0620_w24_qp_43
A paper of Chemistry, 0620
Questions:
6
Year:
2024
Paper:
4
Variant:
3

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This question is about compounds of tin. Tin(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 = % 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 SnO3 2–. State the meaning of the term amphoteric. Write the symbol equation for the reaction between SnO2 and NaOH. Tin is a metal that forms both covalent and ionic compounds. Suggest why this is unusual for a metal. Tin(chloride, SnCl 4, is covalently bonded. A tin atom has four electrons in its outer shell. Complete the dot-and-cross diagram in for a molecule of SnCl 4. Show the outer shell electrons only. Sn Cl Cl Cl Cl Tin(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 Explain, in terms of structure and bonding, why SnCl 4 has a much lower melting point than SnO. Part of the reactivity series is shown. magnesium most reactive tin aluminium copper least reactive When aluminium foil is added to aqueous tin(sulfate, a reaction does not occur even though aluminium is above tin in the reactivity series. Explain why a reaction does not occur. An aqueous solution of tin(sulfate contains Sn2+ ions. Two experiments are carried out. Experiment 1 Copper is added to aqueous tin(sulfate. Experiment 2 Magnesium is added to aqueous tin(sulfate. Write an ionic equation for any reaction that occurs in each experiment. If no reaction occurs, write ‘no reaction’. Experiment 1 Experiment 2 Hydrated tin(nitrate, Sn(NO3)2•20H2O, decomposes when it is heated. State what is meant by the term hydrated. Complete the equation for the decomposition of Sn(NO3)2•20H2O. 2Sn(NO3)2•20H2O → SnO + NO2 + O2 + H2O
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This question is about sulfuric acid, H2SO4. 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 . 25.0 cm3 aqueous sodium hydroxide 0.100 mol / dm3 dilute sulfuric acid conical flask Thymolphthalein is used as an indicator for this titration. State the colour change of thymolphthalein at the end-point of this titration. from to Suggest why universal indicator is not used for this titration. 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 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. 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. Potassium hydrogen sulfate, KHSO4, dissolves in water to form solution X. Solution X contains K+, H+ and SO4 2– ions. Name the type of solution that contains H+ ions. State the observations when the following tests are done. ● A flame test is carried out on X. ● Solid copper(carbonate is added to X. ● Aqueous barium nitrate acidified with dilute nitric acid is added to X. 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 Determine whether Zn or H2SO4 is the limiting reactant. Explain your answer. 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
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This question is about rate of reaction and equilibrium. A student investigates the rate of decomposition of aqueous hydrogen peroxide, H2O2, using manganese(oxide as a catalyst. The equation for the reaction is shown. 2H2O2→ 2H2O+ O2The student uses the apparatus shown in . loosely fitting cotton wool aqueous hydrogen peroxide balance manganese(oxide catalyst container 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. Table 5.1 shows the mass recorded at regular time intervals. Table 5.1 time / s mass / g 50.64 49.80 49.38 49.17 49.07 49.02 48.99 48.97 48.97 48.97 Suggest why the mass decreases as time increases. After a certain time the reaction stops. Explain why the reaction stops. Suggest why it is not possible to use the results in Table 5.1 to determine the exact time when the reaction stops. shows a graph of the mass against time. mass / g time / s The experiment is repeated at a higher temperature. All other conditions remain the same. Explain, in terms of collision theory, why the rate of reaction is higher at a higher temperature. On , sketch the line expected when the experiment is repeated at a higher temperature. Manganese(oxide is the catalyst in this reaction. Explain the meaning of (in manganese(oxide. State how the mass of the catalyst has changed, if at all, at the end of the experiment. 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+ O22NO2∆H = –113 kJ / mol NOand O2are passed into a beaker as shown in . beaker O2NOExplain why the method shown in will not allow the reaction to reach equilibrium. 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. 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
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