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Answer to Chemistry EOY 2007

Answer to Chemistry EOY 2007. Paper 1 1. B 2. D 3. A 4. A 5. D 6. D 7. C 8. B 9. A 10. B. 11. D 12. D 13. B 14. B 15. A 16. C 17. D 18. C 19. D 20. A. 21. B 22. C 23. C 24. D 25. A 26. C 27. D 28. B 29. D 30. D. 1. s. 2. p. 2. s.

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Answer to Chemistry EOY 2007

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  1. Answer to Chemistry EOY 2007 Paper 1 1. B 2. D 3. A 4. A 5. D 6. D 7. C 8. B 9. A 10. B 11. D 12. D 13. B 14. B 15. A 16. C 17. D 18. C 19. D 20. A 21. B 22. C 23. C 24. D 25. A 26. C 27. D 28. B 29. D 30. D

  2. 1 s 2 p 2 s Paper 2, Section A, Question A1 a) i) 6 electrons of carbon ii) The electrostatic attraction between the positively charged protons/nucleus and the negatively charged electrons. iii) iv)

  3. Question A1 (b) (ii) Aluminium fluoride has the highest melting point, followed by hydrogen fluoride, then boron trifluoride. Aluminium fluoridehas a giant ionic structure/ions are held in a giant ionic lattice.Ions are held by strong electrostatic attraction, which requires a large amount of energy to overcome these strong forces. Both hydrogen fluoride and boron trifluoride exists as simple (covalent) molecules/have simple molecular structure. Thus their melting points are much lower than that of aluminium fluoride. But hydrogen fluoride molecules are held by hydrogen bonds whereas boron trifluoride molecules are held by van der Waals forces. Hydrogen bonds are stronger than van der Waals forces/hydrogen bonds require more energy to overcome than that of van der Waals forces. Thus the melting point of hydrogen fluoride is higher than that of boron trifluoride.

  4. A2 • A b) D c) C d) E • A3 • i) Chlorine acts as an oxidizing agent. • ii) Acidified potassium dichromate (VI)/acidified potassium manganate (VII)/bromine/hydrogen peroxide • b) It is both oxidized and reduced. • It is oxidized as the oxidation state of chlorine increases from 0 (in Cl2) to +1 (in NaClO). • It is reduced as the oxidation state of chlorine decreases from 0 (in Cl2) to -1 (in NaCl)

  5. Question A4 • i) The minimum amount of energy that molecules/reactants must possess in order for a chemical reaction to occur. • ii) A large amount of energy is needed to overcome the strong triple covalent bonds between the nitrogen atoms, thus the process has a high activation energy. • iii) By adding an (iron) catalyst

  6. Question A4 b) i) I and II ii) Condition II. Forward reaction results in a decrease in no. of mol of gases/pressure If pressure is increased (from 100 to 200 atm), forward reaction would be favoured/ equilibrium will shift to right to offset the increase in pressure. c) The yield of ammonia would increase.

  7. Volume of gas/dm3 2.40 CH3COOH 0.600 HCl time • Question A5 • Mg + 2H+→ Mg2+ + H2

  8. Question A6 • a) • b) i) Extremely soluble • ii) Citric acid contains –OH and 3 –COOH groups which are able to form hydrogen bonds with water molecules. • c) i) To prevent the volatilemethanol from escaping.

  9. Question A6 c) ii) iii) Concentrated sulphuric acid

  10. Question A7 • Excessive emission of greenhouse gases which leads to global warming, and in turn climate change. • b) Melting of polar ice caps/rising sea levels OR • Destruction of crops/Decrease in crop yield due to • extreme weathers OR • Greater spread of diseases OR • More incidence of hurricane OR

  11. Question A7 Hence empirical formula is CCl2F2

  12. Question B8 • i) CnH2n • ii) Substitution • i) Addition polymerization • ii) • iii) Name: 1-chloro-1-fluoroethene

  13. Question B8 • iv) (1) The aqueous bromine will turn from reddish-brown to colourlessimmediately. • (2) The aqueous bromine remains reddish-brown/no visible change. • Polymer X is non-biodegradable. Thus it accumulates and leads to land pollution/occupy dumping sites. OR • The combustion of polymer X gives out poisonous gas such as hydrogen chloride/hydrogen fluoride which leads to air pollution/acid rain.

  14. Question B8 ( c ) Volume of CO2 produced = 30-20 = 10 cm3 Volume of O2 reacted = 40 – 20 = 20 cm3 Thus the molecular formula is CH4

  15. Question B9 • The mixture containing the aqueous lead (II) nitrate • and aqueous sodium chloride should be stirred/mixed. • Residue on filter paper should be washed with • deionised water. • Stirring would ensure the reactants are completely • mixed so that reaction is complete. OR • Residue should be washed with deionised water to • remove excess reactants as well as sodium nitrate to • ensure that the lead (II) chloride obtained would be • pure.

  16. Question B 9 c) Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 No. of mol of Pb(NO3)2 = 20.0/1000 x 0.500 = 0.0100 1 mol of Pb(NO3)2 produces 1 mol of PbCl2 No. of mol of PbCl2 = 0.0100 Mass of PbCl2 = 0.0100 x (207+3x35.5) = 2.78 g d) The filtrate was heated until it was saturated. e) Test both reagents with aqueous potassium iodide. If bright yellow precipitate is observed, the reagent must be lead (II) nitrate If there is no visible change, then the reagent must be aqueous sodium chloride

  17. Energy Activation energy H = -435 kJ F2 + 2HBr 2HF + Br2 Progress of reaction • B10 EITHER • i) H-F bond. It has the largest bond energy. • ii) Enthalpy change for bond breaking • = 158 + 2(365) = +888 kJ • Enthalpy change for bond making • = 2 (-565) + (-193) = -1323 kJ • Enthalpy change for the reaction • = 888-1323 = -435 kJ • iii)

  18. Question B10 EITHER b) i) Brown/pink solid coated on the iron bookmark. The aqueous copper (II) sulphate turns from blue to light blue/colourless Effervescence of colourless and odourless gas at anode. ii) Cu2+ + 2e- → Cu iii) Graphite If graphite is used, there would not be enough supply of copper (II) ions. After sometime, all the copper (II) ions from aqueous copper (II) sulphate would be depleted and electrolysis would stop. The metal bookmark may not be evenly/completely plated.

  19. Potential energy Activation energy Ethanol + oxygen/air Carbon dioxide + water Progress of reaction • Question B 10 OR • i) Propane has a higher enthalpy change of combustion per kg of fuel burnt. • ii) Ethanol is renewable while propane is not. • Unlike propane which is a gas, ethanol is a liquid at • room temperature and is easier to store. • iii)

  20. Question B 10 OR • Anode: Cu → Cu2+ + 2e- • Cathode: Cu2+ + 2e- → Cu • ii) The concentration of copper (II) sulphate remains unchanged/constant. • For each copper (II) ion that is discharged at the cathode, one copper (II) ion is produced at the anode by the oxidation of one copper atom. OR • At the anode, each copper atom loses (2) electrons/is oxidised to give a copper (II) ion. • At the cathode, each copper (II) ion gains (2) electrons/is reduced/discharged to give a copper atom.

  21. Question B 10 OR • The impure copper (anode) becomes smaller/thinner/size decreases. • Or The pure copper (cathode) becomes thicker/bigger in size.

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