ocr a f325 revision thread

I chose lactic acid for by removal of the other three:-
Benzoic acid is carcinogenic (Unit 4, benzene based compounds tend to be carcinogenic), pyruvic acid is a bitch to get at (unless we want whole factories full of metabolising cells!) and if we used acetic acid then we'd have yummy vinegar sweets. That was my thought process anyway.

did you get the right answer by any chance, can you show me your working.

I just looked at the mark scheme which was
rate = 1.0 × 10–12 /4 = 2.5 × 10–13 (mol dm–3 s–1), and worked backwards. I was really puzzled at first, but after re-reading it, it says the initial rate data in the table is for the formation of methanol, so you have to use that value

Experiment 1's initial rate of formation of HCHO is 1.0 × 10–12, and the stoichiometry of HCHO : O2 is 2 : 0.5 from the equation:

O3(g) + C2H4(g) <-> 2HCHO(g) + ½ O2(g)

-you have to use the initial rate of formation of methanol that's given, and divide it by 4 - ( 2 : 0.5 = 4 : 1 ), to get the initial rate of formation of oxygen.

Oh I see , trick question that , thank you

Is anyone going over the AS / F324 stuff, because F325 supposed to be synoptic?

Could somebody help me June 2012 question 3cii please?
I'm looking at the markscheme and it still isnt making much sense to me

Many thanks

Can someone help me with june 2011 question 8? I'm unsure as to how the copper precipitate is formed in the 2nd ionic equation as well as the equation for step 4. Thanks

Step 2 :
Because it says aq sodium carbonate, is added to neutralise any acid, you know that the CO3^2- must have reacted with H⁺ so:
2H⁺ + CO3^2- --> CO2 + H2O (This is the gas that is formed)

and the rest must've reacted with the Cu²⁺


Cu²⁺ + CO3 ^2- ->> CuCO₃ (This is the ppte that is formed)

For Step 4 you know that CuI and I2 must be formed, so what I did was I wrote that as my products :

____________ --> CuI + I2

(I think this reaction is in the spec)

Then I put the most likely reactants :

Cu2+ + I- --> CuI + I2

and then balanced it (not that it's not 3I-, as then the charges on each sides of the equation don't match.

2Cu2+ + 4I- --> 2CuI + I2

Ahh thanks! What do you mean by the charges balancing out? I originally thought it was 3I- before looking at the mark scheme.

In reactions the total charge on the left hand side has to match the right hand side.

For example the total charge on the left in this reaction is -3; and on the right it's 0, so it can't be right:

Cu2+ + 3I- --> CuI + I2
0 + (3*-1) =/= 0 +0

But on this the total charges on both sided balance to 0:

2Cu2+ + 4I- --> 2CuI + I2

2*2 + 4*-1 = 0 + 0
4 -4 0 + 0
=0 =0


EDIT: though of a better example :

In balancing the oxidation of I- to I2 you first write out the two species and balance the atoms:

2I- --> I2

because there's 2 (one) negative Iodide ions on the left and I2 has no charge, you need to make the equation balance in terms of charge -- you need to either add 2+ on the the left hand side to make both sides equal 0, or add 2- on the right to make both sides equal -2; you do the latter:

2I- --> I2 + 2e-

Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?

Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?

Guys now I know its late but can I ask what everyone got for the practical exams out of 40? Do you think the grade boundaries will be high again this year?

31 unfortunately, I always crack during practicals, not sure why. I hoping that's at least 40 UMS.

Grade boundaries are likely to be high, they are every year, make sure to do your best on the paper.

It's funny how 80% on coursework equates to ~70% UMS and 70% on this paper equates to 80% UMS. They really need to sort out the papers lol.