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Chemistry 0620/41 May/June 2022: question paper and mark scheme

Cambridge IGCSE Chemistry (0620), Paper 4 Theory (Extended), May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0620_s22_qp_410620_s22_ms_410620/41/M/J/22

Paper
Paper 4 Theory (Extended)
Total marks
80
Questions
7
Code on the paper
0620/41/M/J/22
0620_s22_ms_41.pdf

0620/41 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 19 marks

Question

A list of substances is shown. aluminium oxide carbon dioxide chlorine diamond ethanol glucose iron(III) oxide limestone nitrogen oxygen Answer the questions using the list of substances. Each substance may be used once, more than once or not at all. State which of the substances: (a) is a reactant in photosynthesis . …… [1] (b) is the main constituent of bauxite . …… [1] (c) are two products of fermentation …… and . …… [2] (d) is used as a fuel . …… [1] (e) is a gas used to convert iron into steel . …… [1] (f) is a greenhouse gas . …… [1] (g) is a gas that is approximately 78% of clean, dry air . …… [1] (h) is a form of carbon. . …… [1] [Total: 9]

Mark scheme

1(a) carbon dioxide 1 1(b) aluminium oxide 1 1(c) ethanol (1) 2 carbon dioxide (1) 1(d) ethanol 1 1(e) oxygen 1 1(f) carbon dioxide 1 1(g) nitrogen 1 1(h) diamond 1

Question 28 marks

Question

(a) Atoms are made of protons, neutrons and electrons. Atoms of the same element are known as isotopes. (i) Complete the table. particle relative charge relative mass 1 electron 1840 neutron proton +1 [2] 12Mg and 12Mg are isotopes of magnesium. (ii) 24 25 Complete the table to show the numbers of electrons, neutrons and protons in these isotopes of magnesium. number of number of number of isotope electrons neutrons protons 12Mg 24 12Mg 25 [2] (iii) Explain why magnesium ions have a charge of 2+. . …… . …… [1] (b) Mg2+ ions have the electronic structure 2,8. Give the formula of the following particles which have the same electronic structure as Mg2+ ions. ● a cation (positive ion) . …… ● an anion (negative ion) . …… ● an atom . …… [3] [Total: 8]

Mark scheme

2(a)(i) 2 charge relative mass –1 0 1 1 (1) (1) Mark by column 2(a)(ii) 2 12 12 12 (1) 12 13 12 (1) Mark by row 2(a)(iii) (they have) 2 more protons than electrons 1 OR (they have) 2 fewer electrons than protons OR (they have) 12 protons and 10 electrons 2(b) Na+ or Al 3+ (1) 3 F- or O2– or N3– (1) Ne (1)

Question 312 marks

Question

This question is about sodium and compounds of sodium. (a) (i) Describe the bonding in a metallic element such as sodium. You may include a diagram as part of your answer. . …… . …… . …… [3] (ii) Describe how solid sodium conducts electricity. . …… [1] (b) Some properties of sodium chloride are shown: ● melting point of 801 °C ● non-conductor of electricity when solid ● conductor of electricity when molten ● soluble in water. (i) Name the type of bonding in sodium chloride. . …… [1] (ii) Explain why sodium chloride conducts electricity when molten. . …… . …… [1] (c) A student determines the concentration of a solution of dilute sulfuric acid, H2SO4, by titration with aqueous sodium hydroxide, NaOH. step 1 25.0 cm3 of 0.200 mol / dm3 NaOH is transferred into a conical flask. step 2 Three drops of methyl orange indicator are added to the conical flask. step 3 A burette is filled with H2SO4. step 4 The acid in the burette is added to the conical flask until the indicator changes colour. The volume of acid is recorded. This process is known as titration. step 5 The titration is repeated several times until a suitable number of results is obtained. (i) Name the piece of apparatus used to measure exactly 25.0 cm3 of 0.200 mol / dm3 NaOH in step 1. . …… [1] (ii) State the colour change of the methyl orange indicator in step 4. from …… to …… [1] (iii) State how the student decides that a suitable number of results have been obtained. . …… . …… [1] (iv) 20.0 cm3 of H2SO4 reacts with 25.0 cm3 of 0.200 mol / dm3 NaOH. The equation for the reaction is shown. H2SO4 + 2NaOH → Na2SO4 + 2H2O Calculate the concentration of H2SO4 using the following steps. ● Calculate the number of moles in 25.0 cm3 of 0.200 mol / dm3 NaOH. …… mol ● Determine the number of moles of H2SO4 that react with the NaOH. …… mol ● Calculate the concentration of H2SO4. …… mol / dm3 [3] [Total: 12]

Mark scheme

3(a)(i) positive ions / cations (1) 3 sea of electrons / mobile electrons / delocalised electrons (1) attraction between positive ions and electrons (1) 3(a)(ii) electrons move / 1 electrons mobile / electrons flow 3(b)(i) ionic 1 3(b)(ii) ions move / 1 ions are mobile / ions flow 3(c)(i) pipette 1 3(c)(ii) yellow to orange 1 3(c)(iii) at least two results are within 0.2 cm3 or less 1 3(c)(iv) 0.005 / 5 × 10–3 (1) 3 0.0025 / 2.5 × 10–3 (1) 0.125 (1)

Question 410 marks

Question

This question is about compounds of sulfur. (a) Sulfuric acid, H2SO4, is manufactured using the Contact process. This manufacture involves four stages. stage 1 Molten sulfur burns in air to produce sulfur dioxide. stage 2 Sulfur dioxide reacts with oxygen to form sulfur trioxide. stage 3 Sulfur trioxide combines with concentrated sulfuric acid to form oleum, H2S2O7. stage 4 Oleum reacts to form concentrated sulfuric acid. (i) Write a chemical equation for the reaction occurring in stage 1. . …… [1] (ii) State the essential conditions that are necessary for stage 2. Write an equation for the chemical reaction that occurs. . …… . …… . …… . …… [4] (iii) Write a chemical equation for the reaction occurring in stage 3. . …… [1] (iv) Name the substance that reacts with oleum in stage 4. . …… [1] (b) Hydrogen sulfide has the formula H2S. (i) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of hydrogen sulfide. Show outer shell electrons only. H S H [2] (ii) Balance the chemical equation for the reaction of hydrogen sulfide with sulfur dioxide shown. ……H2S + SO2 → ……S + ……H2O[1] [Total: 10]

Mark scheme

4(a)(i) S + O2 → SO2 1 4(a)(ii) (temperature) 450 °C (1) 4 (pressure) 1–2 atmosphere(s) (1) vanadium(V) oxide catalyst (1) 2SO2 + O2 ⇌ 2SO3 (1) 4(a)(iii) SO3 + H2SO4 → H2S2O7 1 4(a)(iv) water 1 4(b)(i) 2 bonding pairs as one dot and cross each (1) 2 2 lone pairs on S (and no additional electrons on Hs) to complete the outer shell on S and both Hs (1) 4(b)(ii) 2H2S + SO2 → 3S + 2H2O 1

Question 519 marks

Question

Ethanoic acid is manufactured by the reaction of methanol with carbon monoxide. An equilibrium mixture is produced. CH3OH(g) + CO(g) CH3COOH(g) (a) State two characteristics of an equilibrium. 1 . …… 2 . …… [2] (b) The purpose of the industrial process is to produce a high yield of ethanoic acid at a high rate of reaction. The manufacture is carried out at a temperature of 300 °C. The forward reaction is exothermic. Use this information to state why the manufacture is not carried out at temperatures: ● below 300 °C . …… ● above 300 °C. . …… [2] (c) Complete the table using only the words increases, decreases or no change. effect on the rate of effect on the equilibrium the forward reaction yield of CH3COOH(g) adding a catalyst no change decreasing the pressure [3] (d) Suggest which of the following metals is a suitable catalyst for the reaction. Give a reason for your answer. aluminium calcium cobalt magnesium potassium suitable catalyst …… reason . …… [2] (e) Ethanoic acid is a member of the homologous series of carboxylic acids. State the general formula of this homologous series. . …… [1] (f) Draw the structure of the carboxylic acid containing three carbon atoms. Show all of the atoms and all of the bonds. [2] (g) When carboxylic acids react with alcohols, esters are produced. The formula of ester X is CH3CH2CH2COOCH3. (i) Name ester X. . …… [1] (ii) Give the name of the carboxylic acid and the alcohol that react together to produce ester X. carboxylic acid …… alcohol …… [2] (h) Ester Y has the following composition by mass: C, 48.65%; H, 8.11%; O, 43.24%. Calculate the empirical formula of ester Y. empirical formula = …… [3] (i) Ester Z has the empirical formula C2H4O and a relative molecular mass of 88. Determine the molecular formula of ester Z. molecular formula = …… [1] [Total: 19]

Mark scheme

5(a) the rate of forward reaction equals the rate of the reverse reaction (1) 2 concentrations of reactants and products are constant (1) 5(b) reaction too slow (1) 2 yield of ethanoic acid too low (1) 5(c) 3 increases (1) decreases (1) decreases (1) 5(d) cobalt (1) 2 transition element (1) 5(e) CnH2n+1COOH 1 5(f) COOH fully displayed (1) 2 whole molecule completely correct (1) 5(g)(i) methyl butanoate 1 5(g)(ii) butanoic acid (1) 2 methanol (1) 5(h) C 48.65 / 12 3 H 8.11 / 1 O 43.24 / 16 OR 4.05:8.11:2.70 (1) fractions shown dividing all by smallest OR 1.5:3:1 OR 3:6:2 (1) C3H6O2 (1) 5(i) C4H8O2 1

Question 611 marks

Question

This question is about zinc and its compounds. (a) Zinc is extracted from its ore which is mainly zinc sulfide, ZnS. The steps for this extraction are shown. step 1 Zinc sulfide is converted into zinc oxide. step 2 The zinc oxide is then reduced to zinc in a furnace. The zinc formed becomes a gas. step 3 The zinc gas is cooled to form molten zinc. (i) Name the ore of zinc, which is mainly zinc sulfide. . …… [1] (ii) Describe how zinc sulfide is converted into zinc oxide in step 1. . …… . …… [1] (iii) Name the reducing agent used in step 2. . …… [1] (iv) Explain why the zinc forms a gas in step 2 inside the furnace. . …… [1] (v) State the name of the physical change occurring when zinc gas is converted into molten zinc. . …… [1] (b) Zinc sulfate crystals, ZnSO4•7H2O, are hydrated. Zinc sulfate crystals are made by reacting zinc carbonate with dilute sulfuric acid. The equation for the overall process is shown. ZnCO3 + H2SO4 + 6H2O → ZnSO4•7H2O + CO2 step 1 Large pieces of solid zinc carbonate are added to dilute sulfuric acid until the zinc carbonate is in excess. This forms aqueous zinc sulfate. step 2 The excess zinc carbonate is separated from the aqueous zinc sulfate. step 3 The aqueous zinc sulfate is heated until a saturated solution is formed. step 4 The saturated solution is allowed to cool and crystallise. step 5 The crystals are removed and dried. (i) In step 1, zinc carbonate is in excess when no more zinc carbonate dissolves. State one other observation that indicates the zinc carbonate is in excess in step 1. . …… [1] (ii) Name a different substance, other than zinc carbonate, that can be added to dilute sulfuric acid to produce aqueous zinc sulfate in step 1. . …… [1] (iii) Step 1 is repeated using powdered zinc carbonate instead of large pieces. All other conditions are kept the same. The rate of reaction increases. Give a reason why the rate of reaction increases. Explain your answer in terms of particles. . …… . …… . …… [2] (iv) Suggest what is observed when the solution is saturated in step 3. . …… . …… . …… [1] (v) The formula of zinc sulfate crystals is ZnSO4•7H2O. Give the formula of the solid formed if the crystals are heated to dryness in step 3. . …… [1] [Total: 11]

Mark scheme

6(a)(i) zinc blende 1 6(a)(ii) heat( zinc sulfide) strongly in air / roast in air 1 6(a)(iii) carbon or carbon monoxide 1 6(a)(iv) the temperature in the furnace is above or higher than the boiling point of zinc 1 OR the boiling point of zinc is below or less than the temperature of the furnace 6(a)(v) condensation / condensing / condense 1 6(b)(i) no bubbles 1 or no fizzing or no effervescence 6(b)(ii) zinc / zinc oxide / zinc hydroxide 1 6(b)(iii) (powder has) larger surface area OR lumps have smaller surface area (1) 2 (powder has) more collisions per unit time / more collision frequency OR lumps have fewer collisions per unit time / less collision frequency (1) 6(b)(iv) crystals (form on glass rod or microscope slide) 1 6(b)(v) ZnSO4 1

Question 711 marks

Question

The Periodic Table can be used to classify elements. (a) Group I elements react with cold water to form alkaline solutions. (i) Place the Group I elements caesium, lithium, potassium, rubidium and sodium in their order of reactivity with water. Put the most reactive element first. most reactive least reactive [1] (ii) Name the alkaline solution formed when caesium reacts with cold water. . …… [1] (b) Group I elements have lower melting points than transition elements. Describe one other difference in the physical properties of Group I elements and transition elements. . …… [1] (c) Group VII elements are known as the halogens. Astatine is below iodine in Group VII. Predict the physical state of astatine at room temperature and pressure. . …… [1] (d) Some Group VII elements react with aqueous solutions containing halide ions. When aqueous chlorine is added to aqueous potassium bromide a reaction occurs. The ionic half-equations for the reaction are shown. Cl 2(aq) + 2e– → 2Cl –(aq) 2Br –(aq) → Br2(aq) + 2e– (i) Describe the colour change of the solution. original colour of potassium bromide solution …… final colour of reaction mixture . …… [2] (ii) Identify the species that is oxidised. Explain your decision. species oxidised …… explanation …… [2] (e) Bromine monochloride, BrCl, is made by the reaction between bromine and chlorine. The chemical equation is shown. Br2(g) + Cl 2(g) → 2BrCl (g) bond bond energy in kJ / mol Br–Br 190 Cl –Cl 242 Br–Cl 218 Calculate the overall energy change for the reaction using bond energies. Use the following steps. ● Calculate the total amount of energy required to break the bonds in 1 mole of Br2(g) and 1 mole of Cl 2(g). …… kJ ● Calculate the total amount of energy released when the bonds in 2 moles of BrCl (g) are formed. …… kJ ● Calculate the overall energy change for the reaction. …… kJ / mol [3] [Total: 11] The Periodic Table of Elements Group I II III IV V VI VII VIII 1 2 H He hydrogen helium Key 1 4 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 7 9 relative atomic mass 11 12 14 16 19 20 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23 24 27 28 31 32 35.5 40 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 16 85 88 89 91 93 96 – 101 103 106 108 112 115 119 122 128 127 131 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 133 137 178 181 184 186 190 192 195 197 201 204 207 209 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 114 116 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium flerovium livermorium – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 139 140 141 144 – 150 152 157 159 163 165 167 169 173 175 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232 231 238 – – – – – – – – – – – The volume of one mole of any gas is 24 dm3 at room temperature and pressure (r.t.p.).

Mark scheme

7(a)(i) from left to right 1 caesium → rubidium→ potassium →sodium → lithium 7(a)(ii) caesium hydroxide 1 7(b) Group I element is less strong / not strong 1 ORA OR Group I element has low(er) density ORA OR Group I element is soft(er) ORA 7(c) solid 1 7(d)(i) colourless (1) 2 orange / brown / yellow (1) 7(d)(ii) Br – (1) 2 loses electron(s) (1) 7(e) 432(1) 3 436(1) – 4(1)

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Chemistry 0620/41 for May/June 2022 is out of 80 marks, across 7 questions.

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