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F.4 Chem MC *40marks* with explanation THX THX THX
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1.A solution may be considered as acidic if it I. turns blue litmus red. II. liberates a colourless gas when mixed with sodium carbonate. III. effervesces with magnesium. IV. undergoes electrolysis, liberating another colourless gas at cathode.(a) I and III only.(b) I, II and III only.(c) II, III... 顯示更多 1.A solution may be considered as acidic if it I. turns blue litmus red. II. liberates a colourless gas when mixed with sodium carbonate. III. effervesces with magnesium. IV. undergoes electrolysis, liberating another colourless gas at cathode. (a) I and III only. (b) I, II and III only. (c) II, III and IV only. (d) I, II, III and IV 2. What is the resultand concentration of the salt solution formed when 100cm3 of 1.2M 更新: sulphuric acid has just been neutralized completely by adding 1.2M potassium hydroxide solution? (a) 0.4M (b) 0.6M (c) 0.8M (d) 1.2M 3.Which of the following process will not release chlorine gas? (a) blow out all the liquid inside. (b) Mixing bleaching solution with vinegar. 更新 2: (c) Electrolysis of molten calcium chloride using carbon electrodes. (d) Electrolysis of brine using carbon electrodes. 4. A sample of hydrogen chloride gas contains trace amount of chlorine gas. Which of the following is the best way to verify the presence of chlorine gas in the sample? 更新 3: (a) Add a few drops of dilute sodium sulphite solution and shake. (b) Dissolve the sample into water and test the pH value of the solution with a pH meter. (c) Test the sample with a piece of filter paper soaked with potassium dichromate solution. 更新 4: (d) Test the sample with a piece of filter paper soaked with potassium iodide solution.
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其他解答:
『扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。』|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。 認證碼：23598675354145061456716738|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。』|||||(1) Please note that in the question, it mentions "acidic" solution instead of "acid". Therefore, a copper(II) sulphate acidified with sulphuric acid can also be considered as an "acidic" solution. However, when acidified copper(II) sulphate solution undergoes electrolysis, owing to the lower position of copper than hydrogen in ECS, copper(II) ions are preferentially discharged and hence NO gas bubbles are given out at the cathode. So option IV is not necessarily incorrect. For options II and III, we may use the fact that H+ ions are present with more amount in acidic solution to explain and option I can be explained based on the definition. So ans = B. (2) Looking at the equation: H2SO4 + 2KOH → K2SO4 + 2H2O We can see that 2 moles of KOH completely neutralize 1 mole of H2SO4. Hence, volume of KOH needed = 200 cm3. Then, resulting solution = 300 cm3 No. of moles of salt (K2SO4) formed = 0.12 moles (no. of moles sulphuric acid originally) Hence, molarity of salt in resulting solution = 0.4 M Ans = A (3) Ans = A Suppose that the liquid is chlorine water (bleach), if all aqueous part (water) is blown out, those remained are salt of sodium hypochlorite (active ingredient of bleach) and no chlorine gas will be evolved. (4) Ans = D In fact, for option A, the solution looks very pale yellow and still it can be decolourized if blown into sodium sulphite solution. However, human visual systems are not that sensitivie to detect the colour change. Therefore it is regarded as wrong answer. For option B, since chlorine and HCl are both acidic in aqueous state, so pH test fails to distinguish the presence. For option C, potassium dichromate solution is also an oxidizing agent that cannot react with chlorine gas which is an oxidizing agent too. Finally for option D, iodide ions are readily oxidized by chlorine to give iodine: 2I- + Cl2 → I2 + 2Cl- And further, those iodine formed dissolves in potassium iodide solution, giving a brown complex: I- + I2 → I3-
F.4 Chem MC *40marks* with explanation THX THX THX
發問:
1.A solution may be considered as acidic if it I. turns blue litmus red. II. liberates a colourless gas when mixed with sodium carbonate. III. effervesces with magnesium. IV. undergoes electrolysis, liberating another colourless gas at cathode.(a) I and III only.(b) I, II and III only.(c) II, III... 顯示更多 1.A solution may be considered as acidic if it I. turns blue litmus red. II. liberates a colourless gas when mixed with sodium carbonate. III. effervesces with magnesium. IV. undergoes electrolysis, liberating another colourless gas at cathode. (a) I and III only. (b) I, II and III only. (c) II, III and IV only. (d) I, II, III and IV 2. What is the resultand concentration of the salt solution formed when 100cm3 of 1.2M 更新: sulphuric acid has just been neutralized completely by adding 1.2M potassium hydroxide solution? (a) 0.4M (b) 0.6M (c) 0.8M (d) 1.2M 3.Which of the following process will not release chlorine gas? (a) blow out all the liquid inside. (b) Mixing bleaching solution with vinegar. 更新 2: (c) Electrolysis of molten calcium chloride using carbon electrodes. (d) Electrolysis of brine using carbon electrodes. 4. A sample of hydrogen chloride gas contains trace amount of chlorine gas. Which of the following is the best way to verify the presence of chlorine gas in the sample? 更新 3: (a) Add a few drops of dilute sodium sulphite solution and shake. (b) Dissolve the sample into water and test the pH value of the solution with a pH meter. (c) Test the sample with a piece of filter paper soaked with potassium dichromate solution. 更新 4: (d) Test the sample with a piece of filter paper soaked with potassium iodide solution.
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1 D (i) turning blue litmus paper to red is a property of acidic solutions (ii)acidic solutions form carbon dioxide when mixed with carbonate ion containing compounds. e.g. 2HCl+ Na2CO3 -------> H2O+CO2+NaCl (iii) magnesium forms hydrogen gas when mixed with acidic solution Mg+2H+ --------> H2+Mg2+ P.S. effervesces means gas bubbles (iv) cathode is where cations gather during electrolysis. In a solution of acid, the only possible cations present are hydrogen ions. Hydrogen ions, being very ready to accept electrons and the only kind of cations present, receive electrons and form hydrogen gas. 2. no solution Only the concentration of potassium hydroxide is given but not the volume(amount). Therefore, the 4 answers may be possible 3. A (i) I don't know what you mean by blowing out all the liquids in a solution so it may be possible, after considering the other points. (ii) chlorine gas will be produced when bleaching solution is mixed with acidic solutions as they contain hydrogen ions. Bleaching solutions contain hypochlorite ions (OCl-) HOCl+HCl -------> H2O+Cl2 (iii) In molten calcium chloride, only calcium ions and chloride ions are present, therefore, calcium metal and chlorine gas are formed around the electrodes. (iv) In brine solution, although the anions present include hydroxide ions and chloride ions, as the concentration of chloride ions is very high, chlorine will be produced around the anode. 4. D (i) chlorine will oxidize sodium sulphite solution to sodium sulphate solution, BUT there is NO visible change as the two solutions are both COLORLESS (ii) This method can only test for the presence of acidic gas. Chlorine will produce an acidic solution when mixed with water but hydrogen chloride gas have the same property when it is mixed with water as hydrochloric acid will be formed. (iii) Chlorine will not react with potassium dichromate as the two solutions are both oxidizing agent. (iv) potassium iodide will react with chlorine as chlorine will displace the iodide ions, forming potassium chloride. The solution will turn to dark when chlorine is present due to the formation of iodine 2007-06-19 18:16:24 補充： Sorry, the answer should be 2A 2007-06-19 18:16:34 補充： Sorry, I couldn't see "neutralize completely" in no. 2100cm3 of 1.2M H2SO4 (0.12 mol) requires 0.24 mol of KOHTherefore, 200cm3 of 1.2M KOH is needed2KOH H2SO4 --- K2SO4 2H2OTherefore, 0.12mol of salt is formedConcentration of salt solution=0.12/ [(100 200)/1000]M =0.4M其他解答:
『扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。』|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。」|||||扶正除奸聯盟只為除去不義之士，無意傷及無辜。現為閣下補回失去之分數，不便之處敬請見諒。 認證碼：23598675354145061456716738|||||『扶正除奸聯盟只為除去不義之士，無意傷及無辜。 現為閣下補回失去之分數，不便之處敬請見諒。』|||||(1) Please note that in the question, it mentions "acidic" solution instead of "acid". Therefore, a copper(II) sulphate acidified with sulphuric acid can also be considered as an "acidic" solution. However, when acidified copper(II) sulphate solution undergoes electrolysis, owing to the lower position of copper than hydrogen in ECS, copper(II) ions are preferentially discharged and hence NO gas bubbles are given out at the cathode. So option IV is not necessarily incorrect. For options II and III, we may use the fact that H+ ions are present with more amount in acidic solution to explain and option I can be explained based on the definition. So ans = B. (2) Looking at the equation: H2SO4 + 2KOH → K2SO4 + 2H2O We can see that 2 moles of KOH completely neutralize 1 mole of H2SO4. Hence, volume of KOH needed = 200 cm3. Then, resulting solution = 300 cm3 No. of moles of salt (K2SO4) formed = 0.12 moles (no. of moles sulphuric acid originally) Hence, molarity of salt in resulting solution = 0.4 M Ans = A (3) Ans = A Suppose that the liquid is chlorine water (bleach), if all aqueous part (water) is blown out, those remained are salt of sodium hypochlorite (active ingredient of bleach) and no chlorine gas will be evolved. (4) Ans = D In fact, for option A, the solution looks very pale yellow and still it can be decolourized if blown into sodium sulphite solution. However, human visual systems are not that sensitivie to detect the colour change. Therefore it is regarded as wrong answer. For option B, since chlorine and HCl are both acidic in aqueous state, so pH test fails to distinguish the presence. For option C, potassium dichromate solution is also an oxidizing agent that cannot react with chlorine gas which is an oxidizing agent too. Finally for option D, iodide ions are readily oxidized by chlorine to give iodine: 2I- + Cl2 → I2 + 2Cl- And further, those iodine formed dissolves in potassium iodide solution, giving a brown complex: I- + I2 → I3-
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