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Chemistry 0620/42 October/November 2024: question paper and mark scheme

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

0620_w24_qp_420620_w24_ms_420620/42/O/N/24

Paper
Paper 4 Theory (Extended)
Total marks
80
Time
1 hour 15 minutes
Questions
6
Code on the paper
0620/42/O/N/24
0620_w24_ms_42.pdf

0620/42 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

The formulae of the molecules A to I are shown in Table 1.1. Table 1.1 molecule formula A C 2H 4 B C2H5OH C CO D CO2 E Cl 2 F NO2 G N2 H O2 I SO2 Answer the following questions about the molecules, A to I. Each letter may be used once, more than once or not at all. State which of the molecules A to I: (a) is an element with a triple bond …… [1] (b) is a product of photosynthesis …… [1] (c) is used as a fuel …… [1] (d) turns limewater milky …… [1] (e) undergoes a substitution reaction with alkanes …… [1] (f) is a colourless liquid at r.t.p. …… [1] (g) is unsaturated …… [1] (h) is 21% of clean, dry air …… [1] (i) is a reactant in the Haber process. …… [1] [Total: 9]

Mark scheme

1(a) G 1 1(b) H 1 1(c) B 1 1(d) D 1 1(e) E 1 1(f) B 1 1(g) A 1 1(h) H 1 1(i) G 1

Question 213 marks

Question

Aluminium is manufactured by the electrolysis of aluminium oxide. (a) State the name of the main ore of aluminium. …… [1] (b) Name the substance mixed with aluminium oxide to reduce the operating temperature of the process. …… [1] (c) Explain why the molten mixture in (b) conducts electricity. …… [1] (d) Table 2.1 contains some information about the processes which take place at the anode and the cathode. Table 2.1 anode cathode (i) Complete Table 2.1: ● Write the number of electrons needed to balance the ionic half-equation for the reaction at the anode. ● Write the ionic half-equation for the reaction at the cathode. [3] (ii) State why the process at the anode is an oxidation. …… [1] (iii) Oxygen is formed at the anode. Explain why the main gas given off at the anode is carbon dioxide and not oxygen. …… [2] (e) State why aluminium is used in food containers. …… [1] (f) Aluminium reacts with fluorine to form the ionic compound aluminium fluoride. Complete the dot-and-cross diagram in Fig. 2.1 of the ions in aluminium fluoride. Give the charges on the ions. Al F F F Fig. 2.1 [3] [Total: 13]

Mark scheme

2(a) bauxite 1 2(b) cryolite 1 2(c) (it has) mobile ions 1 2(d)(i) M1 4 3 Al 3+ + 3e- → Al M2 Al 3+ AND e- M3 correct equation 2(d)(ii) electrons are lost (from oxide ions) 1 2(d)(iii) M1 anode / electrode made of carbon or graphite 2 M2 anode / electrode reacts with oxygen / burns / combusts (in oxygen) 2(e) aluminium is resistant to corrosion 1 2(f) M1 eight crosses in second shell of Al 3 M2 seven dots and one cross in second shell of each F M3 ‘3+’ charge on Al ion on correct answer line AND ‘–‘ charge on each F ion on correct answer line

Question 319 marks

Question

Sulfur forms two chlorides, P and Q. Chloride P has the formula S2Cl 2. Chloride Q has the formula SCl 2. (a) Both chlorides are covalently bonded and have low melting points. Suggest, in terms of attraction between particles, why these chlorides have low melting points. …… [2] (b) Chloride P, S2Cl 2, forms when sulfur reacts with chlorine. Write the symbol equation for this reaction. …… [1] (c) Complete the dot-and-cross diagram in Fig. 3.1 of a molecule of chloride Q, SCl 2. Show outer electrons only. Cl S Cl Fig. 3.1 [3] (d) Chloride P is converted to chloride Q by reaction with chlorine in a closed system. The reversible reaction reaches an equilibrium. S2Cl 2(g) + Cl 2(g) 2SCl 2(g) P Q The forward reaction is exothermic. Suggest two changes to the conditions which will result in a decrease in the concentration of chloride Q at equilibrium. 1 …… 2 …… [2] (e) The rate of the forward reaction in (d) is determined by collision theory. The rate of reaction depends upon two factors: ● the frequency of collisions between particles ● the proportion of collisions which have energy greater than or equal to the activation energy. (i) Define the term activation energy. …… [1] (ii) Give the symbol for activation energy. …… [1] (iii) Complete Table 3.1 to show the effect, if any, when the conditions are changed. Use only the words increases, decreases or no change. Table 3.1 effect on the effect on the proportion of collisions change to conditions frequency of collisions which have energy greater than between particles or equal to the activation energy concentration of chlorine is increased temperature is increased a catalyst is added [5] (f) The reaction of chloride P with chlorine is a redox reaction. S2Cl 2(g) + Cl 2(g) 2SCl 2(g) P Q The oxidation number of Cl in chloride P and chloride Q is –1. Use oxidation numbers to explain why: ● sulfur is oxidised in the forward reaction ● chlorine is oxidised in the reverse reaction. [4] [Total: 19] Question 4 starts on the next page.

Mark scheme

3(a) M1 attraction between molecules 2 M2 attraction is weak 3(b) 2S + Cl2 → S2Cl2 1 3(c) M1 S with one dot-cross bonding pair with each Cl 3 M2 four non-bonding dots for S M3 six non-bonding crosses for each Cl 3(d) M1 decrease concentration of S2Cl2 / P and / or Cl2 2 M2 increase temperature 3(e)(i) the minimum energy that colliding particles must have to react 1 3(e)(ii) Ea 1 3(e)(iii) M1 increases M2 no change 5 M3 increases M4 increases M5 increases 3(f) M1 oxidation number of S goes from +1 4 M2 oxidation number of S goes to +2 M3 oxidation number of Cl (goes from –1) to 0 (in Cl2) M4 oxidation numbers of Cl AND S increase

Question 414 marks

Question

Silver bromide, AgBr, is made when aqueous silver ethanoate, CH3COOAg, is added to aqueous sodium bromide, NaBr. The equation for the reaction is shown in equation 1. equation 1 CH3COOAg + NaBr → CH3COONa + AgBr The method includes the following steps. step 1 Add 200.0 cm3 of 0.0500 mol / dm3 CH3COOAg to a beaker. This volume contains 0.0100 mol of Ag+ ions. step 2 Add 50.0 cm3 of aqueous NaBr. This volume contains 0.0100 mol of Br – ions. A precipitate forms. step 3 Filter the mixture. step 4 Dry the solid residue until all the water is removed. step 5 Record the mass of the dry residue. (a) Complete the ionic equation for the reaction by adding the missing state symbols. Ag+(……) + Br –(……) → AgBr(……) [1] (b) Name a different aqueous silver salt which could be used in step 1. …… [1] (c) Use the information in step 2 to calculate the concentration of aqueous NaBr. concentration = …… mol / dm3 [1] (d) State the colour of the precipitate which forms in step 2. …… [1] (e) Use the information in step 1, step 2 and equation 1 to determine the number of moles of AgBr formed. Use this value to calculate the mass of AgBr formed. number of moles of AgBr = …… mass of AgBr = …… g [3] (f) Name the salt dissolved in the filtrate in step 3. …… [1] (g) The recorded mass of the dry residue in step 5 is greater than the mass calculated in (e) because a step is missing from the procedure. (i) Suggest the missing step. …… [1] (ii) Name the substance responsible for the greater mass of the dry residue. …… [1] (h) Barium sulfate can be made by the same method but with different aqueous solutions. (i) Suggest two aqueous solutions which can be added together to make barium sulfate. …… and …… [2] (ii) Write the balanced symbol equation for this reaction. …… [2] [Total: 14]

Mark scheme

4(a) Ag+(aq) + Br–(aq) → AgBr(s) 1 4(b) (silver) nitrate 1 4(c) 0.01 × 1000 / 50 = 0.2(00) (mol / dm3) 1 4(d) cream 1 4(e) M1 mol of AgBr = 0.01(00) 3 M2 Mr of AgBr = 188 M3 188 x 0.01 = 1.88 (g) 4(f) sodium ethanoate 1 4(g)(i) rinsing of residue 1 4(g)(ii) (crystals of) sodium ethanoate 1 4(h)(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 4(h)(ii) M1 any equation with BaSO4 as a product 2 M2 correct equation using soluble reactants

Question 511 marks

Question

Alkenes are manufactured by cracking larger alkane molecules. (a) State the source of the large alkane molecules used in cracking. …… [1] (b) State two conditions needed for cracking large alkane molecules. 1 …… 2 …… [2] (c) When one molecule of dodecane, C12H26, is cracked, three molecules of but-1-ene and one other product are formed. (i) Use molecular formulae to complete the symbol equation for this reaction. C12H26 → …… + …… [2] (ii) Suggest the type of chemical reaction which happens during cracking. …… [1] (d) Propene will undergo polymerisation. (i) Suggest the name of the polymer formed from propene. …… [1] (ii) Draw part of this polymer molecule to show three repeat units. [3] (iii) State the type of polymerisation propene undergoes. …… [1] [Total: 11] Question 6 starts on the next page.

Mark scheme

5(a) petroleum 1 5(b) M1 high temperature 2 M2 catalyst 5(c)(i) C12H26 → 3C4H8 + H2 2 M1 C4H8 M2 correct equation 5(c)(ii) thermal decomposition 1 5(d)(i) poly(propene) 1 5(d)(ii) M1 chain of six C atoms joined by single bonds in a chain 3 M2 three correctly placed CH3 groups M3 correct structure AND continuation bonds 5(d)(iii) addition 1

Question 614 marks

Question

Polyamides and polyesters are polymers. Polyamides can occur naturally or can be manufactured. (a) Part of the structure of a polyamide is shown in Fig. 6.1. O O O O O C C N N C C N N C H H H H Fig. 6.1 (i) On Fig. 6.1, draw a circle around one amide linkage. [1] (ii) Complete Fig. 6.2 to show the structures of the two monomers needed to make the polymer in Fig. 6.1. Show all of the atoms and all of the bonds in the functional groups. Fig. 6.2 [2] (iii) Name the other product formed in this polymerisation. …… [1] (iv) State the term given to natural polyamides. …… [1] (v) Name the type of monomers which are used to make natural polyamides. …… [1] (vi) One of the monomers which forms part of a natural polyamide has three carbon atoms. Complete Fig. 6.3 to show the displayed formula of this monomer. C H Fig. 6.3 [3] (b) PET is a polyester. (i) Name the two types of monomer molecules needed to make polyesters. …… and …… [2] (ii) Draw part of the structure of PET which shows two repeat units. Show all of the atoms and all of the bonds in the linkages. [3] [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

6(a)(i) one amide linkage circled 1 6(a)(ii) M1 two fully displayed carboxylic acid groups on the shaded box 2 M2 two fully displayed amino groups on the unshaded box 6(a)(iii) water 1 6(a)(iv) protein(s) 1 6(a)(v) amino acids 1 6(a)(vi) displayed formula of 2-aminopropanoic acid (3) 3 M1 displayed NH2 or displayed COOH group present M2 displayed NH2 and displayed COOH groups attached to same C atom M3 correct displayed structure of 2-aminopropanoic acid 6(b)(i) M1 dicarboxylic acids 2 M2 diols 6(b)(ii) 3 M1 any one correct fully displayed ester link between any two blocks M2 three fully displayed inter-block ester links with correct orientation M3 fully correct diagram showing two complete repeat units of polyester including continuation bonds

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Where can I find the 0620/42 October/November 2024 mark scheme?

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Chemistry 0620/42 for October/November 2024 is out of 80 marks, with 1 hour 15 minutes to answer it, across 6 questions.

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