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Chemistry 0620/43 May/June 2024: question paper and mark scheme

Cambridge IGCSE Chemistry (0620), Paper 4 Theory (Extended), May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0620_s24_qp_430620_s24_ms_430620/43/M/J/24

Paper
Paper 4 Theory (Extended)
Total marks
80
Code on the paper
0620/43/M/J/24
0620_s24_ms_43.pdf

0620/43 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 17 marks

Question

Name the process used to: (a) produce ammonia from nitrogen . …… [1] (b) produce lead from molten lead(II) bromide . …… [1] (c) separate an insoluble solid from a mixture of an insoluble solid and a solution . …… [1] (d) produce ethanol from ethene . …… [1] (e) identify the components of a mixture of soluble coloured substances . …… [1] (f) separate a mixture of several liquids with different boiling points . …… [1] (g) determine the volume of an acid required to neutralise a given volume of an alkali. . …… [1] [Total: 7]

Mark scheme

1(a) Haber (process) 1 1(b) electrolysis 1 1(c) filtration 1 1(d) catalytic addition 1 1(e) chromatography 1 1(f) fractional distillation 1 1(g) titration 1

Question 25 marks

Question

Complete Table 2.1. Table 2.1 number of number of number of atom or ion protons electrons neutrons 29Cu 63 29 17Cl 37 – 20 30 28 34 [5]

Mark scheme

2 M1 Cu: 29 and 34 5 M2 Cl– : 17 and 18 M3 6430 to left of symbol M4 Zn M5 2+

Question 314 marks

Question

This question is about elements and compounds. (a) Some properties of graphite, oxygen and carbon monoxide are shown in Table 3.1. Table 3.1 conduction of melting point boiling point electricity when / °C / °C solid graphite 3652 4827 good oxygen –218 –183 poor carbon monoxide –199 –191 poor (i) Explain why graphite conducts electricity when solid. . …… . …… [1] (ii) Complete the dot-and-cross diagram in Fig. 3.1 of a molecule of oxygen. Show outer shell electrons only. O O Fig. 3.1 [2] (iii) Deduce the physical state of carbon monoxide at –195 °C. Use the data in Table 3.1 to explain your answer. physical state . …… explanation …… . …… [2] (iv) Explain in terms of structure and bonding why graphite has a much higher melting point than carbon monoxide. . …… . …… . …… . …… [3] (b) Potassium reacts with chlorine to form potassium chloride. Write a symbol equation for this reaction. . …… [2] (c) A dilute aqueous solution of potassium chloride undergoes electrolysis. Oxygen is produced at the anode. (i) State what is meant by the term electrolysis. . …… . …… . …… [2] (ii) Write an ionic half-equation for the production of oxygen at the anode. . …… [2] [Total: 14]

Mark scheme

3(a)(i) mobile electrons 1 3(a)(ii) M1 double bond of two dots and two crosses 2 M2 4 non-bonding electrons on each oxygen atom 3(a)(iii) M1 liquid 2 M2 melting point is below –195 °C and boiling point is above –195 °C or –195 °C is higher than –199 °C / melting point and lower than –191 °C / boiling point or –195 °C is in between melting point / –199 °C and boiling point / –191 °C 3(a)(iv) M1 graphite has a giant covalent structure 3 M2 attraction between molecules in carbon monoxide M3 weaker attraction (between particles) in carbon monoxide ORA 3(b) 2K + Cl2 → 2KCl 2 M1 KCl as only product M2 correct equation 3(c)(i) M1 breakdown by (the passage of) electricity 2 M2 of an ionic compound in molten or aqueous (state) 3(c)(ii) 4OH– → 2H2O + O2 + 4e– 2 M1 any negatively charged OH species losing electrons M2 correct ionic half-equation

Question 410 marks

Question

Dinitrogen tetroxide, N2O4, decomposes into nitrogen dioxide, NO2.The reaction is reversible. N2O4(g) 2NO2(g) colourless gas brown gas Fig. 4.1 shows a gas syringe containing a mixture of dinitrogen tetroxide and nitrogen dioxide. The gas syringe is sealed. The mixture reaches equilibrium and the colour of the mixture of gases is a pale brown. mixture of dinitrogen tetroxide and nitrogen dioxide end sealed plunger Fig. 4.1 (a) Describe a reversible reaction at equilibrium in terms of: ● the rate of the forward reaction and the rate of the reverse reaction . …… ● the concentration of reactants and products. . …… [2] (b) The pressure of the mixture is increased. All other conditions stay the same. The mixture immediately turns darker brown before the position of equilibrium changes. Explain in terms of particles why the mixture immediately turns darker brown. . …… [1] (c) The temperature of the mixture is increased. All other conditions stay the same. The mixture turns darker brown. State what can be deduced about the forward reaction from this information. . …… [1] (d) Sulfur is converted into sulfuric acid, H2SO4, by a series of reactions. Sulfur dioxide, SO2, and oxygen, O2, react to form sulfur trioxide, SO3. The reversible reaction reaches equilibrium. 2SO2(g) + O2(g) 2SO3(g) (i) Complete Table 4.1 using only the words, increases, decreases or no change. Table 4.1 effect on the rate of effect on the the forward reaction equilibrium yield of SO3(g) add a catalyst increase the pressure [4] (ii) Deduce the oxidation number of sulfur in: S …… SO3 …… [2] [Total: 10]

Mark scheme

4(a) M1 (rates are) equal 2 M2 (concentrations are) no longer changing 4(b) more molecules of nitrogen dioxide per unit volume 1 OR same number of nitrogen dioxide molecules in a smaller volume 4(c) (the forward reaction is) endothermic 1 4(d)(i) 4 M1 M3 no increases change M2 M4 increases increases 4(d)(ii) M1 (the oxidation number of sulfur in S) = zero 2 M2 (the oxidation number of sulfur in SO3) = +6

Question 517 marks

Question

(a) Barium sulfate, BaSO4, is an insoluble salt and is made by precipitation. (i) Name two aqueous solutions that produce a precipitate of barium sulfate when they are mixed. 1 . …… 2 . …… [2] (ii) Describe how to produce a pure sample of barium sulfate from the mixture of aqueous solutions in (a)(i). . …… . …… [2] (iii) Write an ionic equation for the precipitation reaction which produces barium sulfate. Include state symbols. . …… [3] (b) Soluble salts are made from dilute acids. Name the dilute acid and one other substance that react together to make copper(II) sulfate. dilute acid …… other substance …… [2] (c) Nitrates decompose when they are heated. When hydrated copper(II) nitrate is heated, oxygen gas is produced. (i) Describe a test for oxygen. test . …… observations …… [1] (ii) Complete the equation for the decomposition of hydrated copper(II) nitrate. 2Cu(NO3)2•3H2O → ……CuO + ……NO2 + O2 + ……H2O[2] (d) Hydrated zinc sulfate gives off water when it is heated. ZnSO4•xH2O(s) → ZnSO4(s) + xH2O(g) A student does an experiment to determine the value of x in ZnSO4•xH2O. step 1 The student weighs a sample of hydrated zinc sulfate. step 2 The student heats the sample of hydrated zinc sulfate. step 3 The student weighs the solid after heating. step 4 The student repeats step 2 and step 3 until the mass of solid after heating is constant. (i) State why the student does step 4. . …… [1] (ii) In an experiment, 0.574 g of ZnSO4•xH2O is heated until the mass is constant. The mass of ZnSO4 that remains is 0.322 g. [Mr: ZnSO4, 161; H2O, 18] Determine the value of x using the following steps. ● Calculate the number of moles of ZnSO4 remaining. …… mol ● Calculate the mass of H2O given off. …… g ● Calculate the number of moles of H2O given off. …… mol ● Determine the value of x. x = …… [4] [Total: 17]

Mark scheme

5(a)(i) M1 named soluble barium salt e.g. barium chloride / barium nitrate 2 M2 named soluble sulfate salt e.g. sodium sulfate / potassium sulfate / ammonium sulfate / sulfuric acid 5(a)(ii) M1 filter 2 M2 wash (residue) with water and dry (residue) between filter papers / in a warm place 5(a)(iii) Ba2+ (aq) + SO42– (aq) → BaSO4(s) 3 M1 BaSO4 as only product M2 Ba2+ + SO42– as only reactants M3 state symbols 5(b) M1 sulfuric (acid) 2 M2 copper(II) carbonate / copper(II) oxide / copper(II) hydroxide 5(c)(i) glowing splint and relights 1 5(c)(ii) M1 2CuO + 4NO2 2 M2 6H2O 5(d)(i) (To ensure that) all the water is given off 1 5(d)(ii) M1 mol ZnSO4 4 (0.322 / 161 =) 0.002(00) M2 mass H2O given off (0.574 – 0.322 =) 0.252 M3 mol H2O given off (0.252 ÷ 18 =) 0.014(0) M4 value of x (0.014 ÷ 0.002 =) 7

Question 613 marks

Question

This question is about iron. (a) Fig. 6.1 shows a blast furnace used to extract iron from its ore. waste gases raw materials: coke, iron ore, limestone A molten iron Fig. 6.1 (i) Name the main ore of iron used in the blast furnace. . …… [1] (ii) Name the substance that enters the blast furnace at A. . …… [1] (iii) Name the reducing agent in the extraction of iron in the blast furnace. . …… [1] (iv) Explain why limestone is added to the blast furnace. Give details of the chemical reactions that are involved. . …… . …… . …… . …… [3] (b) The list shows the properties of some elements. ● act as catalysts ● have low densities ● have low melting points ● form acidic or basic oxides ● form coloured compounds ● form positive or negative ions Iron is a transition metal. Sodium is a Group I metal. State which property from the list: (i) is true for sodium but not iron . …… [1] (ii) is true for iron but not sodium . …… [1] (iii) is true for both sodium and iron . …… [1] (iv) is not true for sodium and not true for iron. . …… [1] (c) Steel consists mainly of iron. Iron rusts when it reacts with water and oxygen. Fig. 6.2 shows magnesium blocks attached to the bottom of a steel boat. The magnesium does not completely cover the steel. The magnesium blocks provide sacrificial protection for the steel. steel boat water containing magnesium dissolved oxygen blocks Fig. 6.2 (i) Explain, in terms of electrons, why magnesium is used for sacrificial protection. . …… . …… [2] (ii) Name a metal that cannot provide sacrificial protection for steel. . …… [1] [Total: 13]

Mark scheme

6(a)(i) hematite 1 6(a)(ii) air 1 6(a)(iii) carbon monoxide 1 6(a)(iv) M1 limestone thermally decomposes (into calcium oxide and carbon dioxide) 3 M2 calcium oxide reacts with silicon(IV) oxide M3 producing slag 6(b)(i) have low densities 1 or have low melting point 6(b)(ii) act as catalysts 1 or form coloured compounds 6(b)(iii) form basic oxides 1 or form positive ions 6(b)(iv) form acidic oxides 1 or form negative ions 6(c)(i) M1 magnesium loses electrons more readily than / in preference to / instead of iron 2 M2 magnesium is more reactive than iron ORA 6(c)(ii) any metal below iron in the reactivity series e.g. copper / silver / gold 1

Question 714 marks

Question

Many organic compounds contain carbon and hydrogen only. (a) (i) An organic compound A has the following composition by mass. C, 83.33%; H, 16.67% Calculate the empirical formula of compound A. empirical formula = …… [3] (ii) Compound B has the empirical formula C2H5 and a relative molecular mass of 58. Determine the molecular formula of compound B. molecular formula = …… [2] (b) Fig. 7.1 shows a section of a polymer formed from an alkene. (i) Identify the functional group in alkenes that reacts when alkenes form polymers. . …… [1] (ii) A section of a polymer is shown in Fig. 7.1. C 2H 5 H C 2H 5 H C C C C H H H H Fig. 7.1 ● Draw the displayed formula of the monomer that forms this polymer. ● Name the monomer used to form this polymer. . …… [3] (c) Alkenes are produced by cracking alkanes. When C12H26 is cracked, the products are ethene and an alkane which form in a 2 : 1 mole ratio. Write a symbol equation for this reaction. C12H26 → …… + …… [2] (d) (i) State the general formula for alcohols. . …… [1] (ii) Draw the displayed formula of one alcohol with the molecular formula C3H8O. Name the alcohol you have drawn. name of alcohol …… [2] [Total: 14] 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 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 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) M1 C 83.33 / 12: H 16.67 / 1 3 or 6.94:16.67 M2 C 6.94 / 6.94: H 16.67 / 6.94 or 1:2.4 or 5:12 M3 C5H12 7(a)(ii) M1 Mr of C2H5 = 29 2 M2 58 / 29 = 2 and C4H10 7(b)(i) C=C 1 7(b)(ii) M1 3 C C any C=C with both carbons having a valency of 4 M2 displayed formula of but-1-ene M3 but-1-ene 7(c) M1 2C2H4 2 M2 C8H18 7(d)(i) CnH2n + 1OH 1 7(d)(ii) M1 displayed formula of propan-1-ol or propan-2-ol 2 M2 propan-1-ol or propan-2-ol

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