Chemistry: Post your doubts here!

[Syeeeer]

Can anyone help me with questions 9 and 11? http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Any help is appreciated

 

[Syeeeer]

Help needed:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Ques 1, 4, 29

For Question 4, you need to find the volume of oxygen that is not used up as well as the volume of carbon dioxide produced.

CH4 + 2O2 ----> CO2 + 2H20 ( Since 20cm3 is used up, 50cm is left (70cm3 - 20cm3) and 10 cm3 of CO2 is produced. Add the total volume = 60cm3 )
C2H6 + 7/2 O2 ----> 2CO2 + 3H20 (Oxygen used: 35cm3, Remaining oxygen: 35cm3, Carbon dioxide produced: 20cm3. Total =55cm3 )
And so on....

 

[Samaani]

No actually. SO2 is formed as a result of sulfur combusting in the internal combustion engine. This is oxidized to SO3 in the atmosphere.

conversion of SO2 to S03 requires a catalyst, it can't happen spontaneously in air

 

[littlecloud11]

conversion of SO2 to S03 requires a catalyst, it can't happen spontaneously in air

Actually, it can. The proportion of SO2 that's converted to SO3 without a catalyst is very small and the rate is slow but the reaction does occur without a catalyst in air during the formation of acid rain. If you still have doubts refer to the reaction here-
http://www.ausetute.com.au/acidrain.html

You mentioned SO forming but sulfur monoxide can only be an intermediate in a reaction and never exist as it is because it is highly unstable and is immediately converted into SO2.

 

[autumnsakura]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Question 39 please Answer is C

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Question 3, no clue... answer is C

 

[NouranAyman]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q 4 , 6 , 8 and 9 pleaseee

 

[sonaly]

can sumone plz helpme with this question
27 Cyanohydrins can be made from carbonyl compounds by generating CN– ions from HCN in the
presence of a weak base.
R
R′
R
R′
C O + HCN
OH
CN
C
CN–
In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases.
Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong
base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3

 

[messi10]

Need help with this MCQ..

How you determine the sign?? Ans is D.. How you determine the sign... How you solved it... please write the calculation.

 

[iKhaled]

Need help with this MCQ..

How you determine the sign?? Ans is D.. How you determine the sign... How you solved it... please write the calculation.

okay here is the explanation to ur question..

you must know that any bond forming is an exothermic reaction and is (-) and any bond breakin reaction is an endothermic reactiona and (+) so we have one X-X bond formed and 6 X-H and that made a total exothermic energy of = -2775 kj/mol and if u notice the sign is negative cuz it is a bond forming reaction so,

- ( x- x ) + (6-(X-H)) = -2775
- ( x-x) + (6(-395) = -2775
-(x-x) + -2370 = -2775
-(x-x) = -2775+2370
x-x = 405 kj/mol which is D

i hope this helped u if u have any further questions pls do ask and i will try my best to help!

 

[messi10]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Question 39 please Answer is C

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Question 3, no clue... answer is C

For Q39, Answer must be B as 1 and 2 are correct (not C), because:
"1. CO is removed by oxidation, the equation is NO + CO => N2 + CO2 (Harmful CO is oxidized from +2 to +4 (whereas N went to 0 from +2)- so, CO is oxidized, NO is oxidizing agent in it, itself reduced)
2. Hydrocarbons are always removed by oxidation
3. NO is not removed by oxidation. It is removed by Reduction, so this option is incorrect."

As for Q3:
"The electronic structures of calcium, krypton, phosphorus and an element X are shown." I don't know where those were shown. May be in that year, data booklet may have any extra page, showing those structures? lol

 

[messi10]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q 4 , 6 , 8 and 9 pleaseee

Okay..

Q4:
Remember: Bond Forming is Exothermic (- H) and Bond Breaking is Endothermic (+H)
P - Br-Br bond is broken: +193 Energy
Q - Cl-Cl bond is formed: -244 Energy
R - C-Cl bond is formed: -344 Energy
S - CH bond is broken: +410 Energy
Arrange this in most negative to most positive you get the answer.

Q6:
I don't know, may be someone else can help me and you.

Q8:
Write equation first, group 2 metal reacts with Cl
X + Cl2 --> XCl2
mass of X = 2.902
mass of XCl = 5.287
mass of Cl2 = 5.287-2.902 = 2.385
mole of Cl2 = 2.385/35.5 = 0.067
1 mole of Cl2 reacts with 1 mole of X ---> you have mole, you have mass of X .. apply "mole = mass/Ar" - so you get Ar that equals 43 that is near to the Sr. So, its answer.

Q9:
i'll do 1 option for you, and you check the rest with the same yourself:
for example for Option A
3.2 divided by Ar of Oxygen (that is 32) and then multiply the answer by 24.. you will get the volume of oxygen.
Correct answer is C
-8.0/(32+32.1)
=0.1248
-0.1248*24
=2.99 dm3 that is approximately 3. Thats your answer.


Can you please tell me how you solved the question 5 of the same paper. I helped you, hope you can help me as well.

thank you!

 

[messi10]

okay here is the explanation to ur question..

you must know that any bond forming is an exothermic reaction and is (-) and any bond breakin reaction is an endothermic reactiona and (+) so we have one X-X bond formed and 6 X-H and that made a total exothermic energy of = -2775 kj/mol and if u notice the sign is negative cuz it is a bond forming reaction so,

- ( x- x ) + (6-(X-H)) = -2775
- ( x-x) + (6(-395) = -2775
-(x-x) + -2370 = -2775
-(x-x) = -2775+2370
x-x = 405 kj/mol which is D

i hope this helped u if u have any further questions pls do ask and i will try my best to help!

Thank you very much, that did help..
I was thinking in ( Reactant - Product = delta Hf ) way.

But then what happened in this question..
In this question"Reactant - Product" should give correct answer with sign, but it give opposite sign answer..

Can you please help here..

 

[messi10]

Can anyone help me with questions 9 and 11? http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Any help is appreciated

For Q9 , I also need help.!!
Please help me with all such questions if you get the answer. lol

For Q11:
1 mole of each substance is reacted.
equation is:
C2H5OH + CH3CO2H => CH3CO2C2H5 + H2O
1-x mole of C2H5OH remained
1-x mole of Ch3CO2H remained
x mole of Ch3CO2C2H5 formed
x mole of H2O formed

Kp = Products/Reactants

4 = (x)(x)/(1-x)(1-x)
4= x^2/(1-x)^2
Taking square root both sides
2 = x/1-x
2 - 2x = x
2 = 3x
x = 2/3
That's your answer.

 

[aiskw1]

I thought conversion from SO2 to SO3 could happen provided that there is NO2 in the atmosphere... represented in the equation

SO2+NO2 ---> NO + SO3

 

[aiskw1]

9 is C because I'll just scan photo...

everything that flows in one route has to equal the one that flows in another direction... Hess's law... the enthalpy change accompanying a chemical change is independent of the route at which the change occurs

 

[aiskw1]

don't mind all the O2's and stuff... they're just there because I like to write everything in my head on paper when I'm solving problems so I don't make any mistakes...

 

[autumnsakura]

For Q39, Answer must be B as 1 and 2 are correct (not C), because:
"1. CO is removed by oxidation, the equation is NO + CO => N2 + CO2 (Harmful CO is oxidized from +2 to +4 (whereas N went to 0 from +2)- so, CO is oxidized, NO is oxidizing agent in it, itself reduced)
2. Hydrocarbons are always removed by oxidation
3. NO is not removed by oxidation. It is removed by Reduction, so this option is incorrect."

As for Q3:
"The electronic structures of calcium, krypton, phosphorus and an element X are shown." I don't know where those were shown. May be in that year, data booklet may have any extra page, showing those structures? lol

I am sorry I accidentally put the wrong answer for Q39. Thanks!!

I have another question:
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w10_qp_13.pdf
Question number 37. Answer is A

 

[ZohaibAsad]

I am sorry I accidentally put the wrong answer for Q39. Thanks!!

I have another question:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_13.pdf
Question number 37. Answer is A

By inspection it's pretty obvious that the gas being referred is Carbon Monoxide.

The remaining job is to checkout which features apply. On a closer look, I would instantly know that CO is in fact colourless, so 3 should be there. If I go into a bit more detail I find out there are in fact lone 'pairs' of electron in CO as 6 electrons are shared in the covalent bond and 2 electrons each are lone on C and O.

Since 1 and 3 are correct, I don't need to think further, from the options there is no option which says '1 and 3' so I deduce that 2 should be there too although personally I haven't read what catalyst is involved in theory.

Hence, A is the answer

 

[iKhaled]

Thank you very much, that did help..
I was thinking in ( Reactant - Product = delta Hf ) way.

But then what happened in this question..
In this question"Reactant - Product" should give correct answer with sign, but it give opposite sign answer..

Can you please help here..

you know that ΔH = ΔHp - ΔHr

so ΔH = (4NO + 6H20 ) - ( 4NH3)
ΔH = (4(90.3) + 6(-241.8) ) - (4(-46.1)
ΔH = -905.2 therefore the answer is C !

knowing the definition of the enthalpy change of formation will help u a lot in this question. i hope i have helped u though!

 

[iKhaled]

need explanation to this question
isn't it supposed to be -176 because..

ΔHf = (2(38) + 2(-214) ) / 2 really confused