Chemistry: Post your doubts here!

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf

Q8

 

[FarahMJ]

Q15. This is a similar to question Q14 of 2006.

Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)
1.32 g is mass of NO2 and O2 gas, the left over mass 0.68 g is XO.

Mass of XO = 0.68g
Moles of XO = 0.68/ (Mr of X + 16)

Mass of X(NO3)2 = 2 g
Moles of X(NO3)2 = 2/(Mr of X + 62)

since moles of X(NO3)2 = moles of XO
2/(Mr of X + 124) = 0.68/ (Mr of X + 16)

We can solve for Mr of X using the normal maths approach, but it might be faster to do trial and error from the Mr of the four options.
Be : 2/(9 + 124) does not equal to 0.68/ (9 + 16)
Ca: 2/(40 + 124) = 0.68/ (40+ 16)
Therefore, answer is Ca.

Q17. At RTP, 1 mol of gas (assume ideal) occupies 24 000 cm3.

mole of O2 = 300 /24 000 = 0.0125 mol

2Ca + O2 --> 2 CaO
2Mg + O2 --> 2 MgO
4K + O2 --> 2K2O
4Na + O2 --> 2 Na2O

0.0125 mol of O2 will react with:
0.025 mol of Ca (1 g)
0.05 mol of Mg (1.2 g)
0.05 mol of K (1.95 g)
0.05 mol of Na (1.15 g)

Thanks a lot! I appreciate all ur help

 

[FarahMJ]

Answer is C.... why? :S

 

[FarahMJ]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_11.pdf

Q8

(X-X) + 6(X-H) = -2775

Since bond breaking is always exothermic (so negative) it becomes:

- [(X-X) + 6(X-H)] = -2775
- [(X-X) + 6(395] = -2775
- [(X-X) + 2370] = -2775

Re arrange the whole thing, you get X-X is +405.0

 

[Abdel Moniem]

Reverse CO(g)+1/2O2-->CO2(g) Delta H=+283KJ/mol
reverse H2o(g)-->H2O(l) Delta H=+44KJ/mol
Leave H2(g)+1/2O2-->H2O(l) Delta H=-286KJ/mol
Add up

 

[sadiaali]

Please anyone Q5
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
The ans is C

 

[Hadi Murtaza]

Please anyone Q5
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
The ans is C

Enthalpy change = Σ(ΔH Products) - Σ(ΔH Reactants)
Enthalpy change = [ (90.3 × 4) + (- 241.8 × 6) ] - [ - 461.4 × 4 ]
Enthalpy change = [ 361.2 - 1450.8 ] + 184.4
Enthalpy change = - 905.2 kJ/mol

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Q4
Q19
Q21 i have difficulty on finding a ciral carbon from rings any tips?
Q33 what about 3? more the concentration lower the activation energy?
Q35 waht about2 won't CO2 and O2 be made with it?
Q39 how do we make out that it reacts with sodium
Help please

 

[pikachu]

Does anybody have explanations for chemistry mcqs?

 

[DeViL gURl B)]

Guys need a piece of advice.. I sorta messed up in my paper 2 .. Like sorta .. So insha Allah I'll get an A ..
But it is still possible to get an A ... It is right?
Btw I just did some silly mistakes in that paper..

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Q4
Q19
Q21 i have difficulty on finding a ciral carbon from rings any tips?
Q33 what about 3? more the concentration lower the activation energy?
Q35 waht about2 won't CO2 and O2 be made with it?
Q39 how do we make out that it reacts with sodium
Help please

Q4. The top number is mass number (proton + neutron), to find the protons, you need to refer to Periodic Table

A. Bi : 83 protons, 126 neutrons
B. Pb : 82 protons, 126 neutrons
C. Po : 84 protons, 126 neutrons
D. Tl : 81 protons, 127 neutrons
So Tl is "unstable" as it does not contain 126 neutrons.

Q19. Form the equation based on the information given.

(NH4)2SO4 + 4O2 --> H2SO4 + 2HNO3 + 2H2O

Q21. The carbons on the "ring" are benzene carbons, they do not have 4 bonds to bond to 4 different groups, so we need not
check them for "chirality".

Q35. CaCo3 --> CaO + CO2

Q39. The hydroxyl (OH) group reacts with sodium to from hydrogen gas.

 

[Metanoia]

Does anybody have explanations for chemistry mcqs?

Could I confirm the date of the chemistry paper?

 

[ashcull14]

JUNE 2004 MCQ 28 PLX HELP URGENT

 

[ashcull14]

Could I confirm the date of the chemistry paper?

10 june

 

[Metanoia]

JUNE 2004 MCQ 28 PLX HELP URGENT

s04qp1

Use the 4 options and trial and error.

Before dehydration, the C=C bonds were actually CH-COH.

Check which of the 4 could possibly give us a tertiary alcohol (resistant to oxidation) before dehydration.

In this case, it was D.

 

[Gehad Mohamed]

Hi can someone please tell me why the product is a carboxylic acid not aldehyde while they mentioned distillation takes place !
Q 25 http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

 

[Hassan Ali Abid]

Hi can someone please tell me why the product is a carboxylic acid not aldehyde while they mentioned distillation takes place !
Q 25 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

Its clearly written that ' The reaction mixture was then boiled under reflux for one hour' so this means that carboxylic acid is formed not the aldehyde.

 

[Gehad Mohamed]

Its clearly written that ' The reaction mixture was then boiled under reflux for one hour' so this means that carboxylic acid is formed not the aldehyde.

But they also said distillation ! So plz tell me under which condition will an aldehyde form?

 

[Hassan Ali Abid]

But they also said distillation ! So plz tell me under which condition will an aldehyde form?

aldehyde is formed under the condition of * distill over * not under heat under reflux. Just read the question carefully ...they have said that the acid formed is separated by distillation.
Did you get it or not ???

 

[Metanoia]

But they also said distillation ! So plz tell me under which condition will an aldehyde form?

Honestly, I do see some flaws in the question set.

I believe they intend for students to interpret in this manner.
1st part of experiment: Reflux to oxidize alcohol to acid.

2nd part of experiment: Distillation to separate mixture of alcohol and acid.