NYGH 2015-S3EOY-IP Chem P1 P2 Ans

1 2015 Sec 3 End-of-Year Exam Chemistry Answers Section A 1 B 2 B 3 D 4 B 5 B 6 D 7 B 8 D 9 B 10 A 11 C 12 A 13 C 14 C 15 A 16 D 17 B 18 B 19 D 20 C 21 B 22 C 23 A 24 B 25 A 26 D 27 B 28 D 29 B 30 D Section A No Answers Marks A1(a) The brown colour slowly spreads upwards until the mixture is homogeneous./Two jars will be uniformly brown/coloured. [1] A1(b) The rate of change increases. OR The change would be faster/Time taken for diffusion to occur will be shorter. [1] A1(c) If the bromine sample was impure, it would boil at a temperature higher than 58.8oC. Impure bromine would also boil over a range of temperatures. [1] [1] A2(a) precipitation [1] A2(b) Ca2+(aq) + CO3 2− (aq)  CaCO3(s) [1] A2(c) filtration [1] A3(a) Ionic [1]

2 No Answers Marks A3(b) Sodium nitride has a giant lattice/ ionic lattice/ crystal lattice structure. There are strong electrostatic forces of attraction between the sodium and nitride ions./oppositely charged ions. [1] [1] A3(c) Sodium nitride conducts electricity in the molten or aqueous state. OR Sodium nitride is soluble in water. Sodium nitride is brittle. [1] A3(d) monobasic, strong [2] A3(e) Na3N(s) + 4HCl(aq)  3NaCl(aq) + NH4Cl(aq) [2] A4(a) methyl orange or any other suitable indicator [3] A4(b) A strong acid is one which is completely ionized in water to form H+ ions. [1] A4(c) MgCO3(s) + H2SO4(aq)  MgSO4(aq) + H2O(l) +CO2(g) 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) [1] [1] A4(d) Concentration of NaOH = 0.400 ÷ (23+16+1) = 0.0100 moldm−3 No. of moles of NaOH used in titration = 0.0100 × 20.50/1000 = 2.05 × 10−4 Mole ratio of NaOH to H2SO4 is 2:1 No. of moles of H2SO4 used in titration with NaOH = 2.05 × 10−4 ÷ 2 = 1.03 × 10−4 [1] [1] [1] A4(e) No. of moles of MgCO3 used = 0.400 ÷ (24+12+16×3) = 4.762 × 10−3 Mole ratio of MgCO3 to H2SO4 is 1:1 No. of moles of H 2SO4 used = 0.400 ÷ (24+12+16×3) = 4.76 × 10−3 [1] [1] A4(f) Total number of moles of H2SO4 = 2.05 × 10−4 ÷ 2 × 100/25.0 + 4.762 × 10−3 = 5.172 × 10−3 [1] [1]

3 No Answers Marks Concentration of H2SO4 = 5.172 × 10−3 ÷ 100/1000 = 0.0517 moldm−3 A4(g) Precipitation. Any aqueous magnesium salt plus any aqueous carbonate. [1] [1] A5(a) B: Ca(OH)2 C: AgCl Accept: chemical names [1] [1] A5(b) NH4 +(aq) + OH−(aq)  NH3(g) + H2O(l) [1] A5(c) NH4Cl and CaCl2 Accept: chemical names [2] A5(d) Calcium sulfate is insoluble in water. OR If the sulfate ion was present, A should be a white suspension / white solid in a colourless solution instead of just a colourless solution. [1] A6(a) concentration of H+ ions, surface area of zinc / particle size of zinc [2] A6(b) experiment A B C D E curve II I IV V III Number of marks awarded = number of correct answers capped at 4 [4] Section B B7 (a) 2Mg + Si → Mg2Si [1]

4 (b)(i) Covalent bonds are found between atoms of silicon and hydrogen while van der Waals’ forces are found between molecules of silane. [1] Van der Waals’ forces between the silane molecules are weak which require less energy to overcome them. [1] [2] (b)(ii) [1] (b)(iii) Silane is soluble in organic solvents but insoluble in water. Silane does not conduct electricity in any state. [2] (c) Number of neutrons = 24 – 12 = 12 [1] Electronic configuration = 2, 8 [1] [2] (d) Number of protons = 48 Number of neutrons = 112 – 48 = 64 Number of electrons = 48 – 2 = 46 [1] [1M] for both numbers of protons and neutrons [2] B8 (a)(i) Relative atomic mass of lead sample A 207 100 8.7208100 5.41207100 0.48206100 7.2204 = ×+×+×+×= [1] (a)(ii) Pb + Cl2 → PbCl2 Let the relative atomic mass of lead sample B be x. From the equation, no. of moles of Pb = no. of moles of PbCl2 [2]

5 208 71341.0 71341.1 25.35 341.1000.1 = = += ×+= x x xx xx The relative atomic mass of lead sample B is 208. [1] Alternative answer: Mass of Cl atoms reacted = 0.341 g [ ]1 mol 10606.935.5 0.341 atoms Cl of moles of No. 3−×== mol 104.8032 10606.9 atoms Pb of moles of No. 3- 3 ×=×= − [ ]1 208 10803.4 000.1 Pb ofA 3r = × = − (a)(iii) Since the relative atomic masses of the two lead samples are different, they have different isotopic compositions. [1] (b)(i) Add solution of the compound to any carbonates/ moderately reactive metals [1] On reaction with carbonates, carbon dioxide is evolved which forms a white precipitate in limewater/ hydrogen is evolved which extinguishes a lighted splint with a ‘pop’ sound. [1] [2] (b)(ii) or [1] (c)(i) Barium carbonate reacts with the hydrochloric acid in the stomach to form soluble barium chloride which is toxic to the [1] [1]

6 body. Accept: Barium carbonate reacts with dilute hydrochloric acid in the stomach to produce gas which causes discomfort. (c)(ii) Ionic equation 1: Reaction of barium hydroxide with dilute nitric acid/ hydrochloric acid Ba(OH)2(s) + 2H+(aq) → Ba2+(aq) + 2H2O(l) [1] Ionic equation 2: Reaction of aqueous barium nitrate/ barium chloride with dilute sulfuric acid Ba2+(aq) + SO4 2‒(aq) → BaSO4(s) [1] [2] B9 Either (a)(i) C H O Mass (%) 40.9 4.6 54.5 Ar 12 1 16 No. of moles 408.312 9.40 = 600.41 6.4 = 406.316 5.54 = Ratio ( )s.f. 4 001.1406.3 408.3 = ( )s.f. 4 351.1406.3 600.4 = ( )s.f. 4 000.1406.3 406.3 = Simplest ratio 3 4 3 Empirical formula = C3H4O3 [1] [2] (a)(ii) H2A(aq) + Na2CO3(aq) → Na2A(aq) + CO2(g) + H2O(l) mol 0.02270 1000 7.2200.1 acidscorbic of moles of Number = ×=a ( )no. wholenearest 761 02270.0 00.4acidascorbic of mass molecular Relative = = [3] [1]

7 Let the molecular formula of ascorbic acid be (C3H4O3)x. ( ) 2 17688 17631641312 = = =×+×+× x x x Molecular formula of ascorbic acid = C6H8O6 (b)(i) mol 0000565.01000 11.300.00500 used hydroxide sodium aqueous of moles of Number =×= [1] (b)(ii) 1 mol of ascorbic acid reacts with 2 mol of sodium hydroxide mol 0.00002825 2 0000565.0 used acidascorbic of moles of Number = = ( )s.f 3 dm mol 0.00113 10000.25 00002825.0 mol/dm in acidascorbic of ionConcentrat 3- 3 = = [2] (b)(iii) mol 0.0008475 1000 7500.00113 bottle the in acidascorbic of moles of No. = ×= ( )s.f. 3 g 149.0 176 0.0008475 bottle the in acidascorbic of moles of Mass = ×= [2] B9 Or (a) XCO3 + 2HNO3 → X(NO3)2 + CO2 + H2O [1] [1] [1] [1] [1] [1] [1]

8 (b) ( ) s.f.) (3 mol 0.0508 1000 1000.5079 added acidnitric of moles of Number s.f. 4 dm mol 5079.063 32.0 mol/dm in acidnitric of ionConcentrat 3-3 = ×= == [2] (c)(i) HNO3 + NaOH → NaNO3 + H2O mol 00268.0 1000 26.800.100 used hydroxide sodium aqueous of moles of Number = ×= 1 mol of nitric acid reacts with 1 mol of aqueous sodium hydroxide mol 00268.0 NaOH withreacted acidnitric of moles of Number = [2] (c)(ii) mol 0.0268 0.25 25000268.0 remained that acidnitric of moles of Number = ×= [2] (d) ( )s.f. 3 mol 0240.0 0268.005079.0 carbonate the withreacted acidnitric of moles of Number = −= [1] (e) 2 mol of nitric acid reacts with 1 mol of XCO3 ( )s.f. 3 mol 0.0120 2 02399.0 XCO of moles of Number 3 = = ( )no. wholenearest 081 0120.0 30.1 XCO of mass molecular Relative 3 = = [2] [1] [1] [1] [1] [1] [1] [1] [1]