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 - 550 = 110 110 cm3of oxygen in excess and left behind. Total volume of gas = 110 + 400 = 500 cm3

17 B aqueous sodium chloride and aqueous silver nitrate (Precipitation reaction)  Sodium nitrate and silver chloride NaCl + AgNO3  AgCl + NaNO3(2 salts) 18 A Use of alkali will result in reaction with ammonium salts in the fertilizers which results production of ammonia gas and loss of nitrogen to environment. Alkali + ammonium salt  salt + water + ammonia gas 19 D Solution X is more acidic than solution Y therefore has a higher concentration of hydrogen ions. 20 C step test observations 1 Dissolve 5 g of P in 10 ml of water. Green solution formed. (Hint of iron (II) ions) Add 2 cm3 of the solution into two test tube. 2 To the first test tube, add aqueous ammonia dropwise. Allow the mixture to stand for 5 minutes. Green precipitate formed in green solution. (QA: Iron(II) ions present) Green precipitate turns into reddish-brown precipitate. (Iron (II) ions oxidised to iron(III) ions) 3 To the second test tube, add an equal volume of nitric acid and followed by aqueous barium nitrate. White precipitate formed in the green solution. (QA: sulfate ions is present) 21 B Oxidation state of underlined element HNO3 : +5 CaSO4 : +6 FeO : -2 NaHCO3 :+4 22 B acidified aqueous potassium manganate(VII): test for reducing agent (RA) – purple to colourless potassium iodide: test for oxidising agent (OA) – colourless to brown Q: Both RA and OA R: OA S: RA T: no RA, no OA Both Q and S contains RA

23 B RA OA Ba(s) + 2H+(aq) → Ba2+(aq) + H2(g) 0 +1 +2 0 Ba acts as a reducing agent: reduces H from +1 in H+ to 0 in H2. Ba is oxidised. 24 C 1 A X atom(2.6) loses gains 2 electrons to form X2- ion. X 2 X burns with hydrogen to form a neutral oxide(H2O).√ 3 X is a strong oxidising agent.√ 4 X is found in Group 2 of the Periodic Table X (group 16) X X2-  electronic configuration of 2.8 (oxide ion) An atom of X  2.6  Gained 2 electrons to form the ion  Proton no: 8  oxygen. Burns with hydrogen to produce water which is a neutral oxide  Oxygen gains electron easily to form oxide ions  strong oxidising agent (gains electron from the other atom)  Group 16 +2 5 D 26 C Halogen displacement more reactive halogen displaces less reactive halogen Reactivity: F2 (pale-yellow)>Cl2 (yellow-green)>Br2(red-brown) P: Chlorine unable to displace F2 from NaF , chlorine remains (yellow-green) Q: Chlorine displaces bromine from LiBr to form bromine gas. (red-brown) Water: bromine gas dissolves in water (reddish brown solution)

27 A Fe + CuSO4  FeSO4 + Cu Displacement of metal. Iron is more reactive than copper metal. Iron will displace copper from copper(II) sulfate solution to form reddish-brown solids (copper metal) on the surface of the container. Solution will turn from blue to pale green. 28 B Key concepts Displacement of metal  More reactive metal displaces less reactive metal from its solution Carbonate very reactive metals do not undergo thermal decomposition (typical of Group 1 metal carbonate)  The metals form cations Q2+, R2+, S+ and T2+.  When a piece of solid Q is placed in a solution of T2+ ions, a layer of solid T is formed on the piece of solid Q. [Q > T]  No reaction when a piece of solid Q is placed in a solution containing R2+ ions. [ Q < R]  Only the carbonate of S does not undergo thermal decomposition. [ S is group 1 carbonate and does not decoompose on heating]  T reacts with hydrochloric acid to produce hydrogen gas. [T is above H] S> R> Q> T 29 B Positive enthalpy change which products has higher energy level than reactant  endothermic process A combustion of fuels {exothermic} B Melting (endothermic-energy taken in) C reaction of acid with metals (exothermic) D reaction of alkali metals with water( exothermic) 30 C Enthalpy change = Energy taken in for bond-breaking in reactants – energy given out during bond-forming 2C- H C≡C 2 H-H 6C-H C-C Enthalpy change = + (2 x413 + 839 + 2 x432 ) – ( 6x413 + 347) = 2529 – 2825 = -296 kJ/mol

31 A No of mole of ammonia formed = no of mole of nitrogen A 1.0 mol (NH4)3PO4 [ 1 x 3 ] B 1.0 mol (NH4)2SO4 [ 1 x 2] C 2.0 mol NH4Cl[ 2 x 1] D 2.0 mol NH4NO3 [ 2 x1] A  3 mole of ammonia B  2 mole of ammonia C  2 mole of ammonia D  2 mole of ammonia 32 D The reaction is reversible N2 + 3H2 ⇌ 2NH3 33 B Cobalt(II) nitrate solution, ions Co2+, NO3-, H+ , OH- Cobalt ions preferentially discharged at platinum foil cathode, Co2+(aq) + 2e  Co(s) (gain electrons / reduced) Hydroxide ions preferentially discharged at graphite rod anode, 4OH-(aq)  O2(g) + 2H2O(l)+4e (lose electrons/ oxidised) Graphite rod reacts with o