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1.A student converted lead(II) nitrate(Pb(NO3)2) to lead(II) sulphate(PbSO4) by the experiment described below:3.31g of pure lead(II) nitrate were dissolved in distilled water and an excess of sodium sulphate solution (Na2SO4) was added. The white precipitate formed was filtered, washed, and heated to dryness.... 顯示更多 1.A student converted lead(II) nitrate(Pb(NO3)2) to lead(II) sulphate(PbSO4) by the experiment described below: 3.31g of pure lead(II) nitrate were dissolved in distilled water and an excess of sodium sulphate solution (Na2SO4) was added. The white precipitate formed was filtered, washed, and heated to dryness. Its mass was found to be 2.90g. (a)Explain why (i)an excess of sodium sulphate solution was added. (ii)it was necessary to wash the precipitate. (iii)the precipitate was dried. (b)From the above information, calculate the theoretical mass of lead(II) sulphate that could be obtained. (c)Suggest why the amount of lead(II) sulphate obtained was less than the theoretical value. 2.Two iron rods A and B are partly plated with silver(Ag) and zinc(Zn)respectively. They are then separately placed in a gel containing potassium hexacyanoferrate(III)(銹蝕指示劑) and phenolphthalein(酚酞). After some time, what would be observed around each rod? Explain the observations. 3.Y is a compound containing carbon, hydrogen and oxygen only. When 0.15g of Y is completely burnt in oxygen, it gives 0.22g of carbon dioxide and 0.09g of water. (a)Determine the simplest formula(實驗式) of Y from the experimental data given above. Show your steps. (b)If the relative molecular mass of Y is 60, show that the molecular formula(分子式)of Y is C2H4O2. Show your steps.
1. (a) (i) This is to ensure that all lead(II) ions precipitate. (ii) This is to remove the solution which sticks on the lead(II) sulphate solid. (iii) This is to remove the water which sticks on the lead(II) sulphate solid. (b) Pb(NO3)2(s) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) Mole ratio Pb(NO3)2 : PbSO4 = 1 : 1 Molar mass of Pb(NO3)2 = 207 + 2(14 + 16x3) = 331 g/mol No. of moles of Pb(NO3)2 used = 3.31/331 = 0.01 mol Theoretical no. of moles of PbSO4 formed = 0.01 mol Molar mass of PbSO4 = 207 + 32 + 16x4 = 303 g/mol Theoretical mass of PBSO4 formed = 0.01 x 303 = 3.03 g (c) Some lead(II) sulphate is lost in the process. 2. A: The gel near the iron end turns blue, and that near the silver plated end turns pink. Iron is more reactive than silver, and thus iron is oxidized to iron(II) ions. The iron(II) ions turns the potassium hexacyanoferrate(III) (rust indicator) to blue. Fe(s) → Fe2+(aq) + 2e- Electrons move from iron to silver. On the surface of silver, oxygen in water is reduced to hydroxide ions, which turn the phenolphthalein to pink. 2H2O(l) + O2(g/aq) + 4e- → 4OH-(aq) B: The gel near the zinc plated end gives no observable change, and that near the iron end turns pink. Zinc is more reactive than zinc, and thus zinc is oxidized to zinc iron. This gives no observable change. Zn(s) → Zn2+(aq) + 2e- Electrons move from zinc to iron. On the surface of iron, oxygen in water is reduced to hydroxide ions, which turn the phenolphthalein to pink. 2H2O(l) + O2(g/aq) + 4e- → 4OH-(aq) 3, (a) Mass fraction of C in CO2 = 12/(12 + 16x2) = 12/44 Mass of H in H2O = (1x2)/(1x2 + 16) = 2/18 In 0.15 g of Y: Mass of C = 0.22 x (12/44) = 0.06 g Mass of H = 0.09 x (2/18) = 0.01 g Mass of O = 0.15 - (0.06 + 0.01) = 0.08 g Mole ratio C : H : O = 0.06/12 : 0.01/1 : 0.08/16 = 1 : 2 : 1 Simplest formula = CH2O (b) Let (CH2O)n be the molecular formula of Y. n(12 + 1x2 + 16) = 60 n = 2 Molecular formula of Y = (CH2O)2 = C2H4O2 =
其他解答:
救命`T^T (請以英文回答` )
發問:
1.A student converted lead(II) nitrate(Pb(NO3)2) to lead(II) sulphate(PbSO4) by the experiment described below:3.31g of pure lead(II) nitrate were dissolved in distilled water and an excess of sodium sulphate solution (Na2SO4) was added. The white precipitate formed was filtered, washed, and heated to dryness.... 顯示更多 1.A student converted lead(II) nitrate(Pb(NO3)2) to lead(II) sulphate(PbSO4) by the experiment described below: 3.31g of pure lead(II) nitrate were dissolved in distilled water and an excess of sodium sulphate solution (Na2SO4) was added. The white precipitate formed was filtered, washed, and heated to dryness. Its mass was found to be 2.90g. (a)Explain why (i)an excess of sodium sulphate solution was added. (ii)it was necessary to wash the precipitate. (iii)the precipitate was dried. (b)From the above information, calculate the theoretical mass of lead(II) sulphate that could be obtained. (c)Suggest why the amount of lead(II) sulphate obtained was less than the theoretical value. 2.Two iron rods A and B are partly plated with silver(Ag) and zinc(Zn)respectively. They are then separately placed in a gel containing potassium hexacyanoferrate(III)(銹蝕指示劑) and phenolphthalein(酚酞). After some time, what would be observed around each rod? Explain the observations. 3.Y is a compound containing carbon, hydrogen and oxygen only. When 0.15g of Y is completely burnt in oxygen, it gives 0.22g of carbon dioxide and 0.09g of water. (a)Determine the simplest formula(實驗式) of Y from the experimental data given above. Show your steps. (b)If the relative molecular mass of Y is 60, show that the molecular formula(分子式)of Y is C2H4O2. Show your steps.
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最佳解答:1. (a) (i) This is to ensure that all lead(II) ions precipitate. (ii) This is to remove the solution which sticks on the lead(II) sulphate solid. (iii) This is to remove the water which sticks on the lead(II) sulphate solid. (b) Pb(NO3)2(s) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) Mole ratio Pb(NO3)2 : PbSO4 = 1 : 1 Molar mass of Pb(NO3)2 = 207 + 2(14 + 16x3) = 331 g/mol No. of moles of Pb(NO3)2 used = 3.31/331 = 0.01 mol Theoretical no. of moles of PbSO4 formed = 0.01 mol Molar mass of PbSO4 = 207 + 32 + 16x4 = 303 g/mol Theoretical mass of PBSO4 formed = 0.01 x 303 = 3.03 g (c) Some lead(II) sulphate is lost in the process. 2. A: The gel near the iron end turns blue, and that near the silver plated end turns pink. Iron is more reactive than silver, and thus iron is oxidized to iron(II) ions. The iron(II) ions turns the potassium hexacyanoferrate(III) (rust indicator) to blue. Fe(s) → Fe2+(aq) + 2e- Electrons move from iron to silver. On the surface of silver, oxygen in water is reduced to hydroxide ions, which turn the phenolphthalein to pink. 2H2O(l) + O2(g/aq) + 4e- → 4OH-(aq) B: The gel near the zinc plated end gives no observable change, and that near the iron end turns pink. Zinc is more reactive than zinc, and thus zinc is oxidized to zinc iron. This gives no observable change. Zn(s) → Zn2+(aq) + 2e- Electrons move from zinc to iron. On the surface of iron, oxygen in water is reduced to hydroxide ions, which turn the phenolphthalein to pink. 2H2O(l) + O2(g/aq) + 4e- → 4OH-(aq) 3, (a) Mass fraction of C in CO2 = 12/(12 + 16x2) = 12/44 Mass of H in H2O = (1x2)/(1x2 + 16) = 2/18 In 0.15 g of Y: Mass of C = 0.22 x (12/44) = 0.06 g Mass of H = 0.09 x (2/18) = 0.01 g Mass of O = 0.15 - (0.06 + 0.01) = 0.08 g Mole ratio C : H : O = 0.06/12 : 0.01/1 : 0.08/16 = 1 : 2 : 1 Simplest formula = CH2O (b) Let (CH2O)n be the molecular formula of Y. n(12 + 1x2 + 16) = 60 n = 2 Molecular formula of Y = (CH2O)2 = C2H4O2 =
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