SKSS 4E Chem Prelim Answers

MARK SCHEME for SKSS 2024 4E Chemistry Prelim Paper 1 & 2 Setter: Ms Kwok Honey 1 PAPER 1 [40 marks] 1 2 3 4 5 6 7 8 9 10 D C B B A B B D C B 11 12 13 14 15 16 17 18 19 20 D C C A C C C C C B 21 22 23 24 25 26 27 28 29 30 A D A D D D B A D D 31 32 33 34 35 36 37 38 39 40 C D D B C B C C D B PAPER 2 Section A [70 marks] 1 [This question mainly assesses students’ memory work.] [accept if correct chemical formula is written each time] (a) ethanol/water [1] [accept if both are written] (b) calcium hydroxide [1] (c) sulfur dioxide [1] (d) aluminium nitrate [1] (e) ammonia [1] (f) methane [1] 2 [This question is similar to the Specimen Paper Q4] (a) [1m for each correct answer; max. of 2m] Any TWO of the following answers: • forms/gives coloured compounds • higher density • higher melting and boiling point [reject: good catalyst, variable oxidation states as these are not physical properties] (b) [1m for all correct number of electrons and protons; 1m for all correct number of neutrons; max. of 2m] Cr24 52 Cr24 53 number of electrons 24 24 number of neutrons 28 29 number of protons 24 24 (c) (i) 2Cr2O3(s) + 3C(s) → 4Cr(s) + 3CO2(g) [1] [reject if the coefficients are not in the simplest form]

MARK SCHEME for SKSS 2024 4E Chemistry Prelim Paper 1 & 2 Setter: Ms Kwok Honey 2 [Note to marker: All state symbols must be written correctly to be awarded 1m if students choose to include in the balanced chemical equation.] (ii) Amphoteric oxide can react with both acids and bases while acidic oxide can only react with bases. [1] Any ONE of the following equations: • Cr2O3 + 6HCl → 2CrCl3 + 3H2O OR with any other acids • CO2 + Ca(OH)2 → CaCO3 + H2O OR with any other bases [1] [Note to marker: It is not within the syllabus for students to write the chemical equation of chromium(III) oxide with a base as complex ions are formed.] (d) • crystal dissolves [1] • (idea of collision) particles collide / particles bounce off each other [1] • (idea of diffusion) particles move further apart / particles move/diffuse from higher concentration to lower concentration / movement of particles down a concentration gradient [reject the word ‘spread’ to describe diffusion as this word is already seen in the question] [1] (e) [1m for every 2 correct order of arrangement; max. of 2m] (most reactive) sodium, lanthanum, nickel, mercury (least reactive) 3 (a) [1] (b) (i) H H │ │ I―C―C―Cl │ │ H H [1] (ii) ∆Hbond breaking = 614 + 4(413) + ? = (2266 + ?) kJ ∆Hbond forming = 240 + 4(413) + 348 + 328 = 2568 kJ ∆Hbond breaking − ∆Hbond forming = −94 2266 + ? – 2568 = −94 [1m with correct working] ? = 208 kJ/mol [1m with correct unit] (iii) [This part of the question is similar to Specimen Paper, Q9(d)] [1m for showing energy of reactants is more than products; 1m for showing Ea and correctly labelled with single-headed arrow pointing in the correct direction (upwards); 1m for indicating correct chemical formula of product (allow ecf from (b)(i)); max. of 3m] I Cl      • • • • • • •

MARK SCHEME for SKSS 2024 4E Chemistry Prelim Paper 1 & 2 Setter: Ms Kwok Honey 3 progress of reaction (c) (i) Substituition [accept minor spelling error] [1] (ii) C2H6 + ICl → C2H5Cl + HI [1] 4 (a) Volume of CO2 present in clean, dry air = 0.04 100 × 480 = 0.192 dm3 No. of moles of CO2 = 0.192 24 = 0.008 mol. [1] No. of molecules of CO2 = 0.008 × 6.02 × 1023 = 4.816 × 1021 = = 4.82 × 1021 (3 s.f.) [1] (b) (i) C H mass / g 85.7 14.3 Ar 12 1 No. of mol 85.7 12 = 7.142 14.3 1 = 14.3 ratio 7.142 7.142 = 1 14.3 7.142 = 2.00 [1] ∴ empirical formula = CH2 (ii) Relative molecular mass of CH2 = 12 + 1 + 1 = 14 n × 14 = 128.25 n = 9.16 ≈ 9 ∴ molecular formula = (CH2)9 = C9H18 [1] (iii) Equation: 2C9H18 + 27O2 → 18CO2 + 18H2O [1] Note to marker: There are two solutions to this part of the question. Solution 1 Solution 2 No. of moles of C9H18 present = 1000 (12×9)+(1×18) = 7.9365 Mole ratio = C9H18 : CO2 = 2 : 18 =7.9365: 71.42857 Vol. of CO2 = 71.42857 × 24 = 1710 dm3 (3 s.f.) No. of moles of C9H18 present = 1000 128.25 = 7.79727 Mole ratio = C9H18 : CO2 = 2 : 18 =7.79727: 70.175 Vol. of CO2 = 70.175 × 24 = 1680 dm3 (3 s.f.) [1] [1] (c) (i) Any ONE of the following answers: • Desertification of fertile land would lead to the amount of food that can be produced globally to decrease. [1] C2H4 + I―Cl energy Ea C2H4ICl ∆H = −94 kJ/mol

MARK SCHEME for SKSS 2024 4E Chemistry Prelim Paper 1 & 2 Setter: Ms Kwok Honey 4 • High temperatures from more frequent and severe heat waves can be fatal. • Ocean warming can cause commercially -important fish population to be depleted. • Melting of polar ice caps can cause sea levels to rise and permanently flood coastal areas. [reject: cause climate change / melt ice caps / cause death] (ii) Carbon dioxide, a greenhouse gas , traps heat within the Earth’s atmosphere. [1] This leads to the increase in the average temperature of the Earth’s surface. [1] (d) 6CO2 + 6H2O →C6H12O6 + 6O2 5 (a) (i) Peroxodisulfate ions act as an oxidising agent. [No mark is awarded unless explanation is correct.] It causes iodide ions to be oxidised to iodine due to an increase in oxidation state of iodine from −1 to 0. [1] (ii) [data analysis: inference] Peroxodisulfate ions: Comparing experiment 1 and 2 / 2 and 3 , rate of reaction increases by twice/doubles when concentration of   peroxodisulfate ions doubles with the same concentration of iodide ions at 0.02 mol/dm3. [1] Iodide ions: Comparing experiment 1 and 4 / 4 and 5, rate of reaction increases by twice/doubles when concentration of   iodide ions doubles with the same concentration of peroxodisulfate ions at 0.008 mol/dm3. [1] [reject if the experiment numbers and concentrations are not quoted] (iii) The presence of a catalyst provides an a lternative pathway of lowering/decreasing activation energy, allowing more colliding particles to have energy greater than or equal to activation energy. [1] This increases the frequency/rate of effective collisions and the rate of reaction. [1] (b) (i) [data analysis: inference and deduction, supported by scientific explanation] There are only 4 drops of halogenoalkanes used in experiment 1 as compared to 8 drops of halogenoalkanes used in experiment 2. [1] [reject if number of drops is not quoted] Lesser amount of reacting particles present per unit volume/in the same volume, resulting in lower frequency/rate of effective collision s hence slower rate of reaction. [1] (ii) [data analysis: describing trend] The more reactive the halogen, the slower the rate of reaction between a halogenoalkane and water. [1] OR The less reactive the halogen, the faster the rate of reaction between a halogenoalkane and water. (iii) At lower temperature, reactant particles have less kinetic energy and move slower. [1] There are less reactant particles possessing energy that is greater than or equal to activation energy. [1] This decreases the frequency/rate of effective collisions and the rate of reaction. [1]

MARK SCHEME for SKSS 2024 4E Chemistry Prelim Paper 1 & 2 Setter: Ms Kwok Honey