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   5** 操 Past Paper 正確方法      温書 點先唔會眼訓?     【中文】5** 卷一至卷五心得      那麼,台灣學測英文卷的難度是?

2018-04-19

【CHEM】2018 DSE00 Chem Mock 1B


按照 卷評局題型出卷
難度和 2017 相若




Paper 1B       version 1.3.1    (20/4/2018, 11pm)
Paper 1A        
Marking         version 3.5    (21/4/2018, 1pm)
All the changes are marked as red colour
1B 修改記錄  (draft to official 3.0)thanks to everyone, the mistakes in drafted version are amended. 



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3.       這是 2018 1B 解答區。

請在下面留言區發問 =〕

其他年份
2017          (2017 解答)
2016          (2016 解答區)
2015          (2015 解答區)

30 則留言:

  1. 無中文..sad(不過係自己學校廢 無得賴你地)

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    1. sorry la, making paper is really time-consuming, if I have time, I also want to make a chinese version ;(

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  2. Q9 Al2O3 is insoluble in water! How will there be a colourless liquid of it at room temperature... -.-

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    1. It is insoluble in water does not mean that it cannot have liquid state. It just melts into liquid by heating.

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    2. I know, but it is assumed that the experiments are done in room temperature.

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    3. The question is already amended ;)

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  3. Q4d What do you mean by interconverted?

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  4. Q1bii NaCl(s) can in fact be collected by this method. All the water vaporize and go into the conical flask upon heating. The NaCl, having a boiling point much higher than water, will stay in the round bottomed flask and be collected there.

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  5. Q10c Other condition remain unchanged = number of moles unchanged (i.e. will lead to lower concentration), or concentration unchanged

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  6. Q12 How come 3-methylpentane-1,2,3-"di"ol will even exist?

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  7. Q10d
    The method of titrating H+ against NaOH is in fact rather unsatisfactory. H+ is in fact the catalyst of this reaction, and during quenching, one of the possible ways is to add excess NaHCO3 to remove all the H+, so no H+ will be left for titration.
    Also, NaOH reacts with other reactants and products apart from H+. It reacts with I2 to form (IO3)- + I-, and it also reacts with Ch3COCH2I to form CH3COCH2OH.
    So, the method of titrating I2 against thiosulphate or MnO4- is much more preferrable.

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    1. thanks for your reminding, I will include it in MS as additional information

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  8. Q13 Is my method ok?
    1. H2, Pt catalyst, heat (to remove double bond)
    2. Br2/CCl4, heat or light (substitute Br to the bottom)
    3. NaOH (aq) (change the Br to OH)
    4. KMnO4/H+, heat under reflux (alcohol to ketone)
    The problem is the yield can be quite low due to step 2 (many possible products can be formed by different position and number of Br substituted), and step 3 (the Cl on top may be changed to OH as well)

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    1. Can anyone explain the method stated in the marking scheme?
      (also it should be K2Cr2O7/H+, not H2)

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    2. I think you method is also acceptable, because the yield of product is not a concern if you can correctly draw the structure.

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  9. Q2a
    Can I say:
    The protective layer of Al2O3 is impermeable to water and oxygen, so corrosion of the aluminium inside can be prevented. However, Fe2O3 is permeable to water and oxygen, so the corrosion of iron under the layer of Fe2O3 can still corrode, so corrosion resistance can't be improved.

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    1. Yes, you are correct if you can tell the main difference of Al and Fe

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  10. Q10c (@ MS v3.1)
    The reaction rate is slower when concentration is lower, and the total number of moles of reactants are unchanged, so the time to reach absorbance = 0 should be longer for volume doubled.
    Also for the curves of 10b and 10c, they are cases for autocatalyst. The last bit should still concave upwards a bit as the reactant concentration drops to a very low level.

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  11. Q4b (MS)
    Missing electron

    Q12a (MS)
    Incorrect positions of OH groups

    Q13 (step 1)
    Should be -OH at C3?

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  12. Q6biii Can you show the calculations?

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    1. Q6biii The calculations seems to be based on 100cm3 solution, but only 25cm3 of NaOH is used in each titration...

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    2. i just cannot calculate the answer stated in the marking scheme...

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    3. 0.0257 x 0.235 x 3 ÷ 0.025 x 4 = 2.90 M
      which 3 is the ratio and 4 is dilution factor

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  13. 12c when you convert A to B simply using hydrogenation,isn't that both C=C double bond will be broken down into single bond instead of breaking only one side of it???

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