2020 YIJC Prelim Exam Paper 1 (Suggested Answers)

2020 JC2 H2 Chemistry 9729 Preliminary Examination 1 H2 CHEMISTRY 9729 Paper 1 Multiple Choice Questions 1 2 3 4 5 6 7 8 9 10 B A D C A C B A B C 11 12 13 14 15 16 17 18 19 20 C D D D B A B A D C 21 22 23 24 25 26 27 28 29 30 B B D D D A A B D C 1 Answer: B At ground state, the electronic configuration of C is 1s2 2s2 2p2.  The occupied orbital of the highest energy is the 2p orbital which is dumb-bell in shape.  The occupied orbital of the lowest energy is the 1s orbital which is spherical in shape.  There are 2 electrons in the p orbitals and 4 electrons in the s orbitals (1s and 2s).  The electrons are present in 3 different energy levels. 2 Answer: A The angle of deflection is proportional to charge/mass of the particle. The angle of deflection of X+ is smaller than that of P +. Since both particles have the same charge, X+ must have a larger mass than P+. 3 Answer: D C2H4: trigonal planar, 120o HCHO: trigonal planar, 120o NH4+: tetrahedral, 109.5o NH3: trigonal pyramidal, 107o 4 Answer: C C2H2: 5  bonds C2H5OH: 8  bonds CH3CHO: 6  bonds CH3CO2H: 7  bonds 5 Answer: A Since the flasks are identical, the volumes of the 3 gases are the same. The masses of the 3 gases are also the same. pV = nRT and n = m/Mr pV = (m/Mr)RT Since V, m and R are constant, Mr  T/P Mr(X) = k(t/p); Mr(Y) = 0.5k(t/p); Mr(Z) = 2k(t/p) 6 Answer: C Amount of CO2 at r.t.p. = 500 ÷ 24000 = 0.020833 mol In 1 molecule of CO2, there are 3 atoms. Amount of atoms = 3 0.020833 = 0.0625 mol Number of atoms = 0.0625 6.02 1023 = 3.76 1022 7 Answer: B H2S(g) + 1 2O2(g)  S(s) + H2O(g) ΔH = [ΔHf(products)] – [ΔHf(reactants)] ΔH = [0 + (243.0)] – [(20.5) + 0] = 222.5 kJ mol1 8 Answer: A ΔGo = ΔHo −TΔSo = (+61) – 298 (−30×10−3) = + 69.9 kJ mol−1  ΔGo is positive.  A positive ΔGo indicates that dissolution is not spontaneous, hence silver chromate is almost insoluble in water.  Since ΔS is negative, at a higher temperature, ΔG tends towards a more positive value. Dissolution becomes less spontaneous at higher temperature.

2020 JC2 H2 Chemistry 9729 Preliminary Examination 2 9 Answer: B  At low [substrate], the graph shows a straight line (similar to y = mx graph) shows that the rate is directly proportional to [substrate]. Hence the order of reaction is first order with respect to [substrate].  At high [substrate], the graph shows a horizontal line (similar to y = k graph) shows that the rate is independent of [substrate]. Hence the order of reaction is zero with respect to [substrate].  The enzyme-substrate is an intermediate of the reaction and hence should not appear in the rate equ ation since rate equation is expressed in terms of the concentrations of the reactants. 10 Answer: C 4X  2