Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.
Alright, Bishnu, typing my response to you.
For the first question, it's from 2011, and if you see the syllabus (http://theallpapers.com/papers/CIE/AS_and_ALevel/Chemistry (9701)/9701_y11_sy.pdf) we see that in Nitrogen and Sulphur, the section of sulphur contains the following points:
(g) explain why atmospheric oxides of nitrogen are pollutants, including their catalytic role in the oxidation of atmospheric sulfur dioxide
(h) describe the formation of atmospheric sulfur dioxide from the combustion of sulfur contaminated carbonaceous fuels
(i) state the role of sulfur dioxide in the formation of acid rain and describe the main environmental consequences of acid rain
(j) state the main details of the Contact process for sulfuric acid production
(k) understand the industrial importance of sulfuric acid
(l) describe the use of sulfur dioxide in food preservation
On the other hand, our current syllabus only has 2 points:
a) describe the formation of atmospheric sulfur dioxide from the
combustion of sulfur-contaminated fossil fuels
b) state the role of sulfur dioxide in the formation of acid rain and describe
the main environmental consequences of acid rain
Clearly, we don't need to know as much about sulphur as 2011 candidates needed to. However, I'll still try to answer this. We know it isn't B cause with it it's only as an acid that leaves behind a salt of Magnesium Sulphate. It isn't A either cause although not explicit in the syllabus I've noticed that CIE do expect us to know that ethanol is dehydrated to ethene, so like no oxidation going on there.
With propanenitrile we know that sulphuric acid turn nitriles to carboxylic acids, so we have HYDROLYSIS going on there. This leaves us with D and even if we don't get why D, we did eliminate other options. But if you think about D, it's NaBr. Even in our current syllabus, in the topic of Group VII, we read
describe and explain the reactions of halide ions with:
(ii) concentrated sulfuric acid
With the bromide ion in NaBr, we know that the bromide ion is continously oxidized from Br- to Br2 as . How it reacts with NaBr as an acid is that H2SO4 reacts with NaBr to form HBr and HNaSO4, which means it donates a proton, so that's that. Everythig is explained here http://www.chemguide.co.uk/inorganic/group7/halideions.html
For the second question just see that the last subshell to be filled is the p subshell and since it's COMPLETELY filled, remember that it is a nobel gas, so answer is D. You don't have to bother about anything else.
For question 23, there are ester bonds going on to an OH and COOH will be there (confirmed both by the COOC in the diagram and the fact that conc. sulfuric acid is being used). It can't be A for that reason, nor C. The debate now is between B and D. If you look at D, the left molecule, CH3COOH has only COOH, so it must in all cases react with the OH of the molecule on the right, HOCH2COOH. If that does happen, our product ends up looking as CH3COOCH2COOH, but this isn't a ring at all, so we can imagine that THIS molecule may react with HOCH2COOH again, but that ends up as CH3COOCH2COOCH2COOH, and this just shows that it keeps going on and on in a straight chain. On the other hand, if you look at B, we have just one TYPE of molecule but of course multiple of these in the whole material. So let's say two of these molecules react:
HOCH2COOH + HOCH2COOH
The red reacts with the red part and the blue with the blue. What do you end up with? You end up with a cyclic compound, that I can't draw here, but looks EXACTLY like the diagram. COOCH2COOCH2C It should look like this with the yellow C being the same (in a ring).
For question 21, you should recall instantly that one of the key differences between alcohols and carboxylic acids is that an alcohol reacts ONLY with a metal (sodium) to form a alcoxide ion (-1 charge), as it is not that strong of a oxidizing agent, whereas a carboxylic acid reacts like any acid with metals (sodium) alkalis (NaOH) metal carbonates and so on, to form the carboxylate ion (-1 charge). So when J is reacted with only sodium, the total -1 charges are 3, whereas with an alkali, it is -1. Since with an alkali it's only -1, we know that there's only ONE carboxylic acid group in J, automatically leaving us with TWO alcohol groups. From the 4 options, only C meets this condition. A has 2 carboxylic acid groups and 1 alochol, B has 1 each and an alkanal, D has no carboxylic acid, just an alcohol, ester, ether, and alkanal group.
For question 19, this bothered me too a bit when I first came across it. However, we should know that CO does NOT react spontaneously in air to form CO2, otherwise CO poisoning wouldn't really be a thing. B is something that I haven't heard of, same goes for D. C however is something we are expected to know from the Nitrogen and Sulphur chapter. NO2 does act as a catalyst (this actually came in my CIE finals paper 2 too) with SO2 and turns it into SO3 while itself beocming NO, which means that SO2 reduces NO2 to NO.
36, catalytic converters are meant to turn NO and CO to N2 and CO2. You should know that. 1 is correct, quite obviously. 3 is definitely incorrect cause better conditions for combustion means greater chance of N2 forming NO or NO2 (or NO going to NO2) again, which kind of defeats the purpose. Similarly, 2 should be incorrect cause we want as much time as possible to be available for our substances to be in contact with the catalyst. Increasing the rate of flow would decrease the efficiency of the process.
12, I hate these kind of questions, but in our book, Cambridge International AS and A Level Chemistry Coursebook 2nd Edition, on page 166, it says that barium burns with a green color. So yeah, sometimes CIE like to give us completely rote-learning based questions like this too. You just have to know this.
I hope I helped and I didn't tally my answer with the mark schemes so let me know if I'm wrong about anyone.
It's correct! Thanks a lot bro!
I have the components C A D for my As level should I give a retake ....?
Depends upon your targeted score
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