTRODUCTION / CAUSES OF RADIOCTIVITY Radioactivity is the spontaneous disintegration/decay of an unstable nuclide. 2 A nuclide is an atom with defined mass number (number of protons and neutrons), atomic number and definite energy.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8573057802025281, "ocr_used": true, "chunk_length": 1991, "token_count": 492}}
{"text": "G. COMPREHENSIVE REVISION QUESTIONS A: INTRODUCTION / CAUSES OF RADIOCTIVITY Radioactivity is the spontaneous disintegration/decay of an unstable nuclide. 2 A nuclide is an atom with defined mass number (number of protons and neutrons), atomic number and definite energy. Radioactivity takes place in the nucleus of an atom unlike chemical reactions that take place in the energy levels involving electrons. A nuclide is said to be stable if its neutron: proton ratio is equal to one (n/p = 1) All nuclide therefore try to attain n/p = 1 by undergoing radioactivity. Examples (i)Oxygen nuclide with 168 O has 8 neutrons and 8 protons in the nucleus therefore an n/p = 1 thus stable and do not decay/disintegrate. (ii)Chlorine nuclide with 3517 Cl has 18 neutrons and 17 protons in the nucleus therefore an n/p = 1.0588 thus unstable and decays/disintegrates to try to attain n/p = 1. (ii)Uranium nuclide with 23792 U has 206 neutrons and 92 protons in the nucleus therefore an n/p = 2.2391 thus more unstable than 23592 U and thus more readily decays / disintegrates to try to attain n/p = 1. (iii) Chlorine nuclide with 3717 Cl has 20 neutrons and 17 protons in the nucleus therefore an n/p = 1.1765 thus more unstable than 3517 Cl and thus more readily decays / disintegrates to try to attain n/p = 1. (iv)Uranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an n/p = 1.5543 thus more stable than 237 92U but also readily decays / disintegrates to try to attain n/p = 1. All unstable nuclides naturally try to attain nuclear stability with the production of: (i)alpha(α) particle decay The alpha (α) particle has the following main characteristic: i)is positively charged(like protons) ii) has mass number 4 and atomic number 2 therefore equal to a charged Helium atom ( 42He2+) iii) have very low penetrating power and thus can be stopped /blocked/shielded by a thin sheet of paper.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8438238361037846, "ocr_used": true, "chunk_length": 1918, "token_count": 513}}
{"text": "(iii) Chlorine nuclide with 3717 Cl has 20 neutrons and 17 protons in the nucleus therefore an n/p = 1.1765 thus more unstable than 3517 Cl and thus more readily decays / disintegrates to try to attain n/p = 1. (iv)Uranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an n/p = 1.5543 thus more stable than 237 92U but also readily decays / disintegrates to try to attain n/p = 1. All unstable nuclides naturally try to attain nuclear stability with the production of: (i)alpha(α) particle decay The alpha (α) particle has the following main characteristic: i)is positively charged(like protons) ii) has mass number 4 and atomic number 2 therefore equal to a charged Helium atom ( 42He2+) iii) have very low penetrating power and thus can be stopped /blocked/shielded by a thin sheet of paper. iv) have high ionizing power thus cause a lot of damage to living cells.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8404107894085623, "ocr_used": true, "chunk_length": 898, "token_count": 241}}
{"text": "(iv)Uranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an n/p = 1.5543 thus more stable than 237 92U but also readily decays / disintegrates to try to attain n/p = 1. All unstable nuclides naturally try to attain nuclear stability with the production of: (i)alpha(α) particle decay The alpha (α) particle has the following main characteristic: i)is positively charged(like protons) ii) has mass number 4 and atomic number 2 therefore equal to a charged Helium atom ( 42He2+) iii) have very low penetrating power and thus can be stopped /blocked/shielded by a thin sheet of paper. iv) have high ionizing power thus cause a lot of damage to living cells. v) a nuclide undergoing α-decay has its mass number reduced by 4 and its atomic number reduced by 2 Examples of alpha decay 210 84 Pb -> x 82 Pb + 42He 2+ 210 84 Pb -> 206 82 Pb + 42He 2+\n3 226 88 Ra -> 222 y Rn + 42He 2+ 226 88 Ra -> 222 86 Rn + 42He 2+ x y U -> 23490 Th + 42He 2+ 238 92 U -> 23490 Th + 42He 2+ x y U -> 23088 Ra + 2 42He 2+ 238 92 U -> 23088 Ra + 2 42He 2+ 210 84 U -> xy W + 10 α 210 84 U -> 17064 W + 10 α 210 92U -> xy W + 6 α 210 92U -> 18680W + 6 α (ii)Beta (β) particle decay The Beta (β) particle has the following main characteristic: i)is negatively charged(like electrons) ii)has no mass number and atomic number negative one(-1) therefore equal to a fast moving electron (0 -1e) iii) have medium penetrating power and thus can be stopped /blocked/shielded by a thin sheet of aluminium foil. iv) have medium ionizing power thus cause less damage to living cells than the α particle.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7512156148867314, "ocr_used": true, "chunk_length": 1600, "token_count": 500}}
{"text": "iv) have high ionizing power thus cause a lot of damage to living cells. v) a nuclide undergoing α-decay has its mass number reduced by 4 and its atomic number reduced by 2 Examples of alpha decay 210 84 Pb -> x 82 Pb + 42He 2+ 210 84 Pb -> 206 82 Pb + 42He 2+\n3 226 88 Ra -> 222 y Rn + 42He 2+ 226 88 Ra -> 222 86 Rn + 42He 2+ x y U -> 23490 Th + 42He 2+ 238 92 U -> 23490 Th + 42He 2+ x y U -> 23088 Ra + 2 42He 2+ 238 92 U -> 23088 Ra + 2 42He 2+ 210 84 U -> xy W + 10 α 210 84 U -> 17064 W + 10 α 210 92U -> xy W + 6 α 210 92U -> 18680W + 6 α (ii)Beta (β) particle decay The Beta (β) particle has the following main characteristic: i)is negatively charged(like electrons) ii)has no mass number and atomic number negative one(-1) therefore equal to a fast moving electron (0 -1e) iii) have medium penetrating power and thus can be stopped /blocked/shielded by a thin sheet of aluminium foil. iv) have medium ionizing power thus cause less damage to living cells than the α particle. v) a nuclide undergoing β -decay has its mass number remain the same and its atomic number increase by 1 Examples of beta (β) decay 1.23 x Na -> 2312Mg + 0 -1e 23 11 Na -> 2312Mg + 0 -1e 2. 234 x Th -> y91 Pa + 0 -1e 234 90 Th -> y91 Pa + 0 -1e 3. 20770Y -> x y Pb + 30 -1e 20770Y -> 207 73Pb + 30 -1e 4. x y C -> 147N + 0 -1e\nC -> 147N + 0 -1e 5.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6546417179834527, "ocr_used": true, "chunk_length": 1333, "token_count": 503}}
{"text": "234 x Th -> y91 Pa + 0 -1e 234 90 Th -> y91 Pa + 0 -1e 3. 20770Y -> x y Pb + 30 -1e 20770Y -> 207 73Pb + 30 -1e 4. x y C -> 147N + 0 -1e\nC -> 147N + 0 -1e 5. 1 x n -> y1H + 0 -1e 1 0 n -> 11H + 0 -1e 6. 42He -> 411H + x 0 -1e 42He -> 411H + 2 0 -1e 7. 22888Ra -> 22890Th + x β 22888Ra -> 22892Th + 4 β 8. 23290Th -> 21282Pb + 2 β + x α 23290Th -> 21282Pb + 2 β + 5 α 9. 23892U -> 22688 Ra + x β + 92U -> 22688 Ra + 2 β + 3 α 10. 21884Po -> 20682Pb + x β + 84Po -> 20682Pb + 4β + 3 α (iii)Gamma (y) particle decay The gamma (y) particle has the following main characteristic: i)is neither negatively charged(like electrons/beta) nor positively charged(like protons/alpha) therefore neutral. ii)has no mass number and atomic number therefore equal to electromagnetic waves. iii) have very high penetrating power and thus can be stopped /blocked/shielded by a thick block of lead.. iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure.. v) a nuclide undergoing y -decay has its mass number and its atomic number remain the same. Examples of gamma (y) decay • 3717Cl -> 3717Cl + y • 146C -> 146C + y The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.665395372489194, "ocr_used": true, "chunk_length": 1242, "token_count": 459}}
{"text": "iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure.. v) a nuclide undergoing y -decay has its mass number and its atomic number remain the same. Examples of gamma (y) decay • 3717Cl -> 3717Cl + y • 146C -> 146C + y The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide. 5 radioactive nuclide sheet of paper aluminium foil thick block of lead (radiation source) (block α-rays) (block β-rays) block y-rays) α-rays β-rays y-rays The sketch diagram below illustrates the effect of electric /magnetic field on the three radiations from a radioactive nuclide Radioactive disintegration/decay naturally produces the stable 20682Pb nuclide /isotope of lead.Below is the 238 92 U natural decay series. Identify the particle emitted in each case\n6 Write the nuclear equation for the disintegration from : (i)238 92 U to 23490 T 238 92 U -> 23490 T + 4 2 He 2+ 238 92 U -> 23490 T + α (ii)238 92 U to 222 84 Rn 238 92 U -> 22284 Rn + 4 4 2 He 2+ 238 92 U -> 22284 Rn + 4α 230 90 Th undergoes alpha decay to 222 86 Rn. Find the number of α particles emitted. Write the nuclear equation for the disintegration. Working 230 90 Th -> 222 86 Rn + x 4 2 He Method 1 Using mass numbers 230 = 222 + 4 x => 4 x = 230 - 222 = 8 x = 8 / 4 = 2 α Using atomic numbers 90 = 86 + 2 x => 2 x = 90 - 86 = 4 x = 4 / 2 = 2 α Nuclear equation 230 90 Th -> 222 86 Rn + 2 4 2 He\n7 214 82 Pb undergoes beta decay to 214 84 Rn. Find the number of β particles emitted. Write the nuclear equation for the disintegration.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7352782366925897, "ocr_used": true, "chunk_length": 1577, "token_count": 485}}
{"text": "Working 238 92 U -> 206 82 Pb + x 0 -1 e + y 4 2 He Using Mass numbers only 238 = 206 + 4y => 4y = 238 - 206 = 32 y = 32 = 8 α 4 Using atomic numbers only and substituting the 8 α(above) 238 92 U -> 206 82 Pb + 8 4 2 He + x 0 -1 e 92 = 82 + 16 + - x => 92 – (82 + 16) = - x x = 6 β Nuclear equation 238 92 U -> 206 82 Pb + 6 0 -1 e + 8 4 2 He 298 92 U undergoes alpha and beta decay to 214 83 Bi. Find the number of α and β particles emitted. Write the nuclear equation for the disintegration. Working 298 92 U -> 210 83 Bi + x 4 2 He + y 0 -1 e Using Mass numbers only 298 = 214 + 4x => 4x = 298 - 214 = 84 y = 84 = 21 α 4 Using atomic numbers only and substituting the 21 α (above) 238 92 U -> 214 83Bi + 21 4 2 He + y 0 -1 e 92 = 83 + 42 + - y\n8 => 92 – (83 + 42) = - x x = 33 β Nuclear equation 298 92 U -> 210 83 Bi + 21 4 2 He + 33 0 -1 e B:NUCLEAR FISSION AND NUCLEAR FUSION Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods: a) nuclear fission b) nuclear fusion. a)Nuclear fission Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy. Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6666266437964552, "ocr_used": true, "chunk_length": 1325, "token_count": 461}}
{"text": "Working 298 92 U -> 210 83 Bi + x 4 2 He + y 0 -1 e Using Mass numbers only 298 = 214 + 4x => 4x = 298 - 214 = 84 y = 84 = 21 α 4 Using atomic numbers only and substituting the 21 α (above) 238 92 U -> 214 83Bi + 21 4 2 He + y 0 -1 e 92 = 83 + 42 + - y\n8 => 92 – (83 + 42) = - x x = 33 β Nuclear equation 298 92 U -> 210 83 Bi + 21 4 2 He + 33 0 -1 e B:NUCLEAR FISSION AND NUCLEAR FUSION Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods: a) nuclear fission b) nuclear fusion. a)Nuclear fission Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy. Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors. The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction Examples of nuclear equations showing nuclear fission 10 n + 235 b U -> 9038 Sr + c 54Xe + 310 n + a 10 n + 2713 Al -> 2813 Al + y + a 10 n + 28a Al -> b11 Na + 42 He a0 n + 147 N -> 14b C + 10 n + 11 H -> 21 H + a 10 n + 235 92 U -> 95 42 Mo + 139 57 La + 210 n + 7 a b) Nuclear fusion Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy. Very high temperatures and pressure is required to overcome the repulsion between the atoms. Nuclear fusion is the basic chemistry behind solar/sun radiation.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7508213585237962, "ocr_used": true, "chunk_length": 1660, "token_count": 490}}
{"text": "The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction Examples of nuclear equations showing nuclear fission 10 n + 235 b U -> 9038 Sr + c 54Xe + 310 n + a 10 n + 2713 Al -> 2813 Al + y + a 10 n + 28a Al -> b11 Na + 42 He a0 n + 147 N -> 14b C + 10 n + 11 H -> 21 H + a 10 n + 235 92 U -> 95 42 Mo + 139 57 La + 210 n + 7 a b) Nuclear fusion Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy. Very high temperatures and pressure is required to overcome the repulsion between the atoms. Nuclear fusion is the basic chemistry behind solar/sun radiation. 9 Two daughter atoms/nuclides of Hydrogen fuse/join to form Helium atom/nuclide on the surface of the sun releasing large quantity of energy in form of heat and light. 21H + 21H -> abHe + 11H + a -> 32He 21H + 21H -> a + 11 H 4 11H -> 42He + a 147H + a -> 178O + 11 H C: HALF LIFE PERIOD (t1/2) The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount. It is usually denoted t 1/2. The rate of radioactive nuclide disintegration/decay is constant for each nuclide. The table below shows the half-life period of some elements. Element/Nuclide Half-life period(t 1/2 ) 238 92 U 4.5 x 10 6 C 88 Ra 15 P 84 Po 0.0002 seconds The less the half life the more unstable the nuclide /element. The half-life period is determined by using a Geiger-Muller counter (GM tube) .A GM tube is connected to ratemeter that records the count-rates per unit time.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7917217391304349, "ocr_used": true, "chunk_length": 1750, "token_count": 500}}
{"text": "The table below shows the half-life period of some elements. Element/Nuclide Half-life period(t 1/2 ) 238 92 U 4.5 x 10 6 C 88 Ra 15 P 84 Po 0.0002 seconds The less the half life the more unstable the nuclide /element. The half-life period is determined by using a Geiger-Muller counter (GM tube) .A GM tube is connected to ratemeter that records the count-rates per unit time. 10 This is the rate of decay/ disintegration of the nuclide. If the count-rates per unit time fall by half, then the time taken for this fall is the half-life period. Examples a)A radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute. Calculate the half-life period of the substance. If t 1/2 = x then 240 --x-->120 –x-->60 –x--->30 From 240 to 30 =3x =6 hours =>x = t 1/2 = ( 6 / 3 ) = 2 hours b) The count rate of a nuclide fell from 200 counts per second to 12.5 counts per second in 120 minutes. Calculate the half-life period of the nuclide. If t 1/2 =x then 200 --x-->100 –x-->50 –x--->25 –x--->12.5 From 200 to 12.5 =4x =120 minutes =>x = t 1/2 = ( 120 / 4 ) = 30 minutes c) After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube. Calculate the half-life period of the nuclide. If t 1/2 = x then 240 --x-->120 –x-->60 –x--->30 –x--->15 From 240 to 15 =4x =6 hours =>x = t 1/2 = ( 6 / 4 )= 1.5 hours d) Calculate the mass of nitrogen-13 that remain from 2 grams after 6 halflifes if the half-life period of nitrogen-13 is 10 minutes.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7348944553898115, "ocr_used": true, "chunk_length": 1540, "token_count": 487}}
{"text": "e) What fraction of a gas remains after 1hour if its half-life period is 20 minutes? If t 1/2 = x then: then 60 /20 = 3x 1 --x--> 1/2 –2x--> 1/4 –3x---> 1/8\n11 After the 3rd half-life 1/8 of the gas remain f) 348 grams of a nuclide A was reduced to 43.5 grams after 270days.Determine the half-life period of the nuclide. If t 1/2 = x then: 348 --x-->174 –2x-->87 –3x--->43.5 From 348 to 43.5=3x =270days =>x = t 1/2 = ( 270 / 3 ) = 90 days g) How old is an Egyptian Pharaoh in a tomb with 2grams of 14C if the normal 14C in a present tomb is 16grams.The half-life period of 14C is 5600years. If t 1/2 = x = 5600 years then: 16 --x-->8 –2x-->4 –3x--->2 3x = ( 3 x 5600 ) = 16800years h) 100 grams of a radioactive isotope was reduced 12.5 grams after 81days.Determine the half-life period of the isotope. If t 1/2 = x then: 100 --x-->50 –2x-->25 –3x--->12.5 From 100 to 12.5=3x =81days =>x = t 1/2 = ( 81 / 3 ) = 27 days A graph of activity against time is called decay curve. A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original\n12 (i)From the graph show and determine the half-life period of the isotope. From the graph t 1/2 changes in activity from: ( 100 – 50 ) => ( 20 – 0 ) = 20 minutes ( 50 – 25 ) => ( 40 – 20 ) = 20 minutes Thus t ½ = 20 minutes (ii)Why does the graph tend to ‘O’?", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.662359030781926, "ocr_used": true, "chunk_length": 1379, "token_count": 502}}
{"text": "If t 1/2 = x then: 100 --x-->50 –2x-->25 –3x--->12.5 From 100 to 12.5=3x =81days =>x = t 1/2 = ( 81 / 3 ) = 27 days A graph of activity against time is called decay curve. A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original\n12 (i)From the graph show and determine the half-life period of the isotope. From the graph t 1/2 changes in activity from: ( 100 – 50 ) => ( 20 – 0 ) = 20 minutes ( 50 – 25 ) => ( 40 – 20 ) = 20 minutes Thus t ½ = 20 minutes (ii)Why does the graph tend to ‘O’? Smaller particle/s will disintegrate /decay to half its original. There can never be ‘O’/zero particles D: CHEMICAL vs NUCLEAR REACTIONS Nuclear and chemical reaction has the following similarities: (i)-both involve the subatomic particles; electrons, protons and neutrons in an atom (ii)-both involve the subatomic particles trying to make the atom more stable. (iii)-Some for of energy transfer/release/absorb from/to the environment take place. 13 Nuclear and chemical reaction has the following differences: (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom. Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy. Chemical reactions produce or absorb small quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter. Chemical reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter. (v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size) The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area) E: APPLICATION AND USES OF RADIOCTIVITY.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8456699452417409, "ocr_used": true, "chunk_length": 2034, "token_count": 498}}
{"text": "(iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter. Chemical reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter. (v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size) The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area) E: APPLICATION AND USES OF RADIOCTIVITY. The following are some of the fields that apply and use radioisotopes; a)Medicine: -Treatment of cancer to kill malignant tumors through radiotherapy. -Sterilizing hospital /surgical instruments /equipments by exposing them to gamma radiation. b) Agriculture: If a plant or animal is fed with radioisotope, the metabolic processes of the plant/animal is better understood by tracing the route of the radioisotope. c) Food preservation: X-rays are used to kill bacteria in tinned food to last for a long time. d) Chemistry: To study mechanisms of a chemical reaction, one reactant is replaced in its structure by a radioisotope e.g. During esterification the ‘O’ joining the ester was discovered comes from the alkanol and not alkanoic acid. During photosynthesis the ‘O’ released was discovered comes from water. e) Dating rocks/fossils: The quantity of 14C in living things (plants/animals) is constant. 14 When they die the fixed mass of 14C is trapped in the cells and continues to decay/disintegrate. The half-life period of 14C is 5600 years . Comparing the mass of 14C in living and dead cells, the age of the dead can be determined. F: DANGERS OF RADIOCTIVITY. All rays emitted by radioactive isotopes have ionizing effect of changing the genetic make up of living cells. Exposure to theses radiations causes chromosomal and /or genetic mutation in living cells. Living things should therefore not be exposed for a long time to radioactive substances. One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors. Those who work in these reactors must wear protective devises made of thick glass or lead sheet.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9087646972107706, "ocr_used": true, "chunk_length": 2198, "token_count": 483}}
{"text": "Living things should therefore not be exposed for a long time to radioactive substances. One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors. Those who work in these reactors must wear protective devises made of thick glass or lead sheet. Accidental leakages of radiations usually occur In 1986 the Nuclear reactor at Chernobyl in Russia had a major explosion that emitted poisonous nuclear material that caused immediate environmental disaster In 2011, an earthquake in Japan caused a nuclear reactor to leak and release poisonous radioactive waste into the Indian Ocean. The immediate and long term effects of exposure to these poisonous radioactive waste on human being is of major concern to all environmentalists. G: SAMPLE REVISION QUESTIONS The figure below shows the behaviour of emissions by a radioactive isotope x. Use it to answer the question follow\n15 (a) Explain why isotope X emits radiations. (1mk) -is unstable //has n/p ratio greater/less than one (b) Name the radiation labeled T (1mk) alpha particle (c) Arrange the radiations labeled P and T in the increasing order of ability to be deflected by an electric filed. (1mk) T -> P a) Calculate the mass and atomic numbers of element B formed after 21280 X has emitted three beta particles, one gamma ray and two alpha particles. Mass number = 212 – (0 beta+ o gamma + (2 x 4 ) alpha = 204 Atomic number = 80 – (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79 b)Write a balanced nuclear equations for the decay of 21280 X to B using the information in (a) above. 21280 X -> 20479B + 242He + 3 0-1 e + y Identify the type of radiation emitted from the following nuclear equations. (i) 146 C -> 147N + ……… β - Beta (ii) 11 H + 10 n -> 21H + …… y -gamma (iii) 23592 U -> 9542Mo + 13957La + 10 n +…… 7 β – seven beta particles\n16 (iv) 23892 U -> 23490Th + … … α-alpha (v) 146 C + 11 H -> 157N + …… y-gamma X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8195725769861675, "ocr_used": true, "chunk_length": 1999, "token_count": 513}}
{"text": "Mass number = 212 – (0 beta+ o gamma + (2 x 4 ) alpha = 204 Atomic number = 80 – (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79 b)Write a balanced nuclear equations for the decay of 21280 X to B using the information in (a) above. 21280 X -> 20479B + 242He + 3 0-1 e + y Identify the type of radiation emitted from the following nuclear equations. (i) 146 C -> 147N + ……… β - Beta (ii) 11 H + 10 n -> 21H + …… y -gamma (iii) 23592 U -> 9542Mo + 13957La + 10 n +…… 7 β – seven beta particles\n16 (iv) 23892 U -> 23490Th + … … α-alpha (v) 146 C + 11 H -> 157N + …… y-gamma X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams. If the half-life period isotope is 25 days, calculate the initial mass X of the radio isotope. Number of half-lifes = ( 100 / 25 ) = 4 20g -----> 40g ----> 80g-----> 160g -----> 320g Original mass X = 320g Radium has a half-life of 1620 years. (i)What is half-life? The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7060363569851887, "ocr_used": true, "chunk_length": 1151, "token_count": 384}}
{"text": "Number of half-lifes = ( 100 / 25 ) = 4 20g -----> 40g ----> 80g-----> 160g -----> 320g Original mass X = 320g Radium has a half-life of 1620 years. (i)What is half-life? The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time\n17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6648190406085144, "ocr_used": true, "chunk_length": 1430, "token_count": 510}}
{"text": "The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time\n17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2\n18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6653596520575443, "ocr_used": true, "chunk_length": 1470, "token_count": 527}}
{"text": "Determine the half life of X2\n18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow. 126 C --step 1-->127 N --step 2--> 1211Na (a)126 C and 146 C are isotopes. What does the term isotope mean? Atoms of the same element with different mass number /number of neutrons. (b)Write an equation for the nuclear reaction in step II 127 N -> 1211Na + 0 -1e (c)Give one use of 146 C Dating rocks/fossils: Study of metabolic pathways/mechanisms on plants/animals Study the graph of a radioactive decay series for isotope H below. (a) Name the type of radiation emitted when isotope (i) H changes to isotope J. Alpha-Mass number decrease by 4 from 214 to 210(y-axis) atomic number decrease by 2 from 83 to 81(x-axis) (ii) J changes to isotope K Beta-Mass number remains 210(y-axis) atomic number increase by 1 from 81 to 82(x-axis). 19 (b) Write an equation for the nuclear reaction that occur when isotope (i)J changes to isotope L 21081 J -> 21084L + 3 0 -1e (i)H changes to isotope M 21483 H -> 20682M + 3 0 -1e + 2 4 2He Identify a pair of isotope of an element in the decay series K and M Have same atomic number 82 but different mass number K-210 and M-206 a)A radioactive substance emits three different particles. Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7855895473924952, "ocr_used": true, "chunk_length": 1712, "token_count": 502}}
{"text": "Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical reactions produce or absorb smaller quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass /mass defect. Chemical reactions are not accompanied by a loss in mass. (v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperature/pressure/purity/particle size/ surface area b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows. 20 Identify the particles emitted in steps I and III (2mks) I - α-particle III - β-ray ii)Write the nuclear equation for the reaction which takes place in (a) step I 21483Bi -> 21081Bi + 4 2 He (b) step 1 to 3 21483Bi -> 21081Bi + 4 2 He + 2 0 -1 e (c) step 3 to 5 21082Pb -> 20682Pb + 4 2 He + 2 0 -1 e (c) step 1 to 5 21483Bi -> 20682Pb + 2 4 2 He + 3 0 -1 e The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times. Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8101903644419538, "ocr_used": true, "chunk_length": 1790, "token_count": 494}}
{"text": "Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks) d) Give one use of radioactive isotopes in medicine (1mk) 14.a)Distinguish between nuclear fission and nuclear fusion. (2mks) Describe how solid wastes containing radioactive substances should be disposed of. (1mk) b)(i)Find the values of Z1 and Z2 in the nuclear equation below Z1 1 94 140 1 U + n -> Sr + Xe + 2 n 92 0 38 Z2 0 iii)What type of nuclear reaction is represented in b (i) above? A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom, Write a balanced decay equation for the above change 1 mark If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if the half-life of the element is 30years 2 marks Define the term half-life. The diagram below shows the rays emitted by a radioactive sample\n22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements. (a)Which letter represent the : Explain.", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8280197969746681, "ocr_used": true, "chunk_length": 1733, "token_count": 467}}
{"text": "The diagram below shows the rays emitted by a radioactive sample\n22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow.\n\nThe letters are not the actual symbols of the elements.\n\n(a)Which letter represent the : Explain.\n\n(i)shortest lived nuclide L-has the shortest half life (ii)longest lived nuclide\n23 P-Is stable (iii) nuclide with highest n/p ratio L-has the shortest half life thus most unstable thus easily/quickly decay/disintegrate (iv) nuclide with lowest n/p ratio P-is stable thus do not decay/disintegrate (b)How long would it take for the following: (i)Nuclide A to change to B 10 years (half life of A) (ii) Nuclide D to change to H 27days +162000years+70000years+16days 232000 years and 43 days (iii) Nuclide A to change to P 27days +162000years+70000years+16days 232000 years and 43 days Study THE END", "metadata": {"source": "KCSE-FORM-4-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8531029255056193, "ocr_used": true, "chunk_length": 1203, "token_count": 319}}
{"text": "12.0.0 GAS LAWS (15 LESSONS) (a)Gas laws 1. Matter is made up of small particle in accordance to Kinetic Theory of matter: Naturally, there are basically three states of matter: Solid, Liquid and gas: (i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density. (ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space. (iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container. 2.Gases are affected by physical conditions. There are two physical conditions: (i)Temperature (ii)Pressure 3. The SI unit of temperature is Kelvin(K). Degrees Celsius/Centigrade(oC) are also used. The two units can be interconverted from the relationship: oC + 273= K K -273 = oC Practice examples 1. Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K\n2. Convert the following into degrees Celsius/Centigrade(oC).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8248348589475939, "ocr_used": true, "chunk_length": 1437, "token_count": 406}}
{"text": "Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K\n2. Convert the following into degrees Celsius/Centigrade(oC). (i) 10 K K -273 = oC substituting: 10 – 273 = -263 oC (ii) (i) 1 K K -273 = oC substituting: 1 – 273 = -272 oC (iii) 110 K K -273 = oC substituting: 110 – 273 = -163 oC (iv) -24 K K -273 = oC substituting: -24 – 273 = -297 oC The standard temperature is 273K = 0 oC. The room temperature is assumed to be 298K = 25oC 4. The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used. The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2). The standard pressure is the atmospheric pressure. Atmospheric pressure is equal to about: (i)101325 Pa (ii)101325 Nm-2 (iii)760 mmHg (iv)76 cmHg (v)one atmosphere. 5. Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure. Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7555968547409836, "ocr_used": true, "chunk_length": 1449, "token_count": 456}}
{"text": "Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6. Boyles law states that “the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ” Mathematically:\nVolume α 1 (Fixed /constant Temperature) Pressure V α 1 (Fixed /constant T) ie PV = Constant(k) P From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved. Graphically therefore a plot of volume(V) against pressure (P) produces a curve. V P Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line V 1/P For two gases then P1 V1 = P2 V2 P1 = Pressure of gas 1 V1 = Volume of gas 1 P2 = Pressure of gas 2 V2 = Volume of gas 2\nPractice examples: 1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled. Working P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2 V2 = 102300 x 25 = 12.5cm3 (102300 x 2) 2. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7792014437175728, "ocr_used": true, "chunk_length": 1550, "token_count": 442}}
{"text": "Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa? P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2 V2 = 101325 x 60 = 67.55 cm3 90000 The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination. 7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ” Mathematically: Volume α Pressure (Fixed /constant pressure) V α T (Fixed /constant P) ie V = Constant(k) T From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled. Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure. The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas. This temperature is called absolute zero. It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)\nFor two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7833720300873586, "ocr_used": true, "chunk_length": 1781, "token_count": 505}}
{"text": "It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)\nFor two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1. 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened. V1 = V2 substituting 500 = V2 T1 T2 (0 +273) (-4 +273) = 500 x (-4 x 273) = 492.674cm3 (0 + 273) The capacity of cylinder (500cm3) is less than new volume(492.674cm3). 7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder. 2. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens. V1 = V2 substituting 40000 = V2 T1 T2 (25 +273) (30 +273) = 40000 x (30 x 273) = 40671.1409cm3 (25 + 273) The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3). The tyre thus bursts. 3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7372387344199426, "ocr_used": true, "chunk_length": 1490, "token_count": 467}}
{"text": "3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened. V1 = V2 substituting 80 = V2 T1 T2 (25 +273) (-30 +273) = 80 x (-30 x 273) = 65.2349cm3\n(25 + 273) The capacity of balloon (80cm3) is more than new volume (65.2349cm3). The balloon thus remained intact. 8. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion. The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas. The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion. Examples 1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2. 2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2. 3. Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8282556659027247, "ocr_used": true, "chunk_length": 1848, "token_count": 511}}
{"text": "A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that: “the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure” Mathematically R α 1 and since density is proportional to mass then R α 1 √ p √ m For two gases then: R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas. √M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas. Since rate is inverse of time. i.e. the higher the rate the less the time: For two gases then: T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse. √M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas. Practice examples: 1. It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.835348162475822, "ocr_used": true, "chunk_length": 1128, "token_count": 319}}
{"text": "Molar mass CO2 = 44g Molar mass HCl = 36.5g T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0 T HCl √ molar mass HCl T HCl √ 36.5 T HCl = 200seconds x √ 36.5 = 182.1588 seconds √ 44.0 3. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse. Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7174145738987239, "ocr_used": true, "chunk_length": 694, "token_count": 249}}
{"text": "Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0) Molar mass CO2 = 44.0 Molar mass CO = 28.0 Method 1 25cm3 CO takes 25seconds 75cm3 takes 75 x25 = 75seconds 25\nT CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0 T CO √ molar mass CO 75 √ 28.0 T CO2 =75seconds x √ 44.0 = 94.0175 seconds √ 28.0 Method 2 25cm3 CO2 takes 25seconds 1cm3 takes 25 x1 = 1.0cm3sec-1 25 R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0 R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0 R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1 √ 44.0 0.7977cm3 takes 1 seconds 75cm3 takes 75cm3 = 94.0203seconds 0.7977cm3\n13.0.0 THE MOLE-FORMULAE AND CHEMICAL EQUATIONS (40 LESSONS) Introduction to the mole, molar masses and Relative atomic masses 1. The mole is the SI unit of the amount of substance. 2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles. 3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6617669700674726, "ocr_used": true, "chunk_length": 1299, "token_count": 513}}
{"text": "3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”. The Avogadros Constant contain 6.023 x10 23 particles. i.e. 1mole = 6.023 x10 23 particles = 6.023 x10 23 2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24 0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22 0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21 3. The mass of one mole of a substance is called molar mass. The molar mass of: (i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g. Molar mass of carbon(C)= relative atomic mass = 12.0g 6.023 x10 23 particles of carbon = 1 mole =12.0 g Molar mass of sodium(Na) = relative atomic mass = 23.0g 6.023 x10 23 particles of sodium = 1 mole =23.0 g Molar mass of Iron (Fe) = relative atomic mass = 56.0g 6.023 x10 23 particles of iron = 1 mole =56.0 g (ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule. The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7017628205128206, "ocr_used": true, "chunk_length": 1248, "token_count": 428}}
{"text": "Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g. Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g 6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g\n2 2 Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g 6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g 6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g (ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7104511264950942, "ocr_used": true, "chunk_length": 762, "token_count": 257}}
{"text": "It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound 3.Practically the empirical formula of a compound can be determined as in the following examples. To determine the empirical formula of copper oxide (a)Method 1:From copper to copper(II)oxide Procedure. Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3). Sample results Mass of crucible(M1) 15.6g Mass of crucible + copper before heating(M2) 18.4 Mass of crucible + copper after heating(M3) 19.1 Sample questions 1. Calculate the mass of copper powder used. Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g\n7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8121982731455222, "ocr_used": true, "chunk_length": 1548, "token_count": 431}}
{"text": "Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g\n7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3. Calculate the number of moles of: (i) copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from brown to black Explanation Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide 7. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above. Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesuin oxide and magnesium nitride. This causes experimental mass errors. (b)Method 2:From copper(II)oxide to copper\n8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.783025219850463, "ocr_used": true, "chunk_length": 1660, "token_count": 502}}
{"text": "(b)Method 2:From copper(II)oxide to copper\n8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below; jgthungu@gmail.com52HEATCopper(II)oxideHydrogen /Laboratory gas /Ammonia gas/Carbon(II)Oxidegas from a generatorExcess hydrogenburningDetermining empirical formula from copper(II)oxide to copper Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator. When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating. Continue passing the gases until the glass tube is cool. Turn off the gas generator. Carefully remove the porcelain boat form the combustion tube. Reweigh (M3). Sample results Mass of boat(M1) 15.6g Mass of boat before heating(M2) 19.1 Mass of boat after heating(M3) 18.4 Sample questions\n9 9 1. Calculate the mass of copper(II)oxide used. Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8204921343223303, "ocr_used": true, "chunk_length": 1482, "token_count": 433}}
{"text": "Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3. Calculate the number of moles of: (i) Copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from black to brown Explanation Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal. 7. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above. Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7669730077965791, "ocr_used": true, "chunk_length": 1252, "token_count": 396}}
{"text": "Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8. Write the equation for the reaction that would take place when the reducing agent is: (i) Hydrogen CuO(s) + H2(g) -> Cu(s) + H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) (ii)Carbon(II)oxide CuO(s) + CO (g) -> Cu(s) + CO2(g) (Black) (brown) (colourless gas, form white ppt with lime water ) (iii)Ammonia 3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) 9. Explain why the following is necessary during the above experiment; (i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide. Air combine with hydrogen in presence of heat causing an explosion (ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped. Hot metallic copper can be re-oxidized back to copper(II)oxide (iii) A stream of excess carbon (II)oxide gas should be ignited to burn Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas. 10. State two sources of error in this experiment. (i)All copper(II)oxide may not be reduced to copper. (ii)Some copper(II)oxide may be blown out the boat by the reducing agent. 4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8591936372348534, "ocr_used": true, "chunk_length": 1489, "token_count": 406}}
{"text": "4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)\n11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2\n12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6913690975132719, "ocr_used": true, "chunk_length": 1193, "token_count": 398}}
{"text": "(Cu = 63.5, 16.0)\n11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2\n12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O\n13 13 Hydrated salt has five/5 molecules of water of crystallizations 4.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6793482844160876, "ocr_used": true, "chunk_length": 2233, "token_count": 791}}
{"text": "Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2\n12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O\n13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6976883875618053, "ocr_used": true, "chunk_length": 2002, "token_count": 689}}
{"text": "What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2\n12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O\n13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7233355344290568, "ocr_used": true, "chunk_length": 1858, "token_count": 607}}
{"text": "What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O\n13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7548389257250017, "ocr_used": true, "chunk_length": 1701, "token_count": 524}}
{"text": "The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number. (ii) Relative empirical formula x n = Relative formular mass where n is a whole number. Practice sample examples 1. A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen. If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6\n14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7658126918971568, "ocr_used": true, "chunk_length": 1723, "token_count": 488}}
{"text": "If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6\n14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2. 4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7169653653870449, "ocr_used": true, "chunk_length": 1314, "token_count": 417}}
{"text": "4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 5.28 = 1.44g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 2.16 = 0.24g√ 18 Element Carbon Hydrogen\n15 15 Symbol C H Moles present = mass Molar mass 1.44g 12 0.24g√ 1 Divide by the smallest value 0.12 0.12 0.24 0.12 Mole ratios 1 2√ Empirical formula is CH2√ The molecular formular is thus determined : n = Relative formular mass = 84 = 6√ Relative empirical formula 14 The molecular formula is (CH2 ) x 6 = C6H12. √ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7520376569346842, "ocr_used": true, "chunk_length": 1142, "token_count": 382}}
{"text": "√ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0) Mass of N in A = 5.2% x 0.085 = 0.00442 g Mass of C in A = 12 x 0.224 = 0.0611g 44 Mass of H in A = 2 x 0.0372 = 0.0041g 18 Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O) => 0.0611g + 0.0041g = 0.0652g (Mass of C,H) 0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g Element Nitrogen Carbon Hydrogen Oxygen Symbol N C H O Moles present = mass Molar mass 0.00442 g 14 0.0611g 12 0.0041g 1 0.0154 g 16 Divide by the smallest value 0.00032 0.00509 0.0041g 0.00096\n16 16 0.00032 0.00032 0.00032 0.00032 Mole ratios 1 16 13 3 Empirical formula = C16H13NO3 (d)Molar gas volume The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is: (i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p). i.e.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6560504201680672, "ocr_used": true, "chunk_length": 1190, "token_count": 487}}
{"text": "22.4dm3 -> 6.0 x1023 2.24dm3 -> 2.24 x 6.0 x1023 22.4 =6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2\n18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6763300699300699, "ocr_used": true, "chunk_length": 1375, "token_count": 531}}
{"text": "= 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2\n18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7235301563192204, "ocr_used": true, "chunk_length": 1374, "token_count": 489}}
{"text": "0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2\n18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7414643040815373, "ocr_used": true, "chunk_length": 1465, "token_count": 501}}
{"text": "Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2\n18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2\n19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6941876829192354, "ocr_used": true, "chunk_length": 1515, "token_count": 576}}
{"text": "All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2\n19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6891265597147951, "ocr_used": true, "chunk_length": 816, "token_count": 309}}
{"text": "Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2\n19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy\n20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6928742619397593, "ocr_used": true, "chunk_length": 1958, "token_count": 738}}
{"text": "Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2\n19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy\n20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6893821050189468, "ocr_used": true, "chunk_length": 1930, "token_count": 731}}
{"text": "1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy\n20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7215171218032077, "ocr_used": true, "chunk_length": 1538, "token_count": 543}}
{"text": "If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy\n20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)\n21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7209363970792929, "ocr_used": true, "chunk_length": 1943, "token_count": 695}}
{"text": "Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)\n21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy. Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7362982789206686, "ocr_used": true, "chunk_length": 1397, "token_count": 499}}
{"text": "Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7467618040142984, "ocr_used": true, "chunk_length": 1303, "token_count": 442}}
{"text": "Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3. (a)What volume of Oxygen was used during the reaction.(1mk) Volume of Oxygen used =100-25 =75cm3√ (P was completely burnt) (b)Determine the molecular formula of the hydrocarbon(2mk) CxHy + O2 -> xCO2 + yH2O 15cm3 : 75cm3 15 15 1 : 3√ => 1 atom of C react with 6 (3x2)atoms of Oxygen Thus x = 1 and y = 2 => P has molecula formula CH4√ (g) Ionic equations An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products. Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions) An ionic equation is usually derived from a stoichiometric equation by using the following guidelines Guidelines for writing ionic equations 1.Write the balanced stoichiometric equation 2.Indicate the state symbols of the reactants and products 3.Split into cations and anions all the reactants and products that exist in aqueous state. 4.Cancel out any cation and anion that appear on both the product and reactant side. 5. Rewrite the chemical equation. It is an ionic equation.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8168697344028193, "ocr_used": true, "chunk_length": 1810, "token_count": 513}}
{"text": "5. Rewrite the chemical equation. It is an ionic equation. Practice (a)Precipitation of an insoluble salt\n23 23 All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt 1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed. Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.798357616795069, "ocr_used": true, "chunk_length": 1577, "token_count": 506}}
{"text": "Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)\n24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.743467994803678, "ocr_used": true, "chunk_length": 1682, "token_count": 628}}
{"text": "When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)\n24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)\n25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7273536440144648, "ocr_used": true, "chunk_length": 2851, "token_count": 1123}}
{"text": "Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)\n24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)\n25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7205789093946988, "ocr_used": true, "chunk_length": 2736, "token_count": 1088}}
{"text": "Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)\n24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)\n25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7309403487069341, "ocr_used": true, "chunk_length": 2281, "token_count": 901}}
{"text": "(i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)\n25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7439374865889422, "ocr_used": true, "chunk_length": 1804, "token_count": 701}}
{"text": "The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution. Examples 2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. 0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. “2M” is more concentrated than“0.02M”. Preparation of molar solution\n26 26 Procedure Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask. Using a wash bottle add about 200cm3 of distilled water. Stopper the flask. Shake vigorously for three minutes. Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved. Add more water slowly upto exactly the 250 cm3 mark. Sample questions 1.Calculate the number of moles of sodium hydroxide pellets present in: (i) 4.0 g. Molar mass of NaOH = (23 + 16 + 1) = 40g Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40 (ii) 250 cm3 solution in the volumetric flask.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8305604757085022, "ocr_used": true, "chunk_length": 1482, "token_count": 395}}
{"text": "x 10-3moles 1000 1000 Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g 3. Calculate the volume required to dissolve : (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3 Molarity 0.8 (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.8M x 312.5cm3 = C2 x (312.5 + 100) C2 = 0.8M x 312.5cm3 = 0.6061M\n31 31 412.5 (b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution . Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7049304776499515, "ocr_used": true, "chunk_length": 1521, "token_count": 511}}
{"text": "Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products. Knowing the concentration/ volume of one reactant, the other can be determined from the relationship: M1V1 = M2V2 where: n1 n2 M1 = Molarity of 1st reactant M2 = Molarity of 2nd reactant V1 = Volume of 1st reactant V1 = Volume of 2nd reactant n1 = number of moles of 1st reactant from stoichiometric equation n2 = number of moles of 2nd reactant from stoichiometric equation Examples 1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0) Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l) Method 1\n32 32 M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3 n1 n2 1 2 => M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3 5.0 x 2 Method 2 Moles of HCl used = molarity x volume 1000 => 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles 1000 Mole ratio MCO3 : HCl = 1:2 Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles 2 Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000 Volume 5 = 1.25M / 1.25 moledm-3 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6507066291804333, "ocr_used": true, "chunk_length": 1756, "token_count": 676}}
{"text": "0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B Moles of NaOH used = molarity x volume 1000 => 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles 1000 Mole ratio B : NaOH = 1:1\n34 34 Moles B = 0.0044175 /4.4175 x 10-3moles Molar mass B = mass => 0.388 moles 0.0044175 /4.4175 x 10-3moles = 87.8324 gmole-1 X-COOH = 87.8324 where X is an alkyl group X =87.8324- 42 = 42.8324=43 By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43 Molecula formula : CH3CH2 CH2COOH Molecule name : Butan-1-oic acid Molecular structure H H H O H C C C C O H H H H H 5. 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6805217391304348, "ocr_used": true, "chunk_length": 925, "token_count": 349}}
{"text": "10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0) Equation for neutralization NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l) Mole ratio NaOH(aq):HCl(aq)= 1:1 Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles 1000 1000 Excess moles of NaOH(aq)= 0.0425 moles Equation for reaction with ammonium salt 2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l) Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1 Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles 1000 1000 Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 - 0.0425 = 0.1575moles Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles Molar mass (NH4) 2SO4= 132 gmole-1\n35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6951297835728655, "ocr_used": true, "chunk_length": 1331, "token_count": 502}}
{"text": "07875moles Molar mass (NH4) 2SO4= 132 gmole-1\n35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point. Set up of titration apparatus The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings. To reduce errors, titration process should be repeated at least once more. The results of titration are recorded in a titration table as below\n36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8377740709850803, "ocr_used": true, "chunk_length": 1962, "token_count": 469}}
{"text": "The results of titration are recorded in a titration table as below\n36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided. Calculate the average volume of solution used 24.0 + 24.0 + 24.0 = 24.0 cm3 3 As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point. As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value . The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates. Bonus mark is awarded for averaged reading within 0.1 school value as Final answer. Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the: (i)relationship among the mole, molar mass, mole ratios, concentration, molarity. (ii) mathematical application of 1st principles. Very useful information which candidates forget appears usually in the beginning of the question paper as: “You are provided with…” All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point. Candidates are expected to use a non programmable scientific calculator. 37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.853985445802384, "ocr_used": true, "chunk_length": 2193, "token_count": 494}}
{"text": "37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1 Sample results Titration table 1 Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution B used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution B used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7336314847942755, "ocr_used": true, "chunk_length": 1118, "token_count": 373}}
{"text": "How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M Volume 20\n38 38 (b)Sample Titration Practice 2 (Redox Titration) You are provided with: Acidified Potassium manganate(VII) solution A 0.1M of an iron (II)salt solution B 8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C You are required to (i)standardize acidified potassium manganate(VII) (ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O. Procedure 1 Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears. Record your results to complete table 1. Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7233929004196482, "ocr_used": true, "chunk_length": 1405, "token_count": 505}}
{"text": "Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution A were present in the average volume. Assume one mole of B react with five moles of B Mole ratio A : B = 1:5 => Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles 5 5\n39 39 (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 Volume 20 = 0.025 M /moles per litre /moles l-1 Procedure 2 Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6914061501243989, "ocr_used": true, "chunk_length": 1393, "token_count": 496}}
{"text": "Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present inin the average titre. Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.659444342000123, "ocr_used": true, "chunk_length": 1122, "token_count": 450}}
{"text": "Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O. 40 40 (N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0) Molar mass = mass perlitre = 8.5 = 8500 g Moles per litre 1.0 x 10 -3 moles NH4)2SO4FeSO4.xH2O = 8500 284 + 18x =8500 8500 - 284 = 8216 = 18x = 454.4444 18 18 x = 454 (whole number) (c)Sample Titration Practice 3 (Back titration) You are provided with: (i)an impure calcium carbonate labeled M (ii)Hydrochloric acid labeled solution N (iii)solution L containing 20g per litre sodium hydroxide. You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6653735557065927, "ocr_used": true, "chunk_length": 1234, "token_count": 499}}
{"text": "You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M\n41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7275374974316828, "ocr_used": true, "chunk_length": 1256, "token_count": 447}}
{"text": "Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M\n41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7428902647331894, "ocr_used": true, "chunk_length": 1423, "token_count": 485}}
{"text": "Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles\n42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7305390491036633, "ocr_used": true, "chunk_length": 1283, "token_count": 432}}
{"text": "Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles\n42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7044396233237435, "ocr_used": true, "chunk_length": 1398, "token_count": 486}}
{"text": "Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles\n42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6505437188444292, "ocr_used": true, "chunk_length": 1242, "token_count": 473}}
{"text": "Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles\n42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6701729106628243, "ocr_used": true, "chunk_length": 1041, "token_count": 391}}
{"text": "Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles\n43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6512514511260739, "ocr_used": true, "chunk_length": 1475, "token_count": 579}}
{"text": "Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles\n43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7090745387611147, "ocr_used": true, "chunk_length": 1406, "token_count": 505}}
{"text": "From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles\n43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7459546743273437, "ocr_used": true, "chunk_length": 1299, "token_count": 437}}
{"text": "Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1. Sample Table 1 1 2 3 Final burette reading (cm3) 24.0 24.0 24.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.0 24.0 24.0 Sample calculations\n44 44 (a)Calculate the average volume of solution L used (1mk) 24.0 + 24.0 + 24.0 = 24.0cm3 3 (b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0) Molar mass H2X.2H2O= mass per litre => 5.0g/litre = 100g Moles/litre 0.05molesdm-3 H2X.2H2O =100 X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 66 (c) Calculate the number of moles of the dibasic acid H2X.2H2O. Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.653023522936658, "ocr_used": true, "chunk_length": 1276, "token_count": 511}}
{"text": "Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre. Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5\n45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6718500182923507, "ocr_used": true, "chunk_length": 1122, "token_count": 429}}
{"text": "Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5\n45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7080931000986684, "ocr_used": true, "chunk_length": 1157, "token_count": 425}}
{"text": "Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5\n45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6912838839013337, "ocr_used": true, "chunk_length": 1192, "token_count": 455}}
{"text": "Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N. 25cm3 pipette volume -> 0.0065 /6.5 x10-3 moles 250cm3 -> 0.0065 /6.5 x10-3 moles x 250 = 0.065 / 6.5 x10-2 moles 25 Procedure 3\n46 46 Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7004767257719101, "ocr_used": true, "chunk_length": 1139, "token_count": 426}}
{"text": "Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3. Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7046197037420119, "ocr_used": true, "chunk_length": 1177, "token_count": 432}}
{"text": "Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5\n48 48 Sample questions Calculate the average volume of solution A used.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6848866414730324, "ocr_used": true, "chunk_length": 2421, "token_count": 891}}
{"text": "Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5\n48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6870540758676352, "ocr_used": true, "chunk_length": 2124, "token_count": 782}}
{"text": "(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5\n48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6923625243258271, "ocr_used": true, "chunk_length": 1650, "token_count": 588}}
{"text": "47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5\n48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6841950805112562, "ocr_used": true, "chunk_length": 1653, "token_count": 592}}
{"text": "12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6719822303921569, "ocr_used": true, "chunk_length": 256, "token_count": 95}}
{"text": "46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40\n52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.697032722501485, "ocr_used": true, "chunk_length": 1359, "token_count": 509}}
{"text": "Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40\n52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6924228271588859, "ocr_used": true, "chunk_length": 1205, "token_count": 462}}
{"text": "25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre. Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25\n14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.700784782252208, "ocr_used": true, "chunk_length": 1377, "token_count": 495}}
{"text": "Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25\n14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent. It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because: (i) carbon uses all the four valence electrons to form four strong covalent bond. (ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7399453999454, "ocr_used": true, "chunk_length": 1485, "token_count": 504}}
{"text": "(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9063656950532899, "ocr_used": true, "chunk_length": 758, "token_count": 191}}
{"text": "When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane\n3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8341539574853007, "ocr_used": true, "chunk_length": 1474, "token_count": 508}}
{"text": "The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane\n3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane. 2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8317235774696508, "ocr_used": true, "chunk_length": 1456, "token_count": 489}}
{"text": "2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons. 4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons. 5.One member of the alkane differ from the next/previous by a CH2 group. e.g Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms 6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series. 7.A homologous series: (i) differ by a CH2 group from the next /previous consecutively (ii)have similar chemical properties (iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula CnH2n+2. (iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8469348659003831, "ocr_used": true, "chunk_length": 1944, "token_count": 508}}
{"text": "(iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula. Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula. The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature uses the following basic rules/guidelines: 1.Identify the longest continuous carbon chain to get/determine the parent alkane. 5 2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane. Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8405326931166447, "ocr_used": true, "chunk_length": 1839, "token_count": 491}}
{"text": "Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “1” Type of the branch “methyl” hence Molecular formula CH3 CH3 CH CH2 CH3 // CH3 CH (CH3 ) CH2CH3 Structural formula H H H H H C C C C H H H H H C H\n6 H (ii)2,2-dimethylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “2” Type of the branch two“methyl” hence Molecular formular CH3 CH3 C CH2 CH3 // CH3 C (CH3 )2 CH2CH3 CH3 Structural formula H H C H H H H H C C C C H H H H H C H H\n7 (iii) 2,2,3-trimethylbutane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8326965177077008, "ocr_used": true, "chunk_length": 1865, "token_count": 499}}
{"text": "Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2 and 3” Number of branches at carbon “3” Type of the branch three “methyl” hence Molecular formular CH3 CH3 C CH CH3 // CH3 C (CH3 )3 CH2CH3 CH3 CH3 Structural formula H H C H H H H C C C H H H H H C C H H H C H\n8 H (iv) 1,1,1,2,2,2-hexabromoethane Molecular formula CBr3 CBr3 Structural formula Br Br Br C C Br Br Br (v) 1,1,1-tetrachloro-2,2-dimethylbutane CH3 CCl 3 C CH3 // C Cl 3 C (CH3 )2 CH3 CH3 Structural formula Cl Cl C Cl H H H C C C H H H H C H H (c)Occurrence and extraction Crude oil ,natural gas and biogas are the main sources of alkanes: (i)Natural gas is found on top of crude oil deposits and consists mainly of methane. 9 (ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced. (iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first. As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8431366790483997, "ocr_used": true, "chunk_length": 1859, "token_count": 503}}
{"text": "As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point. Uses of different crude oil fractions Carbon atoms in a molecule Common name of fraction Uses of fraction 1-4 Gas L.P.G gas for domestic use 5-12 Petrol Fuel for petrol engines 9-16 Kerosene/Paraffin Jet fuel and domestic lighting/cooking 15-18 Light diesel Heavy diesel engine fuel 18-25 Diesel oil Light diesel engine fuel 20-70 Lubricating oil Lubricating oil to reduce friction. Over 70 Bitumen/Asphalt Tarmacking roads (d)School laboratory preparation of alkanes In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime. Chemical equation: Sodium alkanoate + soda lime -> alkane + Sodium carbonate CnH2n+1COONa(s) + NaOH(s) -> C n H2n+2 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2. 10 Examples 1. Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.823205447515431, "ocr_used": true, "chunk_length": 1641, "token_count": 476}}
{"text": "Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3. Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water Sodium butanoate + soda lime -> propane + Sodium carbonate CH3 CH2CH2COONa(s) + NaOH(s) -> CH3 CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3 4. Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water Sodium pentanoate + soda lime -> butane + Sodium carbonate CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3 Laboratory set up for the preparation of alkanes\n11 (d)Properties of alkanes I. Physical properties Alkanes are colourless gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.819560000950322, "ocr_used": true, "chunk_length": 1718, "token_count": 509}}
{"text": "They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids . The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases. This reduces the volume occupied by a given mass of the compound. Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas\n12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7726237177484079, "ocr_used": true, "chunk_length": 1583, "token_count": 483}}
{"text": "Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas\n12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7207528148730034, "ocr_used": true, "chunk_length": 1273, "token_count": 442}}
{"text": "Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. Methane + Air -> carbon(II) oxide + water (excess air/oxygen) 2CH4(g) + 3O2(g) -> 2CO(g) + 4H2O(l/g) 2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Propane + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8104116778743645, "ocr_used": true, "chunk_length": 1407, "token_count": 490}}
{"text": "Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Propane + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C3H8(g) + 7O2(g) -> 6CO(g) + 8H2O(l/g) ii)Substitution Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light. Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes. During substitution: (i)the halogen molecule is split into free atom/radicals. (ii)one free halogen radical/atoms knock /remove one hydrogen from the alkane leaving an alkyl radical. (iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane. (iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide. (v)substitution stops when all the hydrogen in alkanes are replaced with halogens. Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst. Examples of substitution reactions Methane has no effect on bromine or chlorine in diffused light/dark.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8557998330255269, "ocr_used": true, "chunk_length": 1682, "token_count": 496}}
{"text": "(v)substitution stops when all the hydrogen in alkanes are replaced with halogens. Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst. Examples of substitution reactions Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane explode to form colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades. Chemical equation 1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride\n14 CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g) H H H C H + Cl Cl -> H C Cl + H Cl H H (b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g) H H H C Cl + Cl Cl -> H C Cl + H Cl H Cl (c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g) Cl H H C Cl + Cl Cl -> Cl C Cl + H Cl H Cl (c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g) H Cl Cl C Cl + Cl Cl -> Cl C Cl + H Cl Cl Cl\n15 Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8449830432421712, "ocr_used": true, "chunk_length": 1343, "token_count": 402}}
{"text": "Chemical equation 1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride\n14 CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g) H H H C H + Cl Cl -> H C Cl + H Cl H H (b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g) H H H C Cl + Cl Cl -> H C Cl + H Cl H Cl (c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g) Cl H H C Cl + Cl Cl -> Cl C Cl + H Cl H Cl (c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g) H Cl Cl C Cl + Cl Cl -> Cl C Cl + H Cl Cl Cl\n15 Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades. Chemical equation (a)Ethane + chlorine -> Chloroethane + Hydrogen chloride CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g) H H H H H C C H + Br Br -> H C C H + H Br H H H Br Bromoethane H H H Br H C C H + Br Br -> H C C H + H Br H Br H Br 1,1-dibromoethane H Br H Br H C C H + Br Br -> H C C Br + H Br H Br H Br 1,1,1-tribromoethane H Br H Br H C C Br + Br Br -> H C C Br + H Br H Br Br Br 1,1,1,2-tetrabromoethane H Br H Br H C C Br + Br Br -> Br C C Br + H Br\n16 Br Br Br Br 1,1,1,2,2-pentabromoethane H Br Br Br Br C C Br + Br Br -> Br C C Br + H Br Br Br Br Br 1,1,1,2,2,2-hexabromoethane Uses of alkanes 1.Most alkanes are used as fuel e.g.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7808393502528309, "ocr_used": true, "chunk_length": 1508, "token_count": 583}}
{"text": "In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades. Chemical equation (a)Ethane + chlorine -> Chloroethane + Hydrogen chloride CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g) H H H H H C C H + Br Br -> H C C H + H Br H H H Br Bromoethane H H H Br H C C H + Br Br -> H C C H + H Br H Br H Br 1,1-dibromoethane H Br H Br H C C H + Br Br -> H C C Br + H Br H Br H Br 1,1,1-tribromoethane H Br H Br H C C Br + Br Br -> H C C Br + H Br H Br Br Br 1,1,1,2-tetrabromoethane H Br H Br H C C Br + Br Br -> Br C C Br + H Br\n16 Br Br Br Br 1,1,1,2,2-pentabromoethane H Br Br Br Br C C Br + Br Br -> Br C C Br + H Br Br Br Br Br 1,1,1,2,2,2-hexabromoethane Uses of alkanes 1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes.Butane is used as the Laboratory gas. 2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process. 3.In manufacture of Carbon black which is a component in printers ink. 4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane. (ii) Alkenes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8320851810998807, "ocr_used": true, "chunk_length": 1433, "token_count": 461}}
{"text": "(ii) Alkenes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms. They include: n General/ Molecular formula Structural formula Name 1 Does not exist 2 C2H6 H H H C C H Ethene\n17 CH2 CH2 3 C3H8 H H H H C C C H H CH2 CH CH3 Propene 4 C4H10 H H H H H C C C C H H H CH2 CH CH2CH3 Butene 5 C5H12 H H H H H H C C C C C H H H H CH2 CH (CH2)2CH3 Pentene 6 C6H14 H H H H H H H C C C C C C H H H H H CH2 CH (CH2)3CH3 Hexene 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptene\n18 CH2 CH (CH2)4CH3 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H CH2 CH (CH2)5CH3 Octene 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H CH2 CH (CH2)6CH3 Nonene 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH2 CH (CH2)7CH3 decene Note 1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond. 2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8095635029384003, "ocr_used": true, "chunk_length": 1424, "token_count": 512}}
{"text": "They include: n General/ Molecular formula Structural formula Name 1 Does not exist 2 C2H6 H H H C C H Ethene\n17 CH2 CH2 3 C3H8 H H H H C C C H H CH2 CH CH3 Propene 4 C4H10 H H H H H C C C C H H H CH2 CH CH2CH3 Butene 5 C5H12 H H H H H H C C C C C H H H H CH2 CH (CH2)2CH3 Pentene 6 C6H14 H H H H H H H C C C C C C H H H H H CH2 CH (CH2)3CH3 Hexene 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptene\n18 CH2 CH (CH2)4CH3 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H CH2 CH (CH2)5CH3 Octene 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H CH2 CH (CH2)6CH3 Nonene 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH2 CH (CH2)7CH3 decene Note 1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond. 2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series. e.g Propene differ from ethene by one carbon and two Hydrogen atoms from ethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7968849080532658, "ocr_used": true, "chunk_length": 1216, "token_count": 463}}
{"text": "2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series. e.g Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 4.A homologous series of alkenes like that of alkanes: (i) differ by a CH2 group from the next /previous consecutively (ii)have similar chemical properties\n19 (iii)have similar chemical formula represented by the general formula CnH2n (iv)the physical properties also show steady gradual change 5.The = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound. 6. The = C= C = double bond in alkene can easily be broken to accommodate more two more monovalent atoms. The = C= C = double bond in alkenes make it thus unsaturated. 7. An unsaturated hydrocarbon is one with a double =C=C= or triple – C C – carbon bonds in their molecular structure. Unsaturated hydrocarbon easily reacts to be saturated. 8.A saturated hydrocarbon is one without a double =C=C= or triple – C C – carbon bonds in their molecular structure. Most of the reactions of alkenes take place at the = C = C =bond. (b)Isomers of alkenes Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula. Ethene and propene do not form isomers. Isomers of alkenes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature of naming alkenes uses the following basic rules/guidelines: 1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene. 2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible. 3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3diene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8844673251090365, "ocr_used": true, "chunk_length": 2057, "token_count": 518}}
{"text": "2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible. 3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3diene. 4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e But-2-ene means the double =C = C= is between Carbon “2”and “3” Pent-1,3-diene means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4” 5. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkene. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of double C = C bonds and branches attached to the alkene. 20 7.Position isomers can be formed when the=C = C= double bond is shifted between carbon atoms e.g. But-2-ene means the double =C = C= is between Carbon “2”and “3” But-1-ene means the double =C = C= is between Carbon “1”and “2” Both But-1-ene and But-2-ene are position isomers of Butene 8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e. Butene has the molecular/general formular C4H8 position but can form both But-1ene and But-2-ene as position isomers. 9. Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed. 10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g Butene and 2-methyl propene both have the same general formualr but different branching chain.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8425130782342916, "ocr_used": true, "chunk_length": 1712, "token_count": 467}}
{"text": "Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed. 10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g Butene and 2-methyl propene both have the same general formualr but different branching chain. Practice on IUPAC nomenclature of alkenes Name the following isomers of alkene H H H H H C C C C H But-1-ene H H H H H H H C C C C H But-2-ene H H H H H H H H H C C C C C C H 4-methylhex-1-ene H H H H C H\n21 H H H C H H H H H H H C C C C C C H 4,4-dimethylhex-1-ene H H H H C H H 3. H H C H H H H H H C C C C C H 4,4-dimethylpent -1- ene H H H C H H 4. H H C H H H H H H C C C C C H 5,5-dimethylhex-2- ene H C H H H H C H H H\n22 5. H H C H H H H H C C C C H 2,2-dimethylbut -2- ene H H H C H H 8.H2C CHCH2 CH2 CH3 pent -1- ene 9.H2C C(CH3)CH2 CH2 CH3 2-methylpent -1- ene 10.H2C C(CH3)C(CH3)2 CH2 CH3 2,3,3-trimethylpent -1- ene 11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3 2,3,3,4,4-pentamethylpent -1- ene 12.H3C C(CH3)C(CH3) C(CH3)2 CH3 2,3,4,4-tetramethylpent -2- ene 13. H2C C(CH3)C(CH3) C(CH3) CH3 2,3,4-trimethylpent -1,3- diene 14.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7201440400922653, "ocr_used": true, "chunk_length": 1137, "token_count": 503}}
{"text": "I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene 18. H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene (c)Occurrence and extraction\n23 At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures. Cracking is a major source of useful hydrogen gas for manufacture of ammonia/nitric(V)acid/HCl i.e. Long chain alkane -> smaller/shorter alkane + Alkene + Hydrogen gas Examples 1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas. Chemical equation CH3CH2CH3 (g) -> CH4(g) + CH2=CH2(g) + H2(g) 2.Octane undergo cracking to form hydrogen gas, butene and butane gases Chemical equation CH3(CH2) 6 CH3 (g) -> CH3CH2CH2CH3(g) + CH3 CH2CH=CH2(g) + H2(g) (d)School laboratory preparation of alkenes In a school laboratory, alkenes may be prepared from dehydration of alkanols using: (i) concentrated sulphuric(VI)acid(H2SO4). (a) aluminium(III)oxide(Al2O3) i.e Alkanol --Conc. H2SO4 --> Alkene + Water Alkanol --Al2O3 --> Alkene + Water e.g. 1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene. The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities. It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8094677479109209, "ocr_used": true, "chunk_length": 1429, "token_count": 452}}
{"text": "1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene. The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities. It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities. Chemical equation CH3CH2OH (l) --conc H2SO4/180oC--> CH2=CH2(g) + H2O(l) (b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene. 24 Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration. Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8387636658118506, "ocr_used": true, "chunk_length": 956, "token_count": 282}}
{"text": "24 Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration. Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7219245567613861, "ocr_used": true, "chunk_length": 1465, "token_count": 569}}
{"text": "Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene\n25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7109153222435844, "ocr_used": true, "chunk_length": 1709, "token_count": 686}}
{"text": "Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene\n25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7004746784694827, "ocr_used": true, "chunk_length": 1641, "token_count": 674}}
{"text": "Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene\n25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6947774778733288, "ocr_used": true, "chunk_length": 1432, "token_count": 595}}
{"text": "Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene\n25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks. (i)Using conentrated sulphuric(VI)acid Some broken porcelain or sand should be put in the flask when heating to: (i)prevent bumping which may break the flask. (ii)ensure uniform and smooth boiling of the mixture The temperatures should be maintained at above160oC. At lower temperatures another compound -ether is predominantly formed instead of ethene gas. (ii)Using aluminium(III)oxide\n26 (e)Properties of alkenes I. Physical properties Like alkanes, alkenes are colourles gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.862854294260077, "ocr_used": true, "chunk_length": 1652, "token_count": 445}}
{"text": "This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene. Summary of physical properties of the 1st five alkenes Alkene General formula Melting point(oC) Boiling point(K) State at room(298K) temperature and pressure atmosphere (101300Pa)\n27 Ethene CH2CH2 -169 -104 gas Propene CH3 CHCH2 -145 -47 gas Butene CH3CH2 CHCH2 -141 -26 gas Pent-1ene CH3(CH2 CHCH2 -138 30 liquid Hex-1ene CH3(CH2) CHCH2 -98 64 liquid II. Chemical properties (a)Burning/combustion Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkene + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water. Alkene + Air -> carbon(II) oxide + water (limited air) Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the =C=C= double bond because they have higher C:H ratio. A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8502374881255937, "ocr_used": true, "chunk_length": 1695, "token_count": 463}}
{"text": "A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Ethene + Air -> carbon(IV) oxide + water (excess air/oxygen) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) (b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. 28 Ethene + Air -> carbon(II) oxide + water (limited air ) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) 2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C3H6(g) + 9O2(g) -> 6CO2(g) + 6H2O(l/g) (a) Propene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8491700356718194, "ocr_used": true, "chunk_length": 1682, "token_count": 501}}
{"text": "Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8913112004395731, "ocr_used": true, "chunk_length": 916, "token_count": 223}}
{"text": "Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond. Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane\n29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7924373063488565, "ocr_used": true, "chunk_length": 1258, "token_count": 431}}
{"text": "Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane\n29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds. But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H\n30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7559440211165895, "ocr_used": true, "chunk_length": 1242, "token_count": 499}}
{"text": "But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H\n30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces. One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8026378502752897, "ocr_used": true, "chunk_length": 849, "token_count": 295}}
{"text": "One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane. Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I\n31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7600168040359429, "ocr_used": true, "chunk_length": 1089, "token_count": 429}}
{"text": "Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I\n31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H2C= CH CH=CH2 + 2I2 H2CI CHICHI - CHI H H H H H H H H H C C - C = C -H + 2(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides. Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond. The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7738025610752883, "ocr_used": true, "chunk_length": 1210, "token_count": 471}}
{"text": "The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane. Chemical equation H2C=CH2 + HBr H3 C - CH2 Br H H H H\n32 C = C + H – Br H - C – C - Br H H H H Ethene + Bromine bromoethane 2. Propene reacts with hydrogen iodide to form 2-iodopropane. Chemical equation H3C CH=CH2 + HI H3C CHI - CH3 Propene + Chlorine 2-chloropropane H H H H H H H C C = C + H – Cl H - C – C - C- H H H H Cl H 3. Both But-1-ene and But-2-ene reacts with hydrogen bromide to form 2- bromobutane Chemical equation But-1-ene + hydrogen bromide 2-bromobutane H3C CH2 CH=CH2 + HBr H3C CH2CHBr -CH3 H H H H H H H H H C C - C = C + H – Br H - C- C – C - C- H H H H H H Br H But-2-ene + Hydrogen bromide 2-bromobutane H3C CH= CH-CH2 + HBr H3C CHBrCH2 - CH3 H H H H H H H H H C C = C - C -H + Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen\n33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7810857311195225, "ocr_used": true, "chunk_length": 1351, "token_count": 507}}
{"text": "But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen\n33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water. Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid). During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8368460344820848, "ocr_used": true, "chunk_length": 1179, "token_count": 376}}
{"text": "During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH\n34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.78043964686465, "ocr_used": true, "chunk_length": 1276, "token_count": 491}}
{"text": "Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH\n34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7727677472761035, "ocr_used": true, "chunk_length": 1268, "token_count": 493}}
{"text": "Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH\n34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H\n35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7572319438516621, "ocr_used": true, "chunk_length": 1188, "token_count": 493}}
{"text": "The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H\n35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents. When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green. When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized. Examples 1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H2C=CH2 [O] in H+/K2Cr2O7 HO CH2 - CH2 OH H H H H C = C+ [O] in H+/KMnO4 H - C – C - H H H OH OH Ethene + [O] in H+/KMnO4 ethan-1,2-diol 2. Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8428955271060535, "ocr_used": true, "chunk_length": 1406, "token_count": 440}}
{"text": "Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H3C CH=CH2 [O] in H+/KMnO4 H3C CHOH - CH2OH Propene [O] in H+/KMnO4 propan-1,2-diol H H H H H H H C C = C [O] in H+/KMnO4 H - C – C - C- OH\n36 H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form butan-1,2-diol and butan-2,3-diol Chemical equation I.But-1-ene + [O] in H+/KMnO4 butan-1,2-diol H3C CH2 CH=CH2 + [O] H3C CH2CHOH - CH2OH H H H H H H H H H C C - C = C + [O] H - C- C – C - C- OH H H H H H OH H (v) Hydrolysis. Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound. Alkenes undergo hydrolysis to form alkanols . This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8221305495318306, "ocr_used": true, "chunk_length": 1255, "token_count": 436}}
{"text": "This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H\n37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8138746145940391, "ocr_used": true, "chunk_length": 840, "token_count": 297}}
{"text": "Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H\n37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition\n38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7999218529026689, "ocr_used": true, "chunk_length": 1423, "token_count": 511}}
{"text": "Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H\n37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition\n38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8139519792342635, "ocr_used": true, "chunk_length": 1541, "token_count": 519}}
{"text": "Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition\n38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.856884399876944, "ocr_used": true, "chunk_length": 1321, "token_count": 379}}
{"text": "Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule\n39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8766753016359403, "ocr_used": true, "chunk_length": 1595, "token_count": 440}}
{"text": "Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule\n39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.860576571637427, "ocr_used": true, "chunk_length": 1792, "token_count": 510}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule\n39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8525010501426484, "ocr_used": true, "chunk_length": 1639, "token_count": 480}}
{"text": "Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8691496882322571, "ocr_used": true, "chunk_length": 1199, "token_count": 326}}
{"text": "The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8663864267508654, "ocr_used": true, "chunk_length": 1626, "token_count": 458}}
{"text": "40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:\n41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8471976972881405, "ocr_used": true, "chunk_length": 1736, "token_count": 526}}
{"text": "Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:\n41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8486298494499132, "ocr_used": true, "chunk_length": 1727, "token_count": 524}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:\n41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8474452969802136, "ocr_used": true, "chunk_length": 1709, "token_count": 519}}
{"text": "Polychloroethene molecule can be represented as:\n41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H\n42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8267433831990795, "ocr_used": true, "chunk_length": 1738, "token_count": 556}}
{"text": "It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H\n42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples\n43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8230240746027215, "ocr_used": true, "chunk_length": 1695, "token_count": 548}}
{"text": "PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H\n42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples\n43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8224198077086702, "ocr_used": true, "chunk_length": 1679, "token_count": 546}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H\n42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples\n43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8140977819998789, "ocr_used": true, "chunk_length": 1523, "token_count": 504}}
{"text": "Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples\n43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8350857606223888, "ocr_used": true, "chunk_length": 1007, "token_count": 301}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3\n44 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8275350627351769, "ocr_used": true, "chunk_length": 1703, "token_count": 507}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3\n44 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8136066927508306, "ocr_used": true, "chunk_length": 1448, "token_count": 447}}
{"text": "propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings\n45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8367264669134116, "ocr_used": true, "chunk_length": 1004, "token_count": 295}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings\n45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8809150488002537, "ocr_used": true, "chunk_length": 1124, "token_count": 332}}
{"text": "Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings\n45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer\n46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8540134936272288, "ocr_used": true, "chunk_length": 1774, "token_count": 551}}
{"text": "(ii)ceiling tiles (iii)clothe linings\n45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer\n46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8512520824559378, "ocr_used": true, "chunk_length": 1709, "token_count": 534}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer\n46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 6.Formation of rubber from Latex Natural rubber is obtained from rubber trees.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8460938442362592, "ocr_used": true, "chunk_length": 1615, "token_count": 507}}
{"text": "polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer\n46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 6.Formation of rubber from Latex Natural rubber is obtained from rubber trees. During harvesting an incision is made on the rubber tree to produce a milky white substance called latex. Latex is a mixture of rubber and lots of water. The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;\n47 H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n- H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8362184761012252, "ocr_used": true, "chunk_length": 1689, "token_count": 506}}
{"text": "The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;\n47 H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n- H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher. During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer. H CH3 H H H CH3 H H - C - C - C - C - C - C - C - C - H S H H S H H CH3 S H H CH3 S H - C - C - C - C - C - C - C - C - H H H H H H Vulcanized rubber is used to make tyres, shoes and valves. 7.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H Sulphur atoms make cross link between polymers\n48 CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (c)Test for the presence of – C = C – double bond.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8341345313541821, "ocr_used": true, "chunk_length": 1726, "token_count": 510}}
{"text": "Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H Sulphur atoms make cross link between polymers\n48 CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (c)Test for the presence of – C = C – double bond. (i)Burning/combustion All unsaturated hydrocarbons with a – C = C – or – C = C – bond burn with a yellow sooty flame. Experiment Scoop a sample of the substance provided in a clean metallic spatula. Introduce it on a Bunsen burner. Observation Inference Solid melt then burns with a yellow sooty flame – C = C –, – C = C – bond (ii)Oxidation by acidified KMnO4/K2Cr2O7 Bromine water ,Chlorine water and Oxidizing agents acidified KMnO4/K2Cr2O7 change to unique colour in presence of – C = C –\n49 or – C = C – bond. Experiment Scoop a sample of the substance provided into a clean test tube. Add 10cm3 of distilled water. Shake. Take a portion of the solution mixture. Add three drops of acidified KMnO4/K2Cr2O7 . Observation Inference Acidified KMnO4 decolorized Orange colour of acidified K2Cr2O7 turns green Bromine water is decolorized Chlorine water is decolorized – C = C – – C = C – bond (d)Some uses of Alkenes 1. In the manufacture of plastic 2. Hydrolysis of ethene is used in industrial manufacture of ethanol. 3. In ripening of fruits. 4. In the manufacture of detergents.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8507177666011694, "ocr_used": true, "chunk_length": 1681, "token_count": 487}}
{"text": "In ripening of fruits. 4. In the manufacture of detergents. (iii) Alkynes (a)Nomenclature/Naming\n50 These are hydrocarbons with a general formula CnH2n-2 and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one triple bond to each other and single bonds to hydrogen atoms. They include: n General/ Molecular formula Structural formula Name 1 Does not exist - 2 C2H2 H C C H CH CH Ethyne 3 C3H4 H H C C C H H CH C CH3 Propyne 4 C4H6 H H H C C C C H H H CH C CH2CH3 Butyne 5 C5H8 H H H H C C C C C H H H H CH C (CH2)2CH3 Pentyne 6 C6H10 H H H H Hexyne\n51 H C C C C C C H H H H H CH C (CH2)3CH3 7 C7H12 H H H H H H C C C C C C C H H H H H H H H CH C (CH2)4CH3 Heptyne 8 C8H14 H H H H H H H C C C C C C C C H H H H H H H CH C (CH2)5CH3 Octyne 9 C9H16 H H H H H H H H C C C C C C C C C H H H H H H H H CH C (CH2)6CH3 Nonyne 10 C10H18 H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH C (CH2)7CH3 Decyne Note 1. Since carbon is tetravalent ,each atom of carbon in the alkyne MUST always be bonded using four covalent bond /four shared pairs of electrons including at the triple bond. 2. Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons. 52 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.813053077177903, "ocr_used": true, "chunk_length": 1319, "token_count": 482}}
{"text": "2. Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons. 52 3. One member of the alkyne ,like alkenes and alkanes, differ from the next/previous by a CH2 group(molar mass of 14 atomic mass units).They thus form a homologous series. e.g Propyne differ from ethyne by (14 a.m.u) one carbon and two Hydrogen atoms from ethyne. 4.A homologous series of alkenes like that of alkanes: (i) differ by a CH2 group from the next /previous consecutively (ii) have similar chemical properties (iii)have similar chemical formula with general formula CnH2n-2 (iv)the physical properties also show steady gradual change 5.The - C = C - triple bond in alkyne is the functional group. The functional group is the reacting site of the alkynes. 6. The - C = C - triple bond in alkyne can easily be broken to accommodate more /four more monovalent atoms. The - C = C - triple bond in alkynes make it thus unsaturated like alkenes. 7. Most of the reactions of alkynes like alkenes take place at the - C = C- triple bond. (b)Isomers of alkynes Isomers of alkynes have the same molecular general formula but different molecular structural formula. Isomers of alkynes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature of naming alkynes uses the following basic rules/guidelines: 1.Identify the longest continuous/straight carbon chain which contains the - C = C- triple bond to get/determine the parent alkene. 2. Number the longest chain form the end of the chain which contains the -C = C- triple bond so as - C = C- triple bond get lowest number possible. 3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3diyne. 4.The position indicated must be for the carbon atom at the lower position in the -C = C- triple bond.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8751195057280875, "ocr_used": true, "chunk_length": 1913, "token_count": 495}}
{"text": "3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3diyne. 4.The position indicated must be for the carbon atom at the lower position in the -C = C- triple bond. i.e But-2-yne means the triple -C = C- is between Carbon “2”and “3” Pent-1,3-diyne means there are two triple bonds; one between carbon “1” and “2”and another between carbon “3” and “4”\n53 5. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkyne. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of triple - C = C- bonds and branches attached to the alkyne. 7.Position isomers can be formed when the - C = C- triple bond is shifted between carbon atoms e.g. But-2-yne means the double - C = C- is between Carbon “2”and “3” But-1-yne means the double - C = C- is between Carbon “1”and “2” Both But-1-yne and But-2-yne are position isomers of Butyne. 9. Like alkanes and alkynes , an alkyl group can be attached to the alkyne. Chain/branch isomers are thus formed. Butyne and 2-methyl propyne both have the same general formular but different branching chain. (More on powerpoint) (c)Preparation of Alkynes. Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction Chemical equation\n54 CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g) (d)Properties of alkynes I.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8332186293717146, "ocr_used": true, "chunk_length": 1663, "token_count": 487}}
{"text": "Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction Chemical equation\n54 CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g) (d)Properties of alkynes I. Physical properties Like alkanes and alkenes,alkynes are colourles gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. Ethyne has a pleasant taste when pure. The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st three straight chain alkynes (ethyne,propyne and but-1yne)are gases at room temperature and pressure. The density of straight chain alkynes increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkyne. Summary of physical properties of the 1st five alkenes Alkyne General formula Melting point(oC) Boiling point(oC) State at room(298K) temperature and pressure atmosphere (101300Pa) Ethyne CH CH -82 -84 gas Propyne CH3 C CH -103 -23 gas Butyne CH3CH2 CCH -122 8 gas Pent-1-yne CH3(CH2) 2 CCH -119 39 liquid Hex-1-yne CH3(CH2) 3C CH -132 71 liquid II. Chemical properties (a)Burning/combustion Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8590350586209092, "ocr_used": true, "chunk_length": 1877, "token_count": 503}}
{"text": "Summary of physical properties of the 1st five alkenes Alkyne General formula Melting point(oC) Boiling point(oC) State at room(298K) temperature and pressure atmosphere (101300Pa) Ethyne CH CH -82 -84 gas Propyne CH3 C CH -103 -23 gas Butyne CH3CH2 CCH -122 8 gas Pent-1-yne CH3(CH2) 2 CCH -119 39 liquid Hex-1-yne CH3(CH2) 3C CH -132 71 liquid II. Chemical properties (a)Burning/combustion Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water. 55 Alkyne + Air -> carbon(II) oxide /carbon + water (limited air) Burning of alkynes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the - C = C – triple bond because they have very high C:H ratio. Examples of burning alkynes 1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water. Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g) (b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water. Ethyne + Air -> carbon(II) oxide + water (limited air ) C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g) 2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8152435008092617, "ocr_used": true, "chunk_length": 1507, "token_count": 475}}
{"text": "Examples of burning alkynes 1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water. Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g) (b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water. Ethyne + Air -> carbon(II) oxide + water (limited air ) C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g) 2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Propyne + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H4(g) + 4O2(g) -> 3CO2(g) + 2H2O(l/g) (a) Propyne when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C3H4(g) + 5O2(g) -> 6CO(g) + 4H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound. Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple - C = C- to single C- C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8444349680170575, "ocr_used": true, "chunk_length": 1540, "token_count": 462}}
{"text": "Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple - C = C- to single C- C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single. Chemical equation\n56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H\n57 (ii) Halogenation.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7861890694239292, "ocr_used": true, "chunk_length": 1354, "token_count": 486}}
{"text": "(i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single. Chemical equation\n56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H\n57 (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane. The reaction of alkynes with halogens with alkynes is faster than with alkenes.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7949441079875863, "ocr_used": true, "chunk_length": 1443, "token_count": 511}}
{"text": "Chemical equation\n56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H\n57 (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane. The reaction of alkynes with halogens with alkynes is faster than with alkenes. The triple bond in the alkyne break and form a double then single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases. Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7850704377395153, "ocr_used": true, "chunk_length": 1337, "token_count": 496}}
{"text": "The triple bond in the alkyne break and form a double then single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases. Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon. Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8080869860686374, "ocr_used": true, "chunk_length": 567, "token_count": 181}}
{"text": "Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon. Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I\n58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7084862694370334, "ocr_used": true, "chunk_length": 1197, "token_count": 537}}
{"text": "Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I\n58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6904354026195458, "ocr_used": true, "chunk_length": 1173, "token_count": 546}}
{"text": "Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I\n58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6999942479148692, "ocr_used": true, "chunk_length": 1220, "token_count": 551}}
{"text": "Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I\n58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2 I I I I H C C - C = C -H + 4(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6970298518685616, "ocr_used": true, "chunk_length": 1209, "token_count": 563}}
{"text": "But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2 I I I I H C C - C = C -H + 4(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides. Hydrogen halides reacts with alkyne to form a halogenoalkene then halogenoalkane. The triple bond in the alkyne break and form a double then single bond. 59 The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethyne reacts with hydrogen bromide to form bromoethane. Chemical equation H C = C H + 2HBr H3 C - CH Br2 H H H H C = C + 2H – Br H - C – C - Br H Br Ethyne + Bromine 1,1-dibromoethane 2. Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product ) Chemical equation H3C C = CH + 2HI H3C CHI2 - CH3 Propene + Chlorine 2,2-dichloropropane H H I H H C C = C + 2H – I H - C – C - C- H H H H I H 3.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.774563024937575, "ocr_used": true, "chunk_length": 1082, "token_count": 392}}
{"text": "Ethyne reacts with hydrogen bromide to form bromoethane. Chemical equation H C = C H + 2HBr H3 C - CH Br2 H H H H C = C + 2H – Br H - C – C - Br H Br Ethyne + Bromine 1,1-dibromoethane 2. Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product ) Chemical equation H3C C = CH + 2HI H3C CHI2 - CH3 Propene + Chlorine 2,2-dichloropropane H H I H H C C = C + 2H – I H - C – C - C- H H H H I H 3. Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane Chemical equation But-1-ene + hydrogen bromide 2,2-dibromobutane H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3 H H H H Br H H C C - C = C + 2H – Br H - C- C – C - C- H H H H H H Br H Carbon atom with more Hydrogen atoms gets extra hydrogen\n60 But-2-yne + Hydrogen bromide 2,2-dibromobutane H3C C = C -CH3 + 2HBr H3C CBr2CH2 - CH3 H H H Br H H H C C = C - C -H + 2Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.76316915995397, "ocr_used": true, "chunk_length": 1100, "token_count": 423}}
{"text": "Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane Chemical equation But-1-ene + hydrogen bromide 2,2-dibromobutane H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3 H H H H Br H H C C - C = C + 2H – Br H - C- C – C - C- H H H H H H Br H Carbon atom with more Hydrogen atoms gets extra hydrogen\n60 But-2-yne + Hydrogen bromide 2,2-dibromobutane H3C C = C -CH3 + 2HBr H3C CBr2CH2 - CH3 H H H Br H H H C C = C - C -H + 2Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)\n61 (A) INTRODUCTION.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7457948539560354, "ocr_used": true, "chunk_length": 878, "token_count": 363}}
{"text": "But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)\n61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol\n62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7605870662799442, "ocr_used": true, "chunk_length": 1712, "token_count": 715}}
{"text": "The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)\n61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol\n62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7632691573552386, "ocr_used": true, "chunk_length": 1706, "token_count": 707}}
{"text": "But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)\n61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol\n62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7578034468024498, "ocr_used": true, "chunk_length": 1652, "token_count": 697}}
{"text": "Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol\n62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.771763182737694, "ocr_used": true, "chunk_length": 1483, "token_count": 599}}
{"text": "(iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points. (vi)they show similar and gradual change in their chemical properties. B. ISOMERS OF ALKANOLS. 63 Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines: (i)Like alkanes , identify the longest carbon chain to be the parent name. (ii)Identify the position of the -OH functional group to give it the smallest /lowest position. (iii) Identify the type and position of the side branches. Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8453618992472496, "ocr_used": true, "chunk_length": 1256, "token_count": 349}}
{"text": "Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)\n64 CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH\n65 (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7354020319303338, "ocr_used": true, "chunk_length": 1325, "token_count": 507}}
{"text": "2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)\n64 CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH\n65 (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS. For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation. In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination. Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7961659402811714, "ocr_used": true, "chunk_length": 1459, "token_count": 454}}
{"text": "Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase. (C6H10O5)n (s) + H2O(l) --diastase enzyme --> C12H22O11(aq) (Starch) (Maltose) (ii)Hydrolysis of Maltose to glucose using the enzyme maltase. C12H22O11(aq)+ H2O(l) -- maltase enzyme -->2 C6H12O6(aq) (Maltose) (glucose) (iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase. C6H12O6(aq) -- zymase enzyme --> 2 C2H5OH(aq) + 2CO2(g) (glucose) (Ethanol) At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction. 66 To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol). Methanol is much more poisonous /toxic than ethanol. Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours. School laboratory preparation of ethanol from fermentation of glucose Measure 100cm3 of pure water into a conical flask. Add about five spatula end full of glucose. Stir the mixture to dissolve. Add about one spatula end full of yeast. Set up the apparatus as below. Preserve the mixture for about three days. D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols (a) Role of yeast\n67 Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process. (b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8801670759047722, "ocr_used": true, "chunk_length": 1972, "token_count": 492}}
{"text": "(b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water. More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate. Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) H2O(l) + CO2 (g) + CaCO3(s) -> Ca(HCO3) 2 (aq) (c)Effects on litmus paper Experiment Take the prepared sample and test with both blue and red litmus papers. Repeat the same with pure ethanol and methylated spirit. Sample Observation table Substance/alkanol Effect on litmus paper Prepared sample Blue litmus paper remain blue Red litmus paper remain red Absolute ethanol Blue litmus paper remain blue Red litmus paper remain red Methylated spirit Blue litmus paper remain blue Red litmus paper remain red Explanation Alkanols are neutral compounds/solution that have characteristic sweet smell and taste. They have no effect on both blue and red litmus papers. (d)Solubility in water. Experiment Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water. Repeat the same with pure ethanol and methylated spirit. Observation No layers formed between the two liquids. Explanation Ethanol is miscible in water.Both ethanol and water are polar compounds . 68 The solubility of alkanols decrease with increase in the alkyl chain/molecular mass. The alkyl group is insoluble in water while –OH functional group is soluble in water. As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases. e)Melting/boiling point. Experiment Place pure ethanol in a long boiling tube .Determine its boiling point. Observation Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure. Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9025397043238257, "ocr_used": true, "chunk_length": 2223, "token_count": 486}}
{"text": "Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling. f)Density Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other. This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density). Summary table showing the trend in physical properties of alkanols Alkanol Melting point (oC) Boiling point (oC) Density gcm-3 Solubility in water Methanol -98 65 0.791 soluble Ethanol -117 78 0.789 soluble Propanol -103 97 0.803 soluble Butanol -89 117 0.810 Slightly soluble Pentanol -78 138 0.814 Slightly soluble Hexanol -52 157 0.815 Slightly soluble Heptanol -34 176 0.822 Slightly soluble Octanol -15 195 0.824 Slightly soluble Nonanol -7 212 0.827 Slightly soluble Decanol 6 228 0.827 Slightly soluble\n69 g)Burning Experiment Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit. Observation/Explanation Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting. Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8297342496360743, "ocr_used": true, "chunk_length": 1635, "token_count": 409}}
{"text": "Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes. Chemica equation C2 H5OH(l) + 3O2 (g) -> 3H2O(l) + 2CO2 (g) ( excess air) C2 H5OH(l) + 2O2 (g) -> 3H2O(l) + 2CO (g) ( limited air) 2CH3OH(l) + 3O2 (g) -> 4H2O(l) + 2CO2 (g) ( excess air) 2 CH3OH(l) + 2O2 (g) -> 4H2O(l) + 2CO (g) ( limited air) 2C3 H7OH(l) + 9O2 (g) -> 8H2O(l) + 6CO2 (g) ( excess air) C3 H7OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) 2C4 H9OH(l) + 13O2 (g) -> 20H2O(l) + 8CO2 (g) ( excess air) C4 H9OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) Due to its flammability, ethanol is used; (i) as a fuel in spirit lamps (ii) as gasohol when blended with gasoline (h)Formation of alkoxides Experiment Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7454004309242961, "ocr_used": true, "chunk_length": 1112, "token_count": 438}}
{"text": "Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles\n70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9159360580092288, "ocr_used": true, "chunk_length": 984, "token_count": 231}}
{"text": "Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles\n70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)\n71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7988172124608012, "ocr_used": true, "chunk_length": 2494, "token_count": 825}}
{"text": "Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles\n70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)\n71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8003428805293402, "ocr_used": true, "chunk_length": 2547, "token_count": 842}}
{"text": "If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)\n71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.759567638806224, "ocr_used": true, "chunk_length": 1857, "token_count": 681}}
{"text": "Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)\n71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7537759002811287, "ocr_used": true, "chunk_length": 1817, "token_count": 672}}
{"text": "Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water. Smell the products. Repeat with methanol Sample observations Substance/alkanol Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid Absolute ethanol Sweet fruity smell Methanol Sweet fruity smell Explanation Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory. Alkanol + Alkanoic acid –Conc. H2SO4-> Ester + water Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that\n72 create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters . Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g. Ethanol + Ethanoic acid -> Ethylethanoate + Water Ethanol + Propanoic acid -> Ethylpropanoate + Water Ethanol + Methanoic acid -> Ethylmethanoate + Water Ethanol + butanoic acid -> Ethylbutanoate + Water Propanol + Ethanoic acid -> Propylethanoate + Water Methanol + Ethanoic acid -> Methyethanoate + Water Methanol + Decanoic acid -> Methyldecanoate + Water Decanol + Methanoic acid -> Decylmethanoate + Water During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol. R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O e.g. 1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8911813971819885, "ocr_used": true, "chunk_length": 1791, "token_count": 484}}
{"text": "1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3COOH(l) --Conc. H2SO4 --> CH3COOCH2CH3(aq) +H2O(l) 2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water. Ethanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2CH3(aq) +H2O(l) 3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water. Methanol + Ethanoic acid --Conc. H2SO4 -->Methylethanoate + Water CH3OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO CH3(aq) +H2O(l) 4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water. Methanol + propanoic acid --Conc. H2SO4 -->Methylpropanoate + Water CH3OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. 73 Propanol + Propanoic acid --Conc.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7257631510754081, "ocr_used": true, "chunk_length": 1135, "token_count": 481}}
{"text": "H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. 73 Propanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C3H7OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C3H7(aq) +H2O(l) CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2 CH2CH3(aq) +H2O(l) (j)Oxidation Experiment Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7788510911424904, "ocr_used": true, "chunk_length": 984, "token_count": 339}}
{"text": "Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8256261163920707, "ocr_used": true, "chunk_length": 1436, "token_count": 440}}
{"text": "pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8006321120963271, "ocr_used": true, "chunk_length": 1373, "token_count": 453}}
{"text": "They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8070307196655425, "ocr_used": true, "chunk_length": 1534, "token_count": 503}}
{"text": "Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”. (k)Hydrolysis /Hydration and Dehydration I. Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e. Alkenes + Water - H3PO4 catalyst-> Alkanol Examples (i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol Ethene + water ---60 atm/300oC/ H3PO4 --> Ethanol H2C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2OH(l) This is the main method of producing large quantities of ethanol instead of fermentation (ii) Propene + water ---60 atm/300oC/ H3PO4 --> Propanol CH3C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2OH(l) (iii) Butene + water ---60 atm/300oC/ H3PO4 --> Butanol CH3 CH2 C=CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2 CH2OH(l) II.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7501876430205949, "ocr_used": true, "chunk_length": 1425, "token_count": 486}}
{"text": "Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e. Alkenes + Water - H3PO4 catalyst-> Alkanol Examples (i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol Ethene + water ---60 atm/300oC/ H3PO4 --> Ethanol H2C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2OH(l) This is the main method of producing large quantities of ethanol instead of fermentation (ii) Propene + water ---60 atm/300oC/ H3PO4 --> Propanol CH3C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2OH(l) (iii) Butene + water ---60 atm/300oC/ H3PO4 --> Butanol CH3 CH2 C=CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2 CH2OH(l) II. Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances. Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e Alkanol --Conc. H2 SO4/180oC--> Alkene + Water Examples\n75 1. At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.749063420380915, "ocr_used": true, "chunk_length": 1389, "token_count": 462}}
{"text": "At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene. Propanol ---180oC/ H2SO4 --> Propene + Water CH3 CH2 CH2OH(l) --180oC/ H2SO4 --> CH3CH =CH2 (g) + H2O(l) 3. Butanol undergoes dehydration to form Butene. Butanol ---180oC/ H2SO4 --> Butene + Water CH3 CH2 CH2CH2OH(l) --180oC/ H2SO4 --> CH3 CH2C =CH2 (g) + H2O(l) 3. Pentanol undergoes dehydration to form Pentene. Pentanol ---180oC/ H2SO4 --> Pentene + Water CH3 CH2 CH2 CH2 CH2OH(l)--180oC/ H2SO4-->CH3 CH2 CH2C =CH2 (g)+H2O(l) (l)Similarities of alkanols with Hydrocarbons I. Similarity with alkanes Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7426998686889832, "ocr_used": true, "chunk_length": 1124, "token_count": 397}}
{"text": "This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. CH2 CH2OH(l) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) CH2 CH2OH(l) + 2O2(g) -Limited air-> 2CO (g) + 3H2 O(l) CH3 CH3(g) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) 2CH3 CH3(g) + 5O2(g) -Limited air-> 4CO (g) + 6H2 O(l) II. Similarity with alkenes/alkynes Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond: (i)decolorize acidified KMnO4 (ii)turns Orange acidified K2Cr2O7 to green. Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH). Alkenes are oxidized to alkanols with duo/double functional groups. Examples 1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid. 76 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol. Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7856892253161516, "ocr_used": true, "chunk_length": 1370, "token_count": 508}}
{"text": "Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water. Alkenes decolorizes bromine and chlorine water to form halogenoalkanols Example When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol. Ethene + Bromine water -> Bromoethanol. H2C=CH2 + HOBr -> BrCH2 -CH2OH IV. Differences in melting and boiling point with Hydrocarbons Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne) This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding. Two alkanol molecules form a dimer joined by hydrogen bonding. Example In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen. This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen. Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer. H H H H C C O H H H H H O C C H Hydrogen bonds Covalent bonds\n77 H H Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting. E.USES OF SOME ALKANOLS (a)Methanol is used as industrial alcohol and making methylated spirit (b)Ethanol is used: 1. as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8898847795163585, "ocr_used": true, "chunk_length": 1850, "token_count": 482}}
{"text": "as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION. Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:\n78 Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where: (i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows. (ii) the members have R-COOH/R C-O-H as the functional group.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.88252352736145, "ocr_used": true, "chunk_length": 755, "token_count": 210}}
{"text": "B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION. Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:\n78 Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where: (i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows. (ii) the members have R-COOH/R C-O-H as the functional group. n General /molecular formular Structural formula IUPAC name 0 HCOOH H – C –O - H │ O Methanoic acid 1 CH3 COOH H H – C – C – O - H │ H O Ethanoic acid 2 CH3 CH2 COOH C2 H5 COOH H H H-C – C – C – O – H H H O Propanoic acid 3 CH3 CH2 CH2 COOH C3 H7 COOH H H H H- C - C – C – C – O – H H H H O Butanoic acid 4 CH3CH2CH2CH2 COOH C4 H9 COOH H H H H H - C – C - C – C – C – O – H H H H H O Pentanoic acid 5 CH3CH2 CH2CH2CH2 COOH C5 H11 COOH H H H H H H C - C – C - C – C – C – O – H H H H H H O Hexanoic acid 6 CH3CH2 CH2 CH2CH2CH2 COOH C6 H13 COOH H H H H H H H C C - C – C - C – C – C – O – H H H H H H H O Pentanoic acid\n79 O (iii)they have the same general formula represented by R-COOH where R is an alkyl group. (iv)each member differ by –CH2- group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting point. (vi)they show similar and gradual change in their chemical properties.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8154090597545018, "ocr_used": true, "chunk_length": 1459, "token_count": 510}}
{"text": "(v)they show a similar and gradual change in their physical properties e.g. boiling and melting point. (vi)they show similar and gradual change in their chemical properties. (vii) since they are acids they show similar properties with mineral acids. (B) ISOMERS OF ALKANOIC ACIDS. Alkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines (i)Like alkanes. identify the longest carbon chain to be the parent name. (ii)Identify the position of the -C-O-H functional group to give it the smallest O /lowest position. (iii)Identify the type and position of the side group branches. Practice examples on isomers of alkanoic acids 1.Isomers of butanoic acid C3H7COOH CH3 CH2 CH2 COOH Butan-1-oic acid CH3 H2C C COOH 2-methylpropan-1-oic acid 2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does. 2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n80 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8127294331446147, "ocr_used": true, "chunk_length": 1506, "token_count": 498}}
{"text": "(iii)Identify the type and position of the side group branches. Practice examples on isomers of alkanoic acids 1.Isomers of butanoic acid C3H7COOH CH3 CH2 CH2 COOH Butan-1-oic acid CH3 H2C C COOH 2-methylpropan-1-oic acid 2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does. 2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n80 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS. In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 81 The oxidation converts the alkanol first to an alkanal the alkanoic acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7788775498508012, "ocr_used": true, "chunk_length": 1153, "token_count": 429}}
{"text": "2.Isomers of pentanoic acid C4H9COOH CH3CH2CH2CH2 COOH pentan-1-oic acid CH3 CH3CH2CH COOH 2-methylbutan-1-oic acid\n80 CH3 H3C C COOH 2,2-dimethylpropan-1-oic acid CH3 3.Ethan-1,2-dioic acid O O HOOC- COOH // H - O – C - C – O – H 4.Propan-1,3-dioic acid O H O HOOC- CH2COOH // H - O – C – C - C – O – H H 5.Butan-1,4-dioic acid O H H O HOOC CH2 CH2 COOH H- O – C – C - C – C –O – H H H 6.2,2-dichloroethan-1,2-dioic acid HOOCCHCl2 Cl H – O - C – C – Cl O H (C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS. In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 81 The oxidation converts the alkanol first to an alkanal the alkanoic acid. NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7 General equation: R- CH2 – OH + [O] --H+/KMnO4--> R- CH –O + H2O(l) (alkanol) (alkanal) R- CH – O + [O] --H+/KMnO4--> R- C –OOH (alkanal) (alkanoic acid) Examples 1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7574202496532594, "ocr_used": true, "chunk_length": 1064, "token_count": 432}}
{"text": "In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming. 81 The oxidation converts the alkanol first to an alkanal the alkanoic acid. NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7 General equation: R- CH2 – OH + [O] --H+/KMnO4--> R- CH –O + H2O(l) (alkanol) (alkanal) R- CH – O + [O] --H+/KMnO4--> R- C –OOH (alkanal) (alkanoic acid) Examples 1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid . CH3- CH2 – OH + [O] --H+/KMnO4--> CH3- CH –O + H2O(l) (ethanol) (ethanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (ethanal) (ethanoic acid) 2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid CH3- CH2 CH2 – OH + [O] --H+/KMnO4--> CH3- CH2 CH –O + H2O(l) (propanol) (propanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (propanal) (propanoic acid) Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from: (a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e. Alkenes + Steam/water -- H2PO4 Catalyst--> Alkanol The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7981476306439379, "ocr_used": true, "chunk_length": 1354, "token_count": 481}}
{"text": "CH3- CH2 – OH + [O] --H+/KMnO4--> CH3- CH –O + H2O(l) (ethanol) (ethanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (ethanal) (ethanoic acid) 2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid CH3- CH2 CH2 – OH + [O] --H+/KMnO4--> CH3- CH2 CH –O + H2O(l) (propanol) (propanal) CH3- CH – O + [O] --H+/KMnO4--> CH3- C –OOH (propanal) (propanoic acid) Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from: (a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e. Alkenes + Steam/water -- H2PO4 Catalyst--> Alkanol The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid. Alkanol + Air -- MnSO4 Catalyst/5 atm pressure--> Alkanoic acid Example Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol. 82 CH2=CH2 + H2O -> CH3 CH2OH (Ethene) (Ethanol) This is the industrial large scale method of manufacturing ethanol Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2OH + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanol) (Ethanoic acid) (b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8040654513932025, "ocr_used": true, "chunk_length": 1503, "token_count": 493}}
{"text": "Alkanol + Air -- MnSO4 Catalyst/5 atm pressure--> Alkanoic acid Example Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol. 82 CH2=CH2 + H2O -> CH3 CH2OH (Ethene) (Ethanol) This is the industrial large scale method of manufacturing ethanol Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2OH + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanol) (Ethanoic acid) (b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals. Alkyne + Water -- Mercury(II)sulphate(VI)catalyst--> Alkanal The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid. Alkanal + air/oxygen -- Manganese(II)sulphate(VI)catalyst--> Alkanoic acid Example Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal. CH = CH + H2O --HgSO4--> CH3 CH2O (Ethyne) (Ethanal) This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas. Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2O + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanal) (Oxygen from air) (Ethanoic acid) (D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8429242989139246, "ocr_used": true, "chunk_length": 1592, "token_count": 464}}
{"text": "CH = CH + H2O --HgSO4--> CH3 CH2O (Ethyne) (Ethanal) This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas. Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid. CH3 CH2O + [O] -- MnSO4 Catalyst/5 atm pressure--> CH3 COOH (Ethanal) (Oxygen from air) (Ethanoic acid) (D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS. I.Physical properties of alkanoic acids\n83 The table below shows some physical properties of alkanoic acids Alkanol Melting point(oC) Boiling point(oC) Density(gcm-3) Solubility in water Methanoic acid 18.4 101 1.22 soluble Ethanoic acid 16.6 118 1.05 soluble Propanoic acid -2.8 141 0.992 soluble Butanoic acid -8.0 164 0.964 soluble Pentanoic acid -9.0 187 0.939 Slightly soluble Hexanoic acid -11 205 0.927 Slightly soluble Heptanoic acid -3 223 0.920 Slightly soluble Octanoic acid 11 239 0.910 Slightly soluble Nonanoic acid 16 253 0.907 Slightly soluble Decanoic acid 31 269 0.905 Slightly soluble From the table note the following: (i) Melting and boiling point decrease as the carbon chain increases due to increase in intermolecular forces of attraction between the molecules requiring more energy to separate the molecules. (ii) The density decreases as the carbon chain increases as the intermolecular forces of attraction increases between the molecules making the molecule very close reducing their volume in unit mass. (iii) Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain. (iv) Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e..", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8081076920924115, "ocr_used": true, "chunk_length": 1743, "token_count": 481}}
{"text": "(iii) Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain. (iv) Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e.. 84 jgthungu@gmail.com104Hydrogen bondscovalent bondsR1C O δ-…….….…H δ+O δO δH δ+……..….O δCR2R1 and 2 are extensions of the molecule.For ethanoic acid the extension is made up of CH3 – to make the structure; jgthungu@gmail.com105For ethanoic acid the extension is made up of CH3 – to make the structure;Hydrogen bondscovalent bonds CH3CO δ-…………… H δ+O δO δH δ+…………O δCCH3Ethanoic acid has a higher melting/boiling point than ethanol .This is because ethanoic acid has two/more hydrogen bond than ethanol. II Chemical properties of alkanoic acids The following experiments shows the main chemical properties of ethanoic (alkanoic) acid. (a)Effect on litmus papers Experiment Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid. Sample observations Solution/acid Observations/effect on litmus papers Inference Ethanoic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion\n85 Succinic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Citric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Oxalic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Tartaric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Nitric(V)acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Explanation All acidic solutions contains H+/H3O+(aq) ions.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8747236551215919, "ocr_used": true, "chunk_length": 1725, "token_count": 484}}
{"text": "(a)Effect on litmus papers Experiment Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid. Sample observations Solution/acid Observations/effect on litmus papers Inference Ethanoic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion\n85 Succinic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Citric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Oxalic acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Tartaric acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Nitric(V)acid Blue litmus paper turn red Red litmus paper remain red H3O+/H+(aq)ion Explanation All acidic solutions contains H+/H3O+(aq) ions. The H+ /H3O+ (aq) ions is responsible for turning blue litmus paper/solution to red (b)pH Experiment Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid. Sample observations Solution/acid pH Inference Ethanoic acid 4/5/6 Weakly acidic Succinic acid 4/5/6 Weakly acidic Citric acid 4/5/6 Weakly acidic Oxalic acid 4/5/6 Weakly acidic Tartaric acid 4/5/6 Weakly acidic Sulphuric(VI)acid 1/2/3 Strongly acidic Explanations Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8567421277306364, "ocr_used": true, "chunk_length": 1635, "token_count": 459}}
{"text": "Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid. Sample observations Solution/acid pH Inference Ethanoic acid 4/5/6 Weakly acidic Succinic acid 4/5/6 Weakly acidic Citric acid 4/5/6 Weakly acidic Oxalic acid 4/5/6 Weakly acidic Tartaric acid 4/5/6 Weakly acidic Sulphuric(VI)acid 1/2/3 Strongly acidic Explanations Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water. All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO- Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H+ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e Examples 1. CH3COOH(aq) CH3COO-(aq) + H+(aq)\n86 (ethanoic acid) (ethanoate ion) (few H+ ion) 2. CH3 CH2COOH(aq) CH3 CH2COO-(aq) + H+(aq) (propanoic acid) (propanoate ion) (few H+ ion) 3. CH3 CH2 CH2COOH(aq) CH3 CH2 CH2COO-(aq) + H+(aq) (Butanoic acid) (butanoate ion) (few H+ ion) 4. HOOH(aq) HOO-(aq) + H+(aq) (methanoic acid) (methanoate ion) (few H+ ion) 5. H2 SO4 (aq) SO42- (aq) + 2H+(aq) (sulphuric(VI) acid) (sulphate(VI) ion) (many H+ ion) 6.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7932554389815136, "ocr_used": true, "chunk_length": 1412, "token_count": 495}}
{"text": "Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Inference Ethanoic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Succinic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Citric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Oxalic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Tartaric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion\n87 Nitric(V)acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that burn with “pop” sound/explosion H3O+/H+(aq)ion Explanation Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group -COOH is /are displaced and not in the alkyl hydrocarbon chain. Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e. Examples 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8850063999109579, "ocr_used": true, "chunk_length": 1510, "token_count": 424}}
{"text": "Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e. Examples 1. For a monovalent metal with monobasic acid 2R – COOH + 2M -> 2R- COOM + 2H2(g) 2.For a divalent metal with monobasic acid 2R – COOH + M -> (R- COO) 2M + H2(g) 3.For a divalent metal with dibasic acid HOOC-R-COOH+ M -> MOOC-R-COOM + H2(g) 4.For a monovalent metal with dibasic acid HOOC-R-COOH+ 2M -> MOOC-R-COOM + H2(g) 5 For mineral acids (i)Sulphuric(VI)acid is a dibasic acid H2 SO4 (aq) + 2M -> M2 SO4 (aq) + H2(g) H2 SO4 (aq) + M -> MSO4 (aq) + H2(g) (ii)Nitric(V) and hydrochloric acid are monobasic acid HNO3 (aq) + 2M -> 2MNO3 (aq) + H2(g) HNO3 (aq) + M -> M(NO3 ) 2 (aq) + H2(g) Examples 1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas. Caution: This reaction is explosive. CH3COOH (aq) + Na(s) -> CH3COONa (aq) + H2(g) (Ethanoic acid) (Sodium ethanoate) 2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas. 2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)\n88 (Ethanoic acid) (Calcium ethanoate) 3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7409386088237264, "ocr_used": true, "chunk_length": 1149, "token_count": 459}}
{"text": "hydrogen gas. 2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)\n88 (Ethanoic acid) (Calcium ethanoate) 3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas. HOOC-COOH+ 2Na -> NaOOC - COONa + H2(g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate) Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate. 4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas. HOOC-R-COOH+ Mg -> ( OOC - COO) Mg + H2(g) (ethan-1,2-dioic acid) (magnesium ethan-1,2-dioate) 5.Magnesium reacts with (i)Sulphuric(VI)acid to form Magnesium sulphate(VI) H2 SO4 (aq) + Mg -> MgSO4 (aq) + H2(g) (ii)Nitric(V) and hydrochloric acid are monobasic acid 2HNO3 (aq) + Mg -> M(NO3 ) 2 (aq) + H2(g) (d)Reaction with hydrogen carbonates and carbonates Experiment Place about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7763242610477027, "ocr_used": true, "chunk_length": 1139, "token_count": 420}}
{"text": "Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Inference Ethanoic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Succinic acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Citric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Oxalic acid (i)effervescence, fizzing, bubbles H3O+/H+(aq)ion\n89 (ii)colourless gas produced that forms a white precipitate with lime water Tartaric acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion Nitric(V)acid (i)effervescence, fizzing, bubbles (ii)colourless gas produced that forms a white precipitate with lime water H3O+/H+(aq)ion All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water. Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint. Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water. Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide Examples 1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8895431612412745, "ocr_used": true, "chunk_length": 1749, "token_count": 468}}
{"text": "Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water. Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide Examples 1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + H2O(l) + CO2 (g) (Ethanoic acid) (Sodium ethanoate) 2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. 2CH3COOH (aq) + Na2CO3 (s) -> 2CH3COONa (aq) + H2O(l) + CO2 (g) (Ethanoic acid) (Sodium ethanoate) 3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. HOOC-COOH+ Na2CO3 (s) -> NaOOC - COONa + H2O(l) + CO2 (g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate) 4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas. HOOC-COOH+ 2NaHCO3 (s) -> NaOOC - COONa + H2O(l) + 2CO2 (g) (ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)\n90 (e)Esterification Experiment Place 4cm3 of ethanol acid in a boiling tube. Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame. Pour the mixture into a beaker containing 50cm3 of water. Smell the products.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8078353079010604, "ocr_used": true, "chunk_length": 1436, "token_count": 495}}
{"text": "Warm/heat gently on Bunsen flame. Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid. Sample observations Solution/acid Observations Ethanoic acid Sweet fruity smell Succinic acid Sweet fruity smell Citric acid Sweet fruity smell Oxalic acid Sweet fruity smell Tartaric acid Sweet fruity smell Dilute sulphuric(VI)acid No sweet fruity smell Explanation Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids. Alkanol + Alkanoic acids -> Ester + water Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g. Ethanol + Ethanoic acid -> Ethylethanoate + Water Ethanol + Propanoic acid -> Ethylpropanoate + Water Ethanol + Methanoic acid -> Ethylmethanoate + Water Ethanol + butanoic acid -> Ethylbutanoate + Water Propanol + Ethanoic acid -> Propylethanoate + Water Methanol + Ethanoic acid -> Methyethanoate + Water Methanol + Decanoic acid -> Methyldecanoate + Water Decanol + Methanoic acid -> Decylmethanoate + Water\n91 During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol. R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O Examples 1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.885859885128315, "ocr_used": true, "chunk_length": 1777, "token_count": 496}}
{"text": "H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. Propanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C3H7OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C3H7(aq) +H2O(l) CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)\n92 C. DETERGENTS Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to: (i)dissolve substances which water can not e.g grease ,oil, fat (ii)be washed away after cleaning. There are two types of detergents: (a)Soapy detergents (b)Soapless detergents (a) SOAPY DETERGENTS Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e. Sodium hydroxide + octadecanoic acid -> Sodium octadecanoate + water NaOH(aq) + CH3 (CH2) 16 COOH(aq) -> CH3 (CH2) 16 COO – Na+ (aq) +H2 O(l) Commonly ,soap can thus be represented ; R- COO – Na+ where; R is a long chain alkyl group and -COO – Na+ is the alkanoate ion. In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8232105053553838, "ocr_used": true, "chunk_length": 1539, "token_count": 499}}
{"text": "In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced. Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol Fats/Oils are esters with fatty acids and glycerol parts in their structure; C17H35COOCH2 C17H35COOCH C17H35COOCH2\n93 When boiled with concentrated sodium hydroxide solution NaOH; (i)NaOH ionizes/dissociates into Na+ and OH- ions (ii)fat/oil split into three C17H35COO- and one CH2 CH CH2 (iii) the three Na+ combine with the three C17H35COO- to form the salt C17H35COO- Na+ (iv)the three OH-ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol) C17H35COOCH2 CH2OH C17H35COOCH +NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol Generally: CnH2n+1COOCH2 CH2OH CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO- Na+ + CHOH CnH2n+1COOCH2 CH2OH Ester Alkali Soap glycerol R - COOCH2 CH2OH R - COOCH +NaOH -> 3R-COO- Na+ + CHOH R- COOCH2 CH2OH Ester Alkali Soap glycerol During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out. The soap is then added colouring agents ,perfumes and herbs of choice.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8167330278216012, "ocr_used": true, "chunk_length": 1479, "token_count": 500}}
{"text": "Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol Fats/Oils are esters with fatty acids and glycerol parts in their structure; C17H35COOCH2 C17H35COOCH C17H35COOCH2\n93 When boiled with concentrated sodium hydroxide solution NaOH; (i)NaOH ionizes/dissociates into Na+ and OH- ions (ii)fat/oil split into three C17H35COO- and one CH2 CH CH2 (iii) the three Na+ combine with the three C17H35COO- to form the salt C17H35COO- Na+ (iv)the three OH-ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol) C17H35COOCH2 CH2OH C17H35COOCH +NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol Generally: CnH2n+1COOCH2 CH2OH CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO- Na+ + CHOH CnH2n+1COOCH2 CH2OH Ester Alkali Soap glycerol R - COOCH2 CH2OH R - COOCH +NaOH -> 3R-COO- Na+ + CHOH R- COOCH2 CH2OH Ester Alkali Soap glycerol During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out. The soap is then added colouring agents ,perfumes and herbs of choice. School laboratory preparation of soap Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8005621908769046, "ocr_used": true, "chunk_length": 1417, "token_count": 498}}
{"text": "The soap is then added colouring agents ,perfumes and herbs of choice. School laboratory preparation of soap Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of\n94 /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve. The action of soap Soapy detergents: (i)act by reducing the surface tension of water by forming a thin layer on top of the water. (ii)is made of a non-polar alkyl /hydrocarbon tail and a polar -COO-Na+ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar -COO-Na+ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water. Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is removed from the garment by rinsing with fresh water.The repulsion of the soap head prevent /ensure the droplets do not mix.Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules. Advantages and disadvantages of using soapy detergents Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution. Soapy detergents have the diadvatage in that: (i)they are made from fat and oils which are better eaten as food than make soap.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9014974460544147, "ocr_used": true, "chunk_length": 2120, "token_count": 510}}
{"text": "Advantages and disadvantages of using soapy detergents Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution. Soapy detergents have the diadvatage in that: (i)they are made from fat and oils which are better eaten as food than make soap. (ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate formed when soap reacts with Ca2+ and Mg2+ present in hard water. Chemical equation 2C17H35COO- Na+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2Na+(aq) (insoluble Calcium octadecanote/scum) 2C17H35COO- Na+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2Na+(aq) (insoluble Magnesium octadecanote/scum) This causes wastage of soap. Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive. (b)SOAPLESS DETERGENTS\n95 Soapless detergent usually called detergent is a long chain salt fromed from byproducts of fractional distillation of crude oil.Commonly used soaps include: (i)washing agents (ii)toothpaste (iii)emulsifiers/wetting agents/shampoo Soapless detergents are derived from reacting: (i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8683746536297164, "ocr_used": true, "chunk_length": 1645, "token_count": 488}}
{"text": "Potassium is more expensive than sodium and thus its soap is also more expensive. (b)SOAPLESS DETERGENTS\n95 Soapless detergent usually called detergent is a long chain salt fromed from byproducts of fractional distillation of crude oil.Commonly used soaps include: (i)washing agents (ii)toothpaste (iii)emulsifiers/wetting agents/shampoo Soapless detergents are derived from reacting: (i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent. alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water sulphate(VI) hydroxide alkyl hydrogen sulphate(VI) R –O-SO3H + NaOH -> R –O-SO3- Na+ + H2O Example Step I : Reaction of Octadecanol with Conc.H2SO4 C17H35CH2OH (aq) + H2SO4 -> C17H35CH2-O- SO3- H+ (aq) + H2O (l) octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water Step II: Neutralization by an alkali C17H35CH2-O- SO3- H+ (aq) + NaOH -> C17H35CH2-O- SO3- Na+ (aq) + H2O (l) Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water sulphate(VI) hydroxide hydrogen sulphate(VI) School laboratory preparation of soapless detergent Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8484350028810299, "ocr_used": true, "chunk_length": 1559, "token_count": 488}}
{"text": "Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI) Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water R –OH + H2SO4 -> R –O-SO3H + H2O (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI) Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent. alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water sulphate(VI) hydroxide alkyl hydrogen sulphate(VI) R –O-SO3H + NaOH -> R –O-SO3- Na+ + H2O Example Step I : Reaction of Octadecanol with Conc.H2SO4 C17H35CH2OH (aq) + H2SO4 -> C17H35CH2-O- SO3- H+ (aq) + H2O (l) octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water Step II: Neutralization by an alkali C17H35CH2-O- SO3- H+ (aq) + NaOH -> C17H35CH2-O- SO3- Na+ (aq) + H2O (l) Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water sulphate(VI) hydroxide hydrogen sulphate(VI) School laboratory preparation of soapless detergent Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water. Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent. The action of soapless detergents\n96 The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8412001791312136, "ocr_used": true, "chunk_length": 1508, "token_count": 476}}
{"text": "Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent. The action of soapless detergents\n96 The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e. vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COO-Na+ vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO3- Na+ (long hydrophobic /non-polar alkyl tail) (hydrophilic/polar/ionic head) The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water. The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching. The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment. The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away. Advantages and disadvantages of using soapless detergents Soapless detergents are non-biodegradable unlike soapy detergents. They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death. Soapless detergents have the advantage in that they: (i)do not form scum with hard water. (ii)are cheap to manufacture/buying (iii)are made from petroleum products but soapis made from fats/oil for human consumption. Sample revision questions 1. Study the scheme below Fat/oil KOH Boiling Sodium Chloride Filtration Filtrate Y Residue X\n97 (a)Identify the process Saponification (b)Fats and oils are esters.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.9121821790623668, "ocr_used": true, "chunk_length": 1793, "token_count": 468}}
{"text": "(ii)are cheap to manufacture/buying (iii)are made from petroleum products but soapis made from fats/oil for human consumption. Sample revision questions 1. Study the scheme below Fat/oil KOH Boiling Sodium Chloride Filtration Filtrate Y Residue X\n97 (a)Identify the process Saponification (b)Fats and oils are esters. Write the formula of the a common structure of ester C17H35COOCH2 C17H35COOCH C17H35COOCH2 (c)Write a balanced equation for the reaction taking place during boiling C17H35COOCH2 CH2OH C17H35COOCH +3NaOH -> 3 C17H35COO- Na+ + CHOH C17H35COOCH2 CH2OH Ester Alkali Soap glycerol (d)Give the IUPAC name of: (i)Residue X Potassium octadecanoate (ii)Filtrate Y Propan-1,2,3-triol (e)Give one use of fitrate Y Making paint (f)What is the function of sodium chloride To reduce the solubility of the soap hence helping in precipitating it out (g)Explain how residue X helps in washing. Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat Has a polar /ionic hydrophilic head that dissolves in water. 98 From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment. (h)State one: (i)advantage of continued use of residue X on the environment Is biodegradable and thus do not pollute the environment (ii)disadvantage of using residue X Uses fat/oil during preparation/manufacture which are better used for human consumption. (i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below. Water sample A B C Drops of residue X 15 2 15 Drops of residue X in boiled water 2 2 15 (i)State and explain which sample of water is: I. Soft Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating. II. Permanent hard Sample C .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8724771084171516, "ocr_used": true, "chunk_length": 1822, "token_count": 503}}
{"text": "Soft Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating. II. Permanent hard Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water. III. Temporary hard Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water. (ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n99 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8052749934629131, "ocr_used": true, "chunk_length": 1043, "token_count": 362}}
{"text": "Boiling remove temporary hardness of water. (ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n99 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7451266159531788, "ocr_used": true, "chunk_length": 1412, "token_count": 576}}
{"text": "(ii)Write the equation for the reaction at water sample C. Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n99 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.766888657429326, "ocr_used": true, "chunk_length": 1611, "token_count": 623}}
{"text": "Chemical equation 2C17H35COO- K+ (aq) + CaSO4(aq) -> (C17H35COO- )Ca2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum)\n99 Chemical equation 2C17H35COO- K+ (aq) + MgSO4(aq) -> (C17H35COO- )Mg2+ (s) + K2SO4(aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iii)Write the equation for the reaction at water sample A before boiling. Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7624753848907985, "ocr_used": true, "chunk_length": 1568, "token_count": 615}}
{"text": "Chemical equation 2C17H35COO- K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO- )Ca2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Ca2+(aq) -> (C17H35COO- )Ca2+ (s) + 2K+(aq) (insoluble Calcium octadecanote/scum) Chemical equation 2C17H35COO- K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO- )Mg2+ (s) + 2KHCO3 (aq) (insoluble Calcium octadecanote/scum) Ionic equation 2C17H35COO- K+ (aq) + Mg2+(aq) -> (C17H35COO- )Mg2+ (s) + 2K+(aq) (insoluble Magnesium octadecanote/scum) (iv)Explain how water becomes hard Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness. (v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc. H2SO4 Ice cold water Brown solid A 6M sodium hydroxide Substance B\n100 (a)Identify : (i)brown solid A Alkyl hydrogen sulphate(VI) (ii)substance B Sodium alkyl hydrogen sulphate(VI) (b)Write a general formula of: (i)Substance A.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8056866106634666, "ocr_used": true, "chunk_length": 1228, "token_count": 432}}
{"text": "(v)State two useful benefits of hard water -Used in bone and teeth formation -Coral polyps use hard water to form coral reefs -Snails use hard water to make their shells 2.Study the scheme below and use it to answer the questions that follow. Olive oil Conc. H2SO4 Ice cold water Brown solid A 6M sodium hydroxide Substance B\n100 (a)Identify : (i)brown solid A Alkyl hydrogen sulphate(VI) (ii)substance B Sodium alkyl hydrogen sulphate(VI) (b)Write a general formula of: (i)Substance A. O R-O-S O3 H // R- O - S - O - H O (ii)Substance B O R-O-S O3 - Na+ R- O - S - O - Na+ O (c)State one (i) advantage of continued use of substance B -Does not form scum with hard water -Is cheap to make -Does not use food for human as a raw material. (ii)disadvantage of continued use of substance B. Is non-biodegradable therefore do not pollute the environment (d)Explain the action of B during washing. Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat. Has a polar/ionic hydrophilic head that dissolves in water Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment. (e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B. 101 Product A Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI) H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O (f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8596048751871807, "ocr_used": true, "chunk_length": 1688, "token_count": 480}}
{"text": "(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B. 101 Product A Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI) H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O (f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation. Product A Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O 3.Below is part of a detergent H3C – (CH2 )16 – O - SO3 - K + (a)Write the formular of the polar and non-polar end Polar end H3C – (CH2 )16 – Non-polar end – O - SO3 - K + (b)Is the molecule a soapy or saopless detergent? Soapless detergent (c)State one advantage of using the above detergent -does not form scum with hard water -is cheap to manufacture 4.The structure of a detergent is H H H H H H H H H H H H H\n102 H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO-Na+ H H H H H H H H H H H H H a) Write the molecular formula of the detergent. (1mk) CH3(CH2)12COO-Na+ b) What type of detergent is represented by the formula?", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7866417377779198, "ocr_used": true, "chunk_length": 1401, "token_count": 489}}
{"text": "Product A Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3-Na+ + H2O 3.Below is part of a detergent H3C – (CH2 )16 – O - SO3 - K + (a)Write the formular of the polar and non-polar end Polar end H3C – (CH2 )16 – Non-polar end – O - SO3 - K + (b)Is the molecule a soapy or saopless detergent? Soapless detergent (c)State one advantage of using the above detergent -does not form scum with hard water -is cheap to manufacture 4.The structure of a detergent is H H H H H H H H H H H H H\n102 H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO-Na+ H H H H H H H H H H H H H a) Write the molecular formula of the detergent. (1mk) CH3(CH2)12COO-Na+ b) What type of detergent is represented by the formula? (1mk) Soapy detergent c) When this type of detergent is used to wash linen in hard water, spots (marks) are left on the linen. Write the formula of the substance responsible for the spots (CH3(CH2)12COO-)2Ca2+ / CH3(CH2)12COO-)2Mg2+ D. POLYMERS AND FIBRES Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8067950398225627, "ocr_used": true, "chunk_length": 1417, "token_count": 459}}
{"text": "POLYMERS AND FIBRES Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization. Polymers and fibres are either: (a)Natural polymers and fibres (b)Synthetic polymers and fibres Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include: -proteins/polypeptides making amino acids in animals -cellulose that make cotton,wool,paper and silk -Starch that come from glucose -Fats and oils -Rubber from latex in rubber trees. Synthetic polymers and fibres are man-made. They include: -polyethene\n103 -polychloroethene -polyphenylethene(polystyrene) -Terylene(Dacron) -Nylon-6,6 -Perspex(artificial glass) Synthetic polymers and fibres have the following characteristic advantages over natural polymers 1. They are light and portable 2. They are easy to manufacture. 3. They can easily be molded into shape of choice. 4. They are resistant to corrosion, water, air , acids, bases and salts. 5. They are comparatively cheap, affordable, colourful and aesthetic Synthetic polymers and fibres however have the following disadvantages over natural polymers 1. They are non-biodegradable and hence cause environmental pollution during disposal 2. They give out highly poisonous gases when burnt like chlorine/carbon(II)oxide 3. Some on burning produce Carbon(IV)oxide. Carbon(IV)oxide is a green house gas that cause global warming. 4. Compared to some metals, they are poor conductors of heat,electricity and have lower tensile strength. 5. To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used: 1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment. 2.Production of biodegradable synthetic polymers and fibres that rot away.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8989061110164841, "ocr_used": true, "chunk_length": 2052, "token_count": 477}}
{"text": "To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used: 1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment. 2.Production of biodegradable synthetic polymers and fibres that rot away. There are two types of polymerization: (a)addition polymerization (b)condensation polymerization (a)addition polymerization Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. 104 Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n105 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8925940067106329, "ocr_used": true, "chunk_length": 2010, "token_count": 498}}
{"text": "The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n105 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8644319520816969, "ocr_used": true, "chunk_length": 1273, "token_count": 367}}
{"text": "Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n105 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8595866405022016, "ocr_used": true, "chunk_length": 1797, "token_count": 512}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons\n105 •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8531291895185862, "ocr_used": true, "chunk_length": 1641, "token_count": 480}}
{"text": "Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8675968992248062, "ocr_used": true, "chunk_length": 1204, "token_count": 328}}
{"text": "The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8662117151082358, "ocr_used": true, "chunk_length": 1627, "token_count": 458}}
{"text": "It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8468845559582575, "ocr_used": true, "chunk_length": 1738, "token_count": 528}}
{"text": "Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials\n106 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8466714151072356, "ocr_used": true, "chunk_length": 1732, "token_count": 527}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8472866597004528, "ocr_used": true, "chunk_length": 1710, "token_count": 520}}
{"text": "Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + …\n107 H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8264367816091955, "ocr_used": true, "chunk_length": 1740, "token_count": 557}}
{"text": "It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n109 104 The commercial name of polyphenylethene is polystyrene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8227110460585094, "ocr_used": true, "chunk_length": 1697, "token_count": 547}}
{"text": "PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8221041896830119, "ocr_used": true, "chunk_length": 1681, "token_count": 545}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + …\n108 (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8137574700616208, "ocr_used": true, "chunk_length": 1525, "token_count": 503}}
{"text": "Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)\n109 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8348055790363482, "ocr_used": true, "chunk_length": 1008, "token_count": 300}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:\n110 H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8273781317186608, "ocr_used": true, "chunk_length": 1704, "token_count": 506}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as:\n110 H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.827867744261507, "ocr_used": true, "chunk_length": 1696, "token_count": 512}}
{"text": "It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n111 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8806618261826182, "ocr_used": true, "chunk_length": 1125, "token_count": 332}}
{"text": "Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n111 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8537035867643344, "ocr_used": true, "chunk_length": 1776, "token_count": 551}}
{"text": "(ii)ceiling tiles (iii)clothe linings 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n111 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8509330782088174, "ocr_used": true, "chunk_length": 1711, "token_count": 534}}
{"text": "During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F\n111 C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8440429102241401, "ocr_used": true, "chunk_length": 1620, "token_count": 509}}
{"text": "polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples\n112 Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 5.Formation of rubber from Latex Natural rubber is obtained from rubber trees. During harvesting an incision is made on the rubber tree to produce a milky white substance called latex. Latex is a mixture of rubber and lots of water. The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-\n113 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8358984591830443, "ocr_used": true, "chunk_length": 1691, "token_count": 506}}
{"text": "The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus; H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n-\n113 H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher. During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer. H CH3 H H H CH3 H H - C - C - C - C - C - C - C - C - H S H H S H H CH3 S H H CH3 S H - C - C - C - C - C - C - C - C - H H H H H H Vulcanized rubber is used to make tyres, shoes and valves. 6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers\n114 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8338329940575331, "ocr_used": true, "chunk_length": 1677, "token_count": 494}}
{"text": "6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers\n114 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (b)Condensation polymerization Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water). Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds During condensation polymerization: (i)the two monomers are brought together by high pressure to reduce distance between them. (ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8690574714235753, "ocr_used": true, "chunk_length": 1822, "token_count": 449}}
{"text": "(ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H H- O - C – (CH2 ) 4 – C – O - H + H –N – (CH2) 6 – N – H\n115 (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H H H- O - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + H 2O . Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.829432730335274, "ocr_used": true, "chunk_length": 1567, "token_count": 451}}
{"text": "Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H Cl - C – (CH2 ) 4 – C – Cl + H –N – (CH2) 6 – N – H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O H H Cl - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + HCl . Polymer bond linkage The two monomers each has six carbon chain hence the name “nylon-6,6” The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets. 2.Formation of Terylene\n116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8422650501672242, "ocr_used": true, "chunk_length": 1600, "token_count": 456}}
{"text": "2.Formation of Terylene\n116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H- O - C – C6H5 – C – O - H + H –O – CH2 CH2 – O – H (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H- O - C – C6H5 – C – O – (CH2) 6 – N – H + H 2O . Polymer bond linkage of terylene\n117 Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol. Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring. The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen During the formation of Terylene (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene\n118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.839640671767772, "ocr_used": true, "chunk_length": 1679, "token_count": 497}}
{"text": "O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene\n118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits. 119 Practice questions Organic chemistry 1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture (i) Name and write the formula of the main products Name…………………………………. Formula…………………………………….. (ii) Which homologous series does the product named in (i) above belong? 2. The structure of the monomer phenyl ethene is given below:- a) Give the structure of the polymer formed when four of the monomers are added together b) Give the name of the polymer formed in (a) above 3. Explain the environmental effects of burning plastics in air as a disposal method 4. Write chemical equation to represent the effect of heat on ammonium carbonate 5. Sodium octadecanoate has a chemical formula CH3(CH2)6 COO-Na+, which is used as soap. HC = CH2 O\n120 Explain why a lot of soap is needed when washing with hard water 6. A natural polymer is made up of the monomer: (a) Write the structural formula of the repeat unit of the polymer (b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer were obtained. Determine the number of the monomer molecules in this polymer. (C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH\n121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain\n122 8.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8384188034188035, "ocr_used": true, "chunk_length": 1872, "token_count": 491}}
{"text": "(C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH\n121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain\n122 8. Study the polymer below and use it to answer the questions that follow: (a) Give the name of the monomer and draw its structures (b) Identify the type of polymerization that takes place (c) State one advantage of synthetic polymers 9. Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture of ethanol and pentane 10. (a) What is absolute ethanol? GLUCOSE SOLUTION CRUDE ETHANOL 95% ETHANOL ABSOLUTE ETHANOL G H H H H H C C C C\n123 (b) State two conditions required for process G to take place efficiently 11. (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time\n124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode\n125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8178815228291471, "ocr_used": true, "chunk_length": 1761, "token_count": 481}}
{"text": "(a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time\n124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode\n125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V\n126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U………………………………………………………..", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7887363353686775, "ocr_used": true, "chunk_length": 1674, "token_count": 490}}
{"text": "(ii) Write the equation of a reaction that occurs at the cathode\n125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V\n126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U……………………………………………………….. (II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K\n127 i) Name the following compounds:- I. Product T ………………………… II.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7938315409026798, "ocr_used": true, "chunk_length": 1418, "token_count": 430}}
{"text": "(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K\n127 i) Name the following compounds:- I. Product T ………………………… II. K ……… ii) State one common physical property of substance G iii) State the type of reaction that occurred in step J\n128 iv) Give one use of substance K v) Write an equation for the combustion of compound P vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12) 14. Study the scheme given below and answer the questions that follow:- H2 (g) Ni High temp Polymer Q Polymerization Compound P CH3CH2CH3 CH3CH2CH2ONa + H2 Na(s) Propan-l-ol Step I Propylethanoate CH3CH2COOH Solution T + CO2 (g) Step III Na2CO3(aq) Conc. H2SO4 180oC Step II\n129 (a) (i) Name compound P …………………………………………………………………… (ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3 (b) State one use of polymer Q (c) Name one oxidising agent that can be used in step II ………………………………….. (d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8080376643419485, "ocr_used": true, "chunk_length": 1695, "token_count": 493}}
{"text": "(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne. Explain your answer (h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour. 130 Calculate the minimum volume of air needed for the complete combustion of ethene (Air contains 20% by volume of oxygen) 15. (a) Study the schematic diagram below and answer the questions that follow:- (i) Identify the following: Substance Q .............................................................................................................. Substance R............................................................................................................... Gas P.......................................................................................................................... CH3CH2COOCH2CH2CH3 CH3CHCH2 CH3CH2CH2ONa + Gas P CH3CH2CH2OH X V HCl Step 5 Step 1 R Na H+ Step 3 Q + H2O MnO4 Step 4Ni H2\n131 (ii) Name: Step 1................................................................................................. Step 4................................................................................................. (iii) Draw the structural formula of the major product of step 5 (iv) State the condition and reagent in step 3 16. Study the flow chart below and answer the questions that follow (a) (i) Name the following organic compounds: M……………………………………………………………..…….. L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P\n132 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6592211672281957, "ocr_used": true, "chunk_length": 2091, "token_count": 492}}
{"text": "L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P\n132 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17. a) Give the names of the following compounds: i) CH3CH2CH2CH2OH …………………………………………………………………… ii) CH3CH2COOH ………………………………………………………………… iii) CH3C – O- CH2CH3 …………………………………………………………………… 18. Study the scheme given below and answer the questions that follow;\n133 n i) Name the reagents used in: Step I: ……………………………………………………………………… Step II …………………………………………………………………… Step III ……………………………………………………………………… ii) Write an equation to show products formed for the complete combustion of CH = CH iii) Explain one disadvantage of continued use of items made form the compound formed in step III 19. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3% i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11) ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH\n134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III\n135 21.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7401113227352181, "ocr_used": true, "chunk_length": 1661, "token_count": 466}}
{"text": "Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH\n134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III\n135 21. Give the IUPAC name for each of the following organic compounds; i) CH3 - CH - CH2 - CH3 OH ii)CH3 – CH – CH2 – CH2 - CH3 C2H5 iii)CH3COOCH2CH2CH3 22. The structure below represents a cleansing agent. O R – S – O-Na+ O a) State the type of cleansing agent represented above b) State one advantage and one disadvantage of using the above cleansing agent. 23. The structure below shows part of polymer .Use it to answer the questions that follow. CH3 CH3 CH3   \n136 ― CH - CH2 – CH- CH2 - CH – CH2 ― a) Derive the structure of the monomer b) Name the type of polymerization represented above 24. The flow chart below represents a series of reactions starting with ethanoic acid:- (a) Identify substances A and B (b) Name the process I 25. a) Write an equation showing how ammonium nitrate may be prepared starting with ammonia gas (b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of ammonia (H=1, N=14, O=16) 26. (a) What is meant by the term, esterification? Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O\n137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid\n138 28. The scheme below shows some reactions starting with ethanol.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8352679525131993, "ocr_used": true, "chunk_length": 1736, "token_count": 463}}
{"text": "Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O\n137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid\n138 28. The scheme below shows some reactions starting with ethanol. Study it and answer the questions that follow:- (i) Name and draw the structure of substance Q Q P CH3COONa Ethanol CH3CH2ONa C H2SO4(l) Cr2O7(aq) / H+(aq) Na(s) Step 2 Step 4 CH3CH2OH/H2SO4 Step 3 2-\n139 (ii) Give the names of the reactions that take place in steps 2 and 4 (iii) What reagent is necessary for reaction that takes place in step 3 29. Substances A and B are represented by the formulae ROH and RCOOH respectively. They belong to two different homologous series of organic compounds. If both A and B react with potassium metal: (a) Name the common product produced by both (b) State the observation made when each of the samples A and B are reacted with sodium hydrogen carbonate (i) A (ii) B 30. Below are structures of particles. Use it to answer questions that follow. In each case only electrons in the outermost energy level are shown key P = Proton N = Neutron X = Electron W U V 19P Z Y\n140 (a) Identify the particle which is an anion 31. Plastics and rubber are extensively used to cover electrical wires. (a) What term is used to describe plastic and rubbers used in this way? (b) Explain why plastics and rubbers are used this way\n141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8557330121659599, "ocr_used": true, "chunk_length": 1859, "token_count": 484}}
{"text": "(b) Explain why plastics and rubbers are used this way\n141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams. The halflife period of the isotope is 20days (a) Find the initial mass of the isotope (b) Give one application of radioactivity in agriculture 34. The structure below represents a polymer. Study and answer the questions that follow:- R Conc. R Cleaning agent X H H -C – C - n\n142 (i) Name the polymer above.................................................................................. (ii) Determine the value of n if giant molecule had relative molecular mass of 4956 35. RCOO-Na+ and RCH2OSO3-Na+ are two types of cleansing agents; i) Name the class of cleansing agents to which each belongs ii) Which one of these agents in (i) above would be more suitable when washing with water from the Indian ocean. Explain iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference in their bleaching properties 36. The formula given below represents a portion of a polymer H H H H C C C C O H O H n\n143 (a) Give the name of the polymer (b) Draw the structure of the monomer used to manufacture the polymer\n15.0.0 NITROGEN AND IT'S COMPOUNDS (30 LESSONS) Introduction to oxides of nitrogen Nirogen has a position in second period of group V in the modern periodic table. It has molecular formula N2. It has atomic number 7 and atomic weight 14.08 and its electronic configuration of 2,5. Besides combining with hydrogen and forming NH3 , nitrogen combines with oxygen in different ratios and forms five different oxides.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8420879691321357, "ocr_used": true, "chunk_length": 1791, "token_count": 422}}
{"text": "It has molecular formula N2. It has atomic number 7 and atomic weight 14.08 and its electronic configuration of 2,5. Besides combining with hydrogen and forming NH3 , nitrogen combines with oxygen in different ratios and forms five different oxides. The oxides of nitrogen and the details of the oxygen states of nitrogen and the N:O ratio can be presented in a tabular form as: Name Formula Oxidation state of N Ratio of N:O 1) Nitrogen Oxide or nitrous oxide NO +2 1:1 2) Nitrogen dioxide NO2 +4 1:2 3) Nitrous oxide (also called laughing gas) N2O +1 2:1 All of these oxides of nitrogen are gases, excepting NO and N2O the other oxides are brownish gases. Except the oxides of NO and N2O all the other oxides are acidic. NO and N2O are neutral. The other oxides are prepared in the laboratory using different methods characteristic to each oxide. Example of Oxides of Nitrogen (nitric Oxide - NO) Structure:\nLaboratory Preparation: The oxide is prepared in the laboratory by treating the metallic copper with a moderately concentarted nitric acid (1:1) at room temperature. The reaction is given as : 2Cu + 8 HNO3 ------> Cu(NO3)2 + 2NO + 4H2O The gas is collected by downward displacement of water.The apparatus used is Wolfe's apparatus. The purification is done by absorbing the NO gas in frashly prepared ferrous sulphate solution. Ferrous sulphate absorbs all the NO gas and forms Fe(H2O)5 NO and the solution becomes brown. On heating this solution pure Nitric Oxide is obatined. Physical Properties: NO is not a combustible gas. At high temperature around 1000°C it decomposes into N2 and O2 . 2NO = N2 + O2 at high temperature From the equation above we can see that once the decomposition starts 50% O2 gets evolved and this O2 supports combustion thus making the reaction more violent.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8430030234396324, "ocr_used": true, "chunk_length": 1797, "token_count": 448}}
{"text": "Physical Properties: NO is not a combustible gas. At high temperature around 1000°C it decomposes into N2 and O2 . 2NO = N2 + O2 at high temperature From the equation above we can see that once the decomposition starts 50% O2 gets evolved and this O2 supports combustion thus making the reaction more violent. Chemical properties: 1) NO acts as an oxidising agent, oxidising SO2 in presence of water to give H2SO4: SO2 + 2NO + H2O → H2SO4 + N2O 2) NO acts as a reducing agent, i) reducing an acidified solution of potassium permanganate (pink) to colorless manganous salt. 3KMnO4 + 6H2SO4 + 5NO2 → 3KHSO4 + 3MnSO4 + 2H2O + 5NO3\nii) It can also reduce aqueous solution of I2 to HI 3I2 + 2NO + 4H2O → 2HNO3 + 6HI 3) With halogens NO can form addition compounds as 2NO + Cl2 → 2NOCl (NOCl is nitrosyl chloride) It reacts in the same way with flourine and bromine. 4) With ferrous sulphate NO forms an addition compound as FeSO4 + 5H2O + NO = [Fe(H2O)5NO]SO4 penta aqua nitrosyl iron (II) sulphate This is the famous brown ring test used to identify the nitrate radical or the NO radical. Uses: NO is used to prepare nitric acid. Nitrogen Dioxide - NO2 Structure: Laboratory Preparation: In the laboratory NO2 is prepared by thermal decomposition of Pb(NO3)2. Thus 2Pb(NO3)2 → 2PbO + 4NO2 + O2 Care is taken to ensure the use of dried Pb(NO3)2 as hydrated nitrate salts on heating react violently and explode. Physical Properties: NO2 is a poisonous gas, main source being the exhaust of automobiles. 1) At room temperature it is a deep brown gas. 2) It does not support combustion. 3) It is not combustible. Chemical Properties: 1) With cold water NO2 reacts to give a mixture of HNO2 and HNO3 acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7976197759031489, "ocr_used": true, "chunk_length": 1696, "token_count": 514}}
{"text": "2) It does not support combustion. 3) It is not combustible. Chemical Properties: 1) With cold water NO2 reacts to give a mixture of HNO2 and HNO3 acid. 2NO2 + H2O → HNO2 + HNO3 2) With hot water the reaction is 3NO2 + H2O → 2HNO3 + NO 3) Being acidic it reacts with bases as 2NO2 + KOH → KNO3 + H2O + KNO2 4) It is also a strong oxidising agent. H2S + NO2 → NO + H2S + S 5) With excess oxygen and water NO2 gives HNO3. 4NO2 + O2 + 2H2O → 4HNO3 6) It reacts with concentrated H2SO4 to give nitrosyl hydrogen sulphate 2NO2 + H2SO4 → SO2(OH)ONO + HNO3 Uses: NO2 is used as a fuel in rockets besides being used to prepare HNO3 . Nitrous Oxide - N2O (laughing Gas) Structure:\nLaboratory Preparation: N2O can be prepared in the laboratory by heating NH4NO3 below 200°C to avoid explosion. Sometimes as a safety measure instead of directly using NH4NO3, a mixture of (NH4)2SO4 and NaNO3 are heated to give NH4NO3 which decomposes further to give N2O. NH4NO3 → N2O + H2O (endothermic reaction) Physical Properties: 1) N2O has a faint sweet smell and produces a tickling sensation on the neck when inhaled and makes people laugh hysterically. Excess of inhalation leads to unconsiousness. 2) Unlike other oxides of nitrogen, N2O supports combustion though it does not burn itself. Chemical Properties: 1) At very high temperature N2O decomposes to N2 and O2 2N2O → 2N2 + O2 If a glowing piece of Mg, Cu, or P is introduced in such an environment, these pieces burn brightly due to the O2 produced from decomposition of N2O. 2) With Sodium and potassium N2O reacts to give the corresponding peroxides liberating N2 in the process.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7825828420750217, "ocr_used": true, "chunk_length": 1621, "token_count": 501}}
{"text": "2) Unlike other oxides of nitrogen, N2O supports combustion though it does not burn itself. Chemical Properties: 1) At very high temperature N2O decomposes to N2 and O2 2N2O → 2N2 + O2 If a glowing piece of Mg, Cu, or P is introduced in such an environment, these pieces burn brightly due to the O2 produced from decomposition of N2O. 2) With Sodium and potassium N2O reacts to give the corresponding peroxides liberating N2 in the process. 2N2O + 2Na → Na2O + 2N2. Na2O is sodium peroxide Uses: 1) It is used as propellent gas. 2)Used in combination with oxygen in the ratio N2O : O2 = 1:10 as a mild anaesthetic. 1 16.0.0 SULPHUR AND ITS COMPOUNDS (25 LESSONS) A.SULPHUR (S) Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S2- A. Occurrence. Sulphur mainly occurs : (i) as free element in Texas and Louisiana in USA and Sicily in Italy. (ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near Naivasha in Kenya (iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of the world. B. Extraction of Sulphur from Fraschs process Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in. Digging of tunnels is thus discouraged in trying to extract the mineral. Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters. Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe. The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas. The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8390978869233633, "ocr_used": true, "chunk_length": 1814, "token_count": 510}}
{"text": "Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe. The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas. The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC. A compressed air at 15 atmospheres is forced /pumped through the innermost pipe. The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank. It is about 99% pure. 2 Diagram showing extraction of Sulphur from Fraschs Process C. Allotropes of Sulphur. 1. Sulphur exist as two crystalline allotropic forms: (i)Rhombic sulphur (ii)Monoclinic sulphur Rhombic sulphur Monoclinic sulphur Bright yellow crystalline solid Has a melting point of 113oC Has a density of 2.06gcm-3 Stable below 96oC Has octahedral structure Pale yellow crystalline solid Has a melting point of 119oC Has a density of 1.96gcm-3 Stable above 96oC Has a needle-like structure Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96oC.This is the temperature at which one allotrope changes to the other. ‘Sketch of Octahedral structure ofRhombic sulphurSulphuratomsvalent bondNan-der-waalsforcesu Covalentbonds Sketch of the needle-like structure of monoclinic sulphur\n4 2. Sulphur exists in non-crystalline forms as: (i)Plastic sulphur- Plastic sulphur is prepared from heating powdered sulphur to boil then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur. (ii)Colloidal sulphur- Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is added hydrochloric acid to form a yellow precipitate. D. Heating Sulphur. A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8873124561693879, "ocr_used": true, "chunk_length": 1909, "token_count": 506}}
{"text": "D. Heating Sulphur. A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8. On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily. On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily. This is because the S8 rings break into S8 chain that join together to form very long chains made of over 100000 atoms of Sulphur. The long chains entangle each other reducing their mobility /flow and hence increases their viscosity. On continued further heating to above 160oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous. This is because the long chains break to smaller/shorter chains. At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring of “flowers of sulphur” on the cooler parts. Summary of changes on heating sulphur\n5 Observation on heating Explanation/structure of Sulphur Solid sulphur Heat to 113oC Amber yellow liquid Heat to 160oC Liquid darkens Heat to 444oC Liquid boils to brown vapour Cool to room temperature Yellow sublimate (Flowers of Sulphur) Puckered S8 ring Puckered S8 ring in liquid form (low viscosity/flow easily) Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow) Mixture of S8 ,S6 ,S2 vapour Puckered S8 ring E. Physical and Chemical properties of Sulphur.(Questions) 1. State three physical properties unique to Sulphur Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC. 2. Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8764065088059952, "ocr_used": true, "chunk_length": 1948, "token_count": 486}}
{"text": "Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made. Observations -Sulphur melts then burns with a blue flame Colourless gas produced that has a pungent smell Red litmus paper remains red. Blue litmus paper turns red. Explanation Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e (i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid (ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid Chemical equation\n6 S(s) + O2(g) -> SO2(g) (Sulphur(IV)Oxide gas) 2S(s) + 3O2(g) -> 2SO3(g) (Sulphur(VI)Oxide gas traces) SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid) SO3(g) + H2O(l) -> H2 SO4 (aq) ( sulphuric(VI)acid). 3. Iron filings were put in a test tube containing powdered sulphur then heated on a Bunsen flame. Stop heating when reaction starts. State and explain the observations made. Test the effects of a magnet on the mixture before and after heating. Explain. Observations Before heating, the magnet attracts iron filings leaving sulphur After heating, the magnet does not attract the mixture. After heating, a red glow is observed that continues even when heating is stopped.. Black solid is formed. Explanation Iron is attracted to a magnet because it is ferromagnetic. When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating. Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic. Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ -∆H) Heated powdered heavy metals combine with sulphur to form black sulphides.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8784787262176229, "ocr_used": true, "chunk_length": 1873, "token_count": 506}}
{"text": "When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating. Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic. Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ -∆H) Heated powdered heavy metals combine with sulphur to form black sulphides. Cu(s) + S(s) -> CuS(s) Zn(s) + S(s) -> ZnS(s) Pb(s) + S(s) -> PbS(s) 4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid. 7 (i)State and explain the observation made. Observation Yellow colour of sulphur fades Orange colour of potassium dichromate(VI)paper turns to green. Explanation Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas. The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed. Chemical equation S(s) + 3H2 SO4 (l) -> 3SO2(g) + 3H2O(l) +SO3(g) Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green. (ii)State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated Nitric (V) acid in the above set up. Observation Yellow colour of sulphur fades Colurless solution formed Brown fumes/gas produced. Explanation Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The Nitric (V) acid is reduced to brown nitrogen(IV)oxide gas. Chemical equation S(s) + 6HNO3 (l) -> 6NO2(g) + 2H2O(l) +H2SO4 (l) NB: Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids. 5.State three main uses of sulphur . Sulphur is mainly used in: (i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid. (ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8755690200682985, "ocr_used": true, "chunk_length": 1829, "token_count": 506}}
{"text": "5.State three main uses of sulphur . Sulphur is mainly used in: (i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid. (ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable. (iii)Making gun powder and match stick heads (iv) As ointments to treat fungal infections\n8 6. Revision Practice The diagram below represents the extraction of sulphur by Fraschs process. Use it to answer the questions that follow. (a)Name the substances that passes through: M Superheated water at 170oC and 10 atmosphere pressure L Hot compressed air N Molten sulphur (b)What is the purpose of the substance that passes through L and M? M- Superheated water at 170oC and 10 atmosphere pressure is used to melt the sulphur L- Hot compressed air is used to force up the molten sulphur. (c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow. A B Appearance Bright yellow Pale yellow Density(gcm-3) 1.93 2.08 Melting point(oC) 119 113 Stability Above 96oC Below 96oC I.What are allotropes? Different forms of the same element existing at the same temperature and pressure without change of state. II. Identify allotrope: A. Monoclinic sulphur B . Rhombic sulphur III. State two main uses of sulphur.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8766498359000252, "ocr_used": true, "chunk_length": 1360, "token_count": 341}}
{"text": "Monoclinic sulphur B . Rhombic sulphur III. State two main uses of sulphur. -Manufacture of sulphuric(VI)acid -as fungicide -in vulcanization of rubber to make it harder/tougher/ stronger -manufacture of dyes /fibres L M N\n9 (d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies 24 dm3 at room temperature) Chemical equation S(s) + O2(g) -> SO2(g) Mole ratio S: SO2 = 1:1 Method 1 32.0 g of sulphur -> 24 dm3 of SO2(g) 0.4 g of sulphur -> 0.4 g x 24 dm3 = 0.3 dm3 32.0 g Method 2 Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles Molar mass of sulphur 32 Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3\n10 B.COMPOUNDS OF SULPHUR The following are the main compounds of sulphur: (i) Sulphur(IV)oxide (ii) Sulphur(VI)oxide . (iii) Sulphuric(VI)acid (iv) Hydrogen Sulphide (v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts (i) Sulphur(IV)oxide(SO2) (a) Occurrence Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. (b) School laboratory preparation In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of Method 1:Using Copper and Sulphuric(VI)acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7965303340149967, "ocr_used": true, "chunk_length": 1467, "token_count": 496}}
{"text": "(iii) Sulphuric(VI)acid (iv) Hydrogen Sulphide (v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts (i) Sulphur(IV)oxide(SO2) (a) Occurrence Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. (b) School laboratory preparation In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of Method 1:Using Copper and Sulphuric(VI)acid. Method 2:Using Sodium Sulphate(IV) and hydrochloric acid. 11 (c)Properties of Sulphur(IV)oxide(Questions) 1. Write the equations for the reaction for the formation of sulphur (IV)oxide using: (i)Method 1 Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l) Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l) Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8109845343575867, "ocr_used": true, "chunk_length": 1091, "token_count": 387}}
{"text": "Method 2:Using Sodium Sulphate(IV) and hydrochloric acid. 11 (c)Properties of Sulphur(IV)oxide(Questions) 1. Write the equations for the reaction for the formation of sulphur (IV)oxide using: (i)Method 1 Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l) Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l) Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas. (ii)Method 2 Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g) + 2H2O(l) K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g) + 2H2O(l) BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g) + H2O(l) CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g) + H2O(l) PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g) + H2O(l)\n12 Lead(II)chloride is soluble on heating thus reactants should be heated to prevent it coating/covering unreacted PbSO3(s) 2.State the physical properties unique to sulphur (IV)oxide gas. Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and choking smell which liquidifies easily. It is about two times denser than air. 3. The diagram below show the solubility of sulphur (IV)oxide gas. Explain.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7528657709848392, "ocr_used": true, "chunk_length": 1281, "token_count": 504}}
{"text": "3. The diagram below show the solubility of sulphur (IV)oxide gas. Explain. Sulphur(IV) oxide is very soluble in water. One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a vacuum. If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum. An acidic solution of sulphuric (IV)acid is formed which turns litmus solution red. Chemical equation SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)\n13 4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made. Observations (i)Dry Blue litmus paper remains blue. Dry red litmus paper remains red. (ii) Wet/damp/moist blue litmus paper turns red. Moist/damp/wet red litmus paper remains red. Both litmus papers are then bleached /decolorized. Explanation Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red litmus papers. Wet/damp/moist litmus papers contain water that dissolves /react with dry sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3 (aq)). Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions: H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic. The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas. It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples: (a)Bleaching agent Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8721717799889441, "ocr_used": true, "chunk_length": 1809, "token_count": 500}}
{"text": "Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions: H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic. The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas. It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples: (a)Bleaching agent Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas. This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/ material to form sulphuric(VI)acid. Chemical equations (i)Formation of sulphuric(IV)acid SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Decolorization/bleaching of the dye/removal of atomic oxygen. Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye (coloured) (colourless) Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (aq) + (dye - O) (coloured) (colourless)\n14 Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a dye unlike chlorine that bleaches by oxidation /adding oxygen. The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary. This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g. Old newspapers turn brown on exposure to air on regaining the atomic oxygen. The bleaching through adding oxygen by chlorine gas is permanent. (b)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.857950909271589, "ocr_used": true, "chunk_length": 1735, "token_count": 491}}
{"text": "Old newspapers turn brown on exposure to air on regaining the atomic oxygen. The bleaching through adding oxygen by chlorine gas is permanent. (b)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or; (ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas. Observation: Orange acidified potassium dichromate(VI) turns to green. Explanation: Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) This is a confirmatory test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)\n15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8476276074589129, "ocr_used": true, "chunk_length": 1582, "token_count": 481}}
{"text": "Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)\n15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas. Observation: Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized. Explanation: Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4- ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) This is another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving a yellow residue. (d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8418316945944699, "ocr_used": true, "chunk_length": 1566, "token_count": 493}}
{"text": "(d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl. Observation: Yellow bromine water turns to colourless/ bromine water is decolorized. Explanation:\n16 Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) SO32-(aq) + Br2 (aq) + H2O(l) -> SO42-(aq) + 2HBr(aq) (yellow) (colourless) This can also be used as another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue. (e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8405653876840503, "ocr_used": true, "chunk_length": 1622, "token_count": 501}}
{"text": "(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq) SO32-(aq) + 2Fe3+ (aq) +3H2O(l) -> SO42-(aq) + 2Fe2+(aq) + 2H+(aq) (yellow) (green)\n17 (f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or; (ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Sulphur(IV) oxide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8195442385382709, "ocr_used": true, "chunk_length": 1454, "token_count": 481}}
{"text": "Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g) (brown fumes/gas) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V) acid. Observation: A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid. Explanation: Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either SO32- ,SO42- ,CO32- ions. i.e. Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate\n18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8154653661762274, "ocr_used": true, "chunk_length": 1564, "token_count": 485}}
{"text": "Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate\n18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I. BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2 gas that turns Orange moist filter paper dipped in acidified Potassium dichromate to green. Chemical equation BaSO3(s) +2H+(aq) -> Ba2+ (aq) + SO2(g) + H2O(l) I. BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO2 gas that forms a white precipitate when bubbled in lime water. Chemical equation BaCO3(s) +2H+(aq) -> Ba2+ (aq) + CO2(g) + H2O(l) 5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples. (a)Reduction by burning Magnesium Experiment Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas Observation Magnesium ribbon continues to burn with difficulty. White ash and yellow powder/speck Explanation Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8304664620207693, "ocr_used": true, "chunk_length": 1525, "token_count": 442}}
{"text": "Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar. During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent) Chemical equation SO2(g) + 2Mg(s) -> 2MgO(s) + S(s) (white ash/solid) (yellow speck/powder)\n19 (b)Reduction by Hydrogen sulphide gas Experiment Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas. Or Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl Observation Yellow powder/speck Explanation Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar. A little moisture/water act as catalyst /speeds up the reaction. Chemical equation SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s) (yellow speck/powder) 6.Sulphur(IV)oxide has many industrial uses. State three. (i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbes’ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8889434906883023, "ocr_used": true, "chunk_length": 1817, "token_count": 499}}
{"text": "(i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbes’ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid . The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas. Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water. 20 The solid fumes out on heating to a highly acidic poisonous gas. Chemical equation 2SO2(g) + O2(g) --platinised asbestos--> 2SO3 (g) Method 2; Heating Iron(II)sulphate(VI) heptahydrate When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white. Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.882952622673435, "ocr_used": true, "chunk_length": 1576, "token_count": 442}}
{"text": "Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g) (green solid) (brown solid) Caution On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky. (c) Uses of sulphur(VI)oxide One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid. (iii) Sulphuric(VI)acid(H2SO4) (a) Occurrence Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids; Nitric(V)acid Sulphuric(VI)acid Hydrochloric acid. 21 Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level. (b)The Contact process for industrial manufacture of H2SO4 . I. Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8829377823850412, "ocr_used": true, "chunk_length": 1763, "token_count": 488}}
{"text": "Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air. Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners Chemical equation S(s) + O2(g) --> SO2 (g) II. Burning/roasting sulphide ores in air. Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like: - Lead from Lead(II)sulphide/Galena,(PbS) - Zinc from zinc(II)sulphide/Zinc blende, (ZnS) - Copper from Copper iron sulphide/Copper pyrites, (CuFeS2) On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced. Chemical equation (i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g) (ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g) (ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s) Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely. 22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.856928217090587, "ocr_used": true, "chunk_length": 1546, "token_count": 493}}
{"text": "22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method . In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is : (i) cheaper/less expensive (ii) less easily poisoned by impurities (iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger. The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst. The close “contact” between the reacting gases and catalyst give the process its name. Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas. Chemical equation 2SO2 (g) + O2(g) -- V2O5 --> 2SO2 (g) This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450oC to ensure that: (i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC (ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC. (iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.889726804226277, "ocr_used": true, "chunk_length": 1733, "token_count": 482}}
{"text": "(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower. The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid. Chemical equation H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)\n23 Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid . Chemical equation H2S2O7 (l) + H2O (l) -> 2H2SO4 (l) The acid is stored ready for market/sale. III. Environmental effects of contact process Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere. In the Contact process, about 2% of these gases do not form sulphuric (VI) acid. The following precautions prevent/minimize pollution from Contact process: (i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers. (ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water. This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8778879311238614, "ocr_used": true, "chunk_length": 1741, "token_count": 504}}
{"text": "This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV. Uses of Sulphuric(VI)acid Sulphuric (VI) acid is used: (i) in making dyes and paint (ii)as acid in Lead-acid accumulator/battery (iii) for making soapless detergents (iv) for making sulphate agricultural fertilizers VI. Sketch chart diagram showing the Contact process\n24 (c) Properties of Concentrated sulphuric(VI)acid (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC. (ii) Concentrated sulphuric(VI)acid is very soluble in water. The solubility /dissolution of the acid very highly exothermic. The concentrated acid should thus be diluted slowly in excess water. Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container. (iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions. Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red. (iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution. This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases. (v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8871294851794072, "ocr_used": true, "chunk_length": 1923, "token_count": 509}}
{"text": "(v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes. Observation; Colour( in brown sugar )change from brown to black. Colour (in glucose) change from white to black. 10cm3 beaker becomes very hot. Explanation\n25 Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds. When added to sugar /glucose a vigorous reaction that is highly exothermic take place. The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon. Caution This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi. Chemical equation Glucose: C6H12O6(s) --conc.H2SO4--> 6C (s) + 6H2O(l) (white) (black) Sugar: C12H22O11(s) --conc.H2SO4--> 12C (s) +11H2O(l) (brown) (black) Experiment II; Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm . Observation; Colour change from blue to white. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds. The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate . Chemical equation CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l) (blue) (white) Experiment III; Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8596737259706544, "ocr_used": true, "chunk_length": 1872, "token_count": 508}}
{"text": "Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds. The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate . Chemical equation CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l) (blue) (white) Experiment III; Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid. Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI)solution on the mouth of the boiling tube. Heat strongly. Caution: Absolute ethanol is highly flammable. Observation; Colourless gas produced. 26 Orange acidified potassium dichromate (VI) paper turns to green. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates ethanol to ethene gas at about 170oC. Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green. Chemical equation C2H5OH(l) --conc.H2SO4/170oC --> C2H4 (g) + H2O(l) NB: This reaction is used for the school laboratory preparation of ethene gas Experiment IV; Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Heat gently Caution: This should be done in a fume chamber/open Observation; Colourless gas produced. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8676005159710531, "ocr_used": true, "chunk_length": 1726, "token_count": 455}}
{"text": "Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas. Chemical equation HCOOH(l) --conc.H2SO4 --> CO(g) + H2O(l) NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas Experiment V; Put about 4cm3 of ethan-1,2-dioic/oxalic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water.Heat gently Caution: This should be done in a fume chamber/open Observation; Colourless gas produced. Gas produced forms a white precipitate with lime water. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent. 27 It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds. The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases. Chemical equation HOOCCOOH(l) --conc.H2SO4 --> CO(g) + CO2(g) + H2O(l) NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas. Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution. II. As an Oxidizing agent Experiment I Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately. Observation; Effervescence/fizzing/bubbles Blue solution formed with copper, Green solution formed with Iron Colourless solution formed with Zinc Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8829776347482363, "ocr_used": true, "chunk_length": 2027, "token_count": 503}}
{"text": "Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately. Observation; Effervescence/fizzing/bubbles Blue solution formed with copper, Green solution formed with Iron Colourless solution formed with Zinc Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a green solution. Chemical equation Cu(s) + 2H2SO4(aq) --> CuSO4(aq) + SO2(g) + 2H2O(l) Zn(s) + 2H2SO4(aq) --> ZnSO4(aq) + SO2(g) + 2H2O(l) Fe(s) + 2H2SO4(aq) --> FeSO4(aq) + SO2(g) + 2H2O(l) Experiment II\n28 Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately. Warm. Observation; Black solid charcoal dissolves/decrease Yellow solid sulphur dissolves/decrease Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8683699982677985, "ocr_used": true, "chunk_length": 1757, "token_count": 476}}
{"text": "Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Explanation Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green. Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to Sulphur(IV)oxide . Chemical equation C(s) + 2H2SO4(aq) --> CO2(aq) + 2SO2(g) + 2H2O(l) S(s) + 2H2SO4(aq) --> 3SO2(g) + 2H2O(l) III. As the least volatile acid Study the table below showing a comparison in boiling points of the three mineral acids Mineral acid Relative molecula mass Boiling point(oC) Hydrochloric acid(HCl) 36.5 35.0 Nitric(V)acid(HNO3) 63.0 83.0 Sulphuric(VI)acid(H2SO4) 98.0 333 1.Which is the least volatile acid? Explain Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by Hydrogen bonds making it to have the highest boiling point/least volatile. 2. Using chemical equations, explain how sulphuric(VI)acid displaces the less volatile mineral acids. (i)Chemical equation KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g) NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)\n29 This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3). (ii)Chemical equation KCl(s) + H2SO4(aq) --> KHSO4(s) + HCl(g) NaCl(s) + H2SO4(aq) --> NaHSO4(s) + HCl(g) This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl). (d) Properties of dilute sulphuric(VI)acid.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8160793775483486, "ocr_used": true, "chunk_length": 1596, "token_count": 512}}
{"text": "(i)Chemical equation KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g) NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)\n29 This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3). (ii)Chemical equation KCl(s) + H2SO4(aq) --> KHSO4(s) + HCl(g) NaCl(s) + H2SO4(aq) --> NaHSO4(s) + HCl(g) This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl). (d) Properties of dilute sulphuric(VI)acid. Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated sulphuric (VI) acid is carefully added to about 90cm3 of distilled water. Diluting concentrated sulphuric (VI) acid should be done carefully because the reaction is highly exothermic. Diluting concentrated sulphuric (VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive. On diluting concentrated sulphuric(VI) acid, water ionizes /dissociates the acid fully/wholly into two(dibasic)free H+(aq) and SO42-(aq)ions: H2SO4 (aq) -> 2H+(aq) + SO42-(aq) The presence of free H+(aq)ions is responsible for ; (i)turn litmus red because of the presence of free H+(aq)ions (ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence a strongly acidic solution. (iii)Reaction with metals Experiment: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8367619329315636, "ocr_used": true, "chunk_length": 1591, "token_count": 478}}
{"text": "Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon. Observation: No effervescence/ bubbles/ fizzing with copper Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium Colourless gas produced that extinguishes burning splint with a “pop” sound. Colourless solution formed with Zinc and Magnesium. Green solution formed with Iron Explanation: When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid,\n30 effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen gas. Impure hydrogen gas extinguishes burning splint with a “pop” sound. A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher than hydrogen in the reactivity/electrochemical series. They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc sulphate(VI). . When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place. Copper thus do not react with dilute sulphuric(VI)acid. Chemical/ionic equation Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) --> Mg2+ (aq) + H2(g) Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g) Zn(s) + 2H+(aq) --> Zn2+ (aq) + H2(g) Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g) Fe(s) + H+(aq) --> Fe2+ (aq) + H2(g) NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8520095614258061, "ocr_used": true, "chunk_length": 1592, "token_count": 470}}
{"text": "When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place. Copper thus do not react with dilute sulphuric(VI)acid. Chemical/ionic equation Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) --> Mg2+ (aq) + H2(g) Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g) Zn(s) + 2H+(aq) --> Zn2+ (aq) + H2(g) Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g) Fe(s) + H+(aq) --> Fe2+ (aq) + H2(g) NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals. (ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid (iv)Reaction with metal carbonates and hydrogen carbonates Experiment: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium hydrogen carbonate. Observation: Effervescence/ bubbles/ fizzing. Colourless gas produced that extinguishes burning splint. Colourless solution formed with Zinc carbonate, sodium hydrogen carbonate and sodium carbonate. Green solution formed with Iron(II)Carbonate Blue solution formed with Copper(II)Carbonate Explanation: When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas\n31 extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate (VI) salts is formed.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8512649985459788, "ocr_used": true, "chunk_length": 1707, "token_count": 498}}
{"text": "Green solution formed with Iron(II)Carbonate Blue solution formed with Copper(II)Carbonate Explanation: When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas\n31 extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate (VI) salts is formed. Chemical/ionic equation ZnCO3(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) + CO2(g) ZnCO3(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l) + CO2(g) CuCO3(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) + CO2(g) CuCO3(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) + CO2(g) FeCO3(s) + H2SO4(aq) --> FeSO4(aq) + H2O(l) + CO2(g) FeCO3(s) + 2H+(aq) --> Fe2+ (aq) + H2O(l) + CO2(g) 2NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l) + 2CO2(g) NaHCO3(s) + H+(aq) --> Na+ (aq) + H2O(l) + CO2(g) Na2CO3(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) + CO2(g) NaHCO3(s) + H+(aq) --> Na+ (aq) + H2O(l) + CO2(g) (NH4)2CO3(s) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g) (NH4)2CO3 (s) + H+(aq) --> NH4+ (aq) + H2O(l) + CO2(g) 2NH4HCO3(aq) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g) NH4HCO3(aq) + H+(aq) --> NH4+ (aq) + H2O(l) + CO2(g) NB: Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.6691797880333136, "ocr_used": true, "chunk_length": 1331, "token_count": 599}}
{"text": "(v)Neutralization-reaction of metal oxides and alkalis/bases Experiment I: Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir. Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place of the Copper(II)Oxide. Observation: Blue solution formed with Copper(II)Oxide Colourless solution formed with other oxides Explanation: When a metal oxide is put in a test tube containing dilute sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt. Chemical/ionic equation ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) ZnO(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l)\n32 CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) CuO(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) MgO(s) + H2SO4(aq) --> MgSO4(aq) + H2O(l) MgO(s) + 2H+(aq) --> Mg2+ (aq) + H2O(l) Na2O(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) Na2O(s) + 2H+(aq) --> 2Na+ (aq) + H2O(l) K2CO3(s) + H2SO4(aq) --> K2SO4(aq) + H2O(l) K2O(s) + H+(aq) --> 2K+ (aq) + H2O(l) NB: Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides. Experiment II: Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.749021879021879, "ocr_used": true, "chunk_length": 1260, "token_count": 496}}
{"text": "Chemical/ionic equation ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) ZnO(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l)\n32 CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) CuO(s) + 2H+(aq) --> Cu2+ (aq) + H2O(l) MgO(s) + H2SO4(aq) --> MgSO4(aq) + H2O(l) MgO(s) + 2H+(aq) --> Mg2+ (aq) + H2O(l) Na2O(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) Na2O(s) + 2H+(aq) --> 2Na+ (aq) + H2O(l) K2CO3(s) + H2SO4(aq) --> K2SO4(aq) + H2O(l) K2O(s) + H+(aq) --> 2K+ (aq) + H2O(l) NB: Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides. Experiment II: Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution Observation: Colour of phenolphthalein changes from pink to colourless at the end point. Explanation Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.72994126848272, "ocr_used": true, "chunk_length": 1093, "token_count": 444}}
{"text": "Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution Observation: Colour of phenolphthalein changes from pink to colourless at the end point. Explanation Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only. Colour of the indicator used changes when a slight excess of acid is added to the base at the end point Chemical equation: 2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l) 2KOH(aq) + H2SO4(aq) --> K2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l) 2NH4OH(aq) + H2SO4(aq) --> (NH4)2SO4(aq) + H2O(l) OH-(s) + H+(aq) --> H2O(l)\n33 (iv) Hydrogen sulphide(H2S) (a) Occurrence Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. It is present in rotten eggs and human excreta. (b) Preparation Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid. (c) Properties of Hydrogen sulphide(Questions) 1. Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide. 34 Chemical equation: FeS (s) + 2HCl (aq) -> H2S (g) FeCl2 (aq) 2. State three physical properties unique to Hydrogen sulphide. Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8320480769230769, "ocr_used": true, "chunk_length": 1664, "token_count": 504}}
{"text": "State three physical properties unique to Hydrogen sulphide. Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red. 3. Hydrogen sulphide exist as a dibasic acid when dissolved in water. Using a chemical equation show how it ionizes in aqueous state. H2S(aq) -> H+(aq) + HS-(aq) H2S(aq) -> 2H+(aq) + S2- (aq) Hydrogen sulphide therefore can form both normal and acid salts e.g Sodium hydrogen sulphide and sodium sulphide both exist 4. State and explain one gaseous impurity likely to be present in the gas jar containing hydrogen sulphide above. Hydrogen/ H2 Iron(II)sulphide contains Iron as impurity .The iron will react with dilute hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes with hydrogen sulphide gas. 5. State and explain the observations made when a filter paper dipped in Lead(II) ethanoate /Lead (II) nitrate(V) solution is put in a gas jar containing hydrogen sulphide gas. Observations Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black. Explanation When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed. All sulphides are insoluble black salts except sodium sulphide, potassium sulphide and ammonium sulphides. Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper . The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide. This is the chemical test for the presence of H2S other than the physical smell of rotten eggs.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8926054910632026, "ocr_used": true, "chunk_length": 1620, "token_count": 403}}
{"text": "Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper . The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide. This is the chemical test for the presence of H2S other than the physical smell of rotten eggs. Chemical equations Pb2+(aq) + H2S -> PbS + 2H+(aq) (black) Fe2+(aq) + H2S -> FeS + 2H+(aq) (black) Zn2+(aq) + H2S -> ZnS + 2H+(aq) (black) Cu2+(aq) + H2S -> CuS + 2H+(aq) (black)\n35 2Cu+(aq) + H2S -> Cu2S + 2H+(aq) (black) 6. Dry hydrogen sulphide was ignited as below. (i) State the observations made in flame A Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g) Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water. Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s) 7. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide. (a)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or; (ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing Hydrogen sulphide gas. Observation: Orange acidified potassium dichromate (VI) turns to green. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8301193847722947, "ocr_used": true, "chunk_length": 1639, "token_count": 495}}
{"text": "Observation: Orange acidified potassium dichromate (VI) turns to green. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur. Chemical/ionic equation: 4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l) This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide gas. Flame A Dry Hydrogen sulphide gas\n36 Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a yellow residue. (b)Decolorizes acidified potassium manganate(VII) Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Hydrogen Sulphide gas. Observation: Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized. Yellow solid residue. Explanation: Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO4- ions to green Mn2+ ions leaving a residue as the gas itself is oxidized to sulphur. Chemical/ionic equation: 5H2S(g) + 2MnO4- (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l) (purple) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (c)Decolorizes bromine water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8632788349861521, "ocr_used": true, "chunk_length": 1722, "token_count": 496}}
{"text": "Chemical/ionic equation: 5H2S(g) + 2MnO4- (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l) (purple) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (c)Decolorizes bromine water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow bromine water turns to colourless/ bromine water is decolorized. Yellow solid residue Explanation: Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to sulphur. 37 Chemical/ionic equation: H2 S(g) + Br2 (aq) -> S (s) + 2HBr(aq) (yellow solution) (yellow solid) (colourless) This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas. Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving no yellow residue. (d)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green. Yellow solid solid Explanation: Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8642882404951371, "ocr_used": true, "chunk_length": 1740, "token_count": 493}}
{"text": "Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green. Yellow solid solid Explanation: Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur. Chemical/ionic equation: H2S(aq) + 2Fe3+ (aq) -> S (s) + Fe2+(aq) + 2H+(aq) (yellow solution) (yellow residue) (green) (e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or; (ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur. Chemical/ionic equation: H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)\n38 (yellow residue) (brown fumes) (f)Reduces sulphuric(VI)acid to Sulphur Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or; (ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8504715177774946, "ocr_used": true, "chunk_length": 1652, "token_count": 484}}
{"text": "Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur Chemical/ionic equation: H2S(g) + H2O2 (l) -> 2H2O(l) + S (s) (yellow residue) 8.Name the salt formed when: (i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution: Sodium hydrogen sulphide Chemical/ionic equation: H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq) (ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution: Sodium sulphide Chemical/ionic equation:\n39 H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq) Practice Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below (a)Name the black solid Copper(II)sulphide (b)Identify the cation responsible for the formation of: I. Blue solution Cu2+(aq) II. Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8565428294241854, "ocr_used": true, "chunk_length": 1593, "token_count": 442}}
{"text": "Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green. -Yellow solid residue/deposit is formed with Hydrogen sulphide Chemical/ionic equation: 4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator. ObservationGas continues burning with a blue flame Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g) Hydrogen sulphide Blue solution Brown solution Black solid Green solution\n40 (v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts 1. Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4. 2. Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3. 3. Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8204996646545943, "ocr_used": true, "chunk_length": 1491, "token_count": 473}}
{"text": "The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO4 (aq) -> 2H+(aq) + SO42-(aq) H2SO4 (aq) -> H+(aq) + HSO4 -(aq) All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4) All Hydrogen sulphate (VI) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids. Other Hydrogen sulphate (VI) (HSO4-) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2). 4. Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)\n41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7816565290709501, "ocr_used": true, "chunk_length": 1373, "token_count": 490}}
{"text": "It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)\n41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids. Other Hydrogen sulphate (IV) (HSO3-) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2). 5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts: Experiment: In a clean dry test tube place separately about 1.0g of : Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8082397260743963, "ocr_used": true, "chunk_length": 1337, "token_count": 449}}
{"text": "Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV). -Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI) -Colour changes from blue to white then black in case of Copper (II) sulphate (VI) -Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV) -Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI). Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)\n42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7800936329588016, "ocr_used": true, "chunk_length": 1335, "token_count": 492}}
{"text": "Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)\n42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI). (i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water. Sulphur (IV) oxide does not. (ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7700146024160361, "ocr_used": true, "chunk_length": 1395, "token_count": 510}}
{"text": "(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6. The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound: Experiments/Observations: (a)Using Lead(II)nitrate(V) I. To about 5cm3 of a salt solution in a test tube add four drops of Lead(II)nitrate(V)solution. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- , Cl- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1\n43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8525766077198347, "ocr_used": true, "chunk_length": 1712, "token_count": 473}}
{"text": "Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1\n43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2\n44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8456486758759645, "ocr_used": true, "chunk_length": 1555, "token_count": 420}}
{"text": "To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2\n44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Explanations Using Lead(II)nitrate(V) (i)Lead(II)nitrate(V) solution reacts with chlorides(Cl-), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV). Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8301044043421144, "ocr_used": true, "chunk_length": 1607, "token_count": 495}}
{"text": "Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g) PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g) (iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed; - Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling) - Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling. (iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8128021868504229, "ocr_used": true, "chunk_length": 1211, "token_count": 414}}
{"text": "Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. Using Barium(II)nitrate(V)/ Barium(II)Chloride (i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV). 46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.801603481163567, "ocr_used": true, "chunk_length": 1440, "token_count": 505}}
{"text": "46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g) BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g) (iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.779844894411098, "ocr_used": true, "chunk_length": 1307, "token_count": 472}}
{"text": "Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7762222695740701, "ocr_used": true, "chunk_length": 722, "token_count": 252}}
{"text": "Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1. Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl\n48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.8009899404434214, "ocr_used": true, "chunk_length": 1537, "token_count": 484}}
{"text": "Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl\n48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7713520342229663, "ocr_used": true, "chunk_length": 1539, "token_count": 536}}
{"text": "Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I. Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7679526138745796, "ocr_used": true, "chunk_length": 826, "token_count": 294}}
{"text": "49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I. Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087\n50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7277011663821548, "ocr_used": true, "chunk_length": 1509, "token_count": 549}}
{"text": "Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087\n50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow. (a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.759491854477927, "ocr_used": true, "chunk_length": 1436, "token_count": 474}}
{"text": "Add acidified Barium nitrate(V)/chloride.\n\nWhite precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087\n50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.\n\n(a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.\n\nBa2+ (aq) + CO32- (aq) -> BaCO3(s) Ba2+ (aq) + SO32- (aq) -> BaSO3(s) Ba2+ (aq) + SO42- (aq) -> BaSO4(s)\nst", "metadata": {"source": "KCSE-FORM-3-CHEMISTRY-NOTES.pdf", "file_type": "pdf", "language": "en", "quality_score": 0.7100098233796436, "ocr_used": true, "chunk_length": 1279, "token_count": 469}}