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 Hess’s law, ∆H1 = ∆H2 + (∆H3 + ∆H4 + ∆H5 + ∆H6 + ∆H7) ‒ ∆H8 rearrange, ∆H2 = ∆H1 + ∆H8 ‒ (∆H3 + ∆H4 + ∆H5 + ∆H6 + ∆H7) Hence statement 3 is correct. 7 (A) Calibration methods account for heat loss by assuming a constant heat capacity C of the set-up. In experiment 1, reaction is a neutralisation reaction. Total amount of heat evolved depends on amount of water formed. n(water) = 2.0 x 50/1000 Using heat evolved = heat absorbed and letting C be heat capacity in kJ K‒1. 2.0 x 50/1000 x 57.4 = C x 10.0 -Eqn  For experiment 2, we calculate the enthalpy change of neutralisation for one mole of the weak acid H2C2O4. n(H2C2O4) reacted = 2.0 x 50/1000 x ½ (as NaOH is limiting) Using heat evolved = heat absorbed 2.0 x 50/1000 x ½ x |∆Hrxn| = C x 8.5 -Eqn  Taking Eqn/Eqn, 2.0 x 50/1000 x ½ x |∆Hrxn| 2.0 x 50/1000 x 57.4 = C x 8.5 C x 10.0 Hence |∆Hrxn| = 57.4 x 8.5 x 2 10

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 4 8 (D) Statement Equation ∆GƟ ∆SƟ 1 SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq) ‒ + 2 Cl2(g) + 2I‒(aq) → I2(aq) + 2Cl‒(aq) ‒ ‒ 3 MgCO3(s) → MgO(s) + CO2(g) + + Statement 1 only is correct. 9 (B) Graph of colour intensity against time shows 1 st order characteristics with constant half - lives. Since NO was used in excess, the variable is [Cl 2]. As reaction proceeds, rate decreases proportionally (flatter gradient) with decreasing [Cl 2]. Reaction is 1 st order wrt Cl2. Graph of rate against [NO] 2 shows a straight line with constant positive gradient passing through origin, rate is directly proportional to [NO]2. Hence 2nd order wrt to NO. Statement 1 is correct. Statement 2 test whether students can draw the link b etween orders of reaction and stoichiometric coefficient in the mechanism. 1 st order wrt to Cl 2 and 2nd order wrt NO tells us one molecule of Cl2 and two molecule of NO takes part in the reaction up to and including the slow step of the reaction. Hence, for the given mechanism, when the slow step is the second step, the mechanism will be consistent with the rate equation. Statement 3 is wrong. For an overall 3rd order reaction, units of k = mol dm-3 min-1 [mol dm-3] [mol dm-3] 2 = mol‒2 dm6 min‒1. 10 (D) At constant temperature, decomposition of hydrogen peroxide is a first order reaction, hence rate is directly proportional to [H2O2]. The time taken to use up 10% of initial amount remains constant. After 5 minutes, 90% of initial [H2O2] will remain. Prove: rate = k[H2O2] When [H2O2]=0.10 mol dm‒3, rate = 0.1k. 10% of 0.10 mol dm‒3 is 0.01 mol dm‒3. At a rate of 0.1k, it takes 5 mins to use up 0.01 mol dm‒3 of H2O2. When [H2O2]=1.00 mol dm‒3, rate = k. 10% of 1.00 mol dm ‒3 is 0.10 mol dm ‒3. When the amount to be used up increase by 10x and the rate also increase by 10x, the increases cancel out and the time taken will remain constant at 5 mins. Δ

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 5 11 (B) Human blood is buffered by the H2CO3, HCO3‒ system. The weak base will react with and remove small amounts of H+(aq) in the blood to maintain a relatively constant pH of 7.4. HCO3‒ + H+ → H2CO3 12 (C) For precipitation, ionic product ≥ Ksp. Substitute [IO3‒] = 0.10 mol dm‒3, [CrO42‒] = 0.10 mol dm‒3, [AsO43‒] = 0.10 mol dm‒3 into the corresponding ionic products. AgIO3: [Ag+] [IO3‒] ≥ 3.2 x 10–8 mol2 dm-6 [Ag+] ≥ 3.2 x 10–7 mol dm-3 Ag2CrO4: [Ag+]2 [CrO42‒] ≥ 9.0 x 10–12 mol3 dm-9 [Ag+] ≥ 9.5 x 10–6 mol dm-3 Ag3AsO4: [Ag+]3 [AsO43‒] ≥ 1.0 x 10–22 mol4 dm-12 [Ag+] ≥ 1.0 x 10–7 mol dm-3 The first precipitate is Ag3AsO4 and it appears at the smallest [Ag+] of 1.0 x 10–7 mol dm-3. 13 (A) Rearranging ∆GƟ = −nFEƟ, −∆GƟ/F = nEƟ A quick calculation of EƟ will give the more standard reduction potential values. half-equation −∆GƟ/F (V) EƟ / V UO22+ + e– UO2+ +0.16 +0.16 UO2+ + 4H+ + e– U4+ + 2H2O +0.27 +0.27 U4+ + e– U3+ −0.52 −0.52 U3+ + 3e– U −4.98 −1.66 Mg2+ + 2e– Mg −2.38 Mg metal is a reducing agent that will reduce UO2+ and itself oxidised to Mg2+. Using EcellƟ=EredƟ – EoxdƟ, we can see E cellƟ > 0 (and reaction will be spontaneous) when EredƟ > EMg2+/MgƟ. Since EUO2+/U4+Ɵ > EU4+/U3+Ɵ > EU3+/UƟ > EMg2+/MgƟ, UO2+ will be reduced by Mg metal all the way to U.

Nanyang Junior College 2019 J2 H2 Chemistry Prelim Exam Answers 6 14 (B) Using the equations Q=It and Q=nzF, where n = amount of Cu deposited, z = 2 for Cu2+ + 2e → Cu n = It zF The amount of charge supplied will affect the amount of Cu deposited. The amount of charge, Q, is affected by current used and the length of time for which the current is run. 15 (C) The strongest reducing agent has the lowest 1st + 2nd ionisation energies. IEs decrease down the group and increase across the period. Sr is in group 2, to the le