Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 119192 times)

Offline moon

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #270 on: November 18, 2010, 05:14:01 pm »
thank u soooooooooo much really don't know to thank u and  if don't mind can u explain me june 2010 varient1 Q40 please???
I really appreciate ur help and effort..thanx a lot. :-[ :-[ :-[

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #271 on: November 18, 2010, 05:28:52 pm »
W07 Q 9, 10
W06 Q 4
W04 Q 9, 10, 23

Any help would be humbly appreciated :) Thank you very much :)

Nov 07 p1

9. From the information above we can note that ratio of metal to SO32- is 2 : 1.

So now we calculate the oxidation number of S in SO32- and SO42-. You'll see that it is +4 and +6 respectively.

So we can observe that there is a change of +2 in oxidation number. Or you could have observed that by the 2 electrons emitted. So the metal will on the contrary decrease in oxidation number by 2. But since the ratio of metal was twice compared to the sulfite, change in oxidation number for each metal cation will be -2/2 = -1.

SO it will decrease from +3 to +2. (+3 - 1)

Answer is B

10.
Initial number of moles of NO2 : 4 while initial number of the products are nil.

At equilibrium number of moles of oxygen is given to be 0.8. So that of the other product NO will be 1.6
Now we can determine the number of moles of NO2 at equilibrium : (4 - 2(0.8)) = 2.4

Now Kc will be 1.62 x 0.8 / 2.42

So answer is D

Nov 06 No 4

10CH4 + 20O2 -----> 10CO2 + 20H2O

So residual gas at r.t.p will exclude volume of H2O but include the 50cm3 excess of oxygen. ;)

Hence for CH4, total volume of residual gas will be 60cm3. This eliminates A, B and C.

Answer is D

Nov 04 p1

9. The alkane will release most energy since the others already carry an oxygen atom, meaning that they are already a bit oxidised, while the alkane needs to be completely burned. ;)

So answer is B

23. A mixture of bromine is allowed to react with ethane. ---> Free radical substitution reaction.

I believe the bromine will be in excess, so it can be bromoethane, dibromoethane(2 types), tribromoethane(2 types), tetrabromoethane(2 types), pentabromoethane and heptabromoethane.

So answer is D.

NOTE : AM sorry but i've not had time to check marking schemes, so forgive me if some of the answers are wrong. Do tell me and i'll try my best to correct my mistakes. :-[
« Last Edit: November 18, 2010, 05:43:24 pm by Deadly_king »

Offline Dania

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #272 on: November 18, 2010, 05:32:17 pm »
thank u soooooooooo much really don't know to thank u and  if don't mind can u explain me june 2010 varient1 Q40 please???
I really appreciate ur help and effort..thanx a lot. :-[ :-[ :-[

I also need an explanation to the same question! Anyone?
:)

Offline $!$RatJumper$!$

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #273 on: November 18, 2010, 05:44:03 pm »
Nov 07 p1

9. From the information above we can note that ratio of metal to SO32- is 2 : 1.

So now we calculate the oxidation number of S in SO32- and SO42-. You'll see that it is +4 and +6 respectively.

So we can observe that there is a change of +2 in oxidation number. Or you could have observed that by the 2 electrons emitted. So the metal will on the contrary decrease in oxidation number by 2. But since the ratio of metal was twice compared to the sulfite, change in oxidation number for each metal cation will be -2/2 = -1.

SO it will decrease from +3 to +2. (+3 - 1)

Answer is B

10.
Initial number of moles of NO2 : 4 while initial number of the products are nil.

At equilibrium number of moles of oxygen is given to be 0.8. So that of the other product NO will be 1.6
Now we can determine the number of moles of NO2 at equilibrium : (4 - 2(0.8)) = 2.4

Now Kc will be 1.62 x 0.8 / 2.42

So answer is D

Many thankx.. may god bless you DK :)
I jus dont understand, why did you divide -2/2? as in where did the -2 come from?

Offline Dania

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #274 on: November 18, 2010, 06:52:20 pm »
The answer is A.
:)

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #275 on: November 19, 2010, 03:28:22 am »
Many thankx.. may god bless you DK :)
I jus dont understand, why did you divide -2/2? as in where did the -2 come from?

No problem buddy :)

So first you should realise that this is a redox reaction such that one substance must get oxidised and another one reduced.

We have found that the change in oxidation number of S is +2, i.e it has been oxidised. So there should be another substance which has undergone same oxidation state but in the other direction, i.e which has been reduced. In this case it's the metal M.

So the change in oxidation number of the metal should be -2. Since there are two atoms of it, we divide by 2. ;)

Hope it helps :)

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #276 on: November 19, 2010, 03:46:18 am »
The answer is A.

Both reactions, i.e with bromine and hydrogen under respective conditions will occur at the C=C. Both of them will break the 2 C=C bonds, each with one molecule.

One molecule of H2 is required to break one C=C bond to form the corresponding alkane. Since there are two C=C, two moles of H2 will be required.

The same applies for bromine. ;)

Hope it helps :)

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #277 on: November 19, 2010, 04:06:25 am »
N04 Q10!!!!!!!!!!!!!!!!!!!!!!!! plzzzzzzzzzzzzzzzzzzzzzzzzzz

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #278 on: November 19, 2010, 04:12:26 am »
Nov 04 No 10

2H2O(g) <==> 2H2(g) + O2(g)

Initially there is only pressure of 1 atm due to steam.

                                  P(H2O)                        P(H2)                        P(O2)
Initially                             1                              0                               0

At equilibrium                 1 - 2x                          2x                             x

Since we have been told that 20% of the steam has been converted, this implies that there still remains 80% at equilibrium.

So 1 - 2x = (80/100) x 1 -----> x = 0.1

So the total pressure at equilibrium will be [(1 - 2x) + 2x + x] ----> Replace x = 0.1 to get a total of 1.1 atm.

Now we can find each partial pressure at equilibrium by dividing by total pressure.

P(H2O) = (1 - 2x)/1.1 ----> 0.8/1.1
I'll let you find the other by yourself. ;)

Answer is D.

Jun 10 p11 No 40

Bromoethane has a lower boiling point(39oC) than the solution from which it is being made, so as it is formed, the heat supplied will cause it to evaporate and we'll be able to collect it using the apparatus shown.

Ethanal also has a lower boiling point(21oC) that the reactants producing it. So it can be successfully separated using the apparatus.

1, dibromoethane however has a much higher boiling points(132oC) than the other substances in the reacting mixture. Hence we'll not be getting it through this apparatus.

So answer is B.

Hope it helps :)

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #279 on: November 19, 2010, 04:19:48 am »
thanx man.
J05 Q2 plz
+rep

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #280 on: November 19, 2010, 04:34:55 am »
thanx man.
J05 Q2 plz
+rep

Anytime dude :)

Hmm.................you have been given an equation where 2 moles of Sodium azide forms 2 moles of sodium and 3 moles of nitrogen.

Divide everything by 2 so as to get the equation resulting from one mole of sodium azide. You'll get 1.5 moles of nitrogen and 1 moles of sodium.

But from second equation provided, you'll note that 10 moles of sodium form one mole of N2. So the one mole of sodium formed above will form only 0.1 mole of N2.

Total number of moles of nitrogen produced will be 1.5 + 0.1 = 1.6

Answer is B

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #281 on: November 19, 2010, 04:36:55 am »
ohh Chicken! thanxx man!

Offline $!$RatJumper$!$

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #282 on: November 19, 2010, 05:15:42 am »
Nov 04 No 10

2H2O(g) <==> 2H2(g) + O2(g)

Initially there is only pressure of 1 atm due to steam.

                                  P(H2O)                        P(H2)                        P(O2)
Initially                             1                              0                               0

At equilibrium                 1 - 2x                          2x                             x

Since we have been told that 20% of the steam has been converted, this implies that there still remains 80% at equilibrium.

So 1 - 2x = (80/100) x 1 -----> x = 0.1

So the total pressure at equilibrium will be [(1 - 2x) + 2x + x] ----> Replace x = 0.1 to get a total of 1.1 atm.

Now we can find each partial pressure at equilibrium by dividing by total pressure.

P(H2O) = (1 - 2x)/1.1 ----> 0.8/1.1
I'll let you find the other by yourself. ;)

Answer is D.

Jun 10 p11 No 40

Bromoethane has a lower boiling point(39oC) than the solution from which it is being made, so as it is formed, the heat supplied will cause it to evaporate and we'll be able to collect it using the apparatus shown.

Ethanal also has a lower boiling point(21oC) that the reactants producing it. So it can be successfully separated using the apparatus.

1, dibromoethane however has a much higher boiling points(132oC) than the other substances in the reacting mixture. Hence we'll not be getting it through this apparatus.

So answer is B.

Hope it helps :)

Excellent! :) Thank you very much! +rep

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #283 on: November 19, 2010, 05:21:33 am »
Anytime guys :)

Just ace the paper today ;D

Good luck to you all ;)

Offline cs

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #284 on: November 19, 2010, 06:29:46 am »
Good Luck to all of you.. Thanks for those who helped all my doubts. I am done with my papers already. Thanks Student Forums..