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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 b... | {
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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 ... | {
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(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 equat... | {
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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... | {
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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+... | {
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(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 Substa... | {
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(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 β> H3C β CH2 βO-SO3H Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen su... | {
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Product A Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water H3C-CH2OH + H2SO4 β> H3C β CH2 βO-SO3H + H2O Product B Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water hydrogen sulphate(VI) H3C β CH2 βO-SO3H + NaOH -> H3C β CH2 βO-SO3-Na+ + H2O 3.Below is part of a detergent H3C β (CH2... | {
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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)Natura... | {
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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 bi... | {
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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 t... | {
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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 reduc... | {
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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 + β¦ H H H H H H H H Ethene + Ethene + Ethene + Ethene + β¦ (ii)the double bond joining the ethane molecule break to free rea... | {
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Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C β C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer... | {
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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, ... | {
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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 ... | {
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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)... | {
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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 + β¦ H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + β¦ (ii)the double bond joining the chlo... | {
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Polychloroethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦
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 β C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of... | {
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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 press... | {
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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 br... | {
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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 + β¦ H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + β¦
49 (ii)the double bond joini... | {
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Polyphenylethene molecule can be represented as: H H H H H H H H - C β C - C β C - C β 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 β C )n H C6H5 Examples Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in... | {
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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 ... | {
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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 + β¦ H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + β¦ (ii)the double bond joining the phenylethene molecul... | {
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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 jo... | {
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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... | {
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(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... | {
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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 + β¦ F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + β¦ (ii)the double bond... | {
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polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer - C β C - C β C - C β C - C β C- + β¦ F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C β C )n F F Examples
53 Polytetrafluorothene has a molar mass of 4760.Calculate the numb... | {
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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 β 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 β2β thus; H CH3... | {
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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 β C = C - H During polymerization to synthetic rubber, one double C=C bond is broken t... | {
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(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 m... | {
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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 thu... | {
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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 -... | {
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O O Cl - C β C5H5 β C β Cl + H βO β CH2 CH2 β 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 β C5H5 β C β O β CH2 CH2 β O β H + HCl . Polymer bond linkage of terylene
59 The commercial name of terylene is Polyester /polys... | {
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(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 t... | {
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(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) ... | {
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(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... | {
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(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 ... | {
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(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 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propy... | {
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Lβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 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 οΊCH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦. Step 4 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... | {
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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... | {
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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 c... | {
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(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 t... | {
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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 a... | {
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(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 thu... | {
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(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(Ξ±) particle decay The... | {
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iv) have high ionizing power thus cause a lot of damage to living cells. v) a nuclide undergoing Ξ±-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 + ... | {
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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 Ξ² 22888Ra -> 22892Th + 4 Ξ² 8. 23290Th -> 2128... | {
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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 β’ 3717Cl -> 3717Cl + y β’ 146C -> 146C + y The sketch diagram below shows the penetrating power of... | {
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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 Ξ± 4 Using atomic numbers only and substituting the 8 Ξ±(above) 238 92 U -> 206 82 Pb + 8 4 2 He + x 0 -1 e 92 = 82 + 16 + - x => 92 β (82 + 16) = - x x = 6 Ξ² Nuclear equation 238 92 U -> 206 82 Pb... | {
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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 Ξ± 4 Using atomic numbers only and substituting the 21 Ξ± (above) 238 92 U -> 214 83Bi + 21 4 2 He + y 0 -1 e 92 = 83 + 42 + - y
8 => 92 β (83 + 42) = - x x = 33 Ξ² Nuclear equation 298 92 U -> 210... | {
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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 ... | {
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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... | {
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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 β2x--> 1/4 β3x---> 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.... | {
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If t 1/2 = x then: 100 --x-->50 β2x-->25 β3x--->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)... | {
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(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... | {
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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 r... | {
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Mass number = 212 β (0 beta+ o gamma + (2 x 4 ) alpha = 204 Atomic number = 80 β (-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 fol... | {
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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 mill... | {
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The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> ... | {
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Determine the half life of X2
18 1βx-> 1 /2 βx-> 1 /4 βx-> 1 /8βx-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow. 126 C --step 1-->127 N --step 2--> 1211Na (a)126 C and 146 C are isotopes. What does the term isotope mean? A... | {
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Identify the particle: (i)with the highest mass. Alpha/ Ξ± (ii) almost equal to an electron Beta/ Ξ² 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy le... | {
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Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half β life of the Bismuth isotope II. Original mass of the Bismuth isotope given that the mass that r... | {
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The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( Ξ² )particle/ray R- Alpha (Ξ± )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not bloc... | {
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12.0.0 GAS LAWS (15 LESSONS) (a)Gas laws 1. Matter is made up of small particle in accordance to Kinetic Theory of matter: Naturally, there are basically three states of matter: Solid, Liquid and gas: (i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume ... | {
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Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
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Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6. Boyles law states that βthe volume of a fixed mass of a gas is inversely proportional to the pressure at constant/f... | {
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Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into... | {
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It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two g... | {
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3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened. V1 = V2 substituting 80 = V2 T1 T2 (25 +273) (-30 +273) = 80 x (-30 x 273) = 65.2349cm3
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A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that: βthe rate of diffusion of a gas is ... | {
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Molar mass CO2 = 44g Molar mass HCl = 36.5g T CO2 = β molar mass CO2 => 200 seconds = β 44.0 T HCl β molar mass HCl T HCl β 36.5 T HCl = 200seconds x β 36.5 = 182.1588 seconds β 44.0 3. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffus... | {
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Molar mass O2 = 32g Molar mass Z = x g T O2 = β molar mass O2 => 250 seconds = β 32.0 T Z β molar mass Z 227seconds β x β x = 227seconds x β 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate und... | {
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3. The number of particles in one mole is called the Avogadros Constant. It is denoted βLβ. The Avogadros Constant contain 6.023 x10 23 particles. i.e. 1mole = 6.023 x10 23 particles = 6.023 x10 23 2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24 0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22 0.0065 moles =... | {
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Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g. Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g 6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
2 2 Molar mass chlorine molecule(C... | {
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It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound 3.Practically the empirical formula of a compound can be determined as in the following examples. To determine the empirica... | {
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Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II... | {
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(b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below; jgthungu@gmail.com52HEATCopper(II)oxideHydrogen /La... | {
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Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxy... | {
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Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8. Write the equation for the reaction that would take place when the reducing agent is: (i) Hydrogen CuO(s) + H2(g) -> Cu(s) + H2O(l) (Black) (brown) (colourless liquid form on cooler ... | {
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4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % ... | {
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(Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass o... | {
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Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 β 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by ... | {
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What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 β 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of I... | {
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What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 β 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)su... | {
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The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is ... | {
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If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 β 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The ... | {
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4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO... | {
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β molecular name Hexeneβ/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C Hβ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g o... | {
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22.4dm3 -> 6.0 x1023 2.24dm3 -> 2.24 x 6.0 x1023 22.4 =6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw... | {
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= 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2... | {
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0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1β Molar m... | {
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Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118gβ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232gβ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232gβ 1 Divide by the s... | {
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All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.... | {
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Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2... | {
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Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) ->... | {
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1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)... | {
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If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 2... | {
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