Suggested answers to GCE O Level 2014 Chemistry (5073) Paper 1

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Here are the suggested answers for O Level Chemistry Paper 1, worked out by our O-Level Chemistry Tutor Mr Eric Lee (Mr Lee is also our H2 Chemistry tutor). Detailed explanations will be posted soon. Answers for paper 2 will be posted soon too.

1. B
2. A
3. A (corrected)
4. B
5. A
6. A
7. C
8. D
9. D
10. B
11. C
12. D
13. B
14. A
15. D
16. A
17. B
18. C
19. C
20. C
21. C
22. D
23. C
24. D
25. C
26. A
27. B
28. C
29. A
30. C
31. B
32. B
33. B
34. B
35. D
36. B
37. C
38. B
39. D
40. D

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59 Responses to Suggested answers to GCE O Level 2014 Chemistry (5073) Paper 1

  1. You Do Not Need To Know says:

    Hi, i am skeptical of the answers for questions 1, 3 and 11 respectively. Could you please enlighten me on the indicated questions?

    For question 1 , i chose C complying with the fact that there are 2 dots on the chromatogram. Having B as the answer seems to be a bit strange as the solvent should be positioned at the starting line and not below the starting line Z.

    For question 3 , i chose A. A is acknowledged as the answer in the textbook itself under the qualitative analysis tests. Dilute nitric acid should be used as an acid to acidify the solution instead of sulfuric acid.

    For question 11, i believe that steam is considered as a gas as well. With reference to the equation C6H12O6 + 6O2 > 6H2O + 6CO2 , there are 12 moles of gaseous product in total. As such , it is a firm stance of mine that the answer should be D 288 dm3.

    • Dexter says:

      Qn28. Your workings are correct. However, it would be erroneous to state that the answer is 288dm^3 because the question blatantly states *at room temperature and pressure. Steam turns to liquid water at room temperature and is no longer ingaseous state. Thus, it is not included.

    • You do not need to know me either:) says:

      For question 11 the answer is (C) because the qn asked for the volume of gas at room temperature and pressure thus the water vapour has already cooled down to the liquid state. Therefore, it’s not considered as a gas.

    • Mr Eric Lee says:

      For combustion, the H2O produced is in liquid state, not gaseous state at room temperature and pressure This is a common careless mistake, so hope you can take note in future 🙂

    • Mr Eric Lee says:

      For Q1, the solvent must be BELOW the start line. If it is at the start line the spot will be dissolved in the solvent and there will be no separation. Although there are 2 spots, it is still possible that each spot contains more than one substance as this solvent may not be able to separate them effectively. The best answer is B.

  2. Jessicca Lim says:

    If I’m not wrong Qn 3’s answer is (A) because if you add sulfuric acid. You are introducing sulfate ions. And does not make it fair

    • Mr Eric Lee says:

      You are correct. I misread the question and hence, the careless mistake. The correct answer is A.

  3. Hot boy says:

    I think qns 3 and 28. For qn 3, it’s supposed to be acidified with the corresponding acid for qn 28 the question stated for both gases oxidized not a redox rxn

    • Hot boy says:

      Hence qn 3 answer should be A and qn 28 answer should be A too

      • Mr Eric Lee says:

        Question 28: A is not correct because CO2 cannot be oxidized further. The carbon in CO2 has the oxidation state of +4 already. The answer is C because both carbon monoxide and unburnt petro gases can be further oxidized to carbon dioxide.

  4. Zi Xuan says:

    Please reconsider the answer for question 3. Thank you.

  5. - says:

    Shouldn’t question 3 be A?

  6. Mr Eric Lee says:

    Hi everyone, thanks for the feedback. Question 3 should be A. I misread the question when I was doing the paper, very sorry about that!

    Question 1: the solvent should be below the starting line so that the spot at the starting line will not be dissolved in the solvent, if not there will be no separation into different spots. If there are only 2 substances, there should not be a spot at the starting line. The spot at the starting line may require another solvent to be able to move up the chromatography paper.

    Question 11: for combustion, the H2O produced is in liquid state, not gaseous state.

    Question 28: A is not correct because CO2 cannot be oxidized further. The carbon in CO2 has the oxidation state of +4 already. The answer is C because both carbon monoxide and unburnt petro gases can be further oxidized to carbon dioxide.

  7. Hot boy says:

    Anyone here above 35? I need calculate bell curve for A2…

    • jay says:

      It is not possible to calculate since it involves so many peoples performance… but i got above 35

    • Annonymous says:

      Me. But anyway, how are you gonna calculate? o.O

      • Hot boy says:

        I mean not “calculate”… I need to know how many people do better than me so I can cry… Cuz I got 35…

        • anonymous says:

          35 is good! You don’t need to feel bad about such marks. Considering you get 35, you probably wouldn’t do too bad in Paper 2 either. And no point thinking about things you can’t change. Just look forward. And study for your Bio if you’re taking it. Haha.

        • Mr Eric Lee says:

          You don’t have to cry because you have done well by getting 35. You should be proud of yourself and not be unnecessarily stressed by “bell curves” when you have already did an excellent job. You can’t control the moderation so there is no point in stressing over something you have no control about. The most important thing is you have tried your best under exam conditions and you should be happy that you did pretty well.

  8. Just curious says:

    Hi for Qn14. Isn’t the negative electrode the anode so shouldn’t oxygen be produced there? Because OH- is preferentially discharged

    • Anonymous says:

      The negative electrode is the cathode. This means that copper(II) ions are discharged at the cathode as copper metal.

    • Mr Eric Lee says:

      In electrolysis, the negative electrode is the cathode. The electrode that is connected to the negative terminal of the battery is the cathode. Hence, copper is preferentially discharged.

  9. Xian hui says:

    Hi I thought the answer for Q14 should be (D). Because isn’t the negative electrode the anode. So OH- ions should be discharged. And O2 is formed

    • Mr Eric Lee says:

      No, this is a very common mistake. In electrolysis, the negative electrode is the cathode. The electrode that is connected to the negative terminal of the battery is the cathode. Hence, copper is preferentially discharged.

    • Lol says:

      So I assumed you thought the negative electrode was the anode in paper 2? XD that’s like 3 marks darling.

  10. Hui Xiang says:

    For question 14, isn’t the negative electrode the anode. So should OH- be preferentially discharged. And O2 gas be formed?

    • Mr Eric Lee says:

      The negative electrode is the cathode, not the anode. Cathode is the electrode that is connected to the negative terminal of the battery. Hence, copper is preferentially discharged.

  11. Qns 1!! says:

    May I Know why is Q1 B???

    • Mr Eric Lee says:

      B is the best answer. In MCQ, we always choose the best answer out of the 4 options. It is not necessarily true that there must have been two substances present in the mixture because the solvent used here may not be able to move the spot at Z up the chromatography paper. Also, the spots at X or Y may contain more than 1 substance but the solvent used may not be able to separate them clearly on the paper chromatogram.

      B is the most suitable answer because the solvent needs to be below the start line in order for good separation to occur.

      • Hot boy says:

        But it is also not necessarily true the solvent is below the starting line!!! What if e solvent is just on the starting line o.O

        • Anonymous says:

          If that happens, that means that the mixture will dissolve in the solvent, resulting in an inaccurate result.

        • Mr Eric Lee says:

          The solvent has to be below the start line. If it is on the start line or above the start line, the spot will dissolve in the solvent and there will be no separation.

  12. chen jing ren says:

    Hi i wanna ask a few qns

    Q16) why is it A ?
    Q20) why not B ?
    Q27) why not A ?
    Q30) why not A ?

    Thank you very much .. 🙂

    • Hot Boy says:

      Qn 16 cuz potassium manganate is an oxidizing agent. So it oxidizes the other product and reduces itself.

      Qn 20)How are you sure W and Z are amphoteric? The law of chemistry says that atoms in the same period have the same number of electron shells hence answer is C

      Qn 27)Erm… You must memorize this fact…

    • Anonymous says:

      Carbon monoxide is oxidized by oxygen in a catalytic converter, forming carbon dioxide. Hence, it is not option A.

    • Mr Eric Lee says:

      Q16. A is an oxidizing agent, it is able to oxidize iron(II) ions to iron(III) ions, hence a redox reaction. Option B and D will result in precipitation.

      Q20. Only aluminium, lead and zinc oxides are amphoteric (react with both acids and alkalis). In Period 3, only aluminium forms an amphoteric oxide.

      Q27. A is not correct because it is aqueous, not molten. Once water is present, H+ will be preferentially reduced at the cathode to form H2 gas. This is a very common mistake and you need to take note!

      Q30. Carbon monoxide will be OXIDIZED to carbon dioxide in the catalytic converter. The carbon in CO has an oxidation state of +2 and can be increased to +4 when it is oxidized to carbon dioxide. Hence, the answer is not A.

  13. HEY! says:

    For question 34 , shldnt the answer be C? If its 4.5moles, won’t it be incomplete combustion?

    • Anonymous says:

      For question 34, 2C3H7OH+9O2, forming 6CO2+8H2O
      Dividing both sides by 2, you will obtain 4.5O2 which translates to 4.5 mol of oxygen molecules required.

    • Mr Eric Lee says:

      The correct concept is for complete combustion, the products formed are carbon dioxide and water.

      Try to write and balance the equation for complete combustion of propanol:

      2 C3H7OH + 9 O2 –> 6 CO2 + 8 H2O

      For every 2 moles of propanol, 9 moles of oxygen molecules are needed.
      Hence, for 1 mole of propanol, 4.5 moles of oxygen molecules are needed.

  14. Anonymous says:

    Why is the answer B for Q10? Shouldn’t it be A?

    • Mr Eric Lee says:

      No, the fact that each negative ion is only surrounded by 4 positive ions instead of 8 suggest that the positive ion is doubly charged since each positive ion is surrounded by 8 negative ions. Hence, the answer is B.

  15. Ryan says:

    Is it true paper 2 answers are gonna come out O.O

  16. Potassium loves sodium says:

    Why is question 2 not b?

    • Anonymous says:

      Sodium chloride dissolves in water to form sodium chloride solution. Hence, it is impossible to use filtration to obtain solid NaCl from the solution.

    • Mr Eric Lee says:

      Sodium chloride is soluble in water. In order to obtain solid sodium chloride, you need to use evaporation to dryness (sodium chloride does not decompose upon heating). If the solid decomposes upon heating, you need to use recrystallization. You can only use filtration of the solid is insoluble in water.

      • Potassium lost 1 mark because of table salt says:

        Thank you!
        30 years later, I will invent my own filter paper.
        40 years later, my answer might be accepted.
        🙁

  17. Ryan says:

    Is it true the answers for paper 2 are comin out?

  18. Hot boy says:

    Why is qn 38 B? why not D?

    • Mr Eric Lee says:

      In additional polymerization, the carbon-carbon double bonds of monomers are broken to allow monomers to join together to form the polymer. For Option 2, the CH3 and Cl groups are attached to the same carbon, but they should be attached to different carbon atoms according to the structure of the polymer.

  19. f9 student says:

    Usually,how much must I score to get a b3 or a2 in the o level?

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