2019 DHS Prelim H2 Chem P1 ANS

2019 Y6 Prelims H2 Chemistry 9729 Paper 1 Suggested Solutions  DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 C C B C C D C D A D 11 12 13 14 15 16 17 18 19 20 C C B A D C A B A D 21 22 23 24 25 26 27 28 29 30 D C C C B D B A A B 1 C From Reaction (1) 2NaN3 ≡ 2Na ≡ 3N2 From Reaction (2) 10Na(s) + 2KNO3(s) → K2O(s) + 5Na2O(s) + N2(g) 10Na ≡ 1 N2 Combining reactions (1) and (2), 2NaN3 gives 3N2 directly, while 2Na gives 1/5 N2 Hence, 2NaN3 gives a total of 16/5 N2 1.25NaN3 ≡ 2N2 2 C Moles of sodium percarbonate = 10/1000 x 0.1 = 0.001 mol Moles of CO2 = (48/1000) / 24 = 0.002 mol Since (Na2CO3)x•y(H2O2) ≡ xCO2 x = 2 Moles of KMnO4 = 24/1000 x 0.05 = 0.0012 mol 2KMnO4 ≡ 5 H2O2, Moles of H2O2 = 5/2 x 0.00120 = 0.003 mol Since (Na2CO3)x•y(H2O2) ≡ y H2O2 y = 3 y/x = 3/2 3 B Note that elements in the same group (e.g. O and S) will have the same number of unpaired electrons in their ground state. Group 17 elements have only 1 unpaired electron. Hence, option 1 is incorrect.  1 Electronic configuration of Ti: [Ar]3d24s2 (2 unpaired electrons) Cl: [Ne] 3s23p5 (1 unpaired electron)  2 Electronic configuration of Si: [Ne] 3s23p2 (2 unpaired electrons) O: 1s22s22p4 (2 unpaired electron)  3 Electronic configuration of Ni: [Ar]3d84s2 (2 unpaired electrons) S: [Ne] 3s23p4 (2 unpaired electron) 4 C pV = nRT For Graphs A and B, For a fixed mass of gas at constant T, pV = constant. Since we are plotting a p against pV graph, the graph should be a vertical line instead (i.e. x = constant graph) For Graphs C and D, For a fixed mass of gas at constant p, V = (nR/p)T The graph should be y=mx graph (upward sloping straight line that passes through origin), with gradient = (nR/p) Since I has a higher Mr than J, number of moles of I will be smaller than J and hence graph of I should have a smaller gradient. 5 C Carbonate Ethanoate C O O O 2- CH3 C O O - Nitrate Phenoxide N O O O - In ethanoate ion, the shape with respect to the methyl carbon is tetrahedral. The common feature is the delocalization of electrons. 6 D  1 At constant temperature, Kc remains unchanged. Hence G remains unchanged.  2 Reaction is spontaneous when G < 0. For 0 < Kc < 1, ln Kc is negative and hence G > 0 (non-spontaneous)  3 Adding a catalyst increases rate of both forward and backward reaction to the same extent, hence the position of the equilibrium remains unchanged. Kc remains unchanged and hence G remains unchanged.

Dunman High School 2019 Prelims – H2 Chemistry 9729/01 Solutions  DHS Chemistry Unit Page 2 of 6 7 C N2H5+