2024 Prelims ACJC H2 Chem P1 (Ans)

This document consists of 19 printed pages. Anglo-Chinese Junior College JC2 Preliminary Examination Higher 2 CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9729/01 9 September 2024 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, Centre number and index number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.

2 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over Solutions 2 0.01 mol of an unknown ion G2+ required 17.25 cm3 of 0.23 mol dm-3 acidified KMnO4 to reach the end-point. What is the final oxidation state of element G? A +3 B +4 C +5 D +6 MnO4− + 8H+ + 5e− → Mn2+ + 4H2O Amt of MnO4− = 0.01725 x 0.23 = 3.968 x 10−3 mol Amt of e− = 3.968 x 10−3 x 5 = 1.984 x 10−2 mol 1.984 x 10−2 / 0.01 = 1.98 ≈ 2 1 mol of X2+ loses 2 mol of e−, so final O.S of X is +4. 1 D 6 B 11 C 16 A 21 D 26 B 2 B 7 A 12 C 17 D 22 C 27 B 3 B 8 D 13 D 18 D 23 B 28 B 4 C 9 A 14 A 19 C 24 C 29 B 5 D 10 B 15 D 20 A 25 B 30 D 1 The incomplete combustion of a gaseous hydrocarbon produced 80 cm 3 of carbon dioxide, 40 cm3 of carbon monoxide and 160 cm3 of water vapour. What volume of oxygen was used for combustion of the hydrocarbon? A 40 cm3 B 80 cm3 C 160 cm3 D 180 cm3 CxHy + 9 2 O2 → 2CO2 + CO + 4H2O 40 80 40 160 Comparing volume ratio of gases, 40 × 9 2 = 180 cm3

3 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over 3 Which ion will be deflected the most in an applied electric field? A 79Br+ B 81Br2+ C 81Br+ D 82Br2+ Angle of deflection is directly proportional to charge/mass ratio. 81Br2+ has the largest charge/mass ratio, so it will be deflected the most in an electric field. 4 An unstable ion has • a nucleon number of 219, • 51 more neutrons than electrons • an atomic number of 84, 85, 86, or 87. What could this ion be? A Po2+ B At3+ C Rn4+ D Fr5+ Po2+ At3+ Rn4+ Fr5+ number of protons 84 85 86 87 number of electrons 82 82 82 82 number of neutrons 133 133 133 133 nucleon number 217 218 219 220 5 Which species contains two 𝜋 bonds? 1 BF3NH3 2 CH2CHCH2CH3 3 CH2CHCHO 4 HCO2CH2COCH3 A 1 and 4 only B 2 and 3 only C 2 and 4 only D 3 and 4 only 1 contains a dative bond between the boron atom and nitrogen atom. 2 contains only one 𝜋 bond in the alkene.

4 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over pV p N2 B A C D 3 contains two 𝜋 bonds, one in the alkene and one in the aldehyde. 4 contains two 𝜋 bonds, one in ester and one in ketone. 6 What is the strongest intermolecular force in ethanal, ethylamine and decan-1-ol? ethanal ethylamine decan-1-ol A hydrogen bonds hydrogen bonds induced dipoles B permanent dipoles hydrogen bonds induced dipoles C permanent dipoles permanent dipoles hydrogen bonds D hydrogen bonds permanent dipoles hydrogen bonds Ethanal has permanent dipoles between its molecules as it is a polar molecule. Ethylamine has hydrogen bonding between its molecules due to the lone pair of electrons on oxygen and the hydrogen bonded to oxygen in its molecules. While there is hydrogen bonding between molecules of decan-1-ol, s trongest intermolecular force in decan-1-ol is instantaneous dipole-induced dipoles due to the large electron cloud of the molecule. 7 The volumes and pressures of equal masses of two gases, N 2 and NH3, are separately investigated, at constant temperature. The results are plotted on a graph of pV against p. Both gases behave as ideal gases under the conditions chosen. The result for N2 is given. Which plot shows the result for NH3? Since both gases behave ideally, the pV against p plot for NH3 is also a constant. NH3 has a lower molar mass than N 2. Since equal mass of gas is used, there are more NH3 and given pV = nRT, its pV value will be higher. 8 What can be added to a mixture of MgO and Al2O3 to separate them by filtration?

5 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over 1 water 2 HCl(aq) 3 NaOH(aq) A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 3 only MgO and Al2O3 are both insoluble in water, thus cannot be separated with water and then filtration. Both Al2O3 and MgO are soluble in aqueous HC l as they react with HC l to give soluble products. Hence, they cannot be separated by HCl(aq) and then filtration. Al2O3 is soluble in aqueous NaOH to form NaA l(OH)4 while MgO is insoluble. Hence, NaOH(aq) and then filtration can separate the two oxides. 9 The following table shows the results of two experiments involving Group 17 halides, X− and Y−. experiment deduction halogen Z2 added to X− X2 formed halogen Z2 added to Y− Y2 not formed Which row shows the halogens in decreasing order of oxidising strengths? A Y2, Z2, X2 B Y2, X2, Z2 C X2, Z2, Y2 D X2, Y2, Z2 Z2 can oxidise X−. Thus, Z2 is a stronger oxidising agent than X2. There is no visible reaction between Z2 and Y−, Z2 cannot oxidise Y−. Hence Y2 is a stronger oxidising agent than Z2 Thus, the strongest oxidising agent is Y2, followed by Z2, then X2.

6 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over 10 An Ellingham diagram is a plot of ∆G versus temperature and it can be used to show the stability of compounds at various temperatures. The following Ellingham diagram is for reactions 1 and 2. Reaction 1: W + X → Y Reaction 2: 2W + X → 2Z Which statement is incorrect? A Reaction 1 is favoured at lower temperatures. B The entropy change of reaction 2 is negative. C The enthalpy change of reaction 2 is −220 kJ mol−1 D At 983 K, G of the reaction 2Z → W + Y is zero. A is correct. As temperature decreases, G becomes more negative. B is incorrect. G = H−TS. The gradient of the graph is −S. Since gradient of reaction 2 is negative, S (entropy change) of reaction 2 is positive. C: Since G = H−TS, the y-intercept is H, hence is −220 kJ mol−1. D: The lines for reaction 1 and 2 intersect at 983 K, the G for both reactions have the same value, approximately −370 kJ mol−1. Reaction 1: W + X → Y −370 kJ mol−1 -600 -500 -400 -300 -200 -100 0 0 500 1000 1500 2000 T / K ∆Go / kJ mol−1 983 −220 reaction 1 reaction 2

7 © ACJC2024 9729/Preliminary Examinations/2024 [Turn over Reaction 2: 2W + X → 2Z −370 kJ mol−1 The reaction: 2Z → W + Y is Reaction 1 − Reaction 2 Hence, G of the reaction is −370 −(−370) = 0 11 When an instant cold pack is used, a vigorous reaction occurs, and the temperature falls from 25 °C to 5 °C. What are the correct signs of ∆G and ∆S for this reaction? ∆G ∆S A + + B + − C − + D − − A spontaneous reaction occurred (indicated by the drop in temperature) so ∆G is negative. The reaction is endothermic as the temperature fell. ∆S must be positive