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Questions on AS Chemistry

Can anyone help explain this question? I don’t get why the answer is B... (question 31)
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1 is true because Z is aluminium. Z+ contains 12 electrons. X+ contains 4 electrons.
2 is true. Do the calculation
3 is false. X and Y are isotopes. By definition, they have the same number of protons.
B
I can’t calculate the given answer though. Shouldn’t the calculation be this:
10*0.1552 + 11*0.7448 ?
Original post by helloyoongles
I can’t calculate the given answer though. Shouldn’t the calculation be this:
10*0.1552 + 11*0.7448 ?

No

(10*0.1552 + 11*0.7448)/(0.1552+0.7488)


https://www.wolframalpha.com/input/?i=(10*0.1552+%2B+11*0.7448)%2F0.9


Oooo! I can understand it now. Thank you so much!!
Why is this not C? For a reaction to not have the given pathway, it should not be tertiary. But, if that is the case, both A and C are not tertiary 🤔 DDC49EA3-B6A7-4D7E-A165-4DAB0281B86F.jpg.jpeg
Original post by helloyoongles
Why is this not C? For a reaction to not have the given pathway, it should not be tertiary. But, if that is the case, both A and C are not tertiary 🤔 DDC49EA3-B6A7-4D7E-A165-4DAB0281B86F.jpg.jpeg

??
C involves an alkene
Electrophilic addition
carbocation intermediate

so A
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It would be helpful if anyone could also answer those two questions as well. Thanks in advance!
Original post by helloyoongles
C8AE1EB7-4216-4845-8E0D-68476A717949.jpg.jpeg
It would be helpful if anyone could also answer those two questions as well. Thanks in advance!


M (MgCO3)
P (Mg(OH)2)
Q (BaCO3)
so 3
so B

ratio is 4 x 46 / 32 = 1/0.173
so A
(edited 4 years ago)
Original post by BobbJo
??
C involves an alkene
Electrophilic addition
carbocation intermediate

so A


So A does not have an intermediate? Only when these is an intermediate can the diagram be applied?
Original post by helloyoongles
So A does not have an intermediate? Only when these is an intermediate can the diagram be applied?


A does not have an intermediate
Only reactions which have an intermediate have the energy profile diagram in the question
Original post by BobbJo
M (MgCO3)
P (Mg(OH)2)
Q (BaCO3)
so 3
so B

ratio is 32/(4 x 46) = 1/0.174
so A


For the second one, I am a bit confused. I thought the mass of X released or the mass of NO2 released was n*Mr = 4*36. I know it’s supposed to be 32 divided by that, but why is it not the other way around?
Original post by helloyoongles
For the second one, I am a bit confused. I thought the mass of X released or the mass of NO2 released was n*Mr = 4*36. I know it’s supposed to be 32 divided by that, but why is it not the other way around?

mass of NO2 is 4 x 46

Sorry I corrected my post
Question states Y is oxygen
X is NO2
so ratio is mass of NO2 / mass of O2 hence 4 x 46 / 32 = 1 / 0.173
Original post by BobbJo
mass of NO2 is 4 x 46

Sorry I corrected my post
Question states Y is oxygen
X is NO2
so ratio is mass of NO2 / mass of O2 hence 4 x 46 / 32 = 1 / 0.173


OOO it’s clearer now. Thank you sooo much!!! You’re a life saver! Did you finish your chemistry alevels already? You’re so good at it!
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Why should we keep both A and B closed?
Opening A would increase the volume so pressure will decrease and backwards reaction will be favoured so conc. of ammonia decreases hence we dont open A.

The question says lithium reacts with nitrogen so when we open B the N2 will react and its concentration will decrease so backwards reaction will be favoured and ammonia will decompose decreasing the concentration of ammonia.
Original post by helloyoongles
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Why should we keep both A and B closed?
Original post by apple7171
Opening A would increase the volume so pressure will decrease and backwards reaction will be favoured so conc. of ammonia decreases hence we dont open A.

The question says lithium reacts with nitrogen so when we open B the N2 will react and its concentration will decrease so backwards reaction will be favoured and ammonia will decompose decreasing the concentration of ammonia.


Thank you! So an increase in volume would decrease the pressure.... got it!
I don’t understand these two questions.
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This one as well

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