JPJC Prelim H2 Chemistry P1 (Worked Solution)

2021 JPJC JC1 H2 Chemistry (9729) 1 2021 Preliminary Exam Paper 1 (Worked Solution) JURONG PIONEER JUNIOR COLLEGE 2021 JC2 H2 Chemistry (9729) Preliminary Exam Paper 1 (Worked Solutions) Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 B 6 A 11 C 16 B 21 A 26 D 2 C 7 A 12 B 17 D 22 B 27 C 3 C 8 A 13 C 18 D 23 C 28 B 4 D 9 A 14 B 19 C 24 D 29 A 5 B 10 D 15 C 20 D 25 A 30 B 1 Answer: B The reaction causes the original particle to lose 1 proton, 1 electron and gain 1 neutron. Options A and C both have the same number of electrons and protons before and after the reaction. Option D shows a decrease in number of electrons and protons (36 in Kr to 35 in Br) but the number of neutrons remain unchanged (both particles have 40 neutrons). 2 Answer: C From Figure 1, large increase between the 2nd and 3rd electron removed indicates that there are two valence electrons in element P. Hence element P is in group 2. P+: ns1 From the electronic configuration of P+, the point corresponding to the second IE of element P is C. 3 Answer: C 8A SiCl4 has 4 bond pairs and no lone pairs around Si Î bond angle is 109.5 °. 8B SO2 has 2 bond pairs and 1 lone pair around S Î bond angle is < 120 °. 9C IF2– has 2 bond pairs and 3 lone pairs around I Î bond angle is 180 °. 8D CH3+ has 3 bond pairs and no lone pairs around C Î bond angle is 120 °. 4 Answer: D 9A and B The structures of N2O4 and (CH3CO2H)2 dimers are shown below. 9C The CH3 groups in (CH3CO2H)2 is still tetrahedral around the C, thus the dimer is non- planar. 8D Due to p-p orbital overlap, the electrons are delocalised across the O=N-O bond, making all the nitrogen-oxygen bonds to be of the same strength and thus of the same length.

2021 JPJC JC1 H2 Chemistry (9729) 2 2021 Preliminary Exam Paper 1 (Worked Solution) 5 Answer: B §· § · § · u u u ¨¸ ¨ ¸ ¨ ¸©¹ © ¹ © ¹ r 92.2 4.7 3.1 of Si 28 29 30 28.109100 100 100A 8A Ar of Si = 28.099 8C Ar of Si = 28.668 8D Ar of Si = 28.854 6 Answer: A { 42 33 42 2 2 Since 1 CH 2 O 50 cm of CH requires 100 cm of O for complete reaction. Since CO will be absorbed by the alkaline KOH, it will not be collected. Volume of gas collected volume of O left 150 100 350 cm 7 Answer: A u uu ' u u ' uu ' u uu 'u? y u u uu kJ kJ Total amount of heat evolved 1371 46.0 300Total amount of heat transferred 300 J 1000 300 1371 300 46.0efficiency 100% 100%1000 46.0 1000 1371 m cTcT cT m cT m 8 Answer: A Using ΔG = ΔH – TΔS, the negative gradient of the graph in the Ellingham diagram corresponds to ΔS of the reaction. Reaction II has a gradient of zero, that means ΔS = 0 (reject options B and C) Reaction I has a positive gradient, that means ΔS < 0 Î decrease in disorderedness Reaction III has a negative gradient, that means ΔS > 0 Î increase in disorderedness 9 Answer: A The graph shows that when pressure increases, % products at equilibrium decreases. Î POE shifts left to form less gas