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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 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.
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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 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.
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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 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.
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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) + 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.
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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+ (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.
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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 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.
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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 –> 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?
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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.
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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)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.
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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 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.
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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 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.
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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 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.
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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 + … 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.
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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.
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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, 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.
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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 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).
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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)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.
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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 + … 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.
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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.
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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 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.
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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 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.
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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 + … 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.
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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.
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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 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.
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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 + … 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.
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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 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.
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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/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).
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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 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.
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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 + … 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.
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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.
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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 – 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.
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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 – 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.
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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 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.
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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 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.
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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 -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.
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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.
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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 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.
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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) 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………………………………………………………..
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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 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.
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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 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.
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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 ………………………………………………………….. (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.
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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.
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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. 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.
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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 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.
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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 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.
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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 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.
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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 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.
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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(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.
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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.
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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 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.
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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 • 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.
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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 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.
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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 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.
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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 + 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.
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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 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.
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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 –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’?
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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.
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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 (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.
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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 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.
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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.
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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 milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained.
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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----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow.
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Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow. 126 C --step 1-->127 N --step 2--> 1211Na (a)126 C and 146 C are isotopes. What does the term isotope mean? Atoms of the same element with different mass number /number of neutrons. (b)Write an equation for the nuclear reaction in step II 127 N -> 1211Na + 0 -1e (c)Give one use of 146 C Dating rocks/fossils: Study of metabolic pathways/mechanisms on plants/animals Study the graph of a radioactive decay series for isotope H below. (a) Name the type of radiation emitted when isotope (i) H changes to isotope J. Alpha-Mass number decrease by 4 from 214 to 210(y-axis) atomic number decrease by 2 from 83 to 81(x-axis) (ii) J changes to isotope K Beta-Mass number remains 210(y-axis) atomic number increase by 1 from 81 to 82(x-axis). 19 (b) Write an equation for the nuclear reaction that occur when isotope (i)J changes to isotope L 21081 J -> 21084L + 3 0 -1e (i)H changes to isotope M 21483 H -> 20682M + 3 0 -1e + 2 4 2He Identify a pair of isotope of an element in the decay series K and M Have same atomic number 82 but different mass number K-210 and M-206 a)A radioactive substance emits three different particles. Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element.
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Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical reactions produce or absorb smaller quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass /mass defect. Chemical reactions are not accompanied by a loss in mass. (v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperature/pressure/purity/particle size/ surface area b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows. 20 Identify the particles emitted in steps I and III (2mks) I - α-particle III - β-ray ii)Write the nuclear equation for the reaction which takes place in (a) step I 21483Bi -> 21081Bi + 4 2 He (b) step 1 to 3 21483Bi -> 21081Bi + 4 2 He + 2 0 -1 e (c) step 3 to 5 21082Pb -> 20682Pb + 4 2 He + 2 0 -1 e (c) step 1 to 5 21483Bi -> 20682Pb + 2 4 2 He + 3 0 -1 e The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times. Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II.
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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 – life of the Bismuth isotope II. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks) d) Give one use of radioactive isotopes in medicine (1mk) 14.a)Distinguish between nuclear fission and nuclear fusion. (2mks) Describe how solid wastes containing radioactive substances should be disposed of. (1mk) b)(i)Find the values of Z1 and Z2 in the nuclear equation below Z1 1 94 140 1 U + n -> Sr + Xe + 2 n 92 0 38 Z2 0 iii)What type of nuclear reaction is represented in b (i) above? A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom, Write a balanced decay equation for the above change 1 mark If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if the half-life of the element is 30years 2 marks Define the term half-life. The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements. (a)Which letter represent the : Explain.
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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 ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow.
The letters are not the actual symbols of the elements.
(a)Which letter represent the : Explain.
(i)shortest lived nuclide L-has the shortest half life (ii)longest lived nuclide
23 P-Is stable (iii) nuclide with highest n/p ratio L-has the shortest half life thus most unstable thus easily/quickly decay/disintegrate (iv) nuclide with lowest n/p ratio P-is stable thus do not decay/disintegrate (b)How long would it take for the following: (i)Nuclide A to change to B 10 years (half life of A) (ii) Nuclide D to change to H 27days +162000years+70000years+16days 232000 years and 43 days (iii) Nuclide A to change to P 27days +162000years+70000years+16days 232000 years and 43 days Study THE END
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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 /occupies definite space. It has a very high density. (ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space. (iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container. 2.Gases are affected by physical conditions. There are two physical conditions: (i)Temperature (ii)Pressure 3. The SI unit of temperature is Kelvin(K). Degrees Celsius/Centigrade(oC) are also used. The two units can be interconverted from the relationship: oC + 273= K K -273 = oC Practice examples 1. Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
2. Convert the following into degrees Celsius/Centigrade(oC).
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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
2. Convert the following into degrees Celsius/Centigrade(oC). (i) 10 K K -273 = oC substituting: 10 – 273 = -263 oC (ii) (i) 1 K K -273 = oC substituting: 1 – 273 = -272 oC (iii) 110 K K -273 = oC substituting: 110 – 273 = -163 oC (iv) -24 K K -273 = oC substituting: -24 – 273 = -297 oC The standard temperature is 273K = 0 oC. The room temperature is assumed to be 298K = 25oC 4. The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used. The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2). The standard pressure is the atmospheric pressure. Atmospheric pressure is equal to about: (i)101325 Pa (ii)101325 Nm-2 (iii)760 mmHg (iv)76 cmHg (v)one atmosphere. 5. Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure. Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6.
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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 “the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ” Mathematically:
Volume α 1 (Fixed /constant Temperature) Pressure V α 1 (Fixed /constant T) ie PV = Constant(k) P From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved. Graphically therefore a plot of volume(V) against pressure (P) produces a curve. V P Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line V 1/P For two gases then P1 V1 = P2 V2 P1 = Pressure of gas 1 V1 = Volume of gas 1 P2 = Pressure of gas 2 V2 = Volume of gas 2
Practice examples: 1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled. Working P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2 V2 = 102300 x 25 = 12.5cm3 (102300 x 2) 2. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?
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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 the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa? P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2 V2 = 101325 x 60 = 67.55 cm3 90000 The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination. 7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ” Mathematically: Volume α Pressure (Fixed /constant pressure) V α T (Fixed /constant P) ie V = Constant(k) T From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled. Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure. The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas. This temperature is called absolute zero. It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1.
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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 gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1. 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened. V1 = V2 substituting 500 = V2 T1 T2 (0 +273) (-4 +273) = 500 x (-4 x 273) = 492.674cm3 (0 + 273) The capacity of cylinder (500cm3) is less than new volume(492.674cm3). 7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder. 2. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens. V1 = V2 substituting 40000 = V2 T1 T2 (25 +273) (30 +273) = 40000 x (30 x 273) = 40671.1409cm3 (25 + 273) The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3). The tyre thus bursts. 3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.
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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
(25 + 273) The capacity of balloon (80cm3) is more than new volume (65.2349cm3). The balloon thus remained intact. 8. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion. The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas. The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion. Examples 1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2. 2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2. 3. Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion.
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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: “the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure” Mathematically R α 1 and since density is proportional to mass then R α 1 √ p √ m For two gases then: R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas. √M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas. Since rate is inverse of time. i.e. the higher the rate the less the time: For two gases then: T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse. √M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas. Practice examples: 1. It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.
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Molar mass CO2 = 44g Molar mass HCl = 36.5g T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0 T HCl √ molar mass HCl T HCl √ 36.5 T HCl = 200seconds x √ 36.5 = 182.1588 seconds √ 44.0 3. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse. Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.
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Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0) Molar mass CO2 = 44.0 Molar mass CO = 28.0 Method 1 25cm3 CO takes 25seconds 75cm3 takes 75 x25 = 75seconds 25
T CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0 T CO √ molar mass CO 75 √ 28.0 T CO2 =75seconds x √ 44.0 = 94.0175 seconds √ 28.0 Method 2 25cm3 CO2 takes 25seconds 1cm3 takes 25 x1 = 1.0cm3sec-1 25 R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0 R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0 R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1 √ 44.0 0.7977cm3 takes 1 seconds 75cm3 takes 75cm3 = 94.0203seconds 0.7977cm3
13.0.0 THE MOLE-FORMULAE AND CHEMICAL EQUATIONS (40 LESSONS) Introduction to the mole, molar masses and Relative atomic masses 1. The mole is the SI unit of the amount of substance. 2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles. 3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.
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3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”. The Avogadros Constant contain 6.023 x10 23 particles. i.e. 1mole = 6.023 x10 23 particles = 6.023 x10 23 2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24 0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22 0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21 3. The mass of one mole of a substance is called molar mass. The molar mass of: (i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g. Molar mass of carbon(C)= relative atomic mass = 12.0g 6.023 x10 23 particles of carbon = 1 mole =12.0 g Molar mass of sodium(Na) = relative atomic mass = 23.0g 6.023 x10 23 particles of sodium = 1 mole =23.0 g Molar mass of Iron (Fe) = relative atomic mass = 56.0g 6.023 x10 23 particles of iron = 1 mole =56.0 g (ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule. The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.
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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(Cl2) =relative molecular mass =(35.5x 2)g =71.0g 6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g 6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g (ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
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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 empirical formula of copper oxide (a)Method 1:From copper to copper(II)oxide Procedure. Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3). Sample results Mass of crucible(M1) 15.6g Mass of crucible + copper before heating(M2) 18.4 Mass of crucible + copper after heating(M3) 19.1 Sample questions 1. Calculate the mass of copper powder used. Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3.
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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 Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3. Calculate the number of moles of: (i) copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from brown to black Explanation Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide 7. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above. Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesuin oxide and magnesium nitride. This causes experimental mass errors. (b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible.
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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; [email protected](II)oxideHydrogen /Laboratory gas /Ammonia gas/Carbon(II)Oxidegas from a generatorExcess hydrogenburningDetermining empirical formula from copper(II)oxide to copper Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator. When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating. Continue passing the gases until the glass tube is cool. Turn off the gas generator. Carefully remove the porcelain boat form the combustion tube. Reweigh (M3). Sample results Mass of boat(M1) 15.6g Mass of boat before heating(M2) 19.1 Mass of boat after heating(M3) 18.4 Sample questions
9 9 1. Calculate the mass of copper(II)oxide used. Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3.
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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 oxygen = 2.8 g 3. Calculate the number of moles of: (i) Copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from black to brown Explanation Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal. 7. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above. Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8.
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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 parts ) (ii)Carbon(II)oxide CuO(s) + CO (g) -> Cu(s) + CO2(g) (Black) (brown) (colourless gas, form white ppt with lime water ) (iii)Ammonia 3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) 9. Explain why the following is necessary during the above experiment; (i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide. Air combine with hydrogen in presence of heat causing an explosion (ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped. Hot metallic copper can be re-oxidized back to copper(II)oxide (iii) A stream of excess carbon (II)oxide gas should be ignited to burn Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas. 10. State two sources of error in this experiment. (i)All copper(II)oxide may not be reduced to copper. (ii)Some copper(II)oxide may be blown out the boat by the reducing agent. 4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula.
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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 = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen.
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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 of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4.
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Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.
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What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.
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What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number.
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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 a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number. (ii) Relative empirical formula x n = Relative formular mass where n is a whole number. Practice sample examples 1. A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen. If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2.
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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 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2. 4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only.
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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 CO2 => Molar mass of CO2 12 x 5.28 = 1.44g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 2.16 = 0.24g√ 18 Element Carbon Hydrogen
15 15 Symbol C H Moles present = mass Molar mass 1.44g 12 0.24g√ 1 Divide by the smallest value 0.12 0.12 0.24 0.12 Mole ratios 1 2√ Empirical formula is CH2√ The molecular formular is thus determined : n = Relative formular mass = 84 = 6√ Relative empirical formula 14 The molecular formula is (CH2 ) x 6 = C6H12. √ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed.
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√ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0) Mass of N in A = 5.2% x 0.085 = 0.00442 g Mass of C in A = 12 x 0.224 = 0.0611g 44 Mass of H in A = 2 x 0.0372 = 0.0041g 18 Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O) => 0.0611g + 0.0041g = 0.0652g (Mass of C,H) 0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g Element Nitrogen Carbon Hydrogen Oxygen Symbol N C H O Moles present = mass Molar mass 0.00442 g 14 0.0611g 12 0.0041g 1 0.0154 g 16 Divide by the smallest value 0.00032 0.00509 0.0041g 0.00096
16 16 0.00032 0.00032 0.00032 0.00032 Mole ratios 1 16 13 3 Empirical formula = C16H13NO3 (d)Molar gas volume The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is: (i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p). i.e.
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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 its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products.
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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= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation.
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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-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1.
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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 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2.
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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.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid.
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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(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1.
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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) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.
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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) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2.
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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 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.
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