NYGH 2012-S3BT2-Chem Ans

1    2012 Secondary Three Block Test 2 (Chemistry) Marking Scheme Section A   1 B 6 C 11 D 2 B 7 D 12 A 3 C 8 A 13 C 4 A 9 D 14 D 5 D 10 C 15 C   Section B & C   No Answers Marks B1 (a) Diamond: Macromolecular Tin iodide: crystal lattice/ giant ionic lattice 1 1 (b) Carbon dioxide is a simple molecular compound with weak intermolecular forces of attraction. (1m) Diamond is a giant covalent substance with strong covalent bonds between the atoms of carbon, forming a macromolecular structure. (1m) More energy is needed to break the strong covalent bond. (1m for energy comparison) 3 (c) Copper has a “sea” of delocaliz ed electrons that can help to conduct electricity. 1 Accept Sec 2 dot and cross diagrams. Please note that sec 2 dot and cross diagrams will NOT be accepted for EOY (d) 1 (e) 1 B2(a) 84, 86, 88 1

2    No Answers Marks (b) Average relative molecular mass = 9/16 x 84 + 6/16 x 86 + 1/16 x 88 =85 (1m) The dichloromethane on the fibre found in the crime scene also has an average relative molecular mass of 85 (correct explanation 1m) and so the suspect is the criminal. (1m) 3 (c) 1 B3 Release both gases at the same starting point in a test tube and use moist red litmus paper and moist blue litmus paper to test. (correct method 1m) Ammonia has a relative molecular mass of 36.5 while sulfur dioxide has a higher relative molecular mass of 64 and so ammonia has a higher diffusion rate. (correct explanation for the difference in diffusion rates between ammonia and sulfur dioxide 1m) Ammonia has a higher diffusion rate and so the moist red litmus paper turned blue first. OR Sulfur dioxide has a lower diffusion rate and so the moist blue litmus paper turned red later. (correct observation to determine diffusion rate 1m) 3 For mole concept calculations, accept 6/6.02/6.023 x 1023 B4(a) Empirical formula = PQ 3, molecular formula = (PQ3)n Relative molecular mass from empirical formula =(12 + 3x1) = 15 n= 30/15 = 2 molecular formula = (PQ 3)2 = P2Q6 1 B4(b) No. of moles of gas = 48/24 = 2 moles Mass of gas = 2 x (12x2 + 1x6) = 60 g 1 1

3    No Answers Marks B4(c) Number of particles in 2 moles = 2 x 6.02 x 1023 = 1.204x1024 1 C1(a)(i) No. of moles of antimony = 40.0/122 = 0.3279 mol~ 0.328mol 1 (a)(ii) No. of moles of atoms in oxygen = 7.9/16 = 0.49375 mol 1 (a)(iii) Mole ratio Sb:O =0.3279: 0.49375 =1:1.5 =2:3 Formula is Sb 2O3. 4Sb (s) + 3O2 (g) → 2Sb2O3 (s) OR 4Sb (s) + 3O 2 (g) → Sb4O6 (s) 4 (b) Mass of antimony oxide = 0.2 x (122x2 + 16x3) (1m) = 58.4 g (1m) OR Mass of antimony oxide = 0.2 x (122x4 + 16x6) (1m) = 116.8 g (1m) 2 (c) I agree with the student. (1m) SbxOy is the empirical formula, Sb2xO2y is the molecular formula. (1m) 2 END OF MARKING SCHEME