1. 1 GCSE Questions and Answers Calculations Remember that you can search using “edit”! 6 consecutive GCSE Chemistry papers: 2002-7

2. 2 I IIIIIIVVVIVIIVIII 1H11H1  Atomic Number at the top  Relative Atomic Mass at the bottom 2 He 4 3 Li 7 4 Be 9 5 B 11 6 C 12 7 N 14 8 O 16 9 F 19 10 Ne 20 11 Na 23 12 Mg 24 13 Al 27 14 Si 28 15 P 21 16 S 32 17 Cl 35.5 18 Ar 40 19 K 39 20 Ca 40 21 Sc 45 22 Ti 48 23 V 51 24 Cr 52 25 Mn 55 26 Fe 56 27 Co 59 28 Ni 59 29 Cu 64 30 Zn 65 31 Ga 70 32 Ge 73 33 As 75 34 Se 79 35 Br 80 36 Kr 84 2 AmmoniumNH 4 + HydroxideOH - Nitrate NO 3 - HydrogencarbonateHCO 3 - Hydrogen sulphateHSO 4 - CarbonateCO 3 2- SulphateSO 4 2- Relative Atomic Masses you’ll need.

3. Covalent elements and Compounds 3 Covalent Elements: Hydrogen H 2 Chlorine Cl 2 Bromine Br 2 Iodine I 2 Oxygen O 2 Nitrogen N 2 Helium He Neon Ne Argon Ar Covalent Compounds Water H 2 O Carbon dioxide CO 2 Carbon monoxide CO Sulphur dioxide SO 2 Sulphur trioxide SO 3 Ammonia NH 3 Hydrogen peroxide H 2 O 2 Nitrogen monoxide NO Nitrogen dioxide NO 2 SulphurS 8 PhosphorusP 4

4. To obtain full marks in this question, you must show your working out 5When washing soda crystals, Na 2 CO 3.10H 2 O, are left exposed to the atmosphere they lose water of crystallisation. The longer they are left, the more water is lost. The amount of water of crystallisation remaining can be found in two ways: either by heating to remove all the remaining water or by titration. 2002, Paper 1

5. 5a)2.675g of a sample of crystals were heated to constant mass. The mass of the residue was 1.325g. (i)Why was the sample heated to constant mass? ______________________________ [1] (ii)Calculate the number of moles of anhydrous sodium carbonate in the residue. ______________________________ [2]

6. Consequential marking applies throughout 5a)2.675g of a sample of crystals were heated to constant mass. The mass of the residue was 1.325g. (i)Why was the sample heated to constant mass? To ensure that all water [1] (of crystallisation) was lost. (ii)Calculate the number of moles of anhydrous sodium carbonate in the residue. Na 2 CO 3 = 1.325 = 0.0125 [1] 106 [1]

7. (iii)Calculate the mass of water lost and from this calculate the number of moles of water lost. _________________________________ ______________________________ [3] (iv)From your answers to part (a)(ii) and (iii) above, calculate the value of x in the formula Na 2 CO 3.xH 2 O. _________________________________ ______________________________ [2]

8. (iii)Calculate the mass of water lost and from this calculate the number of moles of water lost. Mass of water lost = (2.765-1.325) g = 1.35g [1] Number of moles lost = 1.35 = 0.075 [1] 18 [1] (iv)From your answers to part (a)(ii) and (iii) above, calculate the value of x in the formula Na 2 CO 3.xH 2 O. Ratio of moles Na 2 CO 3 : H 2 O 0.0125 : 0.075 [1] 1 : 6 [1]

9. b)1.775g of a different sample of washing soda was dissolved in distilled water and made up of a total volume of 250cm 3. 25.0cm 3 of this solution were titrated with 0.08 mol/dm 3 (moles per litre) nitric acid. 31.25cm 3 of acid were required. The equation for the reaction is: Na 2 CO 3 + 2HNO 3 → 2NaNO 3 + H 2 O + CO 2 (i)Calculate the number of moles of nitric acid used in the titration. ______________________________________ ___________________________________ [2]

10. b)1.775g of a different sample of washing soda was dissolved in distilled water and made up of a total volume of 250cm 3. 25.0cm 3 of this solution were titrated with 0.08 mol/dm 3 (moles per litre) nitric acid. 31.25cm 3 of acid were required. The equation for the reaction is: Na 2 CO 3 + 2HNO 3 → 2NaNO 3 + H 2 O + CO 2 (i)Calculate the number of moles of nitric acid used in the titration. Number of moles HNO 3 = 31.25 x 0.08 1000[1] = 2.5 x 10 -3

11. (ii)Calculate the number of moles of sodium carbonate present in the 25.0cm 3 sample. _______________________________________ ____________________________________[2] (iii)Calculate the number of moles of sodium carbonate present in 250cm 3 of solution. _______________________________________ ____________________________________[2]

12. (ii)Calculate the number of moles of sodium carbonate present in the 25.0cm 3 sample. Na 2 CO 3 + 2HNO 3 → 2NaNO 3 + H 2 O + CO 2 mole ratio 1 : 2 [1] Number of moles Na 2 CO 3 = 2.5 x 10 -3 2 = 1.25 x 10 -3 (iii)Calculate the number of moles of sodium carbonate present in 250cm 3 of solution. Number of moles Na 2 CO 3 in 250cm 3 = 1.25 x 10 -3 x 10 [1] = 1.25 x 10 -2 [1]

13. (iv)Using your answer to part (b)(iii) and the mass of sodium carbonate crystals weighed out find the value of x in the formula Na 2 CO 3.xH 2 O. _________________________________ ______________________________[2]

14. (iv)Using your answer to part (b)(iii) and the mass of sodium carbonate crystals weighed out find the value of x in the formula Na 2 CO 3.xH 2 O. Number of moles = moles rfm rfm = mass = 1.775 no. of moles1.25 x 10 -2 = 142 [1] rfm Na 2 CO 3 = 106 rfm x H 2 O = 142 – 106 =36 [1] x = 36 = 2 [1] 18

15. c)Sodium hydrogencarbonate decomposes when it is heated into sodium carbonate according to the equation: 2NaHCO 3 → NaCO 3 + H 2 O + CO 2 1.68g of sodium hydrogencarbonate were placed in a test tube and heated in a Bunsen flame for some time. (i)Calculate the number of moles of sodium hydrogencarbonate used. _______________________________________ ____________________________________ [2]

16. c)Sodium hydrogencarbonate decomposes when it is heated into sodium carbonate according to the equation: 2NaHCO 3 → NaCO 3 + H 2 O + CO 2 1.68g of sodium hydrogencarbonate were placed in a test tube and heated in a Bunsen flame for some time. (i)Calculate the number of moles of sodium hydrogencarbonate used. Number of moles NaHCO 3 = 1.68 84 [1] = 0.02 [1]

17. (ii)Calculate the number of moles of sodium carbonate formed. ___________________________________ ________________________________ [2] (iii)Calculate the mass of sodium carbonate expected to be formed. ___________________________________ ________________________________ [2]

18. (ii)Calculate the number of moles of sodium carbonate formed. Mole ratio NaHCO 3 : Na 2 CO 3 2 : 1[1] number of moles Na 2 CO 3 formed = 0.02 = 0.01 [1] 2 (iii)Calculate the mass of sodium carbonate expected to be formed. Mass Na 2 CO 3 expected = 0.01 x 106 [1] = 1.06g [1]

19. (iv)Calculate the volume of carbon dioxide produced in this experiment. (1 mole of gas occupies 24dm 3 at room temperature and pressure) _____________________________________ __________________________________ [2]

20. (iv)Calculate the volume of carbon dioxide produced in this experiment. (1 mole of gas occupies 24dm 3 at room temperature and pressure) Mole ratio NaHCO 3 : CO 2 2 :1[1] Number of moles CO 2 expected = 0.02 = 0.01 [1] Volume of CO 2 expected = 24 x 0.01 = 0.24dm 3 [1] or 240cm 3

21. 4To obtain full marks in this question, all steps in the calculation must be shown. a)Copper carbonate, CuCO 3 decomposes on heating according to the following equation: CuCO 3 → CuO + CO 2 (Relative atomic masses: C = 12; O = 16; Cu = 64) The following results were obtained in an experiment in which a sample of copper carbonate was to be heated. Mass of empty crucible = 24.21g Mass of crucible and sample of copper carbonate = 27.31g 2003, Paper 1

22. (i)What colour change would you observe on heating the sample of copper carbonate? _______________________________ [2] (ii)Calculate the mass of copper carbonate in the crucible. _______________________________ [1]

23. (i)What colour change would you observe on heating the sample of copper carbonate? Green [1] to black [1] [2] (ii)Calculate the mass of copper carbonate in the crucible. 27.31 – 24.21 = 3.1g [1]

24. (iii)Calculate the mass of solid copper oxide, CuO, which you would expect to obtain from the complete decomposition of this sample of copper carbonate.

25. (iii)Calculate the mass of solid copper oxide, CuO, which you would expect to obtain from the complete decomposition of this sample of copper carbonate. = 0.025 [1] moles of CuCO 3 CuCO 3 : CuO 1:1 [1] 0.025 [1] moles of CuO 0.025 x 80 [1] = 2.0 [1]g

26. (iv)How would you ensure all the copper carbonate had decomposed? _______________________________ [1]

27. (iv)How would you ensure all the copper carbonate had decomposed? Heat to constant mass [1]

28. b)The experiment was carried out, heating the sample for 3 minutes. Not all of the copper carbonate had decomposed in this time. The results obtained are shown below: Mass of empty crucible= 24.21g Mass of crucible and sample of copper carbonate= 27.31g Mass of crucible and solid after heating for 3 minutes= 26.43g

29. (i)What mass of solid remained after heating for 3 minutes? _______________________________ [2] (ii)Calculate the mass of carbon dioxide gas lost in this experiment. _______________________________ [2]

30. (i)What mass of solid remained after heating for 3 minutes? 26.43 – 24.21 [1] = 2.22 [1]g (ii)Calculate the mass of carbon dioxide gas lost in this experiment. 27.31 – 26.43 [1] = 0.88 [1] g

31. (iii)Calculate the number of moles of carbon dioxide gas lost in this experiment. _______________________________ [2] (iv)Calculate the mass of copper carbonate which must have been decomposed in this experiment. __________________________________ _______________________________ [ ]

32. (iii)Calculate the number of moles of carbon dioxide gas lost in this experiment. = 0.02 [1] moles of CO 2 (iv)Calculate the mass of copper carbonate which must have been decomposed in this experiment. CO 2 : CuCO 3 = 1 : 1 0.02 [1] moles of CuCO 3 decomposed 0.02 x 124 = 2.48 [1]g of CuCO 3 decomposed

33. v)Using your answer to parts (a)(ii) and (b)(iv), calculate the percentage of copper carbonate in the original sample which must have been decomposed in this experiment. [2]

34. v)Using your answer to parts (a)(ii) and (b)(iv), calculate the percentage of copper carbonate in the original sample which must have been decomposed in this experiment. x 100 [1] = 80% [2]

35. c)Copper carbonate reacts with sulphuric acid and crystals of hydrated copper sulphate, CuSO 4.5H 2 O, can be obtained. Calculate the percentage of water of crystallisation in hydrated copper sulphate, CuSO 4.5H 2 O. (Relative atomic masses: H=1; O=16; Cu=64) [3]

36. c)Copper carbonate reacts with sulphuric acid and crystals of hydrated copper sulphate, CuSO 4.5H 2 O, can be obtained. Calculate the percentage of water of crystallisation in hydrated copper sulphate, CuSO 4.5H 2 O. (Relative atomic masses: H=1; O=16; Cu=64) [3] CuSO 4.5H 2 O RFM = 250 [1] Mass of water = 18 x 5 = 90 [1] Percentage water = x 100 =36% [1]

37. d)Carbon dioxide reacts with excess sodium hydroxide solution according to the following equation: CO 2 + 2NaOH → Na 2 CO 3 + H 2 O (1 mole of any gas at room temperature and pressure occupies 24 000cm 3 ) 120cm 3 of carbon dioxide gas is passed into 150cm 3 of 0.1mol/dm 3 moles per litre.mol/l sodium hydroxide solution.

38. (i)Calculate the number of moles in 120cm 3 of carbon dioxide gas. _____________________________________ __________________________________ [2]

39. (i)Calculate the number of moles in 120cm 3 of carbon dioxide gas. [1] = 0.005 [1] moles CO 2

40. (ii)Calculate the number of moles of sodium hydroxide needed to react with this amount of carbon dioxide. _____________________________________ __________________________________ [2]

41. (ii)Calculate the number of moles of sodium hydroxide needed to react with this amount of carbon dioxide. CO 2 : NaOH [1] 0.005 moles x 2 = 0.01 moles NaOH needed [1]

42. (iii)Calculate the number of moles in 150cm 3 of 0.1mol/dm 3 sodium hydroxide solution. _____________________________________ __________________________________ [2]

43. (iii)Calculate the number of moles in 150cm 3 of 0.1mol/dm 3 sodium hydroxide solution. [1] = 0.015 [1] moles NaOH [2]

44. (iv)How many moles of sodium hydroxide are left at the end of the reaction? _____________________________________ __________________________________ [2]

45. (iv)How many moles of sodium hydroxide are left at the end of the reaction? Moles NaOH needed to react with CO 2 = 2 x 0.005 = 0.01 Moles NaOH remaining = 0.015 – 0.01 [1] = 0.005 [1]

46. e)Sodium hydroxide can be neutralised by hydrochloric acid according to the equation. NaOH + HCl → NaCl + H 2 O Using your answer to part (d)(iv) calculate the number of 0.5 mol/dm 3 hydrochloric acid which would be required to neutralise the sodium hydroxide left at the end of the reaction. _____________________________________ __________________________________ [4]

47. e)Sodium hydroxide can be neutralised by hydrochloric acid according to the equation. NaOH + HCl → NaCl + H 2 O Using your answer to part (d)(iv) calculate the number of 0.5 mol/dm 3 hydrochloric acid which would be required to neutralise the sodium hydroxide left at the end of the reaction. NaOH : HCl = 1.1 [1] moles HCl required = 0.005 [1] volume HCl required = [1] = 10 [1] cm 3

48. 5To obtain full marks in this question, all steps in the calculation must be shown. a)In 1908 a German chemist called Fritz Haber succeeded in combining nitrogen with hydrogen to form ammonia. N 2 + 3H 2 → 2NH 3 2004, Paper 1

49. Calculate the volume of nitrogen gas, measured at room temperature and pressure, needed to produce 10dm 3 of ammonia.

50. Calculate the volume of nitrogen gas, measured at room temperature and pressure, needed to produce 10dm 3 of ammonia. 1 volume of nitrogen needs 2 volumes of ammonia [1] hence 10dm 3 needs 5dm3/4.99 [1] dm 3

51. b)A concentrated solution of ammonia can be used as a fertiliser. To determine the concentration of the ammonia it was first diluted by measuring 10.0cm 3 and making the volume up to 1dm 3 (1000cm 3 ). A 25.0cm 3 sample of this dilute ammonia solution was then titrated against 0.05 mol/dm 3 (moles per litre) sulphuric acid. The 25.0 cm 3 of diluted ammonia required 12.5cm 3 of the acid for neutralisation. The equation for the titration is 2NH 3 + H 2 SO 4 → (NH 4 ) 2 SO 4

52. (i)Calculate the number of moles of sulphuric acid used in the titration. (ii)Calculate the number of moles of ammonia in the 25.0cm3 sample which reacted with the acid.

53. (i)Calculate the number of moles of sulphuric acid used in the titration. Number of moles of sulphuric acid = = 0.000625[2] (ii)Calculate the number of moles of ammonia in the 25.0cm3 sample which reacted with the acid. From equation (1:2) 1 mole sulphuric acid reacts with 2 moles of ammonia [1] 0.000625 moles acid reacts with 0.00125 moles ammonia [1]

54. (iii)Calculate the concentration of the dilute ammonia solution in mol/dm 3 (moles per litre). (iv)Calculate the concentration of the original concentrated ammonia solution in mol/dm 3 (moles per litre).

55. (iii)Calculate the concentration of the dilute ammonia solution in mol/dm 3 (moles per litre). Concentration of ammonia = = 0.05 [1] mol/dm 3 (iv)Calculate the concentration of the original concentrated ammonia solution in mol/dm 3 (moles per litre). Diluted 100 times [1] original conc = 0.05 x 100 = 5 [1] mol/dm 3

56. (v)Calculate the concentration of the original concentrated ammonia solution in g/dm 3

57. (v)Calculate the concentration of the original concentrated ammonia solution in g/dm 3 RFM of NH3 = 17[1] conc g/dm 3 = 5 x 17[1] = 85[1] g/dm 3

58. c)Solid fertilisers are easier to store, hence fertilisers like solid ammonium chloride are preferred over ammonia solution. To produce ammonium chloride, ammonia is reacted with hydrochloric acid, according to the equation below. NH 3 + HCl → NH 4 Cl

59. What mass of ammonium chloride is formed when 73g of hydrochloric acid are completely neutralised by ammonia? (Relative atomic masses: H=1, N=14, Cl=35.5

60. What mass of ammonium chloride is formed when 73g of hydrochloric acid are completely neutralised by ammonia? (Relative atomic masses: H=1, N=14, Cl=35.5 RFM of HCl = 36.5 [1] moles of HCl = 73/36.5 = 2[1] (ratio 1:1 hence) moles NH 4 Cl = 2[1] RFM of NH 4 Cl = 53.5[1] 2 x 53.5 = 107g[1]

61. d)Another important fertiliser made from ammonia is urea. It contains 20.00% carbon, 6.66% hydrogen, 46.67% nitrogen and 26.67% oxygen. Calculate the formula of urea. (Relative atomic masses: H=1, C=12, N=14, O=16)

62. d)Another important fertiliser made from ammonia is urea. It contains 20.00% carbon, 6.66% hydrogen, 46.67% nitrogen and 26.67% oxygen. Calculate the formula of urea. (Relative atomic masses: H=1, C=12, N=14, O=16) In 100g of compound there are: 20/12 moles C = 1.67 [1] 6.67/1 moles H = 6.67 [1] 46.67/14 moles N = 3.33[1] 26.67/16 moles O = 1.67[1] formula is CH 4 N 2 O [1] accept any correct whole number multiple.

63. e) Car exhaust fumes contain harmful nitrogen monoxide gas. Research has shown that when a stream of ammonia gas is injected into the hot exhaust a reaction occurs which converts the harmful nitrogen monoxide, NO, to nitrogen gas according to the equation below. 6NO + 4NH 3 → 5N 2 + 6H 2 O

64. (i)How many moles of ammonia would be needed to react with 0.6 moles of nitrogen monoxide, NO? ___________________________ [1]

65. (i)How many moles of ammonia would be needed to react with 0.6 moles of nitrogen monoxide, NO? 0.4 moles [1]

66. (ii)The average car emits 0.033 moles of nitrogen monoxide per km. How many moles of ammonia would be needed to convert this to N 2 gas? ___________________________ [2]

67. (ii)The average car emits 0.033 moles of nitrogen monoxide per km. How many moles of ammonia would be needed to convert this to N 2 gas? 6 moles NO: 4 moles NH 3 /0.033moles NO: 0.033/6x4 [1] = 0.022 [1] moles per km

68. (iii)Using your answer to (e)(ii) calculate the mass of ammonia needed to convert 0.033 moles of NO to N 2 gas. ___________________________ [2]

69. (iii)Using your answer to (e)(ii) calculate the mass of ammonia needed to convert 0.033 moles of NO to N 2 gas. 0.022 x 17[1] = 0.374g [1]

70. 2Lead is extracted from the ore galena, PbS. a)The ore is roasted in air to produce lead(II) oxide, PbO. 2PbS(s) + 3O 2 (g) → 2PbO(s) + 2SO 2 (g) (Relative Atomic Masses: Pb=207, S=32, O=16) (i)Calculate the mass of lead(II) oxide, PbO, produced from 2390g of galena, PbS. (Show all steps in your calculations. [5] 2005, Paper 2

71. (i)Calculate the mass of lead(II) oxide, PbO, produced from 2390g of galena, PbS. (Show all steps in your calculations. [5] RFM PbS = 207+32 = 239 [1] Moles PbS = = 10 [1] Moles PbS = 10 [1] RFM PbO = 270+16 = 223 [1] Mass PbO = 10x223 = 2230g or 2.23kg [1]

72. The lead(II) oxide is reduced to lead by heating it with carbon in a blast furnace. PbO(s) + C(s) → Pb(l) + CO(g) The molten lead is tapped off from the bottom of the furnace. (ii)Using your answer to part (a)(i), calculate the mass of lead that would eventually be produced.

73. (ii)Using your answer to part (a)(i), calculate the mass of lead that would eventually be produced. PbO: Pb = 1:1 [1] Moles Pb = 10 [1] Mass Pb = 10 x 270 = 2070g or 2.07kg [1]

74. b)Lead metal forms several oxides. The formula of lead oxide may be represented as Pb x O y. In an experiment to find the formula of a sample of lead oxide, a porcelain dish was weighed and the mass recorded. The porcelain dish was then filled with the lead oxide and reweighed. The mass was again recorded.

75. The dish was placed in a hard-glass tube and heated in a stream of hydrogen gas. The hydrogen reduced all of the lead oxide to a bead of silvery lead metal. The apparatus was allowed to cool and the dish and its contents were reweighed.

76. (i)Calculate the mass of lead metal produced. _____________________________________ [1] (ii)Calculate the mass of oxygen present in the lead oxide. _____________________________________ [1]

77. (i)Calculate the mass of lead metal produced. 27.56 – 21.35 = 6.21g [1] (ii)Calculate the mass of oxygen present in the lead oxide. 28.20 – 27.56 = 0.64g [1]

78. (iii)Using your answers to (i) and (ii), calculate the formula of the sample of lead oxide. (Relative atomic masses: Pb=207, O=16)

79. (iii)Using your answers to (i) and (ii), calculate the formula of the sample of lead oxide. (Relative atomic masses: Pb=207, O=16) Moles Pb = = 0.03 [1] Moles O == 0.04[1] Ratio 3:4 so formula is Pb 3 O 4 [1]

80. c)Titration is a technique used by chemists to find the concentration of a solution. The apparatus used in a titration is shown opposite.

81. (i)Identify the pieces of apparatus A and B. A is a ________________________________ [1] B is a ________________________________ [1] (ii)Describe in detail, stating precautions to ensure safety and accuracy, how you would transfer 25.0cm3 of an alkali into the conical flask using the piece of apparatus A. _____________________________________ _____________________________________ _____________________________________ _____________________________________ [3]

82. (i)Identify the pieces of apparatus A and B. A is a pipette [1] B is a burette [1] (ii)Describe in detail, stating precautions to ensure safety and accuracy, how you would transfer 25.0cm3 of an alkali into the conical flask using the piece of apparatus A. Rinse with deionised water [1] rinse with alkali [1] use safety pipette filler/safety goggles [1] to draw up liquid until bottom of meniscus on line [1] release [1] into conical flask touch tip of pipette to surface of alkai [1] (Max [3])

83. (iii)Describe in detail, stating precautions to ensure accuracy, the steps you would take to prepare the piece of apparatus B for use in a titration. _______________________________ _______________________________ _______________________________ _______________________________ [4]

84. (iii)Describe in detail, stating precautions to ensure accuracy, the steps you would take to prepare the piece of apparatus B for use in a titration. Rinse with deionised water [1] rinse with solution [1] fill burette with solution [1] use funnel [1] ensure jet is filled [1] ensure no air bubbles [1] (Max [4])

85. d)Limewater is calcium hydroxide solution. In a titration to find the concentration of calcium hydroxide in limewater, 25.0cm 3 of limewater required 16.4cm 3 of hydrochloric acid of concentration 0.040 mol/dm 3 for neutralisation. (Relative atomic masses: Ca=40, O=16, H=1) Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O

86. (i)Calculate the concentration of the calcium hydroxide in mol/dm 3 (mol per litre). Answer __________________ mol/dm 3 [4] (ii)Calculate the concentration of the calcium hydroxide in g/dm 3 (grams per litre). Answer __________________ mol/dm 3 [4]

87. (i)Calculate the concentration of the calcium hydroxide in mol/dm 3 (mol per litre). Answer = 0.000656 [1] Moles Ca(OH)2 = 0.000328 [1] x 1000 [1] = 0.01312 mol/dm 3 (ii)Calculate the concentration of the calcium hydroxide in g/dm 3 (grams per litre). Answer Mass = mol x RFM = 0.01312 x 74 [1] = 0.971 [1] g/dm 3

88. 7a)Oxygen forms ozone gas, O 3, in the upper atmosphere according to the equation: 3O 2 (g) → 2O 3 (g) 150m 3 of oxygen reacts completely to form ozone. 2006, Paper 2

89. (i)State Avogadro’s Law _______________________________ _______________________________ _______________________________ [3]

90. (i)State Avogadro’s Law Equal volumes of gas [1] under the same conditions of temperature and pressure [1] contain the same number of particles [1] or moles of ozone = 4.2 moles [3]

91. (ii)Using Avogadro’s Law or otherwise, calculate the volume of ozone gas produced. ____________________________m 3 [2]

92. (ii)Using Avogadro’s Law or otherwise, calculate the volume of ozone gas produced. [2]

93. b)In the laboratory ozone gas can be produced by passing an electrical discharge through dry air. 450cm3 of ozone gas is produced when the temperature is 300K and the pressure is 1 atmosphere. The ozone gas is compressed using a pressure of 8 atmospheres and the temperature is decreased to 200K. Calculate the volume of ozone gas under these new conditions. _______________________________cm 3 [4]

94. b)Calculate the volume of ozone gas under these new conditions. _______________________________ [4]

95. c)Ozone is used in very small amounts in underground railway stations to remove compounds which cause stations to be stuffy. One of the compounds which is formed is formaldehyde CH 2 O. Calculate the percentage by mass of carbon in CH 2 O. ____________________________% [3]

96. c)Calculate the percentage by mass of carbon in CH 2 O. RFM (CH 2 O) = 30 [1] % carbon = [1] = 40% [1] [3]

97. d)1.92g of sulphur dioxide SO 2, reacts completely with ozone to form 2.40g of sulphur trioxide, SO 3. (i)Calculate the number of moles of sulphur dioxide used. _______________________________ [2]

98. d)1.92g of sulphur dioxide SO 2, reacts completely with ozone to form 2.40g of sulphur trioxide, SO 3. (i)Calculate the number of moles of sulphur dioxide used. RFM (SO 2 ) = 64 [1] Moles = = 0.03 [1] [2]

99. (ii)Calculate the mass of ozone which reacts. _______________________________ [1] (iii)Calculate the number of moles of ozone which reacts. _______________________________ [2]

100. (ii)Calculate the mass of ozone which reacts. 2.4 – 1.92 = 0.48g [1] (iii)Calculate the number of moles of ozone which reacts. RFM (O 3 ) = 48[1] Moles = = 0.01 [1] [2]

101. (iv)Calculate the number of moles of sulphur trioxide formed. _______________________________ [2]

102. (iv)Calculate the number of moles of sulphur trioxide formed. RFM (SO 3 ) = 80[1] Moles = = 0.03[1] [2]

103. (v)Using your answers to (i), (iii) and (iv) or otherwise, balance the symbol equation for the reaction. Equation: SO 2 + O 3 → SO 3 [1]

104. (v)Using your answers to (i), (iii) and (iv) or otherwise, balance the symbol equation for the reaction. Equation: SO 2 + O 3 → SO 3 [1] Ratio: SO 2 : O 3 : SO 3 = 0.03 : 0.01 : 0.03 = 3 : 1 : 3 Equation: 3SO + O 3 → 3SO 2 balancing numbers = [1]

105. e)Oxygen gas is prepared by the decomposition of hydrogen peroxide solution using the catalyst manganese(IV) oxygen. 2H 2 O 2 → 2H 2 O + O 2 A solution of hydrogen peroxide is labelled 0.1mol/dm 3 (moles per litre). 25.0cm 3 of this solution is decomposed completely using manganese(IV) oxide.

106. (i)What is meant by the term catalyst? _______________________________ _______________________________ [3]

107. (i)What is meant by the term catalyst? Substance which speeds up/increases the rate of [1] a (chemical) reaction [1] without being used up/chemically unchanged at the end [1] [3]

108. (ii)Calculate the volume of oxygen gas produced in this decomposition. State the units. (1 mole of any gas occupies a volume of 24dm 3 ) __________________________________ [7]

109. (ii)Calculate the volume of oxygen gas produced in this decomposition. State the units. (1 mole of any gas occupies a volume of 24dm 3 ) Moles of H 2 O 2 = Ratio: H2O2 : O2 = 2 : 1 [1] Moles of O 2 = Volume of oxygen = 0.00125 x 24 [1] = 0.03 [1]dm 3 [1] ( or volume of oxygen = 0.00125 x 24000 [1] = 30[1]cm 3 [1]

110. 5Borax is a salt which is hydrated and is used in cleaning agents. The formula may be represented by Na 2 B 4 O 7.xH 2 O. Borax dissolves in water to give a solution which acts as a weak alkali. 2007, Paper 1

111. a)4.775g of Borax were weighed out and made up to a volume of 250cm 3 with deionised water. 25.0cm 3 portions of this solution were titrated against nitric acid of concentration 0.094 mol/dm 3 (moles per litre). The results were recorded in the table below. Initial burette reading (cm 3 ) Final burette reading (cm 3 ) Volume of nitric acid used (titre) (cm 3 ) Rough Titration0.026.9 1 st Accurate Titration0.026.7 2 nd Accurate Titration0.026.5

112. (i)Calculate the average titre. _______________________________ [2] Initial burette reading (cm 3 ) Final burette reading (cm 3 ) Volume of nitric acid used (titre) (cm 3 ) Rough Titration0.026.9 1 st Accurate Titration0.026.7 2 nd Accurate Titration0.026.5

113. (i)Calculate the average titre. 26.2 [2]. If rough used average = 26.7 award [1] [2]

114. (ii)The indicator used was methyl orange. State the colour change of the indicator in this titration. From ____________ to ____________ [2]

115. (ii)The indicator used was methyl orange. State the colour change of the indicator in this titration. From orange or yellow [1] to pink or red [1] (wrong way round) = [1] [2]

116. (iii)Calculate the number of moles of nitric acid used in this titrations. _______________________________ [2]

117. (iii)Calculate the number of moles of nitric acid used in this titrations. Moles = [1] = 0.0025 [1] (2.5 x 10 -3 )

118. The equation for the reaction is: Na 2 B 4 O 7 + 2HNO 3 + 5H 2 O → 2NaNO 3 + 4H 3 BO 3 (iv)Use the equation to deduce the number of moles of Borax which reacted with the nitric acid. __________________________________ _____________________________________ [2]

119. The equation for the reaction is: Na 2 B 4 O 7 + 2HNO 3 + 5H 2 O → 2NaNO 3 + 4H 3 BO 3 (iv)Use the equation to deduce the number of moles of Borax which reacted with the nitric acid. 1 mole borax : 2 moles nitric acid or = 0.00125 [1] (1.25 x 10 -3 ) [2]

120. (v)Calculate the concentration of the Borax in mol/dm 3 (moles per litre) _______________________________ _______________________________ [2]

121. (v)Calculate the concentration of the Borax in mol/dm 3 (moles per litre) conc = [1] = 0.05 [1] mol/dm 3 [2]

122. (vi)From the mass of Borax used, calculate the concentration of Borax in g/dm 3 _______________________________ [1]

123. (vi)From the mass of Borax used, calculate the concentration of Borax in g/dm 3 4.775g in 250cm3 → 4.775 x 4 = 19.1 [1] g/dm 3 [1]

124. (vii)Using your answers to parts (v) and (vi) find the formula mass of the Borax, Na 2 B 4 O 7.xH 2 O, and hence find the value of x. (Relative atomic masses: H = 1; B = 11; O = 16; Na = 23) _______________________________ _______________________________ _______________________________ [3]

125. (vii)Using your answers to parts (v) and (vi) find the formula mass of the Borax, Na 2 B 4 O 7.xH 2 O, and hence find the value of x. (Relative atomic masses: H = 1; B = 11; O = 16; Na = 23) = 382 [1] 382 = 202 [1] + 18x 18x = 180 x = 10 [1] [3]

126. b)When Borax crystals are left in air they lose some of their water of crystallisation. To find the value of x in a sample of hydrated Borax Na 2 B 4 O 7.xH 2 O, which had been left in air for a month, the sample was heated to constant mass. 7.28g of hydrated Borax produced 4.04g of anhydrous Borax. (Relative atomic masses: H = 1; B = 11; O = 16; Na = 23)

127. (i)What is meant by “heated to constant mass”? _______________________________ [2] (ii)Calculate the mass of water lost. _______________________________ [1]

128. (i)What is meant by “heated to constant mass”? heating and weighing [1] repeat until 2 readings the same [1] [2] (ii)Calculate the mass of water lost. 3.25g, 7.28 – 4.04 = 3.24g [1]

129. (iii)Calculate the number of moles of water lost. _______________________________ _______________________________ [2] (iv)Calculate the number of moles of anhydrous Borax. _______________________________ [1]

130. (iii)Calculate the number of moles of water lost. moles = [1] = 0.18 [1] [2] (iv)Calculate the number of moles of anhydrous Borax. Moles = = 0.02 [1]

131. (v)Using your answers to (iii) and (iv) determine the value of x in Na 2 B 4 O 7.xH 2 O. _______________________________ _______________________________ [2]

132. (v)Using your answers to (iii) and (iv) determine the value of x in Na 2 B 4 O 7.xH 2 O. borax : water 0.02 : 0.18 [1] 1:9x = 9[1] [2]

133. c)When anhydrous Borax is heated it decomposes according to the equation: Anhydrous Borax→Sodium metaborate+Boric Oxide Na 2 B 4 O 7 →2NaBO 2 +B2O3B2O3

134. Calculate the mass of sodium metaborate which is produced when 5.05g of anhydrous Borax is heated. (Relative atomic masses: B = 11; O = 16; Na = 23)

135. Calculate the mass of sodium metaborate which is produced when 5.05g of anhydrous Borax is heated. (Relative atomic masses: B = 11; O = 16; Na = 23) Moles of anhydrous borax = = 0.025 [1] Ratio1 borax:2 sodium metaborate 0.025:0.050[1] 0.05x66[1]:= 3.3 [1]g

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