NYGH 2020-S3EOY-IP Chem P1 and P2 Ans

Class Register Number Name 南洋女子中学校 NANYANG GIRLS' HIGH SCHOOL End-of-Year Examination 2020 Secondary Three Mark Scheme CHEMISTRY 45 minutes Paper 1 and 2 1130 - 1215 Wednesday 07 October 2020 1 hour 45 minutes Additional materials: Writing papers 0845 - 1030 READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, register number and class in the spaces at the top of this page and on any separate writing paper used. Section A (40 marks) Answer all questions. Write your answers in the spaces provided on the question paper. Section B (30 marks) Answer all questions including questions B6, B7 and B8 Either or B8 Or. Write your answers on the separate writing paper provided. At the end of the examination, 1. Fasten all your work securely together; 2. Tick B8 Either (for Either) or B8 Or (for Or) in the grid below to indicate which question you have answered; 3. Hand in Section A and Section B separately. INFORMATION FOR CANDIDATES The intended number of marks is given in the brackets [ ] at the end of each question or part question. You are advised to spend no longer than 1 hour on Section A and no longer than 45 minutes on Section B. The use of approved scientific calculators is allowed. Examiner’s Use Section A Section B B6 B7 B8 Either B8 Or Total Setters: KO/ NHK/ OWJ This document consists of 8 printed pages. NANYANG GIRLS’ HIGH SCHOOL [Turn over]

2020 End-of-Year Examination Secondary 3 IP Chemistry 2 Nanyang Girls’ High School Setters: KO/ NHK/ OWJ Chemistry/ Paper 1 & 2 Paper 1 1 C 2 D 3 B 4 A 5 C 6 D 7 C 8 A 9 C 10 C 11 B 12 C 13 C 14 D 15 A 16 C 17 A 18 B 19 C 20 A 21 C 22 A 23 D 24 B 25 C/D 26 D 27 B 28 B 29 D 30 D 1 Substance exists as liquid at 100C, so melting point is lower than 10OC; it exits as a gas at 600C, so its boiling point is lower than 600C, but higher than 100C. C 2 “homogeneous blue colouration” would mean that the particles, copper(II) ions, sulfate ions and water molecules moved around to mix in order to achieve a homogeneous state. D 3 To obtain water, we will need to carry out distillation so that water can be boiled, condensed over and collected at the distillate. B 4 Iodine undergoes sublimation. Potassium chloride is a soluble salt. Barium sulfate is an insoluble salt. Sublimation must be carried out in the first step. Followed by addition of water to dissolve potassium, then filtering to remove barium sulfate before evaporating the potassium chloride solution. A 5 Both NH3 and HCl are gases, NH3 (smaller Mr) will diffuse faster, reaching the moist red litmus paper first, and turning it blue before HCl reaches it to turn it red. C 6 Expt 1 allows the collection and measurement of gases. Expt 2 allows for the change in mass. Expt 3 allows the collection and measurement of gases. Expt 4 does not have any useful measurement of either the change in mass of reactant/product nor collection and measurement of gas. D 7 Potassium nitrate is neutral (pH 7) while potassium hydroxide is a strong alkaline with high pH. C 8 Amphoteric oxides are insoluble in water. They are metallic oxides and can act as acid or bases when reacted with a base or acid respectively. A 9 Options A and B will cause a coating of insoluble lead (II) sulfate, preventing further reaction, leading to very poor yield. Option C: By reacting lead(II) oxide with nitric acid, lead(II) nitrate is formed. Pb2+ (aq) can then form a precipitate with SO42- (aq). Option D has not reaction between lead(II) chloride and nitric acid, so will not be able to form Pb2+(aq) for precipitate with SO42- (aq) from sodium sulfate. C 10 When barium nitrate, was added to sample of polluted water, a white precipitate (of either BaSO4 or BaCO3) was observed. As it did not react with acid, we deduce that the anion is sulfate and not carbonate. When sample of polluted water was added with nitric acid followed by aqueous ammonia, a white precipitate that is soluble in excess C

2020 End-of-Year Examination Secondary 3 IP Chemistry 3 Nanyang Girls’ High School Setter: KO/ NHK/ OWJ Chemistry/ Paper 1 & 2 [Turn Over] aqueous ammonia was observed. We can deduce that the white precipitate is zinc hydroxide. 11 I: NH3 would be formed as well and present in the ionic equation. IV: Ca2+(aq) + SO42-(aq) + 2H+(aq) + 2OH-(aq) ! 2H2O (l) + CaSO4 (s) II and III: reagents and products except CaSO4(s) are aqueous ions in solution, so spectator ions such as NO3-, H+, and Na+ will be cancelled out. B 12 Green precipitate due to formation of Fe(OH)2(s)and white precipitate due to Ca(OH)2(s). Ammonia gas is evolved indicating that NH4+ was present in the solution. C 13 Fe(NO)3 : NaOH 1 : 3 no. of mol of Fe(NO)3 given = 0.02 mol no. of mol of NaOH given = 0.03 mol NaOH is the limiting reagent, so at the end of reaction, there should be excess of Fe(NO)3 (yellow solution) left apart from the formation of Fe(OH)3 (reddish-brown precipitate). C 14 X forms an ion of +2. So, X : H+ 1: 2 no. of mole of HCl = 0.02 mol no. of mole of X = 0.01 mol Mr of X = 0.65/0.01 = 65 D 15 P4O10 + 6 CaO ! 2 Ca3(PO4)2 0.5 mol : 3mol A 16 I: (NH4)2SO4 + 2NaOH ! Na2SO4 + 2H2O + 2NH3 II: 2NH3 + H2SO4 ! (NH4)2SO4 III: H2SO4 + 2NaOH ! Na2SO4 + 2H2O No. of mole of NaOH in (III) =(25.80/1000)x0.100 =2.58 x 103 No. of mole of H2SO4 in (III) = 2.58 x 103 /2 =1.29 x 10-3 mol (This is the excess of H2SO4 from II) No. of mole of H2SO4 given in (II) = (50.0/1000)x0.0500 = 2.5 x 10-3 mol No. of mole of H2SO4 used in (II) = 2.5 x 10-3 - 1.29 x 10-3 = 1.21 x 10-3 mol No. of mole of NH3 in 25 cm3 of (II) = 2 x 1.21 x 10-3 = 2.42 x 10-3 mol No. of mole of NH3 in 250 cm3 of (II) = 10 x 2.42 x 10-3 = 0.0242 mol No. of mole of (NH4)2SO4 (I) = 0.0242 /2 = 0.0121 mol Percentage mass = (0.0121 x 132.1)/5.00 x 100% = 32.0% C 17 No. of mole of Cl2 = [(0.00495/1000) x 500/1000]/71 = 3.4859 x 10-8 mol No. of Cl atoms = (2 x 3.4859 x 10-8) x 6.02 x 1023 = 4.20 x 1016 A

2020 End-of-Year Examination Secondary 3 IP Chemistry 4 Nanyang Girls’ High School Setters: KO/ NHK/ OWJ Chemistry/ Paper 1 & 2 18 CH3SH(g) + 3O2(g) ! CO2(g) + SO2 (g) +2H2O(l) given 20 80 (excess) 0 0 reacted -20 -60 +20 +20 left 0 20 20 20 Total = 60 cm3 B 19 HClO3 (+1) + O.S. of Cl + 3(-2) = 0 O.S. of Cl = +5 C 20 ZnO lost an oxygen atom to C. ZnO undergoes reduction while C (gained an oxygen atom) undergoes oxidation. A 21 No. of mole of Zn = 0.654/65.4 = 0.01 mol No. of mole of e lost from Zn = 0.01 x 2 = 0.020 mol No. of mole of VO2+ = (100/1000) x 0.10 = 0.010 mol VO2+ : e 1 : 2 1 mole of VO2+ gained 2 mole of e O.S. of V in product = +5 + 2e = +3 C 22 HSO3- reacts with H+ to form water and SO2. There is no change in oxidation state, so it is not redox. A 23 In an atom of Z, there would be 17 electrons: 1s22s22p63s23p5 Z is in group 17, will form a singly charged anion. Mass number = 17 + 18 = 35 D 24 Electron will be attracted and pulled towards the negative plate due to electrostatic forces of attraction. The deflection will be greater as electron has a much lower relative mass compared to proton. B 25 CO2 dissolves in water to form a weak acid known as carbonic acid. H2CO3 ⇌ H+ + HCO3- So it provides some electrical conductivity. D 26 Compound exists as liquid at room temperature tells us that it has a low melting point, so it is a simple molecule. P and Q must be non-metals. D 27 Sodium chloride is an ionic compound with strong electrostatic forces of attraction between oppositely charged ions while silicon tetrachloride is a simple molecule with weak intermolecular forces of attraction. B 28 Q shows mass difference, therefore number of moles of HCl used must be the same, so cannot increase [HCl] as that will also increase its number of moles leading to more gases produced. Only increase in temperature. Pressure has no impact