1. 1 NaCl Find LE f Module 3 Practical: Structured Questions 2002-7 Welcome and thanks for visiting. F Scullion. JustChemy.Com

2. 2 2004/P1

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11. 11 2004/P2

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35. 35 2005/P2

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37. 37 Full explanation provided on the next slide

38. 38 Oxidising ability of Concentrated Sulphuric Acid. Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides. [CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE] 2a) KBr (s) + H 2 SO 4(l)  KHSO 4(aq) + HBr (g) However, the concentrated sulphuric acid will oxidise some of the HBr as follows: -1 0 2b) 2HBr + H 2 SO 4  Br 2 + SO 2 + 2H 2 O Therefore, one will observe a Reddish Vapour due to some bromine being present. The SO 2 is and acidic gas. It is also a reducing agent and will, for example, decolourise purple potassium permanganate solution An increase in O.N.

39. 39 SUMMARY for all of the Halides. Oxidising ability of Concentrated Sulphuric Acid. Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides. [CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE] 1) NaCl(s) + H 2 SO 4 (l) = NaHSO 4 (aq) + HCl(g) The concentrated sulphuric acid will not oxidise HCl 2a) KBr(s) + H 2 SO 4 (l) = KHSO 4 (aq) + HBr(g) The concentrated sulphuric acid will oxidise some of the HBr as follows: 2b) 2HBr + H 2 SO 4 + Br 2 + SO 2 + 2H 2 O Therefore, one will observe a Reddish Vapour due to some bromine being present. 3a) KI(s) + H 2 SO 4 (l) à KHSO 4 + HI(g) The concentrated sulphuric acid will oxidise some of the HI as follows: 3b) H 2 SO 4 (l) + 2HI(g) + SO 2 + I 2 + 2H 2 0 3c) H 2 SO 4 (l) + 6HI(g) + S + 3I 2 + 4 H 2 0 3d) H 2 SO 4 (l) + 8HI(g) + H 2 S + 4I 2 + 4 H 2 0 During the reaction one will observe: - · Violet Iodine Vapour being evolved, · The violet vapour cooling and subliming to form dark solid iodine, · A smell of rotten eggs (H 2 S) · Some free yellow sulphur · Some HI(g) which could be identified in the way one shows the presence of HCl. Use concentrated NH 3 solution

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48. 48 2006/P1

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60. 60 Ethanol Hexane Cyclohexane Cyclohexene Increasing carbon / hydrogen ratio = Increasing smoke and soot

61. 61 2006/P2

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66. 66 To make the Bunsen valve in this case, use a stopper with two holes. Place a glass tube through the second hole. Then put a piece of rubber tubing on the end of the glass tubing, cut a vertical slit in the rubber tubing and put a glass plug in there. This valve acts in one direction, It prevents access of air while release of the gas from the flask is permitted

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70. 70 is primary, secondary or tertiary

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73. 73 2007/P1

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75. 75 2007/P1

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86. 86 2007/P2

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89. 89 More detail provided on the next two slides

90. 90 Oxidising ability of Concentrated Sulphuric Acid. Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides. [CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE] 2a) KBr (s) + H 2 SO 4(l)  KHSO 4(aq) + HBr (g) However, the concentrated sulphuric acid will oxidise some of the HBr as follows: -1 0 2b) 2HBr + H 2 SO 4  Br 2 + SO 2 + 2H 2 O Therefore, one will observe a Reddish Vapour due to some bromine being present. The SO 2 is and acidic gas. It is also a reducing agent and will, for example, decolourise purple potassium permanganate solution An increase in O.N.

91. 91 SUMMARY for all of the Halides. Oxidising ability of Concentrated Sulphuric Acid. Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides. [CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE] 1) NaCl(s) + H 2 SO 4 (l) = NaHSO 4 (aq) + HCl(g) The concentrated sulphuric acid will not oxidise HCl 2a) KBr(s) + H 2 SO 4 (l) = KHSO 4 (aq) + HBr(g) The concentrated sulphuric acid will oxidise some of the HBr as follows: 2b) 2HBr + H 2 SO 4 + Br 2 + SO 2 + 2H 2 O Therefore, one will observe a Reddish Vapour due to some bromine being present. 3a) KI(s) + H 2 SO 4 (l)  KHSO 4 + HI(g) The concentrated sulphuric acid will oxidise some of the HI as follows: 3b) H 2 SO 4 (l) + 2HI(g)  SO 2 + I 2 + 2H 2 0 3c) H 2 SO 4 (l) + 6HI(g)  S + 3I 2 + 4 H 2 0 3d) H 2 SO 4 (l) + 8HI(g)  H 2 S + 4I 2 + 4 H 2 0 During the reaction one will observe: - · Violet Iodine Vapour being evolved, · The violet vapour cooling and subliming to form dark solid iodine, · A smell of rotten eggs (H 2 S) · Some free yellow sulphur · Some HI(g) which could be identified in the way one shows the presence of HCl. Use concentrated NH 3 solution

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93. 93 Please see the next slide

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101. 101 That’s it. Thanks for listening. Do consider using the other three parts in this Series: -  Observation & Deduction  Planning  Titrations Good luck with your Practical Examination. F Scullion, JustChemy.Com