Chemistry: Post your doubts here!

[Oliveme]

Please someone explain these paper 2 questions!

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_21.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_21.pdf
Q 3 (a iii) the answer is that activation energy is too high thats why reaction does not take place but i dont get why and how it will be high
Q 3 (c) explain why the shape of the graph is like that shown in the mark scheme.......i dont get it at all!
Q 4 (a iii) Why cant there be another compound apart from A and D as written in the mark scheme? E can show position isomerism......cant it?

3 (a iii) the activation energy is high because it takes a lot of energy to break the Cl - Cl bond and that is why this reaction only takes place up in the atmosphere where it destroys the ozone layer.
3 (c) okay, this needs a lengthy explanation but overall, it's do to with the bong energy and the energy required to break them and then the graph goes down because the energy is released ---> exothermic. and then the graph stabilizes because products are formed and reaction has stopped.
4 (a iii) I'm not so good at organic chemistry sorry.

 

[leosco1995]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_2.pdf

Q2 (d) (iii)

My volume is 18.26 dm^3 while the marking scheme's answer is 9.13 dm^3. But that's because the equation in the MS was multiplied by 2. Is this necessary, and is my answer right or wrong?

 

[Abhishek Solanki]

not good in organic chemistry not able to remeber any reaction hw to recover from this problem

 

[Amy Bloom]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_2.pdf

Q2 (d) (iii)

My volume is 18.26 dm^3 while the marking scheme's answer is 9.13 dm^3. But that's because the equation in the MS was multiplied by 2. Is this necessary, and is my answer right or wrong?

Your answer is wrong. Let me explain.
First thing, your equation from (d)(i) should be correct, i.e:
H2S + 3/2O2 --> SO2 + H2O

If 34.1 g(Mr) of H2S -------------> 1 mole of H2S
Then, 8.65g of H2S ---------------> 0.2537 moles of H2S

From your equation,
1 mole of H2S ----------------> 3/2 moles of O2
0.2537 moles H2S ------------>0.3804 moles of O2

Volume of O2 = 0.3804 x 24 dm^3 = 9.13 dm^3

 

[Amy Bloom]

not good in organic chemistry not able to remeber any reaction hw to recover from this problem

concept maps may be helpful

 

[Amy Bloom]

Please someone explain these paper 2 questions!

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_21.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_21.pdf
Q 3 (a iii) the answer is that activation energy is too high thats why reaction does not take place but i dont get why and how it will be high
Q 3 (c) explain why the shape of the graph is like that shown in the mark scheme.......i dont get it at all!
Q 4 (a iii) Why cant there be another compound apart from A and D as written in the mark scheme? E can show position isomerism......cant it?

I guess question 4 a(iii) remained unanswered. so let me try to help you guys.
A bears optical isomerism because its central atom bears 4 different constituents.
D, on the other hand, has stereoisomerism, where there is one double bond and 2H groups on top and 2CH3 groups down or vice versa, or it can be One H bond and 1 CH3 bond on top and same at bottom.
See, A and D have 2 different types of isomerism: optical and stereo
However for B, has neither optical nor stereo for it has not 4 different constituents and no similar (e.g. 2H and 2 CH3) groups arranged in different ways.
Also C,E,F have isomers but not of different types, same for B.
Hope my answer has been helpful. ^^

 

[leosco1995]

Your answer is wrong. Let me explain.
First thing, your equation from (d)(i) should be correct, i.e:
H2S + 3/2O2 --> SO2 + H2O

If 34.1 g(Mr) of H2S -------------> 1 mole of H2S
Then, 8.65g of H2S ---------------> 0.2537 moles of H2S

From your equation,
1 mole of H2S ----------------> 3/2 moles of O2
0.2537 moles H2S ------------>0.3804 moles of O2

Volume of O2 = 0.3804 x 24 dm^3 = 9.13 dm^3

Thanks. I realized my (very silly) mistake, I wrote 16g for O2 instead of 32 g.

 

[Amy Bloom]

Thanks. I realized my (very silly) mistake, I wrote 16g for O2 instead of 32 g.

There's no need to find the mass of O2 in one mole. Oh! In the equation, its (1.5)O2, not 32.
Have u understood what i posted, or should i explain it to you other wise?
Btw, u're welcome

 

[leosco1995]

There's no need to find the mass of O2 in one mole. Oh! In the equation, its (1.5)O2, not 32.
Have u understood what i posted, or should i explain it to you other wise?
Btw, u're welcome

I got it. Basically, I solved the entire question using grams, and then converted grams into dm^3 in the end:

H2S : O2
34.1 : 68
8.65 : x

x = 12.17 g of Oxygen.

g : dm^3
32 : 24
12.17 : y

y = 9.13 dm^3

 

[Soulgamer]

Guys, I think the marking scheme is wrong here. Check out Question 5.

The third box, the marking key says E only.
But wouldn't the ketone and aldeyde be reduced by H2(Hydrogen)
It should've been C,D and E.

^Anyone, HELP?
Page 122 last post!

 

[Muhammad Asif]

Guys, I think the marking scheme is wrong here. Check out Question 5.

The third box, the marking key says E only.
But wouldn't the ketone and aldeyde be reduced by H2(Hydrogen)
It should've been C,D and E.

dude Hydroen n platinum catalyst i used for hydrogenation of alkenes to saturate the thats just E

 

[Soulgamer]

dude Hydroen n platinum catalyst i used for hydrogenation of alkenes to saturate the thats just E

Three reducing agents in our course. NaBH4, sodium aluminium hydride and H2. All of these reduce carbonyl compounds!

 

[Muhammad Asif]

Three reducing agents in our course. NaBH4, sodium aluminium hydride and H2. All of these reduce carbonyl compounds!

H2 isnt used to reduce ketones and aldehydes......

 

[dealer]

hey guys... i just dont get that what and how to study for chemistry paper 5.. n good tips for that paper... the marking schemes r not clear enough to assure u...

 

[Amy Bloom]

I got it. Basically, I solved the entire question using grams, and then converted grams into dm^3 in the end:

H2S : O2
34.1 : 68
8.65 : x

x = 12.17 g of Oxygen.

g : dm^3
32 : 24
12.17 : y

y = 9.13 dm^3

sounds okay. But if you have a look at your data booklet, it says that molar volume of gas under room conditions is 24 dm^3 per mole. So, preferably, work in moles.
I don't know if examiners will accept if u work in grams but to play safe, work in moles.

 

[Soulgamer]

H2 isnt used to reduce ketones and aldehydes......

Get your facts straight. They are used to reduce them!

 

[Jaf]

Get your facts straight. They are used to reduce them!

But we haven't studied the chemistry of it yet. So for us such a reaction, in effect, does not exist.
I know, I know there's an MCQ which assumes you DO know this happens but there it's a different issue (the other options are absolutely wrong).

This is just one of the things really messed up about CIE. The syllabi expect us to learn oversimplified stuff which doesn't really happen (or does happen, if you're following my argument here!) in the real world. This is just one example. Another example is the oxidation ketones. We've been taught that potassium per manganate does not (more like, can not...) oxidize ketones but fact is, it can and very well does! We're just not required to know it.

 

[raamish]

How do we make a 3-dimensional structure for hydrocabons? Need help quickly please

 

[Soulgamer]

But we haven't studied the chemistry of it yet. So for us such a reaction, in effect, does not exist.
I know, I know there's an MCQ which assumes you DO know this happens but there it's a different issue (the other options are absolutely wrong).

This is just one of the things really messed up about CIE. The syllabi expect us to learn oversimplified stuff which doesn't really happen (or does happen, if you're following my argument here!) in the real world. This is just one example. Another example is the oxidation ketones. We've been taught that potassium per manganate does not (more like, can not...) oxidize ketones but fact is, it can and very well does! We're just not required to know it.

Actually in our class we have very well studied about H2 and its reducing ability. The fact is CIE are too buzzed to even correct the marking scheme, God knows whats up with them.

 

[dealer]

is there some one who can really help me with paper 5..