• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

Messages
603
Reaction score
1,102
Points
153


w12qp12

Q11. This question does seem a bit confusing at the start. We rely on the definition of bond energy as much as possible to make our choices easier.

Bond energy is the energy required to break 1 mole of covalent bonds (with the species in gaseous phase)

Notice that option A is purely a bond breaking reaction, for the other options, there are bond breaking and bond formation, so its sensible to focus on option A.

However, option A, is actually breaking of n covalent bonds.
XYn --> nX + nY

So we need to take energy change of option A and divide by n, to get energy required to break 1 mole of covalent bonds

Its like
CH4 --> C + 4H
we need to divide this energy change by 4 to get the energy for 1 C-H bond.

Q23. It took me a while to understand what the question is describing.
The molecule must be broken in two and
1) contain same carbon atoms
2) both parts must have isomers

So this eliminates D, as 9 carbons molecule cannot form 2 smaller parts with equal carbons.
Isomers from when hydrocarbon have 4 or more carbon, so minimum we must have 8 carbons in the original chain to start with.

The bottom text is the examiners report, if what I typed earlier don't make sense.

"29% of candidates chose the correct answer, C. The most commonly chosen incorrect answer was A, chosen by 41% of candidates. If n=4 (option A) then the molecule C4H10 must be splitting to give C2H6 and C2H4, neither of which have structural isomers. The answer is C because if n=8 the products of cracking are

C4H10 and C4H8, both of which have structural isomers
 
Last edited:
Messages
603
Reaction score
1,102
Points
153
View attachment 44876 View attachment 44877

Q-A number of alcohols with the formula C4H10O are separately oxidised. Using 70 g of the alcohols

a 62 % yield of organic product is achieved.

What mass of product could be obtained?

1-42.2 g of butanone

2-51.6 g of butanoic acid

3-51.6 g of 2-methyl propanoic acid





working required please. especially Q6
thanks in advance :)

Hi there, please include the answers next time so its easier for us to reply.

Q6. The heat of reaction in the equation is to break 6 N-F bonds, so to get the bond energy, we divide it by 6.



Q15. This is an often set question.

Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)

Mass of XO = 5 -3.29 g = 1.52
Moles of XO = 1.52/ (Mr of X + 16)

Mass of X(NO3)2 = 5 g
Moles of X(NO3)2 = 5/(Mr of X + 124)

since moles of X(NO3)2 = moles of XO
5/(Mr of X + 124) = 1.52/ (Mr of X + 16)

We can solve for Mr of X using the equation , or you can substitute the Mr of the four options and do trial and error.


Last question
Mr of alcohol = 74
Moles of alcohol used = 70/74 = 0.946
Moles of alcohol converted into products = 0.946 x 62% = 0.586

1. If 0.586 mol of butanone was obtained, it would weight 0.586 x Mr of butanone = 0.586 x 72 = 42.2 g (statement 1 true)
2. If 0.586 mol of butanoic acid was obtained, it would weight 0.586 x Mr of butanoic acid = 0.586 x 88 = 51.6 g (statement 2 true)
3. If 0.586 mol of 2-methylpropanoic acid was obtained, it would weight 0.586 x 2-methylpropanoic acid = 0.586 x 88 = 51.6 g (statement 3 true)
 
Messages
603
Reaction score
1,102
Points
153

Please include the answers to make it easier for us to reply. :)

w11qp12

Q18.
Old lime mortar produces a gas when reacted with HCl, so it should be CaCO3 (to produce CO2) , instead of CaO and Ca(OH)2.

This leaves us with Options A and D.
A: CaO is harder than CaCO3
D: Ca(OH)2 is softer than CaCO3

So option D is more likely


Q28. cyclohexene is alkene, cyclohexanol is alcohol, so you have to choose the one reactant that reacts with one but not the other for distinguishing tests.

A. Is Tollen's reagent (no reaction with both)
B. Br2 (decoloursises in cyclohexene, remains brown in cyclohexanol)
C. 2,4 -DNPH (no reaction with both)
D. Reduction agent (no reaction with both)

Q35. During reaction where the mixture heats up, Ba(NO3)2 decomposes to BaO. Mg reacts with oxygen in air to form MgO.

Q39.
1. CH3CH2Cl --> CH3CH2OH
2. CH3CO2CH23 (ester undergo alkali hydrolyse) --> CH3CO2- + CH3OH --> CH3COOH + CH3OH
3. CH3CN --> CH3COO- --> CH3COOH
 
Messages
603
Reaction score
1,102
Points
153
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf
hello can you please help me with these questions: 2,8,10,17,21,25,26,31 and 28
your help will greatly appreciated :)

Hi there, please include the answers next time to make it easier for us to reply.

s10qp12
Q2.
A. 1s2 2s2 2p1 ( 1 unpaired e in p orbital)
B. 1s2 2s2 2p6 3s2 3p1 ( 1 unpaired e in p orbital)
C. 1s2 2s2 2p6 3s2 3p3 (3 unpaired e in p orbital)
D. 1s2 2s2 2p6 3s2 3p6 3d1 4s2 (1 unpaired e in d orbital)

Q8.
X + Cl2 --> XCl2

moles of X = 2.92/Mr of X
moles of XCl2 = 5.287/ Mr of XCl2

since 1 mol of X produces 1 mol of XCl2
2.92/Mr of X = 5.287/ Mr of XCl2
2.92/Mr of X = 5.287/ (Mr of X + 71)

Either solve for Mr of X, or substitute Mr of Ba, Ca, Mg and Sr to see when Mr fits the equation.

Q10.
When temp increases, all particles have more energy, so both the forward AND the backward reaction speeds up.
For this particular reaction, although both directions speeds up, the backward reaction speeds up more than the forward reaction (eqm shifts left).

Q17. Both oxides are acidic.
P: 1s2 2s2 2p6 3s2 3p3 (no unpaired e in p orbital)
S: 1s2 2s2 2p6 3s2 3p4 (has 1 paired e in p orbital)

Q21. The first alcohol will not be dehydrated.
To dehydrate, we need to remove a OH from a carbon, and a H from a neighboring carbon.
The neighboring carbon in the 1st alcohol do have have a H that could be removed.

Q25.
This is a similar to
question 22 of June 2013 paper 12, which i explained a video.

Q26.
rate of hydrolysis of primary halogen alkanes (sN2) is dependent on both the concentration of OH- and the halogen alkanes
rate of hydrolysis of tertiary halogen alkanes (sN2) is dependent only on the concentration of the halogen alkanes
since Y is not dependent of concentration of OH-, it is most likely a tertiary halogen alkane.
 
Messages
515
Reaction score
1,447
Points
153
Q 10:- ............ N2 + 3H2 -------------------> 2NH3
initial mole of N2 = 1
initial mole of H2 = 3
initial mole of NH3= 1.98
At equilibrium:-
mole of N2=> 1 - x = 1.64 so x= - 0.64
mole of 3H2=> 3 - 3x = 3-3(-0.64) = 3 + 1.92= 4.92
mole of 2NH3=> 1.98 + 2x = 1.98 + 2( - o.64) = 0.70
then the Kc value will be = (0.70)^2 / ((1.64) x (4.92)^3)

Note:- u have to take care of the -ve sign in the value of x
 
Messages
162
Reaction score
305
Points
28
Messages
515
Reaction score
1,447
Points
153
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
 
Messages
37
Reaction score
97
Points
28
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).
 
Messages
162
Reaction score
305
Points
28
Messages
37
Reaction score
97
Points
28
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.
 
Messages
70
Reaction score
171
Points
43
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
 
Messages
70
Reaction score
171
Points
43
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
 
Messages
1,824
Reaction score
5,326
Points
523
Last edited:
Messages
37
Reaction score
97
Points
28
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)

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
 
Top