Aqa a2 chem5 19th june 2012
Chemistry exam discussion - share revision tips in preparation for GCSE, A Level and other chemistry exams and discuss how they went afterwards.
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Re: Aqa a2 chem5 19th june 2012Whoops, sorry I'm a bit sleepy!(Original post by rommy123)
hey, i think you`ve attached the wrong one, i asked for JUNE 2011 mark scheme
but thanks for the jan 2012 one -
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Re: Aqa a2 chem5 19th june 2012Jeez chem4 was so stupid, it was nothing like the previous exams(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
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!
I got 90% exactly and the grade boundary for that was 87/100!
Plus for chem5 I have a rare coloublindness so I really can't tell the differences between a lot of the complex colours which isn't helping me to remember them :/ good times. -
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Re: Aqa a2 chem5 19th june 2012lol dont blame you, thanks(Original post by clownfish)
Whoops, sorry I'm a bit sleepy!
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Re: Aqa a2 chem5 19th june 2012there's definitely enough time(Original post by Inspire12)
Hi guys, we have 5/6 weeks till this exam, do u think this will b enough time to revise everything and do papers and get a good grade (B/A)? -
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Re: Aqa a2 chem5 19th june 2012When you make it in the laboratory it will be green due to impurities. Makes sense seeing as it is the end point used to determine if you have an aldehyde or a keytone by oxidisng them with dichromate.(Original post by MuffinMonster)
Guys what colour is [Cr(H2O)6]3+?
Book says green on pg 227 (the nelson text book) and the table in the same book says ruby...
However if you make it without any impurities it's actually sort of violety in colour (or ruby). -
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Re: Aqa a2 chem5 19th june 2012Thanks for the help(Original post by wasting-time)
When you make it in the laboratory it will be green due to impurities. Makes sense seeing as it is the end point used to determine if you have an aldehyde or a keytone by oxidisng them with dichromate.
However if you make it without any impurities it's actually sort of violety in colour (or ruby).
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Re: Aqa a2 chem5 19th june 2012(Original post by wasting-time)
When you make it in the laboratory it will be green due to impurities. Makes sense seeing as it is the end point used to determine if you have an aldehyde or a keytone by oxidisng them with dichromate.
However if you make it without any impurities it's actually sort of violety in colour (or ruby).I believe they accept both.(Original post by MuffinMonster)
Guys what colour is [Cr(H2O)6]3+?
Book says green on pg 227 (the nelson text book) and the table in the same book says ruby...
My chemistry teacher emailed AQA for clarification and that's what she was told, at least
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Re: Aqa a2 chem5 19th june 2012basically use equilibria to explain it(Original post by nicolemonica)
Hi, really hope you guys can help me
Can anyone explain how the emf of a cell changes with a change in concentration of the ion/metals, I cannot get my head round it
say if you have:
Zn --> Zn2+ + 2e-
if you lower the conc. of the Zn2+, the equilibrium will shift to oppose this change by moving to the right and increasing the Zn2+ conc. So you will have an even higher conc. of e- making the cell more negative.Last edited by wasting-time; 13-05-2012 at 21:37. -
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Re: Aqa a2 chem5 19th june 2012The guy above is right - you've got to consider cells in terms of a pair of redox equilibria.
The EMF of a cell is the "pushing power" of the charge carriers as they go around the cell. In physics we see that emf (voltage) and current are directly proportional - i.e. the higher the EMF of the cell, the more charge carriers are passing through the cell per second (electrons).
A cell will involve...
Oxidation, eg. X(s) -><- X+(aq) + e-
and
Reduction, e.g. Y2+(aq) + 2e- -><- Y(s)
To look at changes in EMF it is convenient to consider the NEGATIVE electrode. This will be the electrode which contains the most negative E Standard Value - and is always written on the left hand side of a convential cell "diagram".
Here, oxidation is occuring, because this is where electrons are being released into the system. We have already established that the more electrons present, the higher the EMF of the cell. So, by Le Chatelier's Principle, if the concentration of X(s) decreases - which occurs over time - the position of equilibrium in the oxidation shifts to the Left Hand Side (in an attempt to oppose the change, and replace the X(s) which has been used up). This means it favours the REVERSE reaction and FEWER electrons are released into the cell. Fewer electrons = Lower current = Lower Voltage = Lower EMF as the concentration of the products decreases.
Hope this helped.
Last edited by MattRay94; 13-05-2012 at 22:09. -
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Re: Aqa a2 chem5 19th june 2012Just a few points to note - whilst it's true to say that fluoride and chloride ions won't reduce the Conc H2SO4, they will undergo standard acid-base reactions to produce hydrogen fluoride gas (which is so powerful it can etch glass) and HCl(g) which vaporises.(Original post by thegodofgod)
Fluoride and chloride ions won't reduce concentrated sulphuric acid.
Bromide ions reduce the sulphuric acid to sulphur dioxide. In the process, the bromide ions are oxidised to bromine.
Iodide ions reduce the sulphuric acid to a mixture of products including hydrogen sulphide. The iodide ions are oxidised to iodine.
Therefore, reducing ability of the halide ions increases as you go down the Group.
Don't forget that the normal acid-base reactions also occur for NaBr and NaI - so you will detect steamy fumes of HBr and HI in addition to the reduction products.
The sulphur in the H2SO4 (+6 oxidation state) will be reduced to SO2 (+4 oxidation state - a steamy gas in this concentration, although obviously colourless in the high volume of the atmosphere where it contributes to acid rain), yellow sulphur solid (0 oxidation state) and the H2S (-2 oxidation state) already mentioned by the previous poster.
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Re: Aqa a2 chem5 19th june 2012I'm trying not to be a complete idiot but I don't understand step3 3&4, could someone please explain these?(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?
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Re: Aqa a2 chem5 19th june 2012I think I may fail this =/ I don't know any of unit 5! Any advice on where to start? I need 13/120 UMS to get a B or 73/120 UMS to get an A overall. I know grade boundaries vary but roughly what grade is the latter of the two? A C grade?
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Re: Aqa a2 chem5 19th june 201296 UMS = A(Original post by craig12)
I think I may fail this =/ I don't know any of unit 5! Any advice on where to start? I need 13/120 UMS to get a B or 73/120 UMS to get an A overall. I know grade boundaries vary but roughly what grade is the latter of the two? A C grade?
84 UMS = B
72 UMS = C
So a very low C grade
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Re: Aqa a2 chem5 19th june 2012okay so:(Original post by illuminati786)
I'm trying not to be a complete idiot but I don't understand step3 3&4, could someone please explain these?
Cheers.
step 3 - you've worked out how many mols of H3PO4 (which is equal to the mols of P) you have in 25ml so to work out how many mols there are in the the original tank you need to multiply by 10^6 (as 25000dm3/25ml = 10^6).
Step 4 - now you know how many mols of P were in the original tank, as the P was originally added in the form of P4O10, then the mols of P4O10 originally added must be 1/4 times the mols of P (as each mol of P4O10 provided 4 mols of P). So you divide the mols fo P by 4 to get mols of P4O10.
I got 90% exactly and the grade boundary for that was 87/100!
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