(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
38 (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 .


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)
40 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.


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)
40 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.


(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)
40 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.


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)
40 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.


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
41 (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.


(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
41 (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.


(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
43 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?


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
43 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.


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
44 -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.


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
46 •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.


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
46 •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.


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
46 •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.


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
46 •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.


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
47 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.


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
47 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.


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
47 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- + …
48 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).


Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials
47 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- + …
48 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.


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- + …
48 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.


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- + …
48 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 + …
49 (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.


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 + …
49 (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)
50 104 The commercial name of polyphenylethene is polystyrene.


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 + …
49 (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)
50 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.


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 + …
49 (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)
50 104 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.


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)
50 104 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.


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:
51 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.


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:
51 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.


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
52 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.


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
52 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
53 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).


(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
52 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
53 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.


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
52 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
53 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.


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
53 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-
54 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.


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-
54 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
55 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.


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
55 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.


(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
56 (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.


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
57 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.


2.Formation of Terylene
57 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
58 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
59 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic.


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
59 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
61 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
62 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
63 8.


(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
62 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
63 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
64 (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
65 (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
66 (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) (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
65 (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
66 (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
67 (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) Write the equation of a reaction that occurs at the cathode
66 (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
67 (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
68 i) Name the following compounds:- I. Product T ………………………… II.


(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
68 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
69 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
70 (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.


(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
72 (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
73 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.


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
73 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;
74 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
75 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
76 21.


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
75 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
76 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   
77 ― 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
78 (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
79 28. The scheme below shows some reactions starting with ethanol.


Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O
78 (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
79 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-
80 (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
81 (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
82 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.


(b) Explain why plastics and rubbers are used this way
82 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
83 (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
84 (a) Give the name of the polymer (b) Draw the structure of the monomer used to manufacture the polymer
24.0.0 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.


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.


(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.


(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+
3 226 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.


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+
3 226 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
C -> 147N + 0 -1e 5.


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
C -> 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.


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
6 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
7 214 82 Pb undergoes beta decay to 214 84 Rn. Find the number of β particles emitted. Write the nuclear equation for the disintegration.


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
8 => 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.


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
8 => 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.


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.


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.


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
11 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
12 (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’?


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
12 (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.


(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.


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
15 (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
16 (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.


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
16 (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.