Chemistry: Post your doubts here!

[kruti]

q 39
74.00g of butan-2-ol → 44.64 g of butanone

Moles of butan -2 -ol =1
moles of butanone= Mass/mr
=44.64/72= 0.62

now in theory 1 mole of butan 2 ol produces 1 mole of butanone.
but we got 0.62 moles of butanone
hence 62 percent yield

Use the same method for the rest of the reactions and you will gt 62 percent yield for all of them.

q 40
(CH3)3CBr + NaOH → (CH3)3COH + NaBr

(CH3)3CBr is a tertiary haloalkane .
It has three R groups attached to the Carbon bonded to the halogen
Only Tertiary haloalkanes undergo SN1 mechanism for nucleophillic substituition.
This is because tertiary haloalkanes have electron donationg methyl groups attached they can form an intermediate and stable carbocation.
The graph has two humps so this means that an intermediate is formed before the reaction proceeds.
2 and 3 are not tertiary haloalkanes so an intermediate is not formed

Thanks Maria.

 

[kruti]

need help in:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
markscheme
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_ms_12.pdf

q7 a
q18 d
q29 d
q38 c

looking into it.

 

[sj0007]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q 2 I did it 5 times but I keep on getting answer C even though correct answer is A....help
Thanks a bunch...

 

[kruti]

looking into it.

I don't understand ques 7. If u understand it, please let me know.


q18 - the answer is B because:
bromine dissolves in water which is polar so option A is incorrect.
Br vaporises quite easily - option c is wrong,
bromine gas is brown not purple - option d is incorrect.
this leaves us with option B.


q29 - i don't understand either.


q38 - tricky one!
option one form CH^3COOC2H5
OPTION 2 - HCOOCH2CH2OOCH
OPTION 3 - CH3OOCCOOCH3

only the first option gives the product with the stated molecular formula. check for yourself!

 

[not.maria]

for q

I don't understand ques 7. If u understand it, please let me know.


q18 - the answer is B because:
bromine dissolves in water which is polar so option A is incorrect.
Br vaporises quite easily - option c is wrong,
bromine gas is brown not purple - option d is incorrect.
this leaves us with option B.


q29 - i don't understand either.


q38 - tricky one!
option one form CH^3COOC2H5
OPTION 2 - HCOOCH2CH2OOCH
OPTION 3 - CH3OOCCOOCH3

only the first option gives the product with the stated molecular formula. check for yourself!

for q38 the formula is C4H6O4
yet CH^3COOC2H5
has 2 Oxygen atoms
am i right?

 

[kruti]

for q

for q38 the formula is C4H6O4
yet CH^3COOC2H5
has 2 Oxygen atoms
am i right?

I am sorry the answer to the question is C.
I was in a rush typing.
If you see, only options two and three have products with the molecular formula C4H6O4.
I am really sorry mate.

 

[not.maria]

for q

for q38 the formula is C4H6O4
yet CH^3COOC2H5
has 2 Oxygen atoms
am i right?

I got how the correct ans is C
2. HCO2H and HOCH2CH2OH
2 is correct if we use two of HCO2H to react with HOCH2CH2OH.
we will get ester HCO2CH2CH2CO2H with the molecular formula C4H6O4

3. HO2CCO2H and CH3OH
3 is corret if we react two of CH3OH with one of HO2CCO2H
we will get ester with the formula CH3O2CCO2CH3 with the molecular formula C4H6O4

 

[not.maria]

I am sorry the answer to the question is C.
I was in a rush typing.
If you see, only options two and three have products with the molecular formula C4H6O4.
I am really sorry mate.

Got it

I also have no clue about Q7

 

[kruti]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q 2 I did it 5 times but I keep on getting answer C even though correct answer is A....help
Thanks a bunch...

Okay this one's not that tricky, you just need to read the question well. It took me 15 minutes cause I misread the question.
N : P : K
mass: 15 30 15 total - 60g
%by mass (15/60)*100 (30/60)*100 (15/60)*100
= 25 50 25
No. of mol 25/14 50/31 25/39.1
= 1.786 1.613 0.64
= 3 2.5 1
= 6 5 2

Empirical formula = N6P5K2
no. of moles of nitrogen = 6

14g = 5dm3
for 60 g = 60*5/14 = 21.4
so volume = 21.4 dm3

conc. = 6/21.4 = 0.3mol dm-3

 

[not.maria]

I don't understand ques 7. If u understand it, please let me know.


q18 - the answer is B because:
bromine dissolves in water which is polar so option A is incorrect.
Br vaporises quite easily - option c is wrong,
bromine gas is brown not purple - option d is incorrect.
this leaves us with option B.


q29 - i don't understand either.


q38 - tricky one!
option one form CH^3COOC2H5
OPTION 2 - HCOOCH2CH2OOCH
OPTION 3 - CH3OOCCOOCH3

only the first option gives the product with the stated molecular formula. check for yourself!

kruti
q7

find the number of moles of Helium using PV=nRT
Take temperature to be 273 (or nay random temp since it doesnt matter)
Moles of helium= n= PV/RT = 0.0264
Moles of neon is also = 0.0264

now add the moles and the volumes and use in the equation P=nRT/V to get pressue of 8kPa

 

[Metanoia]

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

Can anybody please explain question 35 ans B

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf

Question 34 ans B

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

question 4 ans A and question 31 ans B

Question 4 I get 630 but the answer suggests 655.

THANKS IN ADVANCE!!!

w11qp12

35.
Under the high temperatures, Ba(NO3)2 decomposes to BaO (refer to group II nitrates decomposition).
Mg reacts with oxygen in the air to form MgO.

s12aq12
complex is formed (not sure if in your syllabus)
Ag+ (aq) + 2NH3 (aq)--> [Ag(NH3)2]+ (aq)

1. Dative bond between Ag+ ion and lone pair of N
2. No change in oxidation state of N

w12qp12

Q4. Could you post your workings? Did you view the NO molecule as N=O or single/triple bonds?

Q31. Are you able to post your diagrams on the 3 structures?

 

[kruti]

kruti
q7

find the number of moles of Helium using PV=nRT
Take temperature to be 273 (or nay random temp since it doesnt matter)
Moles of helium= n= PV/RT = 0.0264
Moles of neon is also = 0.0264

now add the moles and the volumes and use in the equation P=nRT/V to get pressue of 8kPa

Thanks you are the best!

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Number 40 , answer is D
why is 2nd statement incorrect?

 

[Hinafatima]

may i get helped in q# 29 and 35 for may june 2013 variant 13 , answers are D and C respectively?

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q7-D
Q10-C
Q15-B
Q-26-C
Q-29-D (how is D chiral :/ )
Q30-C
Q33-C ( bt catalysts do increase the kinetic energy of reactants ryt )
Thanks.

s08qp1

Q7. Have to read the question carefully, its asking for forces between atoms.
This is same question as Q6 of 2013 P12, I can't seem to embed. You can check it online if interested.

Q10.
Heat of reaction
= heat of formation of products - heat of formation of reactions
= 2(-110) -(-940)
=+720

Q15
CaCO3--> CaO + CO2
Start with 12000 million tonnes of CaCO3, calculate moles of CaCO3 and then moles of CO2. Finally, mass of CO2.

Q26.
Increase in the 2 oxygen means that the 2 OH has been oxidized to 2 COOH groups.
The two OH must be primary alcohols. Since is an unbranched butanol, the OH groups are on carbon 1 and 4.

Q29. Look at the carbon on the top left corner. Thats joined to 4 different groups.

Q33. kinetic energy affected by temperature.

 

[Metanoia]

need help in:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
markscheme
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_ms_12.pdf

q7 a
q18 d
q29 d
q38 c

Q7. Focus on individual gases first.

Helium exerted 12kPa of pressure when occupying 5dm^3 at start, what pressure did it exert when it occupies 15 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
12 x 5 = pressure (end) x 15
pressure (end) = 4 kPa from helium


Neon exerted 6kPa of pressure when occupying 10dm^3 at start, what pressure did it exert when it occupies 15 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
6 x 10 = pressure (end) x 15
pressure (end) = 4 kPa from Neon

Total pressure of Ne + He = 8 kPa

Q29.
A. No C joined to 4 different groups.
B. Nucleophillic addition
C. Intermediate is CH3CH2COCNCH2CH3 intermediate
D. First step uses CN-, last step regenerate CN- (check diagram below which i copied from another website)

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
q26 the answer is D but how...
Q34 couldn't understand at all....the answer is D
.

s07qp1

Q26. I keep reworking the question as I kept getting C as the answer. I think you typed the answer wrongly.

You need a tertiary alcohol (not oxidised easily)

For the alcohol to be chiral, smallest number of carbons for the 3 R groups would be 1,2 and 3 respectively. So minimum carbon would be 7.

Q34. Question is asking if the pressure of the left flask would increase more than the right flask when temperature of whole setup is increased.
A. Temp increase, left flask eqm shift right as forward reaction is endothermic. Left flask more gases particles than before, higher pressure than right flask.

B. Temp increase, left flask eqm shift left as backward reaction is endothermic. No change in moles of gases. So change in pressure is same for both flasks.

C. Change in pressure is same for both flasks.

 

[Metanoia]

can someone help me in these 2 questions?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_11.pdf
question 7 and 24

For this paper, you can try the video explanations I made, if still unclear, feel free to clarify here.

 

[Browny]

w11qp12

35.
Under the high temperatures, Ba(NO3)2 decomposes to BaO (refer to group II nitrates decomposition).
Mg reacts with oxygen in the air to form MgO.

s12aq12
complex is formed (not sure if in your syllabus)
Ag+ (aq) + 2NH3 (aq)--> [Ag(NH3)2]+ (aq)

1. Dative bond between Ag+ ion and lone pair of N
2. No change in oxidation state of N

w12qp12

Q4. Could you post your workings? Did you view the NO molecule as N=O or single/triple bonds?

Q31. Are you able to post your diagrams on the 3 structures?

Thanks so much 31 is ok, and 4 i think the question paperanswer is wrong as it should be 630.

 

[AhsanAfzal]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_13.pdf
explain Q38 please