NYGH 2013-S3EOY-Chem P1 & P2 Ans

2013 Sec 3 Chemistry EOY Marking Scheme Paper 1 1  A  11  D  21  B  2  A  12  C  22  A  3  A  13  D  23  B  4  B  14  B  24  A  5  D  15  B  25  C  6  A  16  A  26  B  7  B  17  B  27  D  8  C  18  B  28  A  9  C  19  C  29  B  10  C  20  C  30  B  Paper 2 Question Answers Marks A1 (a) Magnesium carbonate / Lead (II) iodide (b) Potassium hydroxide and Ammonium sulfate (c) Ethanoic acid / Ammonium sulfate 1 2 1 A2 (a) Lewis Dot Diagram (b) Number of moles of acid: 20.0/1000 x 0.0625 = 0.00125mol Ratio of methylamine to acid is 20:1 . 1 1 1

A3 (a) 1 : 1 (b) No. of moles of oxygen = 23.9/16 = 1.494 mol Percentage of Y = 100- 23.9 = 76.1% M r of Y = 76.1/1.494 = 50.9 or 51(accept either) Hence Y is Vanadium. (c) Amphoteric Lead (II) oxide / Zinc oxide / Aluminium oxide / Other acceptable amphoteric oxides (d) (i) There are strong electrostatic forces of attraction between the oppositely charged ions in the crystal lattice. Hence a large amount of heat energy is needed to overcome the forces of attraction. (ii) Substance Z is able to conduct electricity in aqueous state / molten state but not in solid state. In the aqueous and molten state, the crystal lattice is broken down, the ions are no longer held in the fixed lattice structure and can move to conduct electricity. [Accepted: any other suitable properties such as soluble in water but not organic solvents / hard and brittle] 1 1 1 1 1 1 1 1 1 1 A4 Relative abundance for X-96 = 2.8% Ar of X = (51.5% × 90 + 11.2% × 91+17.1% × 92 + 17.4% × 94 +2.8% × 96) = 91.4 Identity of X = Zr (zirconium) (ECF possible for identity of metal. However, examples where ECF marks are not given: i. A r = 88.6, rounded off to 88, then identity of element is wrong. ii. If the relative abundance is not calculated but Zr is obtained due to averaging, only 1 mark given) 1 1 A5 (a) When black solid (CuO) stops dissolving, CuO is in excess. (Not accepted: precipitate or white precipitate observed or unreacted CuO) (b) The mass of copper(II)oxide It is added in excess. (c) When it cooled, th e solubility decreases and thus some of the copper(II) nitrate will reappear as crystals. (d) CuO(s) + 2H + (aq) → Cu2+(aq) + H2O(l) 1 1 1 1 1 1

(e) No. of moles of HNO 3 = 250/1000 x 2 = 0.5 mol 2 mol of HNO3 : 1 mol of Cu(NO3)2 No. of mol of Cu(NO3)2 = 0.5x ½ = 0.25 mol. Mass of Cu(NO3)2 = 0.25 x 188 = 47.0 g (f) By lowering the temperature further / allowing the solution more time to cool, more crystals can be formed due to the further decrease in the solubility. (Not accepted: heat longer, heat until saturated or heat until half of it original volume) 1 1 1 1 A6 (a) Number of mol of NO 2 = 0.12 /24 = 0.005 mol Number of molecules = 0.005 X 6X1023 = 3 X 1021 (b) (i) The molecules are closely packed together in a