ALL CIE CHEMISTRY DOUBTS HERE !!

[elemis]

@ratjumper

For question 10

@CS

For the Second question :

 

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

[komailnaqvi]

Plz anyone explain N08 Q2 P1. Im having probs in these volume related ques and nothing else! Thanks.

[elemis]

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
 

[komailnaqvi]

thanx so much.
N09 Q20

[yasser37]

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

[komailnaqvi]

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.

[komailnaqvi]

P1 Nov02 Q1 Plz. Thanx.
and also in Q24 why cant it be 1-bromobutane???

[Deadly_king]

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

[Deadly_king]

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.

[moon]

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

[Master_Key]

Can someone please explain me nov.2008 q3,33,39 thanx in advance

Which paper and variant?

[Hypernova]

Mr isn't 88.  its supposed to be strontium.

Could someone do it. no explanation, Just the stochiometry calculations needed. Thanks

[yasser37]

November 06
21
37
39

[Deadly_king]

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 C₂H_6-nCl_n

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.

[$!$RatJumper$!$]

W04 Q3 pls