CJC 2019 Paper 2 Promo Answers

9729 / CJC JC1 Promotional Examination 2019 CANDIDATE NAME CLASS 1T CHEMISTRY 9729/02 Paper 2 Structured Questions Wednesday 2 October 2019 1 hour 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. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculat or is expected, where appropriate. A Data Booklet is provided. At the end of examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 10 printed pages. For Examiner’s Use Paper 1 15 Paper 2 Q1 / 8 Q2 / 4 Q3 / 3 Q4 / 5 Q5 / 6 Q6 / 9 35 Paper 3 Q1 /20 Q2 /20 40 Paper 4 Q1 /15 Q2 /15 30 TOTAL 120 OVERALL/% Grade Catholic Junior College JC1 Promotional Examinations Higher 2 Answers

2 [Turn over 9729 / CJC JC1 Promotional Examination 2019 Paper 2 Answer all the questions. Write your answers in the spaces provided. You are advised to spend not more than one hour on this section. 1 Ethanedioic acid, is a dibasic, organic acid with the formula HO 2CCO2H. It is commonly found in many leafy vegetables, fruits, nuts and seeds. It is able to react with a base such as sodium hydroxide, NaOH. (a) FA 1 is a solution containing 5.00 g dm–3 of a similar dibasic, organic acid, HO2C(CH2)nCO2H. When 25.0 cm 3 of FA 1 is titrated against NaOH of co ncentration 0.125 mol dm–3, 17.00 cm3 of NaOH is required. The equation for this reaction is given as follows. HO2C(CH2)nCO2H(aq) + 2NaOH(aq) → Na2(O2C(CH2)nCO2) (aq) + 2H2O(l) (i) Calculate the amount in moles of NaOH required to react with 25.0 cm 3 of the acid solution in FA 1. Amount of NaOH = ( ૚ૠ.૙૙ ૚૙૙૙ x 0.125) = 0.00213 mol [1] (ii) Calculate the amount in moles of the dibasic acid, HO2C(CH2)nCO2H in 25.0 cm3 of FA 1 that has reacted. Amount of acid = (½ x 0.00213) = 0.00106 mol [1] (iii) Calculate the concentration of the acid, HO2C(CH2)nCO2H in mol dm-3 of solution in FA 1. Concentration of acid = ( ૚૙૙૙ ૛૞.૙ x 0.00106) = 0.0425 mol dm–3 [1] (iv) Hence, determine the value of n in the formula of the acid, HO2C(CH2)nCO2H. ૞.૙૙ ࡹ࢘ 0.0425 Mr of acid, HO2C(CH2)nCO2H = ૞.૙૙ ૙.૙૝૛૞ = 117.6 14n + 90.0 = 117.6; n= 2 (whole number) [1]

3 [Turn over 9729 / CJC JC1 Promotional Examination 2019 (b) Ethanedioic acid, HO2CCO2H (or H2C2O4), is also a reducing agent and reacts with an oxidising agent such as acidified potassium manganate( VII), KMnO4. It can be oxidised by acidified KMnO4 to carbon dioxide, when heated to 60°C. When a 25.0 cm 3 sample of ethanedioic acid is titrated against acidified KMnO 4 of concentration 0.0200 mol dm−3, 23.00 cm3 of KMnO4 is required. (i) Derive a balanced half -equation for the oxidation of ethanedioic acid to carbon dioxide in acidic conditions. H2C2O4(aq) → 2CO2(g) + 2H+(aq) + 2e− ………………………………………………………………………………………… … ..[1] (ii) By reference to the relevant half-equation from the Data Booklet for the reduction of acidified MnO 4−, derive an overall balanced equation for the reaction between ethanedioic acid and acidified manganate(VII) ion, MnO4−. MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (1st equation x 5; 2nd equation x 2 and adding) 5H2C2O4(aq) + 2MnO4−(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) [1] (iii) Hence, calculate the amount in moles of the acid, HO2CCO2H (or H2C2O4) reacted and subsequently its concentration in mol dm−3. Amount of MnO4− = ( ૛૜.૙૙ ૚૙૙૙ x 0.0200) = 0.000460 mol Amount of H2C2O4 = ( ૞ ૛ x 0.000460) = 0.00115 mol Concentration of H2C2O4 = ( ૚૙૙૙ ૛૞.૙ x 0.00115) = 0.0460 mol dm−3 [2] [Total: 8]

4 [Turn over 9729 / CJC JC1 Promotional Examination 2019 2 The graph below shows the second ionisation energies of unknown elements A – J, of consecutive proton numbers. The letters are not the atomic symbols of the elements. J, which has the largest Ar, has an atomic number below 20. (a) Write the ground state electronic configuration of element F. Electronic configuration of F: 1s22s22p63s23p1 [1] (b) Generally the second ionisation energy (2nd I.E.) increases across the period. Explain the decrease in 2nd I.E. between element F and G. The second IE for F involves removing an electron from the 3s orbital while the second IE for G involves removing an electron from the 3p orbital. Electron in the 3p orbital is further from and less strongly attracted by the nucleus. [1] (c) Sketch and label all the valence orbitals of F, clearly showing the labelled axes. [2] [Total: 4] 3 Canisters of flammable gas are used as portable fuel, and may contain a few types of short chain hydrocarbons, which are liquefied under high pressure. A canister was connected to a gas syringe and the valve opened to allow some of the gas into the syringe. It was found that 0.300 g of gas took up 144.0 cm 3 at temperature of 24°C and pressure of 1.02 x 105 Pa. Element A B C D E F G H I J 1000 4000 2500 2nd IE / kJ mol−1 3s 3px 3py 3pz

5 [Turn over 9729 / CJC JC1 Promotional Examination 2019 Calculate the average Mr of the gas mixture assuming it behaves ideally. [Total: 3] 4 Nitroglycerin, C 3H5(NO3)3, is a flammable liquid commonly used to manufacture dynamite. Upon ignition, nitroglycerin decomposes to produce nitrogen, oxygen, carbon dioxide and steam. Given: Standard enthalpy change of formation of nitroglycerin(l) / kJ mol–1 –364 Standard enthalpy change of formation of H2O(g) / kJ mol–1 –242 Standard enthalpy change of formation of CO2(g) / kJ mol–1 –394 (a) Write a balanced equation, with state symbols, for the decomposition of 1 mol of liquid nitroglycerin. [1] C3H5(NO3)3(l)  3/2N2(g) + 1/4O2(g) + 3CO2(g) + 5/2 H2O(g) (b) With reference to the above data, calculate the standard enthalpy change of decomposition of 1 mol of nitroglycerin. [1] Hdecomposition = (Hformation products) – (Hformation