Question 1a Extended Only Fluorine forms both ionic and covalent compounds. Magnesium reacts with fluorine to form the ionic compound magnesium fluoride. The electronic structures of an atom of magnesium and an atom of fluorine are shown. What is the formula of magnesium fluoride? [1 mark] Question 1b When copper is reacted with hot concentrated sulfuric acid, sulfur dioxide gas is formed. Balance the chemical equation for this reaction. Cu + .......H2SO4 → CuSO4 + SO2 + .......H2O [2 marks] Question 1c Extended Only When sulfuric acid reacts with ammonia the salt produced is ammonium sulfate. Write the chemical equation for this reaction. [2 marks] Question 1d Extended Only Barium sulfate is an insoluble salt. Barium sulfate can be made from aqueous ammonium sulfate using a precipitation reaction. Write an ionic equation for this precipitation reaction. Include state symbols. [2 marks] Question 1e Extended Only Oxygen is produced by the decomposition of hydrogen peroxide. Manganese(IV) oxide is the catalyst for this reaction. Oxygen can also be produced by the decomposition of potassium chlorate(V), KClO3. The only products of this decomposition are potassium chloride and oxygen. Write a chemical equation for this decomposition. [2 marks] Question 2a Extended Only This question is about titanium. Titanium is a transition metal that is extracted from titanium dioxide in a two-stage industrial process. In the first stage, titanium dioxide reacts with carbon and chlorine to form titanium tetrachloride and carbon monoxide. Write the balanced symbol equation for this reaction. [3 marks] Question 2b Explain how your chemical symbol equation in part (a) demonstrates the law of conservation of mass. [1 mark] Question 2c Identify two hazards associated with Stage 1. [2 marks] Question 2d Separate: Chemistry and Extended Only Calculate, to three significant figures, the percentage by mass of chlorine in titanium tetrachloride. (Ar: Ti = 48 Cl = 35.5) [2 marks] Question 3a Extended Only Magnesium displaces copper from copper sulfate solution. Write the balanced symbol equation for the reaction. You should include state symbols. [2 marks] Question 3b Extended Only State two changes that would be observed during the displacement reaction. [2 marks] Question 3c Extended Only Write the ionic equation for the displacement of copper from copper sulfate by magnesium. You should include state symbols. [2 marks] Question 3d Extended Only Displacement reactions are examples of redox reactions. Explain why the displacement reaction between magnesium and copper sulfate is both reduction and oxidation. [2 marks] Question 4a Extended Only This question is about lead nitrate. Lead nitrate is an ionic compound. It consists of Pb2+ and NO3- ions. What is the chemical formula of lead nitrate? [1 mark] Question 4b Separate: Chemistry and Extended Only Calculate the percentage by mass of oxygen in lead nitrate, Pb(NO3)2. [2 marks] Question 4c Extended Only A displacement reaction occurs between solutions of lead nitrate and potassium iodide, KI, to form solid lead(II) iodide and aqueous potassium nitrate. Write the balanced symbol equation, including state symbols, for this reaction. [3 marks] Question 4d Separate: Chemistry Only Table 1 summarises the solubility of a selection of ionic compounds in water. Table 1 Soluble All nitrates Most common chlorides Most common sulfates Sodium carbonate, potassium carbonate, ammonium carbonate Insoluble Silver chloride, lead chloride Lead sulfate, barium sulfate, calcium sulfate Most common carbonates The displacement reaction between lead nitrate and potassium iodide forms a yellow precipitate. Justify which chemical is responsible for the yellow colour. [2 marks] Question 5a The relative atomic mass, Ar, for each element is shown in the Periodic Table. Define the term relative atomic mass. [2 marks] Question 5b Separate: Chemistry and Extended Only The element gallium has a relative atomic mass of 69.735 and only contains two isotopes. A sample of gallium contained the isotope 69Ga, with a relative abundance of 63.25 %. Calculate the mass number of the other isotope. You must show all your working. [2 marks] Question 5c Extended Only Define the term empirical formula. [1 mark] Question 5d Separate: Chemistry and Extended Only An unknown compound contains carbon, hydrogen and oxygen. It consists of 18% carbon,and 73 % oxygen. Calculate the empirical formula of the unknown compound [3 marks] Question 1a Separate: Chemistry and Extended Only The following method is used to make crystals of hydrated nickel sulphate. An excess of nickel carbonate, 12.0 g, was added to 40 cm3 of sulphuric acid, 2.0 mol/dm3. The unreacted nickel carbonate was filtered off and the filtrate evaporated to obtain the crystals. NiCO3 + H2SO4 → NiSO4 + CO2 + H2O NiSO4 + 7H2O → NiSO4.7H2O Mass of one mole of NiSO4.7H2O = 281 g Mass of one mole of NiCO3 = 119 g i) Calculate the mass of unreacted nickel carbonate. Number of moles of H2SO4 in 40 cm3 of 2.0 mol/dm3 acid = 0.08 Number of moles of NiCO3 reacted = ...................................... Mass of nickel carbonate reacted = ...................................... g Mass of unreacted nickel carbonate = ...................................... g ii) The experiment produced 10.4 g of hydrated nickel sulphate. Calculate the percentage yield. [3] The maximum number of moles of NiSO4.7H2Othat could be formed = ...................................... The maximum mass of NiSO4.7H2O that could be formed = ...................................... g The percentage yield = ...................................... % [3] [6 marks] Question 1b Extended Only In the above method, a soluble salt was prepared by neutralising an acid with an insoluble base. Other salts have to be made by different methods. i) Give a brief description of how the soluble salt, rubidium sulphate could be made from the soluble base, rubidium hydroxide. [3] ii) Suggest a method of making the insoluble salt, calcium fluoride. [3] [6 marks] Question 2a Soluble salts can be made using a base and an acid. Complete this method of preparing dry crystals of the soluble salt cobalt(II) chloride-6-water from the insoluble base cobalt(II) carbonate. step 1 Add an excess of cobalt(II) carbonate to hot dilute hydrochloric acid. step 2 step 3 step 4 [4 marks] Question 2b i) 5.95 g of cobalt(II) carbonate were added to 40 cm3 of hydrochloric acid, concentration 2.0 mol / dm3. Calculate the maximum yield of cobalt(II) chloride-6-water and show that the cobalt(II) carbonate was in excess. CoCO3 + 2HCl → CoCl2 + CO2 + H2O CoCl2 + 6H2O → CoCl2.6H2O maximum yield: number of moles of HCl used = ..................................................................... number of moles of CoCl2 formed = ............................................................... number of moles of CoCl2.6H2O formed = ....................................................... mass of one mole of CoCl2.6H2O = 238 g maximum yield of CoCl2.6H2O = ...................................................................g to show that cobalt(II) carbonate is in excess: number of moles of HCl used = ..................................... (use your value from above) mass of one mole of CoCO3 = 119 g number of moles of CoCO3 in 5.95 g of cobalt(II) carbonate = .............................. ii) Explain how these calculations show that cobalt(II) carbonate is in excess. [5] [1] [6 marks] Question 3a Crystals of sodium sulphate-10-water, Na2SO4.10H2O, are prepared by titration. 25.0 cm3 of aqueous sodium hydroxide is pipetted into a conical flask. A few drops of an indicator are added. Using a burette, dilute sulphuric acid is slowly added until the indicator just changes colour. The volume of acid needed to neutralise the alkali is noted. Suggest how you would continue the experiment to obtain pure, dry crystals of sodium sulphate-10-water. [4 marks] Question 3b Separate: Chemistry and Extended Only Using 25.0 cm3 of aqueous sodium hydroxide, 2.24 mol / dm3, 3.86 g of crystals were obtained. Calculate the percentage yield. 2NaOH + H2SO4 → Na2SO4 + 2H2O Na2SO4 + 10H2O → Na2SO4.10H2O Number of moles of NaOH used = Maximum number of moles of Na2SO4.10H2O that could be formed = Mass of one mole of Na2SO4.10H2O = 322g Maximum yield of sodium sulphate-10-water, in g = Percentage yield = [4 marks] Question 4a Calcium and other minerals are essential for healthy teeth and bones. Tablets can be taken to provide these minerals. Healthy Bones Each tablet contains calcium magnesium zinc copper boron Boron is a non-metal with a macromolecular structure. i) Predict two physical properties of boron. ii) Name another element and a compound that have macromolecular structures. iii) Sketch the structure of one of the above macromolecular substances. [2] [2] [2] [6 marks] Question 4b Describe the reactions, if any, of zinc and copper(II) ions with an excess of aqueous sodium hydroxide. i) Zinc ions Addition of aqueous sodium hydroxide: Excess sodium hydroxide: ii) Copper(II) ions Addition of aqueous sodium hydroxide: Excess sodium hydroxide: [2] [2] [4 marks] Question 4c Extended Only Each tablet contains the same number of moles of CaCO3 and MgCO3. One tablet reacted with excess hydrochloric acid to produce 0.24 dm3 of carbon dioxide at r.t.p. CaCO3 + 2HCl → CaCl2 + CO2 + H2O MgCO3 + 2HCl → MgCl2 + CO2 + H2O i) Calculate how many moles of CaCO3 there are in one tablet. Number of moles CO2 = Number of moles of CaCO3 and MgCO3 = Number of moles of CaCO3 = [3] ii) Calculate the volume of hydrochloric acid, 1.0 mol / dm3, needed to react with one tablet. Number of moles of CaCO3 and MgCO3 in one tablet = Use your answer to (c)(i). Number of moles of HCl needed to react with one tablet = Volume of hydrochloric acid, 1.0 mol / dm3, needed to react with one tablet = [2] [5 marks] Question 5a The soluble salt hydrated lithium sulfate is made by titration from the soluble base lithium hydroxide. The sulfuric acid is added slowly from the burette until the indicator just changes colour. The volume of sulfuric acid needed to just neutralise the lithium hydroxide is noted. Describe how you would continue the experiment to obtain pure dry crystals of hydrated lithium sulfate. [5 marks] Question 5b Separate: Chemistry and Extended Only Using 25.0 cm3 of aqueous lithium hydroxide, concentration 2.48 mol / dm3, 2.20 g of hydrated lithium sulfate was obtained. Calculate the percentage yield, giving your answer to one decimal place. 2LiOH + H2SO4 → Li2SO4 + 2H2O Li2SO4 + H2O → Li2SO4.H2O Number of moles of LiOH used = ....................... Number of moles of Li2SO4.H2O which could be formed = ....................... Mass of one mole of Li2SO4.H2O = 128g Maximum yield of Li2SO4.H2O = ....................... g Percentage yield = .......................% [4 marks] Question 5c Separate: Chemistry and Extended Only An experiment was carried out to show that the formula of the hydrated salt is Li2SO4.H2O. A sample of the hydrated salt was weighed and its mass recorded. It was then heated and the anhydrous salt was weighed. This procedure was repeated until two consecutive masses were the same. This procedure is called ‘heating to constant mass’. i) What is the reason for heating to constant mass? [1] ii) The mass of the hydrated salt is m1 and the mass of the anhydrous salt is m2. Explain how you could show that the hydrated salt has one mole of water of crystallisation per mole of the anhydrous salt. [3] [4 marks] Question 6a Separate: Chemistry and Extended Only The elements in Period 3 and some of their common oxidation states are shown below. Element Oxidation state Na +1 Mg +2 Al +3 Si +4 i) Why do the oxidation states increase from sodium to silicon? ii) After Group(IV) the oxidation states are negative and decrease across the period. Explain why. P -3 S -2 I -1 Ar 0 [1] [2] [3 marks] Question 6b Aluminium sulfide contains two elements. Predict its formula. [1 mark] Question 6c Separate: Chemistry and Extended Only Choose a different element from Period 3 that matches each description. i) It has a similar structure to diamond. ii) It reacts violently with cold water to form a solution pH = 14. iii) It has a gaseous oxide of the type XO2, which is acidic. [1] [1] [1] [3 marks] Question 6d Extended Only Draw a diagram that shows the arrangement of the outer electrons in the ionic compound sodium phosphide. Use o to represent an electron from sodium. Use x to represent an electron from phosphorus. [3 marks] Question 6e Extended Only Sodium reacts with sulphur to form sodium sulfide. 2Na + S → Na2S An 11.5 g sample of sodium is reacted with 10 g of sulfur. All of the sodium reacted but there was an excess of sulfur. Calculate the mass of sulfur left unreacted. i) Number of moles of sodium atoms reacted = [2 moles of Na react with 1 mole of S] ii) Number of moles of sulfur atoms that reacted = iii) Mass of sulfur reacted, in grams = iv) Mass of sulfur left unreacted, in grams = [1] [1] [1] [1] [4 marks] Question 7a Separate: Chemistry and Extended Only Sulfuric acid is made by the Contact process. 2SO2 + O2 ⇌ 2SO3 This is carried out in the presence of a catalyst at 450°C and 2 atmospheres pressure. i) Name the catalyst used. ii) If the temperature is decreased to 300°C, the yield of sulfur trioxide increases. Explain why this lower temperature is not used. [1] [1] [2 marks] Question 7b Separate: Chemistry and Extended Only Sulfuric acid was first made in the Middle East by heating the mineral, green vitriol, FeSO4.7H2O. The gases formed were cooled. FeSO4.7H2O (s) → FeSO4 (s) + 7H2O (g) green crystals yellow powder 2FeSO4 (s) → Fe2O3 (s) + SO2 (g) + SO3 (g) On cooling SO3 + H2O → H2SO4 sulfuric acid SO2 + H2O → H2SO3 sulfurous acid i) How could you show that the first reaction is reversible? [2] ii) Sulfurous acid is a reductant (reducing agent). What would you see when acidified potassium manganate(VII) is added to a solution containing this acid? [2] iii) Suggest an explanation why sulfurous acid in contact with air changes into sulfuric acid. [1] [5 marks] Question 7c Extended Only 9.12 g of anhydrous iron(II) sulfate was heated. Calculate the mass of iron(III) oxide formed and the volume of sulfur trioxide, at r.t.p., formed. 2FeSO4 (s) → Fe2O3 (s) + SO2 (g) + SO3 (g) mass of one mole of FeSO4 = 152 g number of moles of FeSO4 used = _____________ number of moles of Fe2O3 formed = _____________ mass of one mole of Fe2O3 = _____________ g mass of iron(III) oxide formed = _____________ g number of moles of SO3 formed = _____________ volume of sulfur trioxide formed = _____________ dm3 [6 marks] Question 8a Separate: Chemistry and Extended Only Sulfur dioxide, SO2, and sulfur trioxide, SO3, are the two oxides of sulfur. Sulfur trioxide can be made from sulfur dioxide. i) Why is this reaction important industrially? [1] ii) Complete the word equation. sulfur dioxide + .............................................. → sulfur trioxide iii) What are the conditions for this reaction? [1] [2] [4 marks] Question 8b Separate: Chemistry and Extended Only Sulphur dioxide is easily oxidised in the presence of water. SO2 + 2H2O – 2e– → SO42– + 4H+ i) What colour change would be observed when an excess of aqueous sulphur dioxide is added to an acidic solution of potassium manganate(VII)? [2] ii) To aqueous sulphur dioxide, acidified barium chloride solution is added. The mixture remains clear. When bromine is added, a thick white precipitate forms. What is the white precipitate? Explain why it forms. [3] [5 marks] Question 8c Extended Only Sulphur dioxide reacts with chlorine in an addition reaction to form sulphuryl chloride. SO2 + Cl2 → SO2Cl2 8.0 g of sulphur dioxide was mixed with 14.2 g of chlorine. The mass of one mole of SO2Cl2 is 135 g. Calculate the mass of sulphuryl chloride formed by this mixture. Calculate the number of moles of SO2 in the mixture = .................. Calculate the number of moles of Cl2 in the mixture = .................. Which reagent was not in excess? ............................... How many moles of SO2Cl2 were formed = ................... Calculate the mass of sulphuryl chloride formed = ............. g [5 marks] Question 9a Until recently, arsenic poisoning, either deliberate or accidental, has been a frequent cause of death. The symptoms of arsenic poisoning are identical with those of a common illness, cholera. A reliable test was needed to prove the presence of arsenic in a body. In 1840, Marsh devised a reliable test for arsenic Hydrogen is formed in this reaction. Any arsenic compound reacts with this hydrogen to form arsine which is arsenic hydride, AsH3. i) The mixture of hydrogen and arsine is burnt at the jet and arsenic forms as a black stain on the glass. Write an equation for the reaction which forms hydrogen. ii) Draw a diagram which shows the arrangement of the outer electrons in one molecule of the covalent compound arsine. [2] The electron distribution of arsenic is 2 + 8 + 18 + 5. Use x to represent an electron from an arsenic atom. Use o to represent an electron from a hydrogen atom. [2] [4 marks] Question 9b Separate: Chemistry and Extended Only Another hydride of arsenic has the composition below. arsenic 97.4% hydrogen 2.6% i) Calculate the empirical formula of this hydride from the above data. Show your working. ii) The mass of one mole of this hydride is 154 g. What is its molecular formula? iii) Deduce the structural formula of this hydride. [2] [1] [1] [4 marks] Question 9c Separate: Chemistry and Extended Only In the 19th Century, a bright green pigment, copper(II) arsenate(V) was used to kill rats and insects. In damp conditions, micro-organisms can act on this compound to produce the very poisonous gas, arsine. i) Suggest a reason why it is necessary to include the oxidation states in the name of the compound. ii) The formula for the arsenate(V) ion is AsO43–. [1] Complete the ionic equation for the formation of copper(II) arsenate(V). ......Cu2+ + ......AsO43– → .................................. [2] [3 marks]
