2019 SAJC H2 Chem Prelim P1 ANS

1 NAME Class ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examination H2 Chemistry (9729) Paper 1 Multiple Choice 19 Sep 2019 1 hour Additional Materials: Multiple Choice Answer Sheet, Data Booklet READ THESE INSTRUCTIONS: Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. 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. 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. This document consists of 14 printed pages (including this page).

2 1 Which statement about one mole of sodium metal is always true? A It has the same mass as one mole of 12C. B It has the same number of atoms as 18 g of water. C It has the same number of atoms as 1 12 mole of 12C. D It has the same number of atoms as 12 dm3 of fluorine gas at r.t.p. Ans: D 12/24 = 0.5 mol of F2 = 1.0 mol of F atoms = 6.02 x 1023 H atoms 2 0.84 g of an oxide MO of a metal M was dissolved in excess sulfuric acid. 25.0 cm 3 of 0.12 mol dm3 potassium manganate(VII) solution was required to oxidise M2+ to M3+. What is the relative atomic mass of M? A 36.0 B 40.0 C 52.0 D 56.0 Ans: B Amt of MnO4 = 25/1000 x 0.12 = 0.003 mol Amt of e gained by MnO4 = Amt of e lost by M2+ = 5 x 0.003 = 0.015 mol Mr of MO = 0.84/0.015 = 56 Ar of M = 56-16 = 40 3 X and Y are elements found in the first three periods of the Periodic Table. The outermost shell electronic configurations of two species are given as follows: X2: Y2+: What can best be deduced from the above information? A X has a larger proton number than Y. B X has more unpaired electrons than Y at the ground state. C X exists as a gas while Y is a solid at standard condition. D X2 and Y2+ are isoelectronic. Ans: B Since X and Y are elements in the first three periods and their valence electronic configurations consist of s and p subshells, there are likely to be in period 2 or period 3. X has the valence shell electronic configuration ns2 np4 so it is a Group 16 element and it can be in Period 2 or Period 3. X can be O or S. Y has the valence shell electronic configuration of (n+1)s2 and it should be in Period 3. Y can be Mg.

3 Option A: Since Y is Mg, X can have a smaller or larger proton than Y, depending on whether X is O or S Option B: X would have 2 unpaired electrons while Y has no unpaired electrons. Option C: X can be oxygen which is a gas but it may also be sulfur which is a solid at standard condition. Option D: If X is S, S2 is not isoelectronic with Mg2+. 4 In an experiment, a sample of gaseous 87Sr2+ was passed through an electric field. The angle of deflection for 87Sr2+ was observed to be 2. The experiment was repeated with gaseous sample of particle Q. Which of following could be Q? A 74As3 B 19F C 79Se2 D 127Te2 Ans: D Angle of deflection = k(charge/mass) k = 2 x 87/2 = 87 Angle of deflection of 74As3 = 87(3/74) = 3.5o Angle of deflection of 19F = 87(1/19) = 4.6o Angle of deflection of Se = 87(2/79) = 2.2o Angle of deflection of Te2 = 87(2/127) = 1.4o 5 BeCl2 reacts with CH3NH2 to form compound Z (Mr = 142.0). Which of the following statements are correct? 1 The hybridisation state of N in Z is sp3. 2 Hydrogen bonds exist between molecules of compound Z. 3 1 mol of compound Z is formed from 1 mol of BeCl2 and 2 mol of CH3NH2. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 3 only Ans: C 87Sr2+ ─ + Particle Q

4 Option 1 is correct as N has a sp3 hybridisation state and hence tetrahedral shape around N in compound Z. Option 2 is wrong since there is no lone pair on the N after bonded to Be, hence between the molecules, H-bond no longer exist. Option 3 is correct as Be in BeCl2 has only 4 electrons and it can accommodate another 4 electrons to react octet configuration. 6 Which of the following graphs correctly describes the variation of pV with temperature for a fixed amount of an ideal gas? Ans: A pV = nRT. For an ideal gas, pV  T, hence straight line. But since the scale is in oC, pV = 0 only when T = -273 C. 7 The conversion of graphite into diamond is an endothermic reaction. C(graphite)  C(diamond) Which of the following statements are correct? 1 The carbon-carbon bonds in graphite are stronger than that in diamond. 2 The activation energy of the conversion of graphite to diamond is larger than that of the reverse reaction. 3 The enthalpy change of atomisation of diamond is less endothermic than that of graphite. 4 The enthalpy change of combustion of diamond is less exothermic than that of graphite. pV / Pa m3 T / oC A B C D 0

5 A 1 and 2 only B 1 and 3 only C 1, 2 and 3 only D 1, 2 and 4 only Ans: C Option 1 is correct. Since ΔH > 0, C-C bond energy in graphite is higher than the C-C bond energy in diamond. Option 2 is correct. Since ΔH > 0, the Ea of forward reaction is larger than that of backward reaction. Option 3 is correct. ΔHatomof diamond is less endothermic than that of graphite Option 4 is incorrect: ΔHc of diamond is more exothermic than that of graphite 8 Given the following enthalpy changes: ΔH / kJ mol1 Enthalpy change of formation of H2S(g) 20.6 Enthalpy change of formation of H2O(l) 286.0 Enthalpy change of vaporisation of H2O(l) +40.7 What is the enthalpy change (in kJ mol-1) of reaction for the following reaction? H2S(g) + ½O2(g) → H2O(g) + S(s) A 224.7 B 265.4 C 306.6 D -347.3 Ans: A H2S(g) + ½O2(g) → H2O(g) + S(s) H2O(l) + S(s) H2(g) + S(s) + ½O2(g) ΔHr = 286 + 40.7 + 20.6 = 224.7 C(g) C(diamond) C(graphite) ΔHatom ΔHatom CO2(g) ΔHc ΔHc -20.6 -286 +40.7 ΔHr

6 9 Which equation corresponds to the enthalpy change stated? A H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) ∆Hneutralisation Ɵ B Na+(s) + aq  Na+(aq) ∆Hhydration Ɵ (Na+) C Al2O3(s)  2Al3+(g) + 3O2(g) ∆Hlattice energy Ɵ (Al2O3) D O2(g)  2O(g) 2∆Hatomisation Ɵ (O2) Ans: D Option A: H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2H2O(l) 2∆Hneutralisation Ɵ Option B: Na+(g) + aq  Na+(aq) ∆Hhydration Ɵ (Na+) Option C: 2Al3+(g) + 3O2(g)  Al2O3(s) ∆Hlattice energy Ɵ (Al2O3) 10 A chemical plant illegally dumped some radioactive waste in a landfill. This waste composed of two radioactive isotopes X and Y. The half-life of X is 4 days whereas that of Y is 2 days. The authorities found out about this illegal dumping only when the waste had been in the landfill for 8 days. They did an immediate analysis on a sample of the waste and found equal amounts of X and Y. Considering that the decay of radioactive isotopes follows first -order kinetics, what is the initial molar ratio of X to Y if the waste had been in the landfill for 4 days? X : Y A 1 : 2 B 1 : 4 C 2 : 1 D 4 : 1 Ans: A Day 0 1 2 3 4 5 6 7 8 X 4x 2x x Y 16x 8x 4x 2x x X:Y = 2x: 4x = 1x : 2x =1:2 11 Hydrogen reacts with nitrogen monoxide to form nitrogen and steam only. The rate equation for this reaction is rate = k[NO]2[H2]. Which could be the mechanism for this reaction? A slow NO + 2H2  NH2 + H2O NH2 + NO  N2 + H2O

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