Chemistry: Post your doubts here!

[Metanoia]

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

Can someone help me with q4?

W06qp1

Just take 10cm3 of CH4 as example and form a balance eqn

CH4 + 2O2 --> CO2 + 2H2O

We will use up 20 cm3 of O2 and produce 10 cm3 of CO2.

So 50 cm3 of unused O2 and 10 cm3 of CO2 gives us total of 60cm3 of gas for CH4.

So its graph D if we look at CH4.

 

[kingo44]

metanoia can u do may june 2002

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Number 12 ?? answer is D

w11q13

Check post #9228 on pg 462

It is likely that you wrongly included Ag (s) into the Kc expression? If so, remember we don't include solids in the Kc.

 

[ashcull14]

 

[ashcull14]

 

[ashcull14]

 

[ashcull14]

 

[Metanoia]

View attachment 44851

Please check #9659 on page 583

 

[Metanoia]

Just uploaded

Nov 2013 P11
http://www.youtube.com/playlist?list=PL4Jmce8VJnNU4YaGfNrPYv7SBJXnB1IAC

June 2009 P1 (mistakenly typed as 2007 originally)
http://www.youtube.com/playlist?list=PL4Jmce8VJnNVom2AGZXz6fdxSnIl9G-8D

For those who messaged me to ask, I'ld like to clarify that I did not do these videos for my current students.

We actually are taking a different chemistry paper from 9701.

 

[not.maria]

please help me out in:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
18 28 35 39
Thanks

 

[Youssef Tawil]

Please help with these...i will really appreciate it!

Ans B

1 and 2 correct

Anyone please explain these

 

[raynalist]

It involved the breakage of the triple nitrogen bond...hey is the process endothermic??

the process is endothermic.

 

[Youssef Tawil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_12.pdf
Number 11?? Anyone?? Answer is A how???

 

[ziremm]

4)
P - One Br-Br bond is breaking. So enthalpy change is -193J
Q - One Cl-Cl bond is forming. So enthalpy change is +244J
R - One C-Cl bond is forming. So enthalpy change is +340J
S - One C-H bond is breaking. So enthalpy change is -410J.
Answer is therefore D.

6) NH4NO3 is really NH4+ and NO3-.
Oxidation state of N in NH4+:
x + 4 = +1
x = -3

Oxidation state of N in NO3- :
x + (-2 x 3) = -1
x = +5

Oxidation state of N in N2O:
2x + (-2) = 0
x = +1

So changes in oxidation numbers are:
-3 ---> +1 = +4
+5 ---> +1 = -4
Answer is D

21) Here, they're testing your knowledge of what happens in a free radical substitution reaction. A C-H bond is changed to a C-Cl.

In the compound given, work clockwise with the carbon atoms. In the top most carbon atom, any of the hydrogens being replaced will give the same X radical. So this gives one possible X radical. The next two carbons will also have only one possible distinct replacement. However, these replacements will give radicals which are identical to the replacement given by the first carbon. Try doing this on a paper if you don't get how.
Another possible replacement is on any one of the 2 hydrogens on the second-last carbon on the left. The last one is possible on any one of the 3 hydrogens on the last carbon.
Hence a total of 3 possible X radicals are possible. Ans: C

29) butan-2-ol can form only 2 straight chain alkenes plus one alkene with an alkyl side chain. Ans: B

40) in 3 both reactants are gases. They can't be 'heated under reflux'. Ans: B

I think you did a mistake with question 4.. R is -340 so ans is C

 

[FarahMJ]

S11Q12

In this case, the cross multiplication does give us the answer, but have to be careful at A levels, not all species need to satisfy the octet rule.
E.g AlCl3 and SO4 2-


This was what i suggest some posts ago

Understand that whatever moles of C we have at the beginning will eventually be the same number of moles of C in CO2 at the end.

Let x and y be the ratio of Al and C respectively.
AlxCy --->???---> CO2

Working backwards from CO2,
moles of CO2 = 72/24 000 = 0.003 moles = moles of C in AlxCy

Going through the options
A Al 2C3, moles of Al2C3 = 0.144/90 = 0.0016, moles of C = 0.0016 x 3 = 0.0048 (incorrect)
B Al 3C4 , moles of Al3C4 = 0.144/129 = 0.0016, moles of C = 0.0016 x 4 = 0.0045 (incorrect)
C Al 4C3 , moles of Al4C3 = 0.144/144 = 0.001, moles of C = 0.001 x 3 = 0.003 (BINGO!)
D Al 5C3

Why did you multiply by 3? Here : " moles of C = 0.0016 x 3 = 0.0048 (incorrect) "

 

[..sacrifice4Revenge..]

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

 

[ziremm]

http://papers.xtremepapers.com/CIE/... 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

 

[Metanoia]

Why did you multiply by 3? Here : " moles of C = 0.0016 x 3 = 0.0048 (incorrect) "

Because 1 mole of Al2C3 has 3 moles of C.

So 0.0016 moles of Al2C3 has 0.0016 x 3 moles of C.

 

[Metanoia]

Please help with these...i will really appreciate it!

Ans B

1 and 2 correct

What year is this from? So I can label my replies for easier search next time.

Q26.
W: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3

Q39.
The first two statements requires being clear of about the concept that in terms of stability, HI is less stable, followed by HBr, then HCl.

Statement 1 (True)
It is harder for H-Cl to decompose to H2 and Cl2, compared to H-Br.
So for HCl , the reaction is more endothermic.

Statement 2 (True)
It is easier for HBr to form than HI.
So there is a larger ratio of products/reactants for HBr reaction than for HI.
So Kp for HBr is larger than HI

Statement 3. (False)
It is easier to remove an electron from I than Cl.
So IE for I is lower than Cl.

 

[Metanoia]

View attachment 44847

Image copied from the internet

View attachment 44848

If temperatures are too low, majority of particles will not have sufficient energy to overcome the Ea (even though its already lowered by a catalyst).
This is not limited to just the contact process.