NYGH 2012-S3EOY-Chem Ans

2012 Secondary Three End-of-Year Paper 1 and 2 (Chemistry) Paper 1 1 D 11 C 21 D 2 A 12 C 22 A 3 A 13 B 23 B 4 B 14 A 24 A 5 D 15 B 25 D 6 B 16 A 26 B 7 C 17 A 27 C 8 A 18 C 28 A 9 B 19 B 29 D 10 B 20 B 30 B Paper 2 Section A No Answers Marks A1(a) Element V has a lower melting point. Element V has simple molecular structure with weak intermolecular forces of attraction while element W is a metallic element with strong metallic bonds. During melting, less energy is required to overcome the weak intermolecular forces of attraction between molecules in V than strong metallic bonds in W. [3] b(i) It does not conduct electricity in molten state. [1] (ii) Start with 2 marks, deduct 1 mark for the any of the errors listed (max 2 marks). 1. Wrong chemical symbols. (e.g. Uses C l instead of R) 2. Shares wrong number of electrons./ Wrong number of non bonding electrons 3. Non bonding electrons are not paired. 4. Uses symbols other than dot and cross. 5. Brackets Max 1 mark if ionic compound is drawn. [e.c.f. from bi] [2] c(i) Gp II [1] (ii) U is able to conduct electricity in solid state as it is a metal with ‘sea of delocalized electrons’. [2] A2(a) Any excess aqueous nitric acid in the reaction is difficult to separate from the soluble salt product while excess insoluble magnesium hydroxide may be removed easily by filtration. [2] R  RO 1

No Answers Marks b(i) 3 ions [1] (ii) Covalent and ionic bonds [1] A3(a) (i) K2CO3(aq) + 2HNO3 (aq) →2KNO3 (aq) + H2O(l) + CO2(g) [2] (ii) Amount of K2CO3 added = 4.80÷ Molar mass of K2CO3 = 4.80÷138 = 0.03478 mol Amount of HNO3 used = 0.024x 1.67 = 0.04008 mol Mole ratio of K 2CO3 : HNO3 is 1:2. Thus, 0.03478 x 2 = 0.06956 mole HNO3 is required to completely react with 4.80 g of K2CO3. HNO3 is the limiting reagent. [2] (iii) Since HNO3 is the limiting reagent, and mole ratio of HNO3: KNO3 is 1:1, theoretical yield of KNO3 is 0.04008 mole. Percentage yield = mass of actual yield ÷ mass of theoretical yield x 100 1.70 = mass of actual yield÷(0.04008 x 101) Mass of actual yield = 6.88g (to 3 s.f.) [1] (iv) Excess potassium carbonate added dissolved in the solution and thus could not be removed by filtration. [1] b(i) OH-(aq) + H+(aq) → H2O(l) [2] (ii) M1V1 ÷ M2V2 = Mole ratio of KOH is to HNO3 (0.400 x 0.0300) ÷ (concentration of HNO3 x 0.0200) = 1 Concentration of HNO3 = 0.600 mol/dm3 [2] (c) Student B’s method is more suitable. [1] A4(a) Dibasic/ 2 [1] b(i) Every mole of HA will produce 2 mole of ions. Moles of HA that dissociated = 2 moles HA [1] (ii) Every mole of H2B will produce 3 mole of ions. Moles of H2B that dissociated = 2 moles H2B [1] (iii) H A is the stronger acid. [1] (c) 1. The solid may be zinc as reaction did not occur due to the absence of water. 2. Not all metals will