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Chemistry: Post your doubts here!

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Hi, Im having trouble reasoning and accepting the answers for questions 27. 33. 38. and 39. from this past years paper :

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf

Can anyone please clarify the answers (i.e. give reasons) for the questions above?

The answers are C, D, D, B.

For 27. why is the answer not B?
For 33. Why is the answer not B?

Thank you! :) God bless!
for question 27 i am not sure either, but B dont seem to be stable anion when donating acid..!
for 33, activation only govern the rate...because enthaply change can be low yet HIGH activation energy!
for 38, only 1 is inflammable enuf! thou 2 seems inflammable yet the R group of RBr is big !
and for 39 i dont know how to answer
 
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I can't help more than what the markscheme is providing ! if u dont understand anything in any step ask..
 

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(0.20+x)^2 / (0.5-x)^2 =1.44
(0.20+x)^2 = 1.44 ( 0.5-x)^2
now square root both sides and u get
0.2 + x = 1.2 (0.5 - x)
0.2 +x = 0.6 - 1.2 x
1.2x + x =0.6 - 0.2
2.2 x = 0.4
x = 0.4/2.2
x= 0.181818...
x= 0.18
 
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(0.20+x)^2 / (0.5-x)^2 =1.44
(0.20+x)^2 = 1.44 ( 0.5-x)^2
now square root both sides and u get
0.2 + x = 1.2 (0.5 - x)
0.2 +x = 0.6 - 1.2 x
1.2x + x =0.6 - 0.2
2.2 x = 0.4
x = 0.4/2.2
x= 0.181818...
x= 0.18
ohh got my mistake thanks man
 
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CO2 + H2 <=> CO + H20
Intial 0.50 0.50 0.20 0.20
moles

Let moles to calculated be X
Then we form this(equilibrium moles):
(0.50-X) + (0.50-X) <=> (0.20+X) + ( 0.20+X)
Now we have the equil. moles equation. To find the concentration, we divide moles by volume. Given V=1 dm^3 . so divide it by 1 haha :p
Now we use Kc formula. Given by the question Kc is 1.44
So, 1.44= (0.20+x)(0.20+X)/(0.50-X)(0.50-X)
1.44=(0.20+x)^2/(0.50-X)^2
take the denominator to the left and square root both side
cross multiplication and you end up with 0.40=2.2X X=0.18
now we have the X just put them in the equation.C02=H2=0.32
CO=H20=0.38
Hope it helped :)
 
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At a total pressure of 1.0 atm dinitrogen tetraoxide is 50% dissociated at a temperature of 60 degree Celcius. according to the following equation:
N2O4-------------2NO2( reversible reaction)
what is the value of the equilibrium constant Kp, for this reaction at 60 degree Celcius?
A:1/3 atm
B:2/3 atm
C: 4/3 atm
D:2 atm
Common Guyz Someone reply please:)
 
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for q 31 it wud be B??

as when the atoms become bigger the radii increase thus increasing bond length too because 2 radii make up the bond length as bond length is the measurement of length from the nuceus of one atom to the nucleus of the second atom.
and as the atom becomes bigger the attraction between the atoms decreases so the overlapping area of the radii of the atoms decreases. thus increasing the bond length... like this:
a-===-b

A---=---B

where a and b are smaller atoms
and A and B are larger atoms
and - is the part of radius not overlapping
and = is the part of radius overlapping

understood? :)
thnx :)
also q 35 if u can http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
 
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hey everyone al salam allaykom. I have doubts in Q.5 P1 Oct/Nov 2003, I don't know get it the only "big" leap in the ionisation energies is between the 5th and 6th , but how there's only 3 electrons in the outer shell? And in Q.18 why is the answer D , isn't there hydroxide ions in the solution? what about Q. 31 I thought ionic bonds are only between a metal and a non-metal ! Can any one help me out please??
 
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