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

0.167/ 55.5 = 3.009x10-3 moles of iron
0.487/ 324.6 =1.5x10-3 moles of X

Molar ration of Fe:X is 2:1

2Fe + ?Cl2 -----> X
So the compound X has Fe2 in it 324.6 - (55.5 x2)= 213.6
213.6 / 35.5 = 6.02 approx 6
So it is Fe2Cl6

I haven't done any of the old-spec papers this time. I did a lot of them for F324 and it didn't help to be honest, the old-spec papers are a different style and a lot more simple.

We only have two F325 papers (and a specimen) to work from, so I've been looking at all the stretch/challenge boxes in the Heinemann textbook and doing a few unstructured exercises from university textbooks. That will surely be a bigger help for me.

wenever i ave a titration question i got through the same stages
1. MOLES
look for wen they give u a conc and a volume, so in this qu they give u 0.02moldm3 of KMnO4 and the volume 23.45cm3.
Use n=cv and u shuld get 4.69x10-4 mols of KMn04

2.Ratios
By using the equation given u can compare the number of moles. So Mn04 2:5 H2O2.
so divide the answer we got above by 2, thats so u get 1 mole and then times by 5. 1.1725x10-3

1 thing to look out for in the volumes, the answer we've worked out is in 25cm3, so in 250cm3 its 10X as more, 1.1725x10-2. This is the moles of H2O2 in 250cm3.

3.Units
Now they want the answer in g/dm3. So i think about this in terms of units now. Where can i get grams from? mols=mass/Mr

So 1.1725x10-2 /34= 0.39865g
Right at the beginnin of the qu they tell u that u diluted the soultion in 25cm3.
So we ave grams and cm3, divide cm3 by 1000 u get answer in dm3.
25/1000=0.025DM3
Using the units again g/dm3, i see the / meanin divide.
FINALLY!!

0.39865/ 0.025= 15.97g/dm3




Really hope i didnt confuse u further

0.167/ 55.5 = 3.009x10-3 moles of iron
0.487/ 324.6 =1.5x10-3 moles of X

Molar ration of Fe:X is 2:1

2Fe + ?Cl2 -----> X
So the compound X has Fe2 in it 324.6 - (55.5 x2)= 213.6
213.6 / 35.5 = 6.02 approx 6
So it is Fe2Cl6

I'm just assuming 250cm is volume of entire buffer mixture?

0.25 mol dm3 of ethanoate
and 0.125 mol dm-3 ethanoic acid

[H+] = 1.7x10-5 x 0.125 = 0.0000085 -log = 5.07
0.25

wenever i ave a titration question i got through the same stages
1. MOLES
look for wen they give u a conc and a volume, so in this qu they give u 0.02moldm3 of KMnO4 and the volume 23.45cm3.
Use n=cv and u shuld get 4.69x10-4 mols of KMn04

2.Ratios
By using the equation given u can compare the number of moles. So Mn04 2:5 H2O2.
so divide the answer we got above by 2, thats so u get 1 mole and then times by 5. 1.1725x10-3

1 thing to look out for in the volumes, the answer we've worked out is in 25cm3, so in 250cm3 its 10X as more, 1.1725x10-2. This is the moles of H2O2 in 250cm3.

3.Units
Now they want the answer in g/dm3. So i think about this in terms of units now. Where can i get grams from? mols=mass/Mr

So 1.1725x10-2 /34= 0.39865g
Right at the beginnin of the qu they tell u that u diluted the soultion in 25cm3.
So we ave grams and cm3, divide cm3 by 1000 u get answer in dm3.
25/1000=0.025DM3
Using the units again g/dm3, i see the / meanin divide.
FINALLY!!

0.39865/ 0.025= 15.97g/dm3




Really hope i didnt confuse u further

in buffers calculations does the conc of salt=conc of conjgate base?

hey thnx!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
but can u explain what g/dm3 means?!

in buffers calculations does the conc of salt=conc of conjgate base?

OGHHH, YES, SAME!!! And now i am freaking out over this. I don't think i will ever get over yesterday. Never.

Ahh yes, I forgot that. I probably did get over 48, but I'm not resitting F324 so can't get the A*. Doesn't matter, I only need an A!

hey thnx!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
but can u explain what g/dm3 means?!