Chemistry: Post your doubts here!

[Mohammed salik]

You do realize that CH4 -> C4H10 = first 4, right?

Besides, there's no way in hell the first 10 can be gases. Gasoline = octane = most definitely liquid!

That was A Typo Sorry

 

[aaaamfa]

need help in the following paper:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf

Question No:7

with explanations cuz i dont know where to start frm

 

[NinjaInPyjamas]

Cleavage occurs and the double bond is split.

If the side being split is CH2, it forms CO2+H2O. If it's CHR then it forms COOH. If it's CRR' then it forms a Ketone. Get it?

What's CRR'? Can you give a example?

 

[Mohammed salik]

need help in the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf

Question No:7

with explanations cuz i dont know where to start frm

I will start with D
First write down the equation:-
........................ 2P ------ Q + 2R
Initial Moles= 2 (P) 0(Q) 0(R)
It is given in the question x moles of R at Equilibrium, so we could find moles of Q using Mole ratio.For P since P:R is 2:2=1:1 Then P would be Initial moles - the moles of R at Equilibrium.
Equil Moles= (2-x )(P) ( x/2 ) (Q) ( x)(R)
Total (Add them up) = 2-x+(x/2)+x = (2+x/2) so the ans is D
If u Need further explaination Do tell Me!
Hope i helped!

 

[AbbbbY]

What's CRR'? Can you give a example?

etc.

Two alkyl groups on the same carbon will give you the ketone. The C=C will split and C=O will be formed with the same R groups. If instead of either of the R groups there was a H present, it would've formed carboxylic acid. Same, C=O and the H oxidised to OH.

 

[Haya Ahmed]

You do realize that CH4 -> C4H10 = first 4, right?

Besides, there's no way in hell the first 10 can be gases. Gasoline = octane = most definitely liquid!

omg so who is gas who is liquid

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf


People can anyone help me please in Q10 and Q16 ... Thanks ..

 

[NinjaInPyjamas]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf


People can anyone help me please in Q10 and Q16 ... Thanks ..

10. You can rule out B and D since you know to calculate C-H bond energy you need Hf of methane.
CH4 consists of only 2 elements. In C, combustion of H2 will need O2 - this is unnecessary and not needed in the equation since O is not found in CH4.
And in A, Hatom of C + Hatom of H + Hformation of CH4 are all relevant.

16. B - Initially I oxidation state was -1. Afterwards, it was 0, it was oxidized.
Check the oxidation numbers for each option.

 

[NinjaInPyjamas]

I will start with D
First write down the equation:-
........................ 2P ------ Q + 2R
Initial Moles= 2 (P) 0(Q) 0(R)
It is given in the question x moles of R at Equilibrium, so we could find moles of Q using Mole ratio.For P since P:R is 2:2=1:1 Then P would be Initial moles - the moles of R at Equilibrium.
Equil Moles= (2-x )(P) ( x/2 ) (Q) ( x)(R)
Total (Add them up) = 2-x+(x/2)+x = (2+x/2) so the ans is D
If u Need further explaination Do tell Me!
Hope i helped!

can you please write them and explain in detail please

 

[ZaqZainab]

I will start with D
First write down the equation:-
........................ 2P ------ Q + 2R
Initial Moles= 2 (P) 0(Q) 0(R)
It is given in the question x moles of R at Equilibrium, so we could find moles of Q using Mole ratio.For P since P:R is 2:2=1:1 Then P would be Initial moles - the moles of R at Equilibrium.
Equil Moles= (2-x )(P) ( x/2 ) (Q) ( x)(R)
Total (Add them up) = 2-x+(x/2)+x = (2+x/2) so the ans is D
If u Need further explaination Do tell Me!
Hope i helped!

so you are saying i am suppose to do write the equation of A then B then C and then finally D to know whats the answer won't this consume alot of time?

 

[Mohammed salik]

so you are saying i am suppose to do write the equation of A then B then C and then finally D to know whats the answer won't this consume alot of time?

This Is Detailed Explaination, I dont Think it would consume time if u Practice it! Besides You always Have One and Half Minute Per question.

 

[Mohammed salik]

can you please write them and explain in detail please

Plz if you could specify which part of it , it would be lot easier For Me!

 

[Namehere]

As a matter of fact, I do.
Gimme a day. I'll gather up all the individual facesheets, merge them and upload them and send you the link.
Or, if you want specific topics, let me know and I'll send you those.

Thank you so much, that will be soo helpful! Dont waste too much time on circular motion and baby chapters, things like electromagnetism, a bit of oscillations and applications will do!

 

[saadgujjar]

Plz explain q2b

 

[AbbbbY]

Plz explain q2bView attachment 40134

I'll just quote someone else who has explained this super brilliantly yesterday.

I´ll go in for the rescue!

Lets consider the first row, step 1 is the slowest: This means that the hydrogen peroxide and iodide are involved somehow in the rate determining step. Since they are only involved once (directly or indirectly, you´ll see what I mean later on) and they are in the rate equation, then a=1 and b=1, c=0 because there is no hydrogen ions reacting in step 1.

Lets consider the second row, step 2 is the slowest: This means that the IO- and hydrogen ion are involved somehow in the rate determining step, but the IO- ion doesn´t appear in the rate equation! First, it is clear that the hydrogen ion will give us c=1, since it is only involved once and appears on the rate equation. Secondly, in order for the IO- to be produced, the hydrogen peroxide and hydrogen ion will have to react, and since these are involved in the rate equation, it follows that a=1 and b=1 (since each specie was involved only once), though this time the hydrogen peroxide and hydrogen ion were involved indirectly.

Lets finally consider the third row, step 3 is the slowest: This means that HOI, hydrogen ion and iodide are involved somehow in the rate determining step. Lets analyze HOI. We can see that for HOI to be formed, the IO- and hydrogen ion had to react, but the IO- which was needed for reaction 2 was produced by reaction of hydrogen peroxide with iodide from reaction 1! By looking at the rate equation we see that hydrogen peroxide was involved indirectly once to produce HOI and so we can see a=1. Lets continue with the hydrogen ion. We can see that it is in the rate equation and that it is involved twice, one directly from reaction 3 and one indirectly from reaction 2, since hydrogen ions reacted with IO- to form HOI, one of the reactants somehow involved in the rate determining step in reaction 3. Since it is involved twice, it follows that c=2. Finally with iodide we can also see it is in the rate equation. Here the iodide ion has been involed twice, again once directly from reaction 3 and one indirectly from reaction 1 (since iodide was needed in reaction 1 to produce IO-, which was needed in reaction 2 to make HOI, which was needed in reaction 3 - the rate determining step. Since it is involved twice, it also follows that b=2.

Hope it helps. It is a bit difficult to explain these types of questions by simply writing it. Do ask if you are still unsure about something.

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf


People can anyone help me please in Q10 and Q26 ... Thanks ..

 

[Anikaa]

LE = B – A
= –415 – (131 + 908 + 1730) – {244 + 2(–349)}
This is what markscheme says but where did the 244 come from? I am guessing it's the atomisation of chlorine but it's not given in the question

 

[papajohn]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf


People can anyone help me please in Q10 and Q26 ... Thanks ..

Q10,
Correct me if i m wrong
The enthalpy of atomisation is just converting the substance to the atom of gaseous state..
For example
C(s)------> C(g)
For Hydrogen H2(g)----> 2H(g)

To find the CH bond we need to draw the Hess law ..


CH4(g)
. .
. .
4H(g) + C(g)
The total enthalpy change of atomisation is equal to the the enthalpy change of formation of methane..I hope you get it.

 

[Haya Ahmed]

Q10,
Correct me if i m wrong
The enthalpy of atomisation is just converting the substance to the atom of gaseous state..
For example
C(s)------> C(g)
For Hydrogen H2(g)----> 2H(g)

To find the CH bond we need to draw the Hess law ..


CH4(g)
. .
. .
4H(g) + C(g)
The total enthalpy change of atomisation is equal to the the enthalpy change of formation of methane..I hope you get it.

I got it ^^ what about Q26

 

[daredevil]

someone... ANYONE .... please explain NMR to me!! :``) i seem to not at all be able to do nmr questions :/ (