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Chemistry: Post your doubts here!

B

Browny

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mehria

mehria

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Youssef Tawil said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf
Number 3 ?? Answer is B
Number 10?? Answer is A
Click to expand...
Q 3:- first we will find the no. of moles of TlNO3 n NH4VO3
Mole of TlNO3 = (10/1000) x 0.3 = 3 x 10 ^ -3
Mole of NH4VO3 = (20/1000) x 0.1 = 2 x 10^-3
then we will find the actual mole by simply dividing each value of moles by the least value of mole
TlNO3 = (3 x 10^ -3) / (2 x 10^-3) = 1.5 and NH4VO3 = (2 x 10^ -3) / (2 x 10^ -3) = 1
As the mole of TlNO3 is in fractions so we will multiply both the values of moles with 2 to get them in the form of whole numbers
so:-
mole of TlNO3 = 3
mole of NH4VO3 = 2
3 moles of TlNO3 reduce 2 moles of NH4VO3
and 3 moles of TlNO3 will lose 2 x 3 = 6 electrons
2 moles of V+5 will gain 6 electrons so 1 mole of V+5 will gain 3 electrons
and the oxidation state of V will change from +5 to +2

the oxidation number of V in NH4VO3 => +1 + V - 6 = 0 => V=+5
 
kruti

kruti

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NIM said:
help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf
5, 15, 24, 30, 35, 39
Click to expand...
Hi there, I tried the questions. I was unable to find a solution to question 30.
otherwise, in question 5, you have to halve both the enthalpy changes given and then add them because by definition enthalpy change of formation is only for one mole of a substance and in order to make I2, the enthalpy change of the second reaction is necessary.


ques 15 - only chlorides give colourless solution with excess dilute ammonia.

ques 24 - the answer is B because during hydrolysis the ester bond is broken and the part that comes from the alcohol becomes the alcohol again (inbox me if you still don't get it).
 
B

Browny

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kruti

kruti

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NIM said:
guyzz need help!!!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
1, 4, 30, 33
Click to expand...
1)Mass - 0.216g
no. of moles = 0.216/108 = 2x10^-3
1 mol = 6.03x10^23 atoms
so 2x10^-3 mol = (6.03x10^23) x (2x10^-3) atoms
= 8.0 x 10 ^18

40) the answer is C because the two top and bottom groups are the same so statement 1 is wrong and since statement one is wrong, C has to be right. you can check statements two and three for satisfaction.
 
not.maria

not.maria

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kruti said:
Need Help!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
ques 26,27,39,40
Click to expand...
q26
the ans is B since the aldehyde ethanal gets oxidised by ammonical silver nitrate in Tollen's,
while the silver ions, Ag+, gets reduced to silver metal, Ag, hence a silver mirror forms. The ans cannot be D since there is no reaction of a ketone with Fehling's solution.

q27
strobilurn is an ester.So with sulphuric acid a carboxylic acid is formed.With Hydrogen and palladium catalyst hydrogenation of the double bonds occur so we get A
 
not.maria

not.maria

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kruti said:
Need Help!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
ques 26,27,39,40
Click to expand...
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
 
sj0007

sj0007

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Last edited:
kruti

kruti

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37
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Browny said:
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!!!:)(y)
Click to expand...

W11 :
the answer is B because magnesium nitrate isn't produces. only magnesium oxide, barium oxide and nitrogen oxide gases are produced.
W12 QP12 :
i am getting 65 for question 4 which is obvio. wrong so cant help out. question 31 is fairly simple ,Anions are formed by gaining 3 electrons in this case so optio 3 is wrong
 
kruti

kruti

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not.maria said:
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
Click to expand...


Thanks Maria.
 
kruti

kruti

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kruti said:
looking into it.
Click to expand...
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

not.maria

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for q
kruti said:
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!
Click to expand...
for q38 the formula is C4H6O4
yet CH^3COOC2H5
has 2 Oxygen atoms
am i right?
 
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