[kimmey]

(Original post by DoctorVertigo)
how did you get 1.45V? I got 1.17V by using the tables

i got 1.17V aswell

[mack94]

(Original post by DoctorVertigo)
I got 2.5 too! No idea where i went wrong!

If lots of us got 2.5 and 2.5 was a common mistake chances are we get most the marks

[gwengwen]

(Original post by DoctorVertigo)
how did you get 1.45V? I got 1.17V by using the tables

I got +1.17V as well

[DoctorVertigo]

(Original post by mack94)
If lots of us got 2.5 and 2.5 was a common mistake chances are we get most the marks

Whoop!

[amazon123456789]

does any1 hve the paper yet

[Oromis263]

(Original post by DoctorVertigo)
how did you get 1.45V? I got 1.17V by using the tables

I must have got it wrong then.

[DoctorVertigo]

My predictions: A* 86 A 79 B 71

[l.a.s]

(Original post by Oromis263)
I must have got it wrong then.

or the person with 1.17 used the wrong electrode potential for the methanoic acid. (the one with methanal in it and methanoic acid on the left)

[Stir fried Giblets]

(Original post by Oromis263)
This is a repost of my numbers, in case anyone wanted to see them (they were quite far back)

You think rounding it up to 92% is ok?

[walkbesideme]

(Original post by kimmey)
for the Kc one about when it said about increasing temp and pressure and why it would be hard to predict

i said

as temp increases Kc increases and equ shifts to the left because the forward reaction is exo. i then said that as pressure increases the terms on the bottom of the Kc equation increases so this means that the term on the top needs to increase to restore Kc so the equ would shift to the right. so it sort of opposes each other

this make sense to anyone ?

The forward reaction was exothermic therefore as the temperature increases the value of Kc decreases.

As such the increases in pressure favoured the forward reaction as there were fewer moles of gas on the right hand side but conversly an increase in temp favoured the backward reaction therefore because the effect and extent of the changes are not known it is not possible to understand at what position the equilibrium would lie as the effects oppose one another

[fudgesundae]

(Original post by Oromis263)
I must have got it wrong then.

I got 1.45.

There were two electrode potentials for methanoic acid. Can't remember which one was the right one though.

[kimmey]

(Original post by Oromis263)
I must have got it wrong then.

do you remember the first 2 pH calculations the one to do with pKa and the other one after (not the buffer one) ?

[Oromis263]

(Original post by l.a.s)
or the person with 1.17 used the wrong electrode potential for the methanoic acid. (the one with methanal in it and methanoic acid on the left)

I definitely used the one with methanoic acid on the right, but I couldn't argue with 3 against 1! I'm not really bothered now, I only need to get an A for A*, and just over half marks for the A.

On a more important note: No more chemistry for me! I'm free!

[_Browner_]

charge on the planitum complex people what did you get? 2- ?

[Oromis263]

(Original post by Stir fried Giblets)
You think rounding it up to 92% is ok?

Hello miss. I'm sure with all the method that is required to reach the final answer (which you must have done), you will get 4-5 for getting that. How did you find the rest of it?

[pete clark]

(Original post by fudgesundae)
But is a DMFC not a Direct Methanol Fuel Cell? Which is what they were talking about in the Q.

I believe they specified HCOOH, i may be mistaken. ah well, nothing we can do about it now!!

[walkbesideme]

(Original post by Oromis263)
My numerical answers:
-1046kJ/mol for Lattice Enthalpy

Kc = 0.947

pKa = 4.82
pH = 2.71
pH = 4.22

k = 0.0128 dm^3/mol s
-148 for delta G, although apparently, and I think I must have done, I used the wrong number for delta S (despite changing to kJ and using right temp. Oh well, was only 2 marks right?)

1.51V
1.45V

91.8%

I think most of my explain answers were ok, might drop a few marks for missing a couple of points here and there. Overall, an alright paper, in my opinion. There was nothing outrageous on there at least.

Anyone agree with my numbers?

Exactly the same as all my answers - I only assume this is a good thing

[05sykesd]

For the one where it asked about bubbling CO2 through an equilibrium mixture i put that the CO2 will react with the OH- ions on the right of the equation forming HCO3- meaning that OH- concentration decreases so the equilibrium shifts right to minimise the change and restore concentrations

Anyone else put this?

[RTJP]

(Original post by niaghez)
I dont think its that high because so many of us are now upset at the possibility of grade boundaries being hard so i think its safe to say for the majority of us it was a tricky paper so boundaries should be lower than that

I really really really hope you are right ! The paper wasn't especially difficult but it was tricky ! I am.hoping for similar boundaries to June 2011 so maybe about 72 for an A and then hopefully 76/77 for 126/150 which would get me an A* :') The thing that annoys me is that grade boundaries are always on the up but the papers arent getting any easier

[Jenniferann232]

(Original post by Bright)
Yes, I multiplied it by 4.. it was (I think) 5.1X10^-4 XMr
and then 0.6.../0.7... X100

I am such a fail at remembering properly, sorry.

How did you get 0.6..../.... I got 0.408 for my mass:
5x10^-4x (32+16+16+16) as they wanted you to get so3 %