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Aqa chem 4/ chem 5 june 2016 thread

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Hi i was wondering if anybody could help with this question from the jan13 chemistry paper :smile: question 7a. Im struggling to understand what to draw thanks! :smile: ImageUploadedByStudent Room1466449741.429582.jpg


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Original post by Super199
Can someone explain the chelate effect

So when there's an increase in the number of particles from LHS to RHS there is an increase in entropy and disorder
EG A + 2B ---> 3C + 4D
There would be in increase in entropy due to more product particles
Original post by Sexybadman
Hi i was wondering if anybody could help with this question from the jan13 chemistry paper :smile: question 7a. Im struggling to understand what to draw thanks! :smile: ImageUploadedByStudent Room1466449741.429582.jpg


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Snapshot_20160620.jpg

Hopefully you can make out what I've drawn :smile:
what color is the metal aqua ion of [Fe(H20)6] 3+ ? Some say yellow some violet??
Original post by yung7up
what color is the metal aqua ion of [Fe(H20)6] 3+ ? Some say yellow some violet??


I have always learned it as yellow but you could say either


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Original post by yung7up
what color is the metal aqua ion of [Fe(H20)6] 3+ ? Some say yellow some violet??


I say purple but there are many alternatives
Original post by werdo1997
Snapshot_20160620.jpg

Hopefully you can make out what I've drawn :smile:


Thank Alot i really appreciate you taking the time to do that :smile:


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http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JUN13.PDF

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-MS-JUN13.PDF

I spoke too soon when I said I like the titration questions... would someone please send me a worked through example of 8c?? I'm lost :frown: Thank you in advance
How has everyone learned the colours and equations?
Original post by lahigueraxxx
http://filestore.aqa.org.uk/subjects/AQA-CHEM5-QP-JUN13.PDF

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-MS-JUN13.PDF

I spoke too soon when I said I like the titration questions... would someone please send me a worked through example of 8c?? I'm lost :frown: Thank you in advance


Original post by SirRaza97


I thought you couldn't round to 3 and had to say 1:2.9?
Original post by SirRaza97


Omg that makes sense!! Thank you so much, I really appreciate it :smile:
Original post by ravichauhan11
I thought you couldn't round to 3 and had to say 1:2.9?


I don't know but you can't have .9 of a chlorine atom so I rounded it to 3.
Reply 1853
Original post by The Informant
How has everyone learned the colours and equations?


Reading them and remembering them (y)
Remember all of Fe(iii) is brown.
Fe(ii) is green
Cr(iii) is green except [Cr(nh3)6]3+ which is violet
Al(iii) is colourless except Al(H2o)3(oh)3 is a white ppt
Copper is always blue except dark blue in [cu(nh3)4(h2o)2]2+ and yellow green CuCl4^2- and green blue CuCO3
Co is pink then blue in oh- , straw in ammonia.. Brown when oxidised..
That's about it tbh...
Oh but also orange Cr2o7 and yellow CrO4..
And yellow Co3+
Not too much really
And also Cr2+ is blue
And MnO4- is purple


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Original post by Suits101
Yes with E values

Platinum is used if a half-cell does not contain a metal than can act as an electrode, where the metal MUST be a SOLID. This is when Pt is indicated on the outside of conventional cell diagram.

E.g:

Fe2+ + e- -> Fe+
In this example, the metals in this half-cell are not solid because they are ions, hence you do not have a metal that can act as an electrode which is why a platinum electrode is necessary:
Pt | Fe+, Fe2+ ||

Zn2+ + 2e- -> Zn
In this example, there is a metal that can act as an electrode (Zn) because it is a solid, hence a platinum electrode is not necessary: Zn | Zn2+ ||


Awesome. Cleared that one up nicely for me. Sorry for such a late reply I have been revising for Core 3 which is tomorrow morning!!!
Original post by chzm
Pt|H2(g)|OH–(aq),H2O(l)||O2(g)|H2O(l),OH–(aq)|Pt
is this the same thing even if you swap the OH and the H2O over at the end as this is what the mark scheme states


Yes, since oxygen has an oxidation state of -2 in both OH- and H2O, although there should be a vertical line between OH- and H2O because they are in different phases. A comma is used to separate species in the same phase, such as Fe2+(aq), Fe3+(aq).
Original post by ravichauhan11
I say purple but there are many alternatives


Original post by 26december
I have always learned it as yellow but you could say either


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Ok thanks
Reply 1857
Original post by Jpw1097
Yes, since oxygen has an oxidation state of -2 in both OH- and H2O, although there should be a vertical line between OH- and H2O because they are in different phases. A comma is used to separate species in the same phase, such as Fe2+(aq), Fe3+(aq).


thanks, i copied this off the mark scheme though and i think you put a comma even it is (aq) and (l) because essentially theyre both liquid
Reply 1858
H2O2 + 2e– 2OH–
2[Cr(OH)6]3–+ 3H2O2 2CrO42–+ 8H2O + 2OH–

can someone please explain how they got the overall balanced equation? its from question 7b ii in jan 2011 paper

does anyone know what the half equatoin for the oxidation of [Cr(OH)6] to 2CrO4 would be in order to get the above overall equation as im struggling to come up with a balanced one

thanks
Original post by chzm
H2O2 + 2e– 2OH–
2[Cr(OH)6]3–+ 3H2O2 2CrO42–+ 8H2O + 2OH–

can someone please explain how they got the overall balanced equation? its from question 7b ii in jan 2011 paper

does anyone know what the half equatoin for the oxidation of [Cr(OH)6] to 2CrO4 would be in order to get the above overall equation as im struggling to come up with a balanced one

thanks



[Cr(OH)6]^3– --> CrO4^2- + 2H2O + 2H+ + 3e- x2
H2O2 + 2e- --> 2OH- x3

these are the two half equations

3H2O2 + 2[Cr(OH)6]^3– --> 6OH- + 2CrO4^2- + 4H2O + 4H+

The OH- and H+ cancel out to make H2O, so 4H+ + 4OH- --> 4H2O
3H2O2 + 2[Cr(OH)6]^3– --> 2OH- + 2CrO4^2- + 8H2O


Hopefully this helped, if you still don't understand here is a useful link: http://www.chemguide.co.uk/inorganic/redox/equations2.html

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