NYJC 2022 Prelim P1 Answer

J2 Prelims 2022 H2 Chemistry Paper 1 Answers and CommentsPage 1 of 7 NYJC 2022 H2 Chem 9729 P1 Answer Nanyang JC J2 Preliminary Examinations 2022 H2 Chemistry 9729/01 Paper 1 MCQ Answers and Comments Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 C 6 A 11 C 16 D 21 B 26 C 2 C 7 D 12 D 17 A 22 D 27 D 3 B 8 D 13 A 18 C 23 A 28 B 4 D 9 B 14 B 19 C 24 A 29 A 5 B 10 C 15 B 20 D 25 A 30 D 1 C Step 1: Th90 232 + n0 1 → Th90 233 Law of conservation of mass, addition of one neutron increase nucleon number by 1, does not change the number of protons, element remains as Thorium. Steps 2&3: Th90 233 → → U92 233 Decomposition of Q to uranium-233 does not involve addition of any new particles. Since two steps are involved, each should involve the decomposition of a neutron to a proton. n0 1 → p1 1 + e- -1 0 Therefore for each step, the proton number increase by 1, nucleon number remains the same. Hence: Th90 233 → Pa91 233 → U92 233 Neutron number of Pa91 233 = 233 – 91 = 142 2 C Comparing 3rd IEs, Cd2+: [Kr] 4d10 Rb2+: [Ar] 3d10 4s2 4p5 Sr2+: [Ar] 3d10 4s2 4p6 or [Kr] Xe2+: [Kr] 4d10 5s2 5p4 Across each period, IEs increase. The species with the noble gas configuration has the highest IE. Therefore 3rd IE of Rb < 3rd IE of Sr Down the group, IEs decrease . Hence, for ions with noble gas configurations, we’d expect the trend: [Kr] < [Xe] where [Xe] is [Kr] 4d 10 5s2 5p6, for 3 rd IE, a 2+ ion with this configuration is Ba2+. Hence Cd2+ < Ba2+ and Xe2+ < Ba2+. Since Ba2+ < Sr2+, Sr2+ has the highest 3rd IE. 3 B Solid Ga exist as covalent Ga 2 dimers, Ga−Ga molecules. There are weak instantaneous dipoles – induced dipoles attractions between the dimers which account for the low melting point. No strong covalent bonds need to be broken when Ga melts. Hence B is used to explain the low melting point of Ga but C is not. Liquid Ga can be considered as having strong metallic bonding with high boiling point due to the number of delocalised valence electrons present. Hence A is used to explain the boiling point of Ga. 4 D First, assume volume of the tyre is a constant at its maximum as long as internal pressure exceeds external pressure. i.e. the tyre will expand to its full volume and remains at its full volume when p internal > pexternal. Simplifying pV=nRT gives psea level Tsea level = pluggage Tluggage pluggage= psea level × Tluggage Tsea level pluggage= 6.8 × (273 +2) (273 +30) pluggage= 6.17 bar In the luggage hold, difference in pressure = 6.17 – 0.47 = 5.7 bar Hence, within the maximum allowed difference of 6 bar. Tyre will neither deflate nor burst. A is wrong. When temperature decrease from 30°C to 2°, pressure will also decrease as pressure is proportional to temperature. A: 7 B: 7 C: 7 D: 9

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