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# Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

1. (Original post by haydyb123)
thanks man, I'll check it out!
In some past papers they asked for writing a ligand subst. without giving you any info. about charge of the complex!

Just had a quick look and all ligand subti. start off with +2

They re also eqilibrium

correct me if am wrong
2. (Original post by volvicstar)
do we need to know and use the term mole fraction?
I'm not sure, but it does no harm just to learn it anyway
3. did anyone else find the june 2010 disgusting??? the last few questions erhhhhh
and the q about ammonia in equilibrium?!?!??! i still dont kno after looking at the mark scheme how to work out the 4 mark q on it
4. (Original post by haydyb123)
thanks man, I'll check it out!
how do you get to the past papers?
5. (Original post by nestle)
did you find that on the specimen paper for 2 (d). the mark scheme was x10 out?? or is that just me?
i got 0.013 moldm^-3

and the mark scheme got 0.13?

cos its, [c6h5o-] = (1.3x10^-10 x 0.2) / 1.95x10^-9

which = 0.013

??
Yes! Yes, I also got 0.013

Silly specimen paper...
6. (Original post by 786girl)
did anyone else find the june 2010 disgusting??? the last few questions erhhhhh
and the q about ammonia in equilibrium?!?!??! i still dont kno after looking at the mark scheme how to work out the 4 mark q on it
it gives you the moles of the reactants, and the volume of the container, and you need concentrations for the expression of Kc so you need to divide the moles/volume = concentration

it gives you the stoichiometric equation, so you know its 1 mole of N2, and 3 moles of H2 to give 2 moles of NH3

7.2/6 = 1.2 mol dm3 (N2)
12/6 = 2 mol dm3 (H2)

Kc = [NH3]^2/ 1.2x2^3

rearrange it so that NH3 is on the left
so Kc x 1.2 x 2^3 = [NH3]^2
=0.786, but thats squared, so you need to square root it to get just [NH3]
which is 0.876 mol dm3
question asks for moles, so 0.876 x 6 = 5.26 moles
7. This is a question for anyone in the Heinemans textbook, page 157 (enthalpy of neutralisation) what does it mean by scaling the quantities to match the molar quantities needed to form one mole of water...
does it mean if you have something in excess only work the moles of the acid/base that isn't in excess

i.e.

NaOH + HCl -> NaCl + H20
0.5 0.25

Therefore in your equation do you put in 0.25

So if anyone could clear that up, would be awesome!
8. Hi, here some more recently LEGACY QP and MS, has similar questions to the F325 specification. However there may have some question which is not relevant to our specification.

2816-01 - Unifying Concepts in Chemistry questions - Kc, Ka, Equilibrium, rate- determining step, Orders, Buffers
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions - Born-Haber Cycles, Standard enthalpy, Transiton metal questions as well.

You can also find these question paper on the OCR website.
Attached Images
9. 2816-01Jan10.pdf (208.5 KB, 541 views)
10. 2815-06Jan10.pdf (239.9 KB, 276 views)
11. 2815-01Jan10.pdf (300.0 KB, 175 views)
12. 2816_01June10.pdf (202.7 KB, 1463 views)
13. 2815_06June10.pdf (195.6 KB, 863 views)
14. 2815_01June10.pdf (235.4 KB, 973 views)
15. Legacy MS Jan 2010 ALL.pdf (514.6 KB, 10715 views)
16. Attached Files
17. Legacy MS F325 June 2010.zip (211.7 KB, 116 views)
18. (Original post by jlcf)
Hi, here some more recently LEGACY QP and MS, has similar questions to the F325 specification. However there may have some question which is not relevant to our specification.

2816-01 - Unifying Concepts in Chemistry questions - Kc, Ka, Equilibrium, rate- determining step, Orders, Buffers
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions - Born-Haber Cycles, Standard enthalpy, Transiton metal questions as well.

You can also find these question paper on the OCR website.

19. Could someone explain question 6D of the jan 2011 paper?
20. (Original post by volvicstar)
do we need to know and use the term mole fraction?
pretty sure this isnt on our spec
21. Anyone know how to explain the answer from 6B part 2 of jan 2011

i got the 1st mark by predicting the Kw
but i am not sure what to do from there
i dont understand why the MS says square root of kw= h+
22. (Original post by voices1)
Anyone know how to explain the answer from 6B part 2 of jan 2011

i got the 1st mark by predicting the Kw
but i am not sure what to do from there
i dont understand why the MS says square root of kw= h+
Because: Kw = 1x10 -14 mol2 dm-6 and the root of getting there is Kw= [H+][OH-] so essentially [H+]^2

therefore in order to get from Kw to [H+] you need to square root Kw.
23. Is k, the rate constant, affected by pressure or catalyst? Bit confused
24. (Original post by ChubbyRain)
Is k, the rate constant, affected by pressure or catalyst? Bit confused
K, the rate constant, similarly to Kc, is only changed in itself by temperature..
If you think about it, K is multiplied by the concentrations in the rate equation:

Rate = K[A][B]

And so by changing the concentration, the value of the rate is changed by the change in value of [A] and [B], so there is no requirement to change the constant K.

Similarly, a change in pressure effectively changes the concentration, as the number of particles per unit space is changing.. So as in the above situation, K does not change..

K ONLY CHANGES WITH TEMPERATURE.
25. (Original post by ChubbyRain)
Is k, the rate constant, affected by pressure or catalyst? Bit confused
No, every K you'll ever meet is only affected by temperature
26. (Original post by wilsea05)
at the start of the question it says that propanoic is a weak acid, that's all you really have to go on. you should know that nitric is a strong acid too though
so the stronger acid will act as the acid, making the other act as a base.

7b) so 0.02345dm3 of 0.02mol dm3 MnO4 2- was required to neutralise
so 0.02345x0.02 = moles of MnO4 2- =0.000469moles
ratio of MnO4 2- to H2O2 is 2:5, so you times the moles of MnO4 2- by 2.5 to get 0.0011725 - thats in 25cm3, so x10 to get the moles in 250cm3 which is 0.011725moles
n=c x v so c=n/v so 0.011725/0.025= concentration of H2O2 (because you want it undiluted and the question says there was 25cm3 of undiluted H2O2, so you divide by 0.025 not 0.25)= 0.469mol dm3
but you want grams per dm3
so m=n x M so 0.469 x 34 = 15.946 g dm-3

7d
(Original post by haydyb123)
Hey, no worries:
First question: You know the second one is right because if you look at the reactants, you know HNO3 is a strong acid, and propanoic is a weak acid, therefore in relative you know HNO3 is the acid that will donate H+, and CH3CH2COOH will accept H+

Does that make sense?

Now secondly, hold let me find the paper...

Okay so: moles of KMnO4 = 0.000469
because of the molar rations moles of H2O2 = 2.5 x moles of MnO4 = 0.0011725
now this is in 25cm3 so in order to get the 250cm3 sample, times the moles by ten...
but this is still diluted as the solution was made up to 250cm3 from an original 25, so you can times your moles again by 10 giving you: 0.11725 moles x this by 34 to give you g per 250cm3 (3.9865) x this by 4 to get into g per dm3 (15.9 g dm3)

Now for the O2: Volume = Moles x 24
because you know from the equation on the previous page you know that O2 is only produced in the titration, and therefore the moles to use in this equation is your diluted H202 = 0.0011725 x 24 = 0.028 dm3

does that make sense?

Thanks loads guys!!!! I cant believe how simple the acid base pair one is and i got it wrong!
I understand the calculations too so thanks for that
x
27. (Original post by HEC14)
K, the rate constant, similarly to Kc, is only changed in itself by temperature..
If you think about it, K is multiplied by the concentrations in the rate equation:

Rate = K[A][B]

And so by changing the concentration, the value of the rate is changed by the change in value of [A] and [B], so there is no requirement to change the constant K.

Similarly, a change in pressure effectively changes the concentration, as the number of particles per unit space is changing.. So as in the above situation, K does not change..

K ONLY CHANGES WITH TEMPERATURE.
Coool Thanks for clearing that one up! Makes sense now
28. Do we need to know about the structure of a fuel cell (ie. bottom of page 84 in the CGP book) and how to draw it or do we just need to know the pros and cons to a fuel cell?
29. (Original post by Rogercbinboy)
Do we need to know about the structure of a fuel cell (ie. bottom of page 84 in the CGP book) and how to draw it or do we just need to know the pros and cons to a fuel cell?
I thought you just needed to know what occurs at the positive and negative electrode in terms of hydrogen and oxygens half equations, as well as pros and cons of fuel cell etc.

Another general question to all, about ionic charge, in the OCR textbook it says that ionic charge ncreases and ion radius decreases on pg 176 if anyone has the same book, why does the ionic radius decrease from Na+ -- Mg2+ --- Al3+, if you look at the periodic table their ionic radius increases?

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