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. 發問指引 (請在對應的解答區留言）
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3. 這是 2018 1B 解答區。
30 留言
無中文..sad(不過係自己學校廢 無得賴你地)
回覆刪除sorry la, making paper is really time-consuming, if I have time, I also want to make a chinese version ;(
刪除Q9 Al2O3 is insoluble in water! How will there be a colourless liquid of it at room temperature... -.-
回覆刪除It is insoluble in water does not mean that it cannot have liquid state. It just melts into liquid by heating.
刪除I know, but it is assumed that the experiments are done in room temperature.
刪除The question is already amended ;)
刪除Q4d What do you mean by interconverted?
回覆刪除exchange two solution
刪除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.
回覆刪除Q10c Other condition remain unchanged = number of moles unchanged (i.e. will lead to lower concentration), or concentration unchanged
回覆刪除Q12 How come 3-methylpentane-1,2,3-"di"ol will even exist?
回覆刪除typo is already amended ;)
刪除Q10d
回覆刪除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.
thanks for your reminding, I will include it in MS as additional information
刪除Q13 Is my method ok?
回覆刪除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)
Can anyone explain the method stated in the marking scheme?
刪除(also it should be K2Cr2O7/H+, not H2)
I think you method is also acceptable, because the yield of product is not a concern if you can correctly draw the structure.
刪除Q2a
回覆刪除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.
Yes, you are correct if you can tell the main difference of Al and Fe
刪除Q10c (@ MS v3.1)
回覆刪除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.
yes you are right
刪除Q4b (MS)
回覆刪除Missing electron
Q12a (MS)
Incorrect positions of OH groups
Q13 (step 1)
Should be -OH at C3?
Yes, you are right ;)
刪除Q6biii Can you show the calculations?
回覆刪除Q6biii The calculations seems to be based on 100cm3 solution, but only 25cm3 of NaOH is used in each titration...
刪除i just cannot calculate the answer stated in the marking scheme...
刪除AGREE
刪除0.0257 x 0.235 x 3 ÷ 0.025 x 4 = 2.90 M
刪除which 3 is the ratio and 4 is dilution factor
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???
回覆刪除13
回覆刪除