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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Reply 80
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AqsaMx
5
Also with the ligand substitution reaction of Cu(H2O)62+ + 4Cl- > [CuCl4]2- + 6H2O

Why might we observe the formation of a colour other than yellow?
Original post by AqsaMx
Also with the ligand substitution reaction of Cu(H2O)62+ + 4Cl- > [CuCl4]2- + 6H2O

Why might we observe the formation of a colour other than yellow?


I think it's because the system is in equilibrium and therefore the solution is a mix of the hexaaqua ion (blue) and the tetrachloro ion (yellow) which makes the solution appear a green colour.
Reply 82
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AqsaMx
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Oh makes sense! Thank you 😊
Reply 83
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AqsaMx
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image.jpg

Could anyone help me with this problem? Sorry for the scruffy writing, but I was trying to include the equations which apply to the problem also, thank you :smile:
Original post by AqsaMx
But how do we know 1cm3 = 1.333x10-3 and why are you multiplying volume by 24?
Sorry, I'm really confused :frown:


To find the volume of a gas it occupies 24dm3
The volume of the gas had density so if a question is about the density of a gas it'll always involve the 24dm3/24000cm3 - just something to remember really, especially since the question said it was measured at room temperature and pressure is a indication of that.
I can't really explain it well but I think it was that, 1cm3 is the same as gcm-3 of a gas so you make that equal the density then multiple it by 24000
If it was in gdm-3 then you make it equal to 1dm3 and multiply the density by 24
Reply 85
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AqsaMx
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image.jpg
I know VO2 needs to react with something to reform the catalyst, but could anyone help me with what it needs to react with?
Reply 86
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Sahil_
8
Original post by AqsaMx
image.jpg
I know VO2 needs to react with something to reform the catalyst, but could anyone help me with what it needs to react with?


Second Step: 2VO2 + 0.5O2 --> V2O5
This cancels the 2VO2 formed in first step, reforms the catalyst and adds the 0.5O2 to form the overall equation,
Reply 87
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AqsaMx
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Original post by Sahil_
Second Step: 2VO2 + 0.5O2 --> V2O5
This cancels the 2VO2 formed in first step, reforms the catalyst and adds the 0.5O2 to form the overall equation,


But aren't you supposed to form H2SO4 to form the product?
Reply 88
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AqsaMx
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Attachment not found

Hard, redox mole question!!
Serious help needed guys!
image.jpeg40.0KB
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Reply 89
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Sahil_
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Original post by AqsaMx
But aren't you supposed to form H2SO4 to form the product?


No, SO3 is the product, you are trying to get to the overall equation stated in the question.
Original post by AqsaMx
Attachment not found

Hard, redox mole question!!
Serious help needed guys!


Is the answer to part b 2:1 ??


Posted from TSR Mobile
Original post by AqsaMx


Could anyone help me with this problem? Sorry for the scruffy writing, but I was trying to include the equations which apply to the problem also, thank you



My attempt to your question. Answer ended up to be 40.2 and I assumed X to be calcium.
(edited 7 years ago)
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Reply 92
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AqsaMx
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Original post by suibster

My attempt to your question. Answer ended up to be 40.2 and I assumed X to be calcium.


Thank you so much :smile:
Reply 93
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AqsaMx
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image.jpeg
Could anyone explain (c) (d) and (e)
And in general explain his fuel cells work.
Reply 94
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AqsaMx
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Original post by Saywhatyoumean
Is the answer to part b 2:1 ??


Posted from TSR Mobile


Yes it is, I finally worked it out :smile:
Reply 95
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AqsaMx
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I have a qualitative on identifying transition metals through their reactions
Could anyone possibly give me advice for it ? :smile:
Reply 96
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alow
19
Original post by AqsaMx
I have a qualitative on identifying transition metals through their reactions
Could anyone possibly give me advice for it ? :smile:


Learn the reactions and observations associated with them.
Original post by AqsaMx
Yes it is, I finally worked it out :smile:


Ah okay cool, and did you work out the equation? I can't seem to get it aha


Posted from TSR Mobile
Unfortunately, yes :frown:


Posted from TSR Mobile
Reply 99
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AqsaMx
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Original post by Saywhatyoumean
Ah okay cool, and did you work out the equation? I can't seem to get it aha


Posted from TSR Mobile


Yeah I did, but it's at home so I'll post it after college :smile:

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