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Chemistry: Post your doubts here!

Metanoia

Metanoia

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IGCSE13 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf Q19 ansC , Q22 ansA , Q27 ansD , Q36 ansA please help
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s12qp11

Q19.

Picture 3.png

If wondering why not ans A, in the examiners approach, they don't view CO--> CO2 as spontaneous.

Q22. Products formed CH3CH2OH, CH3CH2COONa
% of CH3CH2OH = Mr of CH3CH2OH/(Mr of CH3CH2O- + Mr of CH3CH2COONa) x 100% = 46/(46+96) x 100% = 32.3%

Q27. C-I bond is weakest, so I- will be released quickest to form ppt with Ag+

Q36.

A. Higher pressure means more expensive equipment
B. Eqm shift to right as there are less molecules on right
C. Rate of backward reaction (and also forward reaction) increases because the gases become more concentrated at higher pressure
 
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Metanoia

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ZaqZainab said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q11
Q13 o_O can someone explain?
Q16 why not D? thats how HCl or HI is formed by breaking I-I bond and Cl-Cl bond
Q18
Q21
Q39 which two alkenes i have problems with ring structures
Q40
Click to expand...

S07qp1

Hi, please try to type the answers next to the questions so we can reply much faster. :)

Q11. The 2nd experiment has more moles of products but slower initial speed.
This can happen when we add a diluted solution to the original concentration
1) the number of moles of H2O2 increases --> more O2
2) the concentration of H2O2 decreases (due to dilution) --> slower production of O2

Q13. Focusing on Al atom, the Al is joined to N and 3 H, so 4 single bonds and no lone pairs around Al.

Q16. Question is trying to justify why it is forming of HI is harder (more endothermic).
Although statement D is true, just statement D alone actually meant that forming of HI is easier!

Q18. Fertiliser is NH4NO3

Q21. Need to replace a H of an alkyl group with halogen, need free radical substitution reaction.

Q39. When the OH is removed, a H from a neighboring carbon is also removed.
So the double bond could be between
1) the carbon( originally with the OH) and the carbon at the 6 o'clock position
2) the carbon( originally with the OH) and the carbon at the 1 o'clock position

Q40. Question says the reagent should be able to react with ester group. Of all 3 options, the acid is the one who can hydrolyse ester groups.
 
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Browny

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Metanoia said:
s04p1

Regarding Q28.

Key phrase in the question is "excess NaOH was used"
NaOH + SO3 --> NaHSO3 ---(1)
As there is excess NaOH, the HSO3- (acid) will further react.
NaHSO3+ NaOH --> Na2SO4 + H2O --(2)

Combine eqn (1) and (2), you will have the overall equation.
2NaOH + SO3 -->Na2SO4 + H2O

Q28. Tertiary alcohols can be oxidized by strong oxidizing agents (which is out of the syllabus), reason why they question use the phrase "not oxidized by mild oxidizing agents" is simply to reassure students that we are definitely talking about tertiary alcohols.

Q35 is as explained by ZaqZainab

Q19. Question is referring to NO2, are you asking about NO or NO2?
Click to expand...

NO2, sorry.
 
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Browny

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Metanoia

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GCE As and a level said:
Hey guys
could u plz help me with this :s
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf

DOUBTS:

Q2 ----> ANSWER D

Q3 ----> ANSWER D

Q9 ----> ANSWER B

Metanoia
Click to expand...

s12qp11

Q2. Removing products shift the equilibrium to the right, but does not increase the forward rate of reaction. Please see post #9494 for a sketch explanation.

Q3. Equation can represent both
Heat of formation of water : H2 (g) + 1/2 O2 (g) --> H2O (l) , exothermic
or heat of combustion of H2

Q9. Please check post 9224 on pg 462

FInally, perhaps a simple request to students posting questions here, besides including the answers, it would be more productive to also write down which part of the questions or which of the suggested options you don't understand or agree with. Then it allows those who intend to help to check your understanding.

Something like what browny did here. :)
Browny said:
...

The problems are as follows:

In 5 why can't A be correct as it is symmetrical.
In 13 why can't A(Mg) be the answer as when it dissolves the pH is around 6.5 which suggests hydrolysis.
In 23 why can't D(180) be the answer as the chain is like -C-C-C-C-C- which shows the C-C-C bond to be 180.
In 36 why is 2(NH3 behaves as a base) is correct.
In 39, actually I don't get a single point!

Thanks in advance and please try these questions out if possible as they are confusing and they might even help you out!
Click to expand...
 
GCE As and a level

GCE As and a level

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Metanoia said:
s12qp11

Q2. Removing products shift the equilibrium to the right, but does not increase the forward rate of reaction. Please see post #9494 for a sketch explanation.

Q3. Equation can represent both
Heat of formation of water : H2 (g) + 1/2 O2 (g) --> H2O (l) , exothermic
or heat of combustion of H2

Q9. Please check post 9224 on pg 462

FInally, perhaps a simple request to students posting questions here, besides including the answers, it would be more productive to also write down which part of the questions or which of the suggested options you don't understand or agree with. Then it allows those who intend to help to check your understanding.

Something like what browny did here. :)
Click to expand...
Thnx
but in my doubts i didnt understand a single word of question :p
and i cant chech the 477 pages to search for my doubt :p
THANK YOU!!!!
 
Metanoia

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Browny said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf

Can anybody please explain question 5 ans B, 9 ans B, 12 ans B, 18 ans C, 20 ans A, 38 ans A, please?

Some other problems are in question 11 why can't B be correct, 30 how to draw A and in 31 if a reaction is like from A to B and the energy change is +20 then if the reaction is from B to A is the change -20?:)
Click to expand...


W04qp1

Q5. You have to imagine yourself filling up the boxes with 8 electrons. If you follow the rules, you will end up with a final configuration that has the lowest energy level.

Q9. The short answer is, compared other organic compounds with the same number of carbons, burning a hydrocarbon usually gives use the most energy.
The C-O bonds for the rest require a bit more energy to break, compared to one that has only C-H single bonds.

Q11. No change in oxidation states for any atom, cant be redox reaction.

Q12. A boltzman distribution curve for a higher temperature with have its peak shifted to the right, and in a lower height.

Q18. For all the other options, there are dative bonds formed from nitrogen's lone pair.
A. N: -->C
B. N: -->H+
D. N: --> Ag+

Q38.

Y reflects the idea that ppt is formed faster than CH3CH2Cl which gives off 1 Cl- to precipitate with Ag+
1. Formed faster as it gives of 2 Cl-
2. Formed faster as C-Br bond is weaker, so Br- is released faster
3. Formed faster as C-I bond is weaker, so I- is released faster
 
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Browny

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Metanoia said:
W04qp1

Q5. You have to imagine yourself filling up the boxes with 8 electrons. If you follow the rules, you will end up with a final configuration that has the lowest energy level.

Q9. The short answer is, compared other organic compounds with the same number of carbons, burning a hydrocarbon usually gives use the most energy.
The C-O bonds for the rest require a bit more energy to break, compared to one that has only C-H single bonds.

Q11. No change in oxidation states for any atom, cant be redox reaction.

Q12. A boltzman distribution curve for a higher temperature with have its peak shifted to the right, and in a lower height.

Q18. For all the other options, there are dative bonds formed from nitrogen's lone pair.
A. N: -->C
B. N: -->H+
D. N: --> Ag+

Q38.

Y reflects the idea that ppt is formed faster than CH3CH2Cl which gives off 1 Cl- to precipitate with Ag+
1. Formed faster as it gives of 2 Cl-
2. Formed faster as C-Br bond is weaker, so Br- is released faster
3. Formed faster as C-I bond is weaker, so I- is released faster
Click to expand...

Thanks so much for the help, I just coudn't figure out thee questions !!!(y)(y)(y)

But in 11 isn't the Cr oxidation number changed.
And in 12 there are others with the curve shifted and with a lower maximum like D for example.
And in 18 how can we know the structure of the compunds such as in D, beacause I think structure of that is not in our syllabus.
 
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