NYJC 2022 Prelim P2 Answer

[Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME CHEMISTRY 9729/02 Paper 2 Structured 12 September 2022 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions in the spaces provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /15 2 /10 3 /6 4 /24 5 /8 6 /12 Total /75 This document consists of 20 printed pages.

2 H2 Chemistry 9729/02 NYJC J2/22 PX [Turn Over For Examiner's Use Answer all questions in the spaces provided. 1(a) X and Y are oxides of elements in the third period of the Periodic Table. The oxidation number of the Period 3 element in X and Y is +3 and +5 respectively. X reacts with both aqueous sodium hydroxide and aqueous hydrochloric acid. Y reacts with aqueous sodium hydroxide, but not with aqueous hydrochloric acid. (i) Identify the formulae of X and Y. X Al2O3 ................................ .................. Y P4O10 ................................ ............ [1] (ii) Write equations for the reactions of X and Y with aqueous sodium hydroxide. X: Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4 [1] ................................ .......................... Y: P4O10 + 12NaOH → 4Na3PO4 + 6H2O [1] ................................ ........................ [2] (b) The graph below shows the boiling points of HCl, HBr and HI. (i) Explain the trend of the boiling points of HCl, HBr and HI. HCl, HBr and HI have simple molecular structures with weak instantaneous dipoles- induced dipoles (id-id) between molecules. [1] The size of electron cloud increases from HCl to HBr to H I, the electron cloud gets more easily polarised. Hen ce more energy is required to overcome the stronger id -id, giving rise to increasing boiling points. [1] ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... ..................................................................................................................... [2] 160 180 200 220 240 260 280 300 HF HCl HBr HI boiling point / K HF HCl HBr HI 300 280 260 240 220 200 180 160 [1]

3 H2 Chemistry 9729/02 NYJC J2/22 PX [Turn Over For Examiner's Use (ii) Complete the sketch on page 2 to predict the boiling point of HF. Explain your answer. HF has hydrogen bonding between molecules. Hydrogen bonding is stronger than id- id hence more energy required to overcome hydrogen bonding. HF has the highest boiling point. [1] ......................................................................................................................... ......................................................................................................................... ..................................................................................................................... [2] (c) Zinc nitrate decomposes at 300 °C whereas barium nitrate decomposes at 600 °C. A 20.0 g sample containing a mixture of zinc nitrate and barium nitrate was heated at 350 °C until no further change occurred. A brown gas and another gas that relights glowing splint were evolved. The remaining white solid weighed 11.3 g. (i) Draw the dot-and-cross diagram for zinc nitrate. Balanced charges – 1 mark; correctly drawn nitrate ion – 1 mark [2] (ii) Use data from the Data Booklet, explain why zinc nitrate decomposes at a much lower temperature than barium nitrate. • Ionic radius of Zn2+(0.074 nm) is much smaller than Ba2+(0.135 nm) • Both have same ionic charge +2 hence charge density (thus polarising power) of Zn2+ is higher. • Electron cloud of NO3− is polarised to larger extent in Zn(NO3)2. • N−O bond is weakened to larger extent. • Hence Zn(NO3)2 has a lower thermal stability /greater ease of thermal decomposition and thermal decomposition temperature is lower. 5●: 3 marks; 4●: 2 marks; 2–3● : 1 mark ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... 2 2+ Zn

4 H2 Chemistry 9729/02 NYJC J2/22 PX [Turn Over For Examiner's Use ..................................................................................................................... [3] (iii) Write an equation for the reaction that occurred when the sample is heated at 350 °C. Zn(NO3)2 → ZnO + 2NO2 + ½ O2 [1] .......................................................... [1] Will not accept if it is Ba(NO3)2 as barium nitrate will not decompose at 350 °C. (iv) Calculate the percentage composition by mass of zinc nitrate in the sample. Let mass of barium nitrate = x g mass of zinc nitrate = (20.0 – x) g mass of zinc oxide = (11.3 – x) g amount of ZnO = amount of Zn(NO3)2 • 11.3 20.0 65.4 16.0 65.4 2(14.0) 6(16.0) xx−− =+ + + 11.3 20.0 81.4 189.4 xx−− = 2140.22 – 189.4x = 1628 – 81.4x 108x = 512.22 • x = 4.742 • mass of Zn(NO3)2 = 20.0 – 4.742 = 15.257 g • % by mass = 15.257 20.0 × 100% = 76.28% = 76.3% 4●: 2 marks; 2,3● : 1 mark [2] [Total: 15] .................................................................................................................................... .................................................................................................................................... ................................................................................................................................ [3]

5 H2 Chemistry 9729/02 NYJC J2/22 PX [Turn Over For Examiner's Use 2(a) Oxalic acid, HO2CCO2H, is a weak diprotic acid. HO2CCO2H Ý HO2CCO2− + H+ pKa1 = 1.27 HO2CCO2− Ý −O2CCO2− + H+ pKa2 = 4.28 (i) 10 cm 3 of 0.100 mol dm−3 HO2CCO2H was mixed with 10 cm 3 of 0.150 mol dm−3 HO2CCO2Na. Write an equation to show how this solution is able to maintain pH upon addition of alkali. OH− + HO2CCO2H → HO2CCO2− + H2O [1] .................................................. [1] (ii) 14.0 cm3 of 0.100 mol dm−3 KOH was added to the solution in (a)(i). Calculate the pH of the resultant solution after adding KOH(aq). OH− + HO2CCO2H → HO2CCO2− + H2O I 0.0014 0.001 0.0015 C −0.001 −0.001 +0.001 F 0.0004 0 0.0025 1 mark for the amou