Zhonghua Prelim P1 Ans

4E Chem 2024 Prelim Paper 1 Ans 1 2 3 4 5 6 7 8 9 10 A C A B C A D B C B 11 12 13 14 15 16 17 18 19 20 C B B B A C B A D C 21 22 23 24 25 26 27 28 29 30 B B B C D C A B B C 31 32 33 34 35 36 37 38 39 40 A D B D A B D A D B Qn Explanation 1 A. Carbon dioxide produced when calcium carbonate reacts with dilute hydrochloric acid. Carbon dioxide subsequently reduced when heated with charcoal (carbon) to form carbon monoxide. Excess carbon dioxide which is an acidic oxide will be removed when bubbled through sodium hydroxide. Carbon monoxide is a neutral oxide and will not react with sodium hydroxide. 2 C 1. Fractional distillation to separate two miscible liquids 2. Separation funnel to separate two immiscible liquids 3. Simple distillation to obtain solvent from a solution 3 A. Mr Helium:4 ; carbon dioxide: 44; carbon monoxide: 28; nitrogen: 28; hydrogen; 2 Mr of gas 1 must be lower than gas 2 so that rate of diƯusion of gas 1 is faster than gas 2 into the porous pot, and therefore exert a greater pressure. 4 B. Argon: monoatomic Chlorine, Cl2: diatomic Nitrogen, N2: diatomic Mixture of 1 monoatomic particle and 2 diƯerent diatomic particles. 5 C. Double bond present in -COOH and O=O 6 A. 2Al3+ + 3O2-  Al2O3 2Al  2Al3+ + 6e

7 D Electrical conductivity of metals and ionic compound are as follows Aqueous/molten ionic compound  mobile ions Metals and graphite  mobile electrons 8 B. 9 C Electronic configuration P: 2.4 Q: 2.8.2 R. 2.8.3 S:2.6 Ionic: R2S3 Covalent: PS2 10 B 243X differ from 239X by having 4 more neutrons and therefore higher density. 11 C Graphite structure 12 B From the graph, 100 cm3 (0.1dm3) of 0.600 mol/dm3 reacts with 50 cm3 of potassium iodide. 1 mole of potassium iodide reacts with 1 mole of silver nitrate (mole ratio from eqn) (m1v1)/(m2v2) = n1/n2 conc = (0.1 x 0.6)/0.05 = 1.2 mol/dm3

1 3 B Mass of MO = 1.12g Molar ratio 1 MO : 1 CO2 Mol of CO2 = mol of MO = 0.48 dm3/24 Mr of MO = mass/mol = 1.12/ (0.48/24) Ar of M = Mr of MO – Ar of O = Mr of MO – 16 = [1.12/ (0.48/24)] - 16 14 B A 0.5 mol of C6H12 [ 0.5 x 6 = 3] B 1.5 mol of C4H8 [ 1.5 x 4 = 6] C 3.0 mol of C3H7OH [ 3 x 3 =9] D 6.0 mol of C2H6 [ 6 x 2=12] No of mol in 144 dm3 of carbon dioxide = 144/24 = 6 mole Hint: use the number of moles of the substance and multiply by the number of carbon atoms in the molecule. e.g 1.5 mol of C4H8 = 1.5 x 4 = 6 mol 15 A Purity of 36% = 36 g of calcium bromide present Formula = CaBr2 Mr of CaBr2 = 200 Mr of 2Br- = 160 Mass of Br = 160/200 x 36 = 28.8g 16 C Molar ratio C4H6 : O2 : CO2 2 : 11 : 8 100 : 550 400 100 cm3 of butyne reacts with 550 cm3 of oxygen to produce 400 cm3 of CO2. Excess O2 = 660 -