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Aqa a2 chem5 19th june 2012

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Reply 40
Original post by thegodofgod
Not too bad on the whole, apart from transition metals. All the memorising for the complexes' colours is killing me :frown:

Luckily, though, most of the colours / complexes follow a trend, e.g. Cr (III) complexes are green, Fe (III) complexes are browny, Cu (II) complexes are bluish, etc.

How about you?


There's some patterns in there thankfully, yeah :tongue:

It's just spotting them really, and using them to remember stuff (i.e. neutral complexes are precipitates).

Plus the equations for the period 3 oxides and the definitions for enthalpy change, the equations for catalysis...

Ugh. :frown:
Reply 41
Original post by thegodofgod
e.g. Cr (III) complexes are green, Fe (III) complexes are browny


This is something that I'm confused about. On page 227 of the Nelson Thornes text book, Cr(III) is clearly green, but the table at the top of page 221 says "red-violet" for [Cr(H2O)6]3+ and "very pale violet" for [Fe(H2O)6]3+ and the table on page 243 says that [Cr(H2O)6]3+ is "ruby" and that [Fe(H2O)6]3+ is "purple/yellow/brown". What are we supposed to write on the exam? :confused:
Original post by Glacier
This is something that I'm confused about. On page 227 of the Nelson Thornes text book, Cr(III) is clearly green, but the table at the top of page 221 says "red-violet" for [Cr(H2O)6]3+ and "very pale violet" for [Fe(H2O)6]3+ and the table on page 243 says that [Cr(H2O)6]3+ is "ruby" and that [Fe(H2O)6]3+ is "purple/yellow/brown". What are we supposed to write on the exam? :confused:


For the ruby one, ruby is the real colour, but in a lab you wouldn't get ruby due to impurities; you'd get green. Either one is accepted by AQA.

My CGP revision guide says that [Fe(H2O)6]3+ is yellow :eek:
Reply 43
Original post by wibletg
There's more memory involved but once you've started to remember everything it gets better :tongue:

The only thing that's likely to be a sting in the tail is the synoptic link :frown:


Isn't section B purely based on chemistry 1 and 2?
Reply 44
Original post by thegodofgod
For the ruby one, ruby is the real colour, but in a lab you wouldn't get ruby due to impurities; you'd get green. Either one is accepted by AQA.

My CGP revision guide says that [Fe(H2O)6]3+ is yellow :eek:

Yeah, they do accept both - especially since green Cr3+ is meant to be the end point of an oxidation of an alcohol :tongue:

Fe2+ - They accept yellow or brown or purple :tongue:
I'd stick with the official AQA textbook for the colours though.

Original post by Alpha5
Isn't section B purely based on chemistry 1 and 2?

Nope, not from the only past paper I've done... it was solely about CHEM5, the colours of the transition metal complexes and redox titration using potassium permanganate :smile:
Original post by wibletg
Yeah, they do accept both - especially since green Cr3+ is meant to be the end point of an oxidation of an alcohol :tongue:

Fe2+ - They accept yellow or brown or purple :tongue:
I'd stick with the official AQA textbook for the colours though.


Nope, not from the only past paper I've done... it was solely about CHEM5, the colours of the transition metal complexes and redox titration using potassium permanganate :smile:


June 11 has a question on moles calculations in section B, and a 2 marker on O-Cu-O bond angles in [Cu(C2O4)3]4- :smile:
Reply 46
Original post by thegodofgod
June 11 has a question on moles calculations in section B, and a 2 marker on O-Cu-O bond angles in [Cu(C2O4)3]4- :smile:


Ah right... I've only done one past paper so far though :tongue: (think it was Jan11)

Just an example to show it's not ONLY CHEM1 & 2 :smile:


What's the odds we get thrown a daft paper?
Original post by wibletg
Ah right... I've only done one past paper so far though :tongue: (think it was Jan11)

Just an example to show it's not ONLY CHEM1 & 2 :smile:


What's the odds we get thrown a daft paper?


Pretty high.

We got a bad Jan 11 Unit 1 paper, a bad June 11 Unit 2 paper, a bad Jan 12 Unit 4 paper, so chances are, we'll get a bad June 12 Unit 5 paper :mad:

In all of those papers, there was a lot of difference between the paper we sat and the past papers, and the bloody grade boundaries were still high!
Reply 48
Original post by thegodofgod
Pretty high.

We got a bad Jan 11 Unit 1 paper, a bad June 11 Unit 2 paper, a bad Jan 12 Unit 4 paper, so chances are, we'll get a bad June 12 Unit 5 paper :mad:

In all of those papers, there was a lot of difference between the paper we sat and the past papers, and the bloody grade boundaries were still high!


Hahah, I think the thing that's infuriated me the most is that they appear to be 'normal' papers in the next exam series! :rolleyes:

I was a bit complacent in January though, that's not happening again. Spent a bit too much time concentrating on M3 too.

I've actually spent today doing CHEM5 for about 4 hours writing down as many equations as I can remember on the sheet, just to see what I need to concentrate on revising... that and some old spec CHM5 papers.
Reply 49
There's this question in Jan 2010:

8 (c) A chemical company has a waste tank of volume 25 000dm3. The tank is full of
phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus(V)
oxide to water in the tank.
A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 mol dm–3 sodium
hydroxide solution for complete reaction.
Calculate the mass, in kg, of phosphorus(V) oxide that must have been added to the
water in the waste tank.


This is the mark scheme:
Moles NaOH = 0.0212 × 0.5 = 0.0106
Moles of H3PO4 = 1/3 moles of NaOH (= 0.00353)
Moles of P in 25000 l = 0.00353 × 106 = 3.53× 103
Moles of P4O10 = 3.53 × 103/4
Mass of P4O10 = 3.53 × 103/4 × 284 = 0.251 × 106 g
= 251 kg

I understand all the of the steps except number 2. How do we know moles of H3PO4 is 1/3 NaOH?
Reply 50
Original post by masterhr1
There's this question in Jan 2010:

8 (c) A chemical company has a waste tank of volume 25 000dm3. The tank is full of
phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus(V)
oxide to water in the tank.
A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 mol dm–3 sodium
hydroxide solution for complete reaction.
Calculate the mass, in kg, of phosphorus(V) oxide that must have been added to the
water in the waste tank.


This is the mark scheme:
Moles NaOH = 0.0212 × 0.5 = 0.0106
Moles of H3PO4 = 1/3 moles of NaOH (= 0.00353)
Moles of P in 25000 l = 0.00353 × 106 = 3.53× 103
Moles of P4O10 = 3.53 × 103/4
Mass of P4O10 = 3.53 × 103/4 × 284 = 0.251 × 106 g
= 251 kg

I understand all the of the steps except number 2. How do we know moles of H3PO4 is 1/3 NaOH?


H3PO4 + 3NaOH -------> Na3PO4 + 3H2O

For complete neutralisation.
Reply 51
Original post by wibletg
H3PO4 + 3NaOH -------> Na3PO4 + 3H2O

For complete neutralisation.


Oh of course, thank you.
Reply 52
Original post by masterhr1
Oh of course, thank you.


No problem :tongue:
Reply 53
Does anyone have a transition metal colours grid with colours and formulas? I had an epic one but I lost it :frown:
If someone has one can they upload it please?
Reply 54
Not sure if I'm in the right thread and this sounds like a really simple question, but about practical skills, can anyone possible let me how you work out the concentration of a solution say of potassium iodide if you mix different volumes of it with different volumes of water? if that makes sense?
Original post by timoaf
Not sure if I'm in the right thread and this sounds like a really simple question, but about practical skills, can anyone possible let me how you work out the concentration of a solution say of potassium iodide if you mix different volumes of it with different volumes of water? if that makes sense?


If you're asking about the EMPA, you don't need to worry about that. :smile:
Original post by timoaf
Not sure if I'm in the right thread and this sounds like a really simple question, but about practical skills, can anyone possible let me how you work out the concentration of a solution say of potassium iodide if you mix different volumes of it with different volumes of water? if that makes sense?


You would find the number of moles of it in the original volume, then divide this number of moles by the new total volume of solution to find the new concentration.

Alternatively if it's a more simple case of ratio then just edit the number accordingly :
eg 25cm3 of 0.1 moldm-3 is made up to 100cm3 with water, it has therefore been diluted by a factor of 4 (from 25 to 100), so the new concentration is 0.1/4 = 0.025 moldm-3.

I hope that has helped and was what you were asking :smile:
Reply 57
anyone got the June 2011 mark scheme please? thanks in advance
Original post by rommy123
anyone got the June 2011 mark scheme please? thanks in advance


Here you go :smile:
Reply 59
Original post by clownfish
Here you go :smile:


Thankyouu :biggrin:

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