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Answer to Chemistry EOY 2006

Answer to Chemistry EOY 2006. Paper 1 1. B 2. B 3. D 4. D 5. A 6. B 7. C 8. A 9. C 10. D. 11. C 12. B 13. A 14. C 15. D 16. A 17. C 18. B 19. C 20. C. 21. C 22. A 23. C 24. D 25. A 26. D 27. D 28. C 29. B 30. C. 1. s. 2. p. 2. s. 2. p. 1. s. 2. s.

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Answer to Chemistry EOY 2006

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  1. Answer to Chemistry EOY 2006 Paper 1 1. B 2. B 3. D 4. D 5. A 6. B 7. C 8. A 9. C 10. D 11. C 12. B 13. A 14. C 15. D 16. A 17. C 18. B 19. C 20. C 21. C 22. A 23. C 24. D 25. A 26. D 27. D 28. C 29. B 30. C

  2. 1 s 2 p 2 s 2 p 1 s 2 s Paper 2, Section A, Question A1 a) i) Electron in box diagram for N (7 e-) ii) O2- (O atom gains 2 electron, thus no. of electron = 8 + 2 = 10)

  3. Question A2 b) i) See from tb that SiH4 has a low bp, hence it must be covalent ii) Ammonia has a higher boiling point than methane. Van der Waals' forces exist between methane molecules while hydrogen bonds exist between ammonia molecules. The hydrogen bonds between ammonia molecules are stronger than the Van der Waals’ forces between the methane molecules. Thus more energy is needed to overcome the stronger hydrogen bonds.

  4. A2 • Water has a higher boiling point than ammonia. • Oxygen is more electronegative than nitrogen, hence the hydrogen bond between water molecules is stronger than that between ammonia molecules. • Or Water forms more hydrogen bonds per molecule than ammonia, hence more energy is needed to overcome the greater no. of hydrogen bonds.

  5. Question A2, part (b) • The boiling point of GeH4 is higher than SiH4. • GeH4 has a bigger molecular size /greater number of electrons (per molecule). • Thus the van der Waals' forces between GeH4 molecules would be stronger than that between SiH4 molecules.

  6. Question A1, part c Dotted line is the hydrogen bond, must show delta + and delta – on H and O atoms respectively.

  7. Question A2 • i) Carbon monoxide is formed by the incomplete combustion of carbon-containing fuel. • ii) Carbon monoxide binds irreversibly with haemoglobin, thus depriving blood of oxygen/preventingblood from transporting oxygen. • b) i)

  8. Question A2 b) • Carbon monoxide is oxidised as oxidation state of carbon increases from +2 (in CO) to +4 (in CO2). • Palladium (II) chloride is reduced as oxidation state of palladium decreases from +2 (in PdCl2) to 0 (in Pd) • Quesiton A3 • Rb • Si • Rb • Cl (Recall maximum oxidation state of an element corresponds to its group no.)

  9. Question A4 • Carbon monoxide, hydrogen, methanol (Both reactants and products present as reaction is reversible) • Working • Initial no. of mol of CO = 1; Hence vol = 24 dm3 (show this value on graph) • As reaction is 75.0 % complete, it would mean that 0.75 mol of CO is reacted. • Hence no. of mol of CO left = 1-0.75 = 0.25. • Volume of CO at the end = 0.25x24 = 6 dm3 (Show this value on graph).

  10. Paper 2, Section A, Question A4 b) Volume of gas/dm3 24 6 Time

  11. Question A4 c) Hydrogen is used in excess so as to shift equilibrium to the right/favour forward reaction, thus increasing yield of methanol. Or Carbon monoxide is poisonous. d) Forward reaction is exothermic, thus temperature can be maintained at 300°C. Further heating would cause equilibrium to shift to left/favour backward reaction, decreasing yield of methanol. Thus reaction is maintained at 300°C, a moderate temperature which allows satisfactory yield of methanol and at the same time ensuring that rate of reaction is not too slow.

  12. Question A5, part (a) No. of mol of PbCO3 = 5.00/(207+12+3x16) = 0.0187 No. of mol of HNO3 = 50.0/1000 x 2.00 = 0.100 PbCO3 + 2HNO3 Pb(NO3)2 + H2O + CO2 1 mol of PbCO3 reacts with 2 mol of HNO3 0.0187 mol of PbCO3 react with 0.0374 mol of HNO3 Thus PbCO3 is the limiting reactant/HNO3 is in excess. 1 mol of PbCO3 gives 1 mol of CO2 No. of mol of CO2 = 0.0187. Volume of CO2 = 0.0187 x 24 = 0.449 dm3

  13. Question A5 b) i)

  14. Question A5 • ii) Lead (II) carbonate reacts with sulphuric acid to give insoluble lead (II) sulphate, which coats around lead (II) carbonate, stopping the reaction after a short while. • Ethanoic acid reacts slower than nitric acid. • Ethanoic acid is a weak acid while nitric acid is a strong acid. • Concentration of hydrogen ions in ethanoic acid would be lower than that in nitric acid. • Frequency of effective collisions between the reactant particles would be lower when ethanoic acid is used.

  15. Question A5 • Warm the nitric acid so that temperature is higher than room temperature. • Grind the lumps of lead (II) carbonate into powder. • Question A6 • Butane (Recall L shaped molecule is still straight chain) • i) • ii) The aqueous bromine turns from reddish-brown to colourless immediately.

  16. Question A6 • i) Isomers are compounds which have the same molecular formulae but different structural formulae. • ii) Disagree, C4H8 has 4 or more than 3 isomers.

  17. Question A7 • Greenhouse effect is caused by gases which trap heatradiation/prevent heat radiation from escaping into space, and hence causes global warming. • Carbon dioxide and methane.

  18. Section B, Question B 8 • i) (hydrated) magnesium sulphate • ii) A = potassium hydroxide or KOH • B = hydrogen gas or H2 • C = magnesium chloride or MgCl2 • D = magnesium hydroxide or Mg(OH)2 • iii) 2K + 2H2O  2KOH + H2 • iv) Mg2+ + 2OH- Mg(OH)2 • v) Salt solution E is heated until is saturated. The hot saturated solution is then cooled to allow crystals to form. Filter the mixture to obtain the crystals. • vi) magnesium/magnesium oxide/magnesium carbonate with dilute sulphuric acid

  19. Section B, B8, part (b) React the white powders with dilute sodium hydroxide solution. If the white powder dissolves in sodium hydroxide to give a colourless solution, the solid is zinc oxide. If there is no visible change, the white powder is calcium oxide. Or React the white powder with dilute sulphuric acid. If the white powder dissolves to give a colourless solution, the solid is zinc oxide. If the white powder remains insoluble, the solid is calcium oxide.

  20. Question B9, part (a) • Cathode: 2H+ + 2e- H2 • Anode: 2Cl- Cl2 + 2e- • (Must label cathode and anode clearly!) • The pH of the electrolyte increases. • When silver is used as the anode, it oxidises to give silver ions. The silver ions then react with chloride ions in the electrolyte to give insoluble silver chloride, a white precipitate. • The concentration of mobile ions decreases, thus decreasing the electrical conductivity of electrolyte.

  21. Question B9, part (b) • Bonds broken: 2 mol C-C bonds, 12 C-H bonds, 7 O=O bonds. • Bonds made: 8 mol C=O bonds, 12 O-H bonds • Enthalpy change for bond breaking • = 2(347) + 12 (435) + 7(497) = +9393 kJ • Enthalpy change for bond making • = 8(-803) + 12 (-463) = -11992 kJ • Enthalpy change for combustion of ethane • = 9393 – 11992 = -2599 kJ. (for 2 mol of C2H6, following equation) • No. of mol of 60.0 g of C2H6 = 60.0/(2x12+6x1) = 2.00 • Hence amount of energy liberated by 60.0 g of ethane • = 2599 kJ. (no sign here!)

  22. Potential energy Activation energy H = -2599 kJ 2C2H6 + 7O2 4CO2 + 6H2O Progress of reaction Question B9, part (b), (iii)

  23. Note question says fully oxidised, meaning both -OH groups are reacted. Question B10 EITHER a) i) alcohols (Question asks for homologous series which means family) ii)

  24. Question B10, EITHER b (i) Empirical formula is CH3O

  25. Question B 10 EITHER b) Let the molecular formula be (CH3O)n Mr of (CH3O)n = 62 n x (12 + 3 + 16) = 62 n = 2 Hence molecular formula is C2H6O2 Possible structure:

  26. Question B10 EITHER c) i) addition polymer ii) methyl-2-methylpropenoate (Read from question that the polymer name is poly(methyl-2-methylpropenoate) iii) iv) Perspex does not break easily like glass.

  27. Question B 10 OR • i) Tartaric acid and methanol are heated under reflux/at high temperature in the presence of concentrated sulphuric acid as catalyst. • ii) • iii) any pH value from 3 to 5.5 • iv) Effervescence of colourless and odourless gas.

  28. Question B 10 OR • i) Let the hydrocarbon be CxHy. • CxHy + (x+y/4) O2 xCO2 + y/2 H2O No. of mol of CxHy 1 = No. of mol of CO2 x 10 1 = 10 x No. of mol of CxHy 1 = No. of mol of O2 X + y/4 10 1 Substitute x =1 = 20 1 + y/4 X = 1 y = 4 Hence formula is CH4

  29. B10 b) ii) CCl4(Question says “excess Cl2”, hence all H replaced) c) i) ii) Nylon iii) Kevlar is lighter in weight as compared to earlier bullet-resistant vests which contained steel or ceramic plates that were (bulky) and heavy.

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