Chemistry 0620/43 October/November 2024: question paper and mark scheme

Question

A list of substances is shown. bauxite carbon dioxide cryolite ethane ethanol ethene graphite helium hematite hydrogen silicon(IV) oxide sodium chloride Answer the following questions using only the substances from the list. Each substance may be used once, more than once or not at all. State which substance: (a) is manufactured by fermentation …… [1] (b) is monatomic …… [1] (c) is a reactant in photosynthesis …… [1] (d) is a solvent in the extraction of aluminium …… [1] (e) is an ore of iron …… [1] (f) is manufactured from methane …… [1] (g) is a compound with a giant covalent structure …… [1] (h) is used as a lubricant …… [1] (i) is tested for with limewater. …… [1] [Total: 9]

Question

This question is about electrolysis. (a) State the meaning of the term electrolysis. …… [2] (b) Table 2.1 gives some information about the electrolysis of two electrolytes using graphite electrodes. Table 2.1 anode cathode (positive electrode) (negative electrode) electrolyte observation name of product observation name of product concentrated aqueous bubbles of potassium iodide colourless gas aqueous bubbles of oxygen pink-brown solid copper(II) sulfate colourless gas (i) Complete Table 2.1. [4] (ii) Oxygen is produced at the anode by the electrolysis of aqueous copper(II) sulfate. Write the ionic half-equation for this reaction. …… [2] (c) Aqueous copper(II) sulfate is electrolysed using copper electrodes instead of graphite electrodes. (i) Explain why the mass of the anode decreases during this electrolysis. …… [1] (ii) Name the product formed at the cathode. …… [1] (iii) State what change, if any, is observed in the appearance of the aqueous copper(II) sulfate. …… [1] [Total: 11]

Question

This question is about compounds of tin. (a) Tin(IV) oxide has the formula SnO2. The relative formula mass, Mr, of SnO2 is 151. Calculate the percentage by mass of tin in SnO2. percentage by mass of tin in SnO2 = ……% [1] (b) SnO2 is an amphoteric oxide. SnO2 reacts with aqueous sodium hydroxide, NaOH, to form a sodium salt and water only. The sodium salt contains a negative ion with the formula SnO32–. (i) State the meaning of the term amphoteric. …… [1] (ii) Write the symbol equation for the reaction between SnO2 and NaOH. …… [2] (c) Tin is a metal that forms both covalent and ionic compounds. Suggest why this is unusual for a metal. …… [1] (d) (i) Tin(IV) chloride, SnCl 4, is covalently bonded. A tin atom has four electrons in its outer shell. Complete the dot-and-cross diagram in Fig. 3.1 for a molecule of SnCl 4. Show the outer shell electrons only. Cl Cl Sn Cl Cl Fig. 3.1 [2] (ii) Tin(II) oxide, SnO, is ionically bonded. The melting points of SnCl 4 and SnO are shown in Table 3.1. Table 3.1 melting point / °C SnCl 4 –33 SnO 1080 Explain, in terms of structure and bonding, why SnCl 4 has a much lower melting point than SnO. …… [3] (e) Part of the reactivity series is shown. magnesium most reactive aluminium tin copper least reactive (i) When aluminium foil is added to aqueous tin(II) sulfate, a reaction does not occur even though aluminium is above tin in the reactivity series. Explain why a reaction does not occur. …… [1] (ii) An aqueous solution of tin(II) sulfate contains Sn2+ ions. Two experiments are carried out. Experiment 1 Copper is added to aqueous tin(II) sulfate. Experiment 2 Magnesium is added to aqueous tin(II) sulfate. Write an ionic equation for any reaction that occurs in each experiment. If no reaction occurs, write ‘no reaction’. Experiment 1 …… Experiment 2 …… [2] (f) Hydrated tin(II) nitrate, Sn(NO3)2•20H2O, decomposes when it is heated. (i) State what is meant by the term hydrated. …… [1] (ii) Complete the equation for the decomposition of Sn(NO3)2•20H2O. 2Sn(NO3)2•20H2O → ……SnO + ……NO2 + O2 + ……H2O [2] [Total: 16]

Question

This question is about sulfuric acid, H2SO4. (a) Dilute sulfuric acid and aqueous sodium hydroxide can be used to prepare sodium sulfate crystals using a method that involves titration. The apparatus for titration is shown in Fig. 4.1. dilute sulfuric acid conical flask 25.0 cm3 aqueous sodium hydroxide 0.100 mol / dm3 Fig. 4.1 Thymolphthalein is used as an indicator for this titration. (i) State the colour change of thymolphthalein at the end-point of this titration. from …… to …… [2] (ii) Suggest why universal indicator is not used for this titration. …… [1] (b) 25.0 cm3 of aqueous sodium hydroxide, NaOH, of concentration 0.100 mol / dm3 is neutralised by 20.0 cm3 of dilute sulfuric acid, H2SO4. The equation for the reaction is shown. 2NaOH + H2SO4 → Na2SO4 + 2H2O Calculate the concentration of H2SO4 using the following steps. ● Calculate the number of moles of NaOH used. …… mol ● Determine the number of moles of H2SO4 that react with the NaOH. …… mol ● Calculate the concentration of H2SO4. …… mol / dm3 [3] (c) A student is provided with an aqueous solution of sodium sulfate. Describe how to prepare a pure sample of sodium sulfate crystals from this solution. …… [3] (d) Potassium hydrogen sulfate, KHSO4, can be prepared by a reaction between aqueous potassium hydroxide and dilute sulfuric acid. Water is the only other product. Write a symbol equation for this reaction. …… [1] (e) Potassium hydrogen sulfate, KHSO4, dissolves in water to form solution X. Solution X contains K+, H+ and SO42– ions. (i) Name the type of solution that contains H+ ions. …… [1] (ii) State the observations when the following tests are done. ● A flame test is carried out on X. ● Solid copper(II) carbonate is added to X. ● Aqueous barium nitrate acidified with dilute nitric acid is added to X. [5] (f) 0.325 g of Zn is added to dilute sulfuric acid which contains 0.0100 moles of H2SO4. The equation for this reaction is shown. Zn + H2SO4 → ZnSO4 + H2 (i) Determine whether Zn or H2SO4 is the limiting reactant. Explain your answer. …… [2] (ii) In another experiment, 48.0 cm3 of hydrogen gas, H2, is produced. The experiment is carried out at room temperature and pressure, r.t.p. Calculate the number of molecules in 48.0 cm3 of H2 gas measured at r.t.p. The value of the Avogadro constant is 6.02 × 1023. …… molecules [2] [Total: 20]

Question

This question is about rate of reaction and equilibrium. A student investigates the rate of decomposition of aqueous hydrogen peroxide, H2O2, using manganese(IV) oxide as a catalyst. The equation for the reaction is shown. 2H2O2(aq) → 2H2O(l) + O2(g) The student uses the apparatus shown in Fig. 5.1. loosely fitting cotton wool manganese(IV) oxide catalyst aqueous hydrogen peroxide container balance Fig. 5.1 The student: ● adds the catalyst to the aqueous hydrogen peroxide ● replaces the container on the balance ● starts a stop-watch ● records the mass at regular time intervals. (a) Table 5.1 shows the mass recorded at regular time intervals. Table 5.1 time / s mass / g 0 50.64 30 49.80 60 49.38 90 49.17 120 49.07 150 49.02 180 48.99 210 48.97 240 48.97 270 48.97 (i) Suggest why the mass decreases as time increases. …… [1] (ii) After a certain time the reaction stops. Explain why the reaction stops. …… [1] (iii) Suggest why it is not possible to use the results in Table 5.1 to determine the exact time when the reaction stops. …… [1] (b) Fig. 5.2 shows a graph of the mass against time. mass / g 0 time / s Fig. 5.2 The experiment is repeated at a higher temperature. All other conditions remain the same. (i) Explain, in terms of collision theory, why the rate of reaction is higher at a higher temperature. …… [3] (ii) On Fig. 5.2, sketch the line expected when the experiment is repeated at a higher temperature. [2] (c) Manganese(IV) oxide is the catalyst in this reaction. (i) Explain the meaning of (IV) in manganese(IV) oxide. …… [2] (ii) State how the mass of the catalyst has changed, if at all, at the end of the experiment. …… [1] (d) Nitrogen monoxide gas, NO, and oxygen gas, O2, react to produce nitrogen dioxide gas, NO2, at room temperature. The reaction can reach equilibrium. The equation is shown. 2NO(g) + O2(g) 2NO2(g) ∆H = –113 kJ / mol NO(g) and O2(g) are passed into a beaker as shown in Fig. 5.3. NO(g) O2(g) beaker Fig. 5.3 (i) Explain why the method shown in Fig. 5.3 will not allow the reaction to reach equilibrium. …… [1] (ii) The apparatus is changed and equilibrium is reached. The temperature of the equilibrium system is then increased and the position of equilibrium shifts to the left. Explain why the position of equilibrium shifts to the left. …… [1] (iii) The pressure of the equilibrium system is then increased. State the direction, if any, in which the position of equilibrium shifts. Explain your answer. direction …… explanation …… [2] [Total: 15]

Question

This question is about hydrocarbons. (a) State the meaning of the term hydrocarbon. …… [1] (b) Propene, C3H6, can be made from long-chain alkanes such as dodecane. Dodecane contains 12 carbon atoms. (i) Deduce the molecular formula of dodecane. …… [1] (ii) Name the type of reaction that occurs when long-chain alkanes are converted into shorter chain alkenes. …… [1] (c) Propene is an unsaturated hydrocarbon. Propene reacts with bromine. (i) State the meaning of the term unsaturated. …… [1] (ii) Write the molecular formula of the product formed when propene reacts with bromine. …… [1] (d) A styrene molecule is represented as shown in Fig. 6.1. H C 6H 5 C C H H Fig. 6.1 (i) The molecular formula of styrene is C8H8. Determine the empirical formula of styrene. …… [1] (ii) Styrene can be polymerised into poly(styrene). State the type of polymerisation that occurs when styrene is converted into poly(styrene). …… [1] (iii) Draw the structure of one repeat unit of poly(styrene). Include all of the atoms and all of the bonds. The C6H5 group should be represented as C6H5. [2] [Total: 9] 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 20 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.).