NYJC 2022 Prelim P1 Answer

J2 Prelims 2022 H2 Chemistry Paper 1 Answers and CommentsPage 1 of 7 NYJC 2022 H2 Chem 9729 P1 Answer Nanyang JC J2 Preliminary Examinations 2022 H2 Chemistry 9729/01 Paper 1 MCQ Answers and Comments Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 C 6 A 11 C 16 D 21 B 26 C 2 C 7 D 12 D 17 A 22 D 27 D 3 B 8 D 13 A 18 C 23 A 28 B 4 D 9 B 14 B 19 C 24 A 29 A 5 B 10 C 15 B 20 D 25 A 30 D 1 C Step 1: Th90 232 + n0 1 → Th90 233 Law of conservation of mass, addition of one neutron increase nucleon number by 1, does not change the number of protons, element remains as Thorium. Steps 2&3: Th90 233 → → U92 233 Decomposition of Q to uranium-233 does not involve addition of any new particles. Since two steps are involved, each should involve the decomposition of a neutron to a proton. n0 1 → p1 1 + e- -1 0 Therefore for each step, the proton number increase by 1, nucleon number remains the same. Hence: Th90 233 → Pa91 233 → U92 233 Neutron number of Pa91 233 = 233 – 91 = 142 2 C Comparing 3rd IEs, Cd2+: [Kr] 4d10 Rb2+: [Ar] 3d10 4s2 4p5 Sr2+: [Ar] 3d10 4s2 4p6 or [Kr] Xe2+: [Kr] 4d10 5s2 5p4 Across each period, IEs increase. The species with the noble gas configuration has the highest IE. Therefore 3rd IE of Rb < 3rd IE of Sr Down the group, IEs decrease . Hence, for ions with noble gas configurations, we’d expect the trend: [Kr] < [Xe] where [Xe] is [Kr] 4d 10 5s2 5p6, for 3 rd IE, a 2+ ion with this configuration is Ba2+. Hence Cd2+ < Ba2+ and Xe2+ < Ba2+. Since Ba2+ < Sr2+, Sr2+ has the highest 3rd IE. 3 B Solid Ga exist as covalent Ga 2 dimers, Ga−Ga molecules. There are weak instantaneous dipoles – induced dipoles attractions between the dimers which account for the low melting point. No strong covalent bonds need to be broken when Ga melts. Hence B is used to explain the low melting point of Ga but C is not. Liquid Ga can be considered as having strong metallic bonding with high boiling point due to the number of delocalised valence electrons present. Hence A is used to explain the boiling point of Ga. 4 D First, assume volume of the tyre is a constant at its maximum as long as internal pressure exceeds external pressure. i.e. the tyre will expand to its full volume and remains at its full volume when p internal > pexternal. Simplifying pV=nRT gives psea level Tsea level = pluggage Tluggage pluggage= psea level × Tluggage Tsea level pluggage= 6.8 × (273 +2) (273 +30) pluggage= 6.17 bar In the luggage hold, difference in pressure = 6.17 – 0.47 = 5.7 bar Hence, within the maximum allowed difference of 6 bar. Tyre will neither deflate nor burst. A is wrong. When temperature decrease from 30°C to 2°, pressure will also decrease as pressure is proportional to temperature. A: 7 B: 7 C: 7 D: 9

J2 Prelims 2022 H2 Chemistry Paper 1 Answers and CommentsPage 2 of 7 NYJC 2022 H2 Chem 9729 P1 Answer Bicycle tyres are generally engineered to work even at 0°C and below. (You can cycle in winter!) 5 B Using pV T = nR, we can see the gradient of a plot of pV against T gives nR. For gas D, nR = 498.6/600 = 0.831. Amount of gas D is 0.831/8.31 = 0.1 mol. Since gradient for gas E is roughly double, amount of gas E is 0.2 mol. From calculations of amount, 34.02 g of SiCl 4 and 12.82 g of SO 2 both gives 0.2 mol. 3.100 g of P and 0.2000 g of H2 both gives 0.1 mol. Hence C and D are wrong. E is a gas at 300 K. Hence 0.2 mol of SO 2 (gas at rtp) is more likely to be gas E than 0.2 mol of SiCl4 (liquid at rtp). NB: 0.2 mol of SO2 has a pV < 498.6 due to negative deviation from ideal gas behaviour. Presence of significant instantaneous dipole – induced dipole attractions cause real gases to exert a smaller pressure than expected. 6 A A is correct. 1 mol of CO 2(g) is formed from 1 mol of C (graphite) which is a solid at standard conditions and 1 mol of O2 gas. B is wrong. ∆ Hneut is for 1 mol of water formed between the reaction of an acid and an alkali. C is wrong. The equation given is for ∆Hʅhyd, enthalpy of hydration. D is wrong. The equation given is for 1 st EA + 2nd EA of sulfur. 7 D ∆Gʅ ∆Sʅ 1 SO3(l) + H2O(l) → H2SO4(aq) − + 2 Cl2(g) + 2I−(aq) → I2(aq) + 2Cl−(aq) − − 3 MgCO3(s) → MgO(s) + CO2(g) + + 8 D A cannot prove order wrt CN −. As rxn proceeds, [CN−] remains relatively constant as it is in huge excess. To deduce order wrt to CN−. we look at how rate changes as [CN−] changes. Since [CN−] does not change, any change in rate is not due to CN−. B is the opposite of A. It would allow us to determine order wrt CN −. However, the graph of [(1 -bromoethyl)benzene] is plotted instead. We cannot determine order wrt CN− when the rate appears to be zero. C is wrong. When the same concentrations of [CN −] and [(1 -bromoethyl)benzene] are used, the dec rease in concentrations must follow the same shape i.e. since we know the graph of [(1 -bromoethyl)benzene] against time follows 1 st order kinetics with constant half-lives, the graph of [CN −] against time must follow the same overall first order kinetics with constant half-lives. D is therefore the correct answer. Another way to understand why the shape of the graph of [CN −] against time will show constant half -lives is to recognise the mechanism as S N1. In the slow step, [(1 - bromoethyl)benzene] decreases following first order kinetics. Any carbocation formed immediately reacts with CN− in the fast step. Hence the decrease in [CN −] will mirror the decrease in [(1-bromoethyl)benzene]. 9 B The operating conditions of Haber process is 500 °C and 60 atm in the presence of Fe catalyst. A mid-high temperature is used to increase rate of reaction. Very high temperatures shift POE to the left, favouring backward endothermic reaction to absorb heat energy and resulting in poor yield. Low temperatures cause the equilibrium to be established too slowly. A relatively high pressure of 60 atm is used to shift POE to the right to reduce total amount of gas particles. 500 atm is too high resulting in the need for very thick steel containers and high costs.

J2 Prelims 2022 H2 Chemistry Paper 1 Answers and CommentsPage 3 of 7 NYJC 2022 H2 Chem 9729 P1 Answer 10 C pKw = pH + pOH A: pH = 1. [H3O+] = 0.1 mol dm−3. [HCl] = 0.1 mol dm−3 B: pH = 1. [H3O+] = 0.1 mol dm−3. [HCl] = 0.1 mol dm−3 C: pH = 13. pOH = 13.5 – 13 = 0.5 [NaOH] = 10-0.5 = 0.316 D: pH = 13. pOH = 14.5 – 13 = 1.5 [NaOH] = 10-1.5 = 0.0316 Hence, C has the highest concentration of ions. 11 C First, calculate the pKa values. Ka pKa A 6.3  10−3 2.2 B 2.0  10−5 4.7 C 6.3  10−8 7.2 D 2.0  10−10 9.7 To determine Ka2, use the H−H equation. pH = pKa2+ lg [HPO4 2-] [H2PO4 -] At maximum buffer capacity, [HPO 42−] = [H2PO4−], pH = pKa2. Hence, we need to find the point in the graph where [HPO 42−] = [H2PO4−]. From graph, when [HPO 42−] = [H 2PO4−], pH ~ 7.2. 12 D CH3COCO2H + OH− → CH3COCO2− + H2O At equivalence point, a weakly alkaline solution where pH > 7 is formed. Hence answer is D. 13 A Chiral Carbons Plane of Symmetry 1 Br CH3 Yes No 2 Br Br Yes No 3 C CC CH3 H CH3 H No No All three molecules rotate plane -polarised light as all does not have a plane of symmetry. 14 B A is correct. This can be seen from step 2. The monomers , CH2=CH2, are added together to form the polymer , (CH3)3C−O−(CH2CH2)nCH2CH2•, using free radicals (CH3)3C−O•. B is wrong. The formation of free radical in step 1 involves homolytic fission where the two electrons of the O−O bond goes back to each O atom, (CH3)3C−O•. C is correct. In a propagation step, a free radical, (CH3)3C−O• reacts with a molecule , CH2=CH2 to form a free radical intermediate