Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 122920 times)

elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #240 on: November 18, 2010, 06:52:44 am »
@ratjumper

For question 10

@CS

For the Second question :

\frac{[NO_2]^2}{[N_2O_4]} 

Imagine there is 1 mole of N2O4 hence there must be 2 moles of NO2

Therefore the partial pressures of each gas are 1/3 (N2O4) and 2/3 for NO2

inputting these into the initial equation at the top gives 4/3



Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #241 on: November 18, 2010, 07:42:06 am »
Plz anyone explain N08 Q2 P1. Im having probs in these volume related ques and nothing else! Thanks.

elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #242 on: November 18, 2010, 08:27:19 am »
Plz anyone explain N08 Q2 P1. Im having probs in these volume related ques and nothing else! Thanks.

CS2(g) + 3O2(g) -----> CO2(g) + 2SO2(g)

10 cm3 of CS2 was used along with 50cm3 of O2. From the above equation we can see that only 30 cm3 of O2 is needed.

Thus, 20 cm3 of O2 is in excess and will remain unreacted at the end.

Since only 10cm3 of CS2 was used it is the LIMITING AGENT. Hence, only 10cm3 of CO2 is produced along with 20cm3 of SO2.

Thus, 20 + 10 + 20 = 50cm3

Hence, answer must be C or D

However, when the acidic gases (SO2 and CO2) are passed through NaOH they will be absorbed leaving only the excess O2 behind.

Hence, after addition of NaOH only 20cm3 of gases are left.

ANSWER = C
 



Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #243 on: November 18, 2010, 09:28:19 am »
thanx so much.
N09 Q20

Offline yasser37

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #244 on: November 18, 2010, 09:30:59 am »

in June 10 paper 11
can someone explain questions 13 , 22 ,28 , and 29

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #245 on: November 18, 2010, 09:36:10 am »
Q.22 J10 p11
Both Alcohols are primary alcohols and so both will give same results with KmnO4, Na, and PCl5 while if you notice with Conc.H2so4 it is meant dehydration and hence two different reactions will take place one alcohol will go through dehydration while the other will not. hence
making B the Ans.

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #246 on: November 18, 2010, 10:52:01 am »
P1 Nov02 Q1 Plz. Thanx.
and also in Q24 why cant it be 1-bromobutane???

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #247 on: November 18, 2010, 10:55:19 am »
Q.22 J10 p11
Both Alcohols are primary alcohols and so both will give same results with KmnO4, Na, and PCl5 while if you notice with Conc.H2so4 it is meant dehydration and hence two different reactions will take place one alcohol will go through dehydration while the other will not. hence
making B the Ans.


Good explanation buddy :)

+rep

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #248 on: November 18, 2010, 11:08:38 am »
P1 Nov02 Q1 Plz. Thanx.
and also in Q24 why cant it be 1-bromobutane???

Nov 02 p1
1. Write the balanced equations for the reactions.

CH4 + 2O2 -----> CO2 + 2H2O

C2H6 + 7/2O2 -----> 2CO2 + 3H2O

From 1st equation, 10 cm3 of methane will produce 10 cm3 of CO2 which is the only product absorbed by the alkali.

From 2nd equation, 10 cm3 of ethane will produce 20 cm3 of CO2.

So, in all 30 cm3 pf CO2 is produced.

Answer is C

24. It cannot be 1-bromobutane since upon oxidation the primary alcohol resulting from the first reaction will be oxidised into a carboxylic acid which does not have the same formula. ;)

« Last Edit: November 18, 2010, 11:10:13 am by Deadly_king »

Offline moon

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #249 on: November 18, 2010, 12:39:18 pm »
Can someone please explain me nov.2008 q3,39 i need help in these questions can't gt them at all I would really appreciate in help.....thanx
« Last Edit: November 18, 2010, 02:44:14 pm by moon »

Offline Master_Key

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #250 on: November 18, 2010, 12:40:42 pm »
Can someone please explain me nov.2008 q3,33,39 thanx in advance

Which paper and variant?

Offline Hypernova

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #251 on: November 18, 2010, 01:17:59 pm »
Mr isn't 88.  its supposed to be strontium.

Could someone do it. no explanation, Just the stochiometry calculations needed. Thanks
Ya Allah! Make useful for me what You taught me and teach me knowledge that will be useful to me.

Offline yasser37

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #252 on: November 18, 2010, 01:24:06 pm »
November 06
21
37
39

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #253 on: November 18, 2010, 01:57:33 pm »
November 06
21
37
39

Nov 06 p1

21. This one, you need to do it step by step. Find the number of chloroethanes when n=1, 2, 3 and 4 in the formula C2H6-nCln

When n=1 ----> C2H5Cl : You can have only one type.

When n=2 ----> C2H4Cl2 : You can have 2 types, one when both chlorine atoms are found on the same carbon and two, when each carbon atoms in the compound carries one chlorine.

When n=3 ----> C2H3Cl3 : You can have 2 types, one when all three chlorine atoms are found on the same carbon and two when one carbon carries 2 chlorine while the other carbon carries one chlorine.

When n=4 -----> C2H2Cl4 : Again 2 types. One when one carbon carries 3 Cl while the other carries one Cl. and two when both carbon atoms carries 2 Cl.

So in all that makes 7 types.

Answer is C

37. To have two organic products upon oxidation, you need to have two double bonds so that when both breaks two different compounds are obtained.

So number two is rejected since only one product will be formed. ;)

Now, to have a ketone, you need to have a secondary alcohol formed upon oxidation of the alkenes.

The answer is D

This is because the double bonds are arranged in such a way that upon oxidation you'll be getting two different compounds each carrying a ketone.

The third option however will produce a compound carrying two ketone groups while the other compound will form a carboxylic acid. So only compound 1 is appropriate. :)

39. Decolourise KMnO4 ----> he's talking about alkenes.

So it's going to be a dehydration process. Therefore number 3 being a carboxylic acid will not be dehydrated.

Both number 1 and 2 are alcohols and will be dehydrated by sulfuric acid to form alkenes.

Hence answer is B

NOTE : I've not had time to confirm my answers with the marking scheme. Do check it out and let me know if there are mistakes. ;)
« Last Edit: November 18, 2010, 02:15:13 pm by Deadly_king »

Offline $!$RatJumper$!$

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #254 on: November 18, 2010, 02:04:53 pm »
W04 Q3 pls
:)