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Chemistry: Post your doubts here!

SomeStudent

SomeStudent

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AbbbbY

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Student12 said:
Some Doubts..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_ms_1.pdf
Questions 1 , 3 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_ms_1.pdf Questions 8 , 9 , 26

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_1.pdf Questions 1, 9 , 10

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_ms_1.pdf Questions 1,2,6,15,31

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_ms_1.pdfQuestions 1,2,4,30

I know these are aloot in one go.. but i need help! =)
Click to expand...


ON07 Q10

2 : 2 : 1
Start: 4 : 0 : 0
Equilibrium: 2.4 : 1.6 : 0.8 [If you don't understand how these concentration values have been reached, ask!]

[Products]/[Reactants]
([0.8][1.6]^2)/[2.4]^2


MJ08

Q1

6CaO + P4O10 -> 2Ca3(PO4)2

So, D

2-

14g per 5dm3
so
2.8g per dm3

100g contains 15g N
2.8g will contain 0.42g of N

Conc: 0.42g/dm3
divide it by Ar to get mol/dm3
Hence, 0.03. So, A.

6-

15-

CaCO3 -> CaO + CO2

1200 ton = 1200/100 = 12mol [in terms of tonnes]
12*44 = 528 tonnes

Hence, B

31-

C2H5OH + 3O2 -> 2CO2 + 3H2O

Since 1 is wrong, it has to be C.

Just to check,
C2H2 + 5/2O2 -> 2CO2 + H2O

CH3CHO + 5/2O2 -> 2CO2 + 2H2O
And again, C.
 
Mairaxo

Mairaxo

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hela said:
PLEASE HELP
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf Q6 C Q7 D Q26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf Q 26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf Q 8 D Q25 D
Click to expand...

Oct 2008 - Q26 - We need to form the reagents C6H5CH2COOH and C2H5OH. In all options the reagents form C6H5CH2OH xept in c
 
Mairaxo

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hela said:
PLEASE HELP
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf Q6 C Q7 D Q26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf Q 26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf Q 8 D Q25 D
Click to expand...
Oct 2010 - Q8 - Ca(S) --> Ca(g) is +177
then add ionisation energies 1 and 2 wich are 590 and 1150
then Ca+2(g) --> Ca+2(aq) is -1565
so 117+590+1150-1565 = 352

Q25- a is cracking wich needs 500 degree. b needs reflux in ethanol so high temp needed. c also needs heating. only d doesnt need to be heated.
 
Mairaxo

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iKhaled said:
can someone tell me how to solve these type of questions ? question 4 oct/nov 2009 paper 12

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Click to expand...
a and d have no dipole. CO2 is linear shaped and so both oxygen atoms cancel out. the same goes for option a. all chlorine molecules cancel eachothers polarity. in b only oxygen is electronegative and it gets -ve charge. in c also there is a dipole but weaker as the chlorine atoms form a -ve charge in the opposite direction so slightly reduce the overall dipole. Hence its b. Just think of it as forces acting at a point. If they act in the same direction the overal force is large. If they act in opposite directions then its less :)
 
Rahma Abdelrahman

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zackle09 said:
can someone pls tell me how to do this?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf

question 2
Click to expand...

You have to try with each one, write the BALANCED chemical equation, then check whether the mole ratios found give 0.2 mol of the hydrocarbon.
First find moles of CO2 and of H2O… à 35.2/44 =0.8 à Carbondioxide
14.4/18 =0.8à Water
Then,
C2H4 + 3O2 à 2CO2 + 2H2O , ratio of the hydrocarbon to CO2 is 1:2 , so this will give 0.4 mol of the hydrocarbon .

But if u try with C4H8 You will get a ratio of 1:4 which will give 0.2 mol of hydrocarbon
 
Z

zackle09

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Rahma Abdelrahman said:
You have to try with each one, write the BALANCED chemical equation, then check whether the mole ratios found give 0.2 mol of the hydrocarbon.
First find moles of CO2 and of H2O… à 35.2/44 =0.8 à Carbondioxide
14.4/18 =0.8à Water
Then,
C2H4 + 3O2 à 2CO2 + 2H2O , ratio of the hydrocarbon to CO2 is 1:2 , so this will give 0.4 mol of the hydrocarbon .

But if u try with C4H8 You will get a ratio of 1:4 which will give 0.2 mol of hydrocarbon
Click to expand...

ohh i get it thanks alot! :)
 
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