EJC 2020 JC2 Prelim H2 Chemistry Paper 1 Worked Solution

2020 JC2 Prelim Exam H2 Chemistry 9729 Paper 1 Worked Solution 1 angle of deflection q m  ion 7Li+ 9Be2+ 16O2– 14N3– q m . 1 7 0 143   . 2 9 0 222   . 2 16 0 125   . 3 14 0 215    B 2 1 The electronic configuration of Cr is [Ar] 3d5 4s1. The five d -electrons occupy each d-orbital singly. 2 The electronic configuration of Mn 2+ is [Ar] 3d5. The five d -electrons occupy each d-orbital singly. 3 The electronic configuration of both Fe2+ and Co3+ are [Ar] 3d6.  C 3  D 4 1 Al2O3 is amphoteric. 2 Charge density of A l3+ is higher than Na+, which gives it higher polarising power, hence imparting higher degree of covalent character to the Al–O bond. 3 latt qqH rr     Since 3A Naqq l and 3A Narr l     latt 2 3 latt 2A O Na OHH   l  Hlatt of Al2O3 is more exothermic.  A 5 mpV nRT pV RT M m RT RTp V M M        5 3 1 gradient 3 10 Pa 8.31 500 1.5 1000 g m 20.8 g mol RT M M M       Neon has a molar mass closest to this.  B 6  24 3Co SO 20.3 0.05 mol406n  A  3 24 3Co SOCo 2nn   and  2 2443 Co SOSO 3nn    32 2443ions Co SOCo SO 5 0.25 moln n n n     23 ionsno. of ions 1.5 10nL    B  3 24 3Co SOCo 2 2 0.05 0.1 molnn      C 2 58.9mass % of Co 100% 29.3%406    D 4 3 16.0mass % of O 100% 47.3%406     C 7 4BrO 20.0 reacted 0.02 0.0004 mol1000n     2NH OH 80.0 reacted 0.01 0.0008 mol1000n     42BrO 2NH OH 6 e  Every 1 mol of NH2OH lose 3 mol of e–  2O.N. of N in NH OH 3 1 2 1    O.N. of N in product 1 3 2    Product is NO (O.N. of N = +2)  A 8 1 Both processes refers to the enthalpy change to form one mole of gaseous metal atoms from one mole of the metal in its standard state. 2  sublimation fusion vapourisation ,m.p. b.p.H H H H      l The energy required to raise the temperature of the molten metal from its melting point to its boiling point is not taken into account. 3 Xenon can be a solid at temperatures below its melting point.  A 9 rS 0 since there is a decrease in the number of gaseous particles. rG rH rTS rH  rTS can be either positive or negative pending on T (which is always >0). rG r0, when HT  rS  C 10 Given that   rate k EF . A Gradient of graph = instantaneous rate, should decrease as reaction proceeds, since E and F are being consumed. B [G] which is the product concentration should increase as reaction proceeds. Since [E]0 = [F]0 and 1E ≡ 1F, therefore [E] = [F]. So the rate equation can be rewritten        22 rate k k k  E F E F . Plot of [ E][F] against t is the same as plot of [E]2 against t, for a second order reaction with rate equation   2 rate k E . Since plot of [E] against t is a curve. C Plot of [ E]2 against t will not be a straight line. D Plot of [E]