NYJC_H1_CHEM_P1_P2_Ans

Nanyang Junior College J2 H1 Chemistry 2017 1 2017 J2 H1 Chemistry Prelim Answers Paper 1 Answer Key 1 B 6 A 11 A 16 D 21 A 26 C 2 C 7 B 12 B 17 A 22 D 27 B 3 D 8 C 13 C 18 C 23 D 28 A 4 D 9 B 14 A 19 A 24 C 29 B 5 C 10 B 15 B 20 D 25 D 30 C

Nanyang Junior College J2 H1 Chemistry 2017 2 Paper 2 Section A Answers 1 (a) (i) The carbon-carbon single bond in the ethane molecule consists of 1  bond. The carbon=carbon double bond in ethane molecule consists of 1  bond and 1  bond. A  bond is weaker than a  bond due to less effective overlap, hence C=C bond is less than twice of C-C bond. (ii) (iii) Ethanol and butanol differs in the size of the non-polar hydrocarbon chain. Butanol is insoluble in water even though it can form hydrogen bonds with water. Its predominantly forms dispersion forces with water due to its long, non-polar hydrocarbon chain. The energy released during formation of these dispersion forces is not enough to overcome the hydrogen bonds between water molecules and the dispersion forces between butanol. (b) (i) (ii) The delocalised  electron cloud results in stability, so the loss of this aromatic character is not energetically favored. Instead, benzene tends to undergo substitution reactions so that its  electron cloud remains intact to maintain aromatic stability. (iii) Benzene do not undergo oxidation. Methylbenzene is oxidized by heating with KMnO 4, H2SO4 (aq)

Nanyang Junior College J2 H1 Chemistry 2017 3 (c) Bond energy of C-H bond = 410 kJ mol-1 Bond energy of C-Cl bond = 340 kJ mol-1 As fluorine atom is smaller than chlorine, bond length of C-F bond is shorter than C-Cl bond. Therefore C-F bond is expected to be stronger than C-Cl bond, hence they do not break easily to form free fluorine atoms to attack the ozone layer. 2 (a) (i) CrO 42– + 8H+ + 3e  Cr3+ + 4H2O (ii) Amount of electrons gained by 0.0150 mole of CrO 42– = 0.0450 mol Amount of electrons lost by 0.0225 mole of SO2 = 0.0450 mol Therefore each mole of SO 2 lost = (0.0450/0. 0225) = 2 mol of electron Original oxidation number of sulfur = +4 New oxidation number of sulfur = +4 + 2 = +6 (iii) SO 42– (b) (i) 2- 27 2- 2 + 2 4 [Cr O ]Kc = [CrO ] [H ] (ii) Amount of PbCrO4 initiallly = 20.00/ (207.2+52.0 +16.0x4) = 0.06188 mol [CrO42–] initially = 0.06188/(100/1000) = 0.6188 ≈ 0.619mol dm-3 (iii) [CrO 42–] at equilibrium = 0.6188 / 5 = 0.1237 = 0.124 mol dm-3 [Cr2O72–] at equilibrium = 0.6