NYJC 2023 H2 Chemistry 9729 P4 Answer

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 1 H2 Chemistry Prelim Exam Answers Paper 4 Answers 1 (a) (i) Final burette reading / cm3 Initial burette reading / cm3 Volume of FA 2 added / cm3 25.40 25.40 (ii) Volume of FA 2 = 25.40 + 25.40 2 = 25.40 cm3 (b) (i) nNaOH = 25.40 1000 x 0.115 = 0.002921 = 0.00292 mol (ii) Mass of 1 dm3 of FA 1 = 1.75 x 1000 = 1750 g Mass of H3PO4 in 1 dm3 of FA 1 = 8.40 100 x 1750 = 147 g Concentration of H3PO4 in FA 1 = 147 g dm–3 Concentration of H3PO4 in FA 1 = 147 ÷ 98.0 = 1.50 mol dm–3 (iii) Concentration of H3PO4 in FA 3 = 10.0 250 x 1.50 = 0.0600 mol dm–3 Amount of H3PO4 in 25.0 cm3 of FA 3 = 25.0 1000 x 0.0600 = 0.00150 mol (iv) 34 NaOH H PO n n = 0.002921 0.00150 = 1.947  2 (nearest whole number)  Major product is Na2HPO4 H3PO4(aq) + 2NaOH(aq)  Na2HPO4(aq) + 2H2O(l)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 2 (c) final burette reading 25.60 cm3 initial burette reading 1.35 cm3 volume added 24.25 cm3  The initial burette reading made by student A was 0.05 cm3 greater than the true value but the volume added was exactly 24.25 cm3. The final burette reading was also 0.05 cm3 greater than the true value.  The initial burette reading made by student B was 0.05 cm3 less than the true value and the actual volume added was exactly 24.15 cm3. The final burette reading was 0.05 cm3 greater than the true value.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 3 2 (a) (i) The temperature, T1, of the hot water at t = 4.0 min is 63.0 °C. t / min T / ºC 0.0 32.0 1.0 32.0 2.0 32.0 3.0 32.0 4.0 –– 5.0 43.0 5.5 42.5 6.0 42.0 6.5 41.5 7.0 41.0 7.5 40.5 8.0 40.0 (ii)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 4 (iii) Minimum temperature, T2, at t = 4.0 min is 32.0 °C. Maximum temperature, T3, at t = 4.0 min is 44.0 °C. Temperature rise for 50 cm3 of cold water in beaker A, (T3 – T2) is 12.0 °C. Temperature fall for 50 cm3 of hot water from the 250 cm3 beaker, (T1 – T3) is 19.0 °C. (iv) Heat lost by hot water = Heat gained by cold water + Heat gained by beaker A Heat gained by beaker A = Heat lost by hot water – Heat gained by cold water = (50.0 x 4.18 x 19.0) – (50.0 x 4.18 x 12.0) = 1463 = 1460 J (v) Heat absorbed by beaker A = Cbeaker A x (T3 – T2) Heat capacity of beaker A = 1463 12.0 = 121.9 = 122 J oC–1 (b) (i) mass of boiling tube + FA 4 / g mass of empty boiling tube / g mass of boiling tube + residual FA 4 / g mass of FA 4 added / g 9.963 initial temperature of water / oC minimum temperature obtained / oC temperature fall, T / oC 6.0 (ii) Mass of NH4Cl = 9.963 53.5 = 0.1862 = 0.186 mol (iii) Heat absorbed by solution = 100 x 4.3 x 6.0 = 2580 J (iv) Total change in heat energy = 2580 + (121.9 x 6.0) = 3311 J 