2019 NYJC Prelim P1 P2 P3 P4 Answers

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 1 H2 Chemistry Prelim Exam Answers (Strictly for Internal Circulation only) Paper 1 Answer Key Paper 1 Worked Solutions 1 (A) Charge of nucleon A = (+⅔) + 2(-⅓) = 0 Charge of nucleon B = 2(+⅔) + (-⅓) = +1 Hence, A is a neutron, B is a proton. Statement 1 is correct. Statement 2: A neutrino does not have a charge. Hence the charge of C must be -1. C will be attracted to the positive plate. Statement 2 is correct. Statement 3: Ne10 22 → Na11 22 is correct. When a neutron breaks up into a proton, mass number remains constant; proton number increase by 1. 2 (B) The ratio of O 2: Cl‒ is 3:2. In chlorates (or any oxoanions), O exist as oxides with OS -2. Hence each O is oxidised from -2 in chlorates to 0 in oxygen gas. Total number of electrons lost by three O2 molecules = 3 x 2 x 2 = 12 Total number of electrons gained by two Cl‒ = 12 Number of electrons gained by each Cl‒ = 6 Change in OS of Cl = -6 Final OS – Initial OS = -6 Initial OS = -1 – (-6) = +5 1 A 11 B 21 D 2 B 12 C 22 B 3 B 13 A 23 A 4 B 14 B 24 C 5 D 15 C 25 A 6 C 16 D 26 D 7 A 17 A 27 C 8 D 18 C 28 C 9 B 19 A 29 C 10 D 20 C 30 B

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 2 3 (B) Cations with higher charge and smaller ionic radius have higher charge density, will polarise anion electron cloud more to form a more covalent bond. For ionic compounds of Group 14 halides, we look for cation with smaller 2+ charge and larger ionic radius (further down the group), hence Pb2+ (PbF2). For covalent compounds, we look for cation with larger 4+ charge and smaller ionic radius (up the group), hence Sn4+ (SnCl4). (Note: Although n ot tested here, students should know that anions with larger electron cloud, are more easily polarised, form more covalent bonds. Hence chlorides form more covalent bonds than fluorides.) 4 (B) NN C NC O H O H 120 105 118 180 120 There are no sp3 hybridised carbon atom with 4 bond pairs and 0 lone pairs to give a 109° bond angle. 5 (D) At constant temperature and pressure, V  n. The initial 5 cm3 is due to air particles. The final reading of 81 cm3 includes the volume due to the air particles. After subtracting off 5 cm 3, the volume due to the volatile liquid only is 76 cm3. Mr = mRT pV Mr = 0.253 x 8.31 x (273+97) 101 325 x 76 x 10-6 Mr = 101.0

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 3 6 (C) This question test how well students know the definitions of the different enthalpy changes. Statement 1: ∆H7 ≡ LE(CaCO3) and ∆H8 ≡ ‒∆Hsol(CaCO3) From ∆Hsol = ∑∆Hhyd ‒ LE, LE = ∑∆Hhyd ‒ ∆Hsol = ∆Hhyd (Ca2+) + ∆Hhyd (CO32–) ‒ (‒∆H8) Hence statement 1 is wrong. Statement 2: ∆H7 ≡ 1st + 2nd IE of Ca = 590 + 1150 = 1740 Hence statement 2 is correct. Statement 3: ∆H2 ≡ ∆Hatm(Ca) From He