Chemistry: Post your doubts here!

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf
Q5

For 5, use bonds broken - bonds formed.

 

[Alice123]

For 5, use bonds broken - bonds formed.

i did.. 2900-2400

 

[h4rriet]

i did.. 2900-2400

Your bonds formed is wrong.

 

[Alice123]

Your bonds formed is wrong.

so cud u please show the correct one??

 

[h4rriet]

so cud u please show the correct one??

(2 x H-O) + (4 x C=O).

 

[Saad (سعد)]

As-Salaamu 'alaykum.

Q7 P12 M/J/12, and Q11 P13 M/J/12, please.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_13.pdf

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21, Q28, Q38 Q36

21. The functional groups that react with Na are CO2H and OH, so three of any of these should be present in the compound. B and D are ruled out. NaOH will react with CO2H, so there should be 1 CO2H and 2 OH groups.
28. The energy profile shows a reaction that has 2 steps, one endothermic and another exothermic. That means in the reaction a bond is broken and another bond is formed.
38. Find the molecular masses of butanone, butanoic acid and 2-methyl propanoic acid. Then use actual yield/calculated yield x 100 and see if you get 60 % for all.
36. Work this out using Le Chatelier's principle. A higher pressure obviously means a higher cost. SO3 is on the side of the fewer molecules, and high pressures favour the sides of fewer molecules. A high pressure will automatically result in an increased rate because collisions will increase, and that'll increase the chance of successful collisions.

 

[cute97]

If your teacher made you understand in a different way ,then it doesn't matter I guess right!

haahhaah noo she didnt explain it she missed many things anyway good luck for everyone

 

[h4rriet]

As-Salaamu 'alaykum.

Q7 P12 M/J/12, and Q11 P13 M/J/12, please.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_13.pdf

وعليكم السلام
7 and 11:
Draw ICE charts, i.e. initial moles, change in moles and equilibrium moles, for ALL the options (A, B, C and D). Then see which one has x + 2/x for its final moles.
For example, 7 D:

We divide the equilibrium moles by 2 because the question says that the final moles of R are x, so we need to make the final moles of R = x.

 

[Alice123]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_13.pdf
Q11 Q23 n Q36

 

[Aries_95]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_13.pdf
Q11 Q23 n Q36

Q.11:
In Hydrazine, the N-N is a single bond and not a triple bond. Simply because it has a high activation energy ie B.
Q.23:
The product produced as a result of oxidation produces orange ppt with 2,4 dinitrophenyl hydrazine means that it contains a carbonyl carbon. ie it is either a ketone or an aldehyde. Since it does not give a positive test with tollens reagent, it is a ketone.
Ketones are produced by oxidation of secondary alcohol. Butan-2-ol is a secondary alcohol hence B is the answer.
Q.36:
the first two options are correct. NaCl dissolves in water and produces Na and Cl ions and the pH is 7 ie neutral. Since 1 and 2 are right, B is right.

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Q11 n Q15

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf
Q5
Jiyad Ahsan

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21, Q28, Q38 Q36


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q28

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_13.pdf
Q5,
combustion is exothermic right?
C-H = 410
C -(triple-bond)- C = 840
O-H = 460
C=O = 740
O=O = 496
acha g, now all you have to is put all the exothermic things one side minus the endothermic on the other side, bond breaking is endothermic and bond forming is exothermic
so we have
endothermic - exothermic, since exothermic is negative

[2 (C-H) + ( C -(triple-bond)- C ) + 2.5 (O=O)] - [4 (C=O) + 2 (O-H)]
[2(410) + (840) + 2.5 (496)] - [4 (740) +2 (460)]
2900 - 3880
-980
so answer is A

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_13.pdf
Q11 Q23 n Q36

11. They're asking why hydrazine doesn't burn SPONTANEOUSLY. Meaning, all by itself. Any reaction that isn't spontaneous is because it needs a large amount of energy for the reaction to start.
36. NaCl is a neutral salt, so the solution it gives will have a pH value of 7. MgCl2 is very slightly acidic. AlCl3 is acidic.

 

[stella jazz]

can any one help me with q 6 and 15 http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
also q 1 and 40 of http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
answers of any of them would be appreciated -thanks

 

[cute97]

PLEASE HELP ME
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21, Q22, Q23
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q16, Q20, Q28
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Q23, Q17
THANKS IN ADVANCE

 

[h4rriet]

can any one help me with q 6 and 15 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
also q 1 and 40 of http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
answers of any of them would be appreciated -thanks

6. pV=nRT. V=m/ρ.
15. Construct a balanced equation for the decomposition of CaCO3. Then use mole ratios.

 

[h4rriet]

can any one help me with q 6 and 15 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
also q 1 and 40 of http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
answers of any of them would be appreciated -thanks

1. Just multiply the moles by Avogrado's constant.
40. The compound is an ester. When ester are hydrolysed with NaOH, the alcohol is given and the sodium salt of carboxylic acid is given.

 

[h4rriet]

PLEASE HELP ME
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Q21, Q22, Q23

21. The functional groups that react with Na are CO2H and OH, so three of any of these should be present in the compound. B and D are ruled out. NaOH will react with CO2H, so there should be 1 CO2H and 2 OH groups.
22. Find the molecular mass of the products of the hydrolysis of ethyl propanoate. Then find the % by mass.
23. The compound is an ester and will get hydrolysed by the acid.

 

[h4rriet]

\
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Q16, Q20, Q28

16. This is something you have to memorise. Cl2 + cold, dilute NaOH will give a ClO- ion (oxidation state = +3) and Cl2 + hot, concentrated NaOH will give a ClO3- ion (oxidation state = +5).
20. Find the molecular masses of the products of propyl propanoate. The find the % by mass of each compound.
28. -NH2 from NH3 substitutes one Br atom. NH2 is a nucleophile too, and has to be close to the other C-Br bond so that it can attack it. 1,5 is closer than 1,4.

 

[Dr.MMM]

An organic compound J reacts with sodium to produce an organic ion with a charge of –3.
J reacts with NaOH(aq) to produce an organic ion with a charge of –1.
What could be the structural formula of J?
A HO2CCH(OH)CH2CO2H
B HO2CCH(OH)CH2CHO
C HOCH2CH(OH)CH2CO2H
D HOCH2COCH2CHO