The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

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

Scroll to see replies

Original post by CoventryCity
How many raw marks would I need for 131 UMS going off past papers?


Similar position to me then, I need roughly 130 for an A overall... it's going to be tough one!
Original post by CoventryCity
How many raw marks would I need for 131 UMS going off past papers?


according to Jan 11...about 67 (80+ was full UMS)

according to June 10...just over 71 (90+ was full UMS)
Reply 1122
Avatar for BA1
BA1
6
Original post by haydyb123
Similar position to me then, I need roughly 130 for an A overall... it's going to be tough one!


Good luck with that! That's quite a lot of pressure!
Original post by raj16

Original post by raj16
surely all the half cells involve ions?
what about the Iodine + thiosulphate, and manganate + hydrogen peroixde??


Yeah, all half cells involve ions but some might be:
Cu^2+ + 2e- --> Cu
And Cu exists as a solid, so you use a copper rod in one of the half cells.
I suggest you learn those equations too.
Original post by haydyb123
That's where your cheeky H+ come in handy :wink:

So you know from oxidation numbers Br is +5 and is going to gain 5 electron as it is being reduced to form Br2

So therefore you need 5 electrons provided by Br- so

BrO3^- + 5Br^- + 6H^+ -> 3Br2 + 3 H2O

Charges balances because charges on the left cancel out :smile:
do you see now?


I think so... But couldn't you just get that from oxidation numbers?
Br is reduced from +5 to 0 so you need 5 electrons
Br is also oxidised from -1 to 0, which gives up 1 electron

Electrons must balance, so you need 5Br- and then balance the rest of the equation from there.

Or have I picked the wrong example to show the importance of balancing charges? XD
(edited 12 years ago)
Original post by haydyb123
Similar position to me then, I need roughly 130 for an A overall... it's going to be tough one!


Mines for an A* :eek:
I think I need around 100 UMS for an A
Reply 1126
Avatar for BA1
BA1
6
Original post by sportycricketer
What are you doing to revise?? I've done all old past papers and the new ones as well :s-smilie: read through my notes a couple of times as well


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.
Original post by BA1
I can't get an A*, I had 78/90 in F324 (which is just under 90%) and I'm not bothering to resit it as it clashes with my holiday dates end of next week :P I also did badly on the A2 practicals so had less than 90% in total for them. I just need a grade A for university so I'm not bothering to try and get the A*.

PS: That's why you should try and do well in the AS modules, to take the pressure off A2! I'm walking into an A2 maths exam next week needing 23/100 raw marks for a grade A, or 90/100 for an A* lmao!


You can still get an A* provided that your practical ums is more than 48
Original post by haydyb123
Hey I have a question but don't have the mark scheme for it I have the correct answer but I have no idea how to get there using maths rather than just working it out trial and error:

a 167 mg sample of iron reacts with a stream of dry chlorine to form 487mg of solid X the molar mass of Solid X was determned to be 324.6 g mol-1

Calculate the molecular formula of X


I did this question the other day: the answer is Fe2Cl6 :smile: the MS suggested using mole ratios.
Reply 1129
Avatar for BA1
BA1
6
Original post by loki276
You can still get an A* provided that your practical ums is more than 48


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!
Reply 1130
Avatar for raj16
raj16
Original post by Medifield

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 x 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 :biggrin:
(edited 12 years ago)
Original post by student777
I think so... But couldn't you just get that from oxidation numbers?
Br is reduced from +5 to 0 so you need 5 electrons
Br is also oxidised from -1 to 0, which gives up 1 electron

Electrons must balance, so you need 5Br- and then balance the rest of the equation from there.

Or have I picked the wrong example to show the importance of balancing charges? XD


It's simply a way of checking you answer essentially, if your charges don't balance either side you know your equation can't be correct because Reduction = Oxidation
Ok i think i know every friking colour regarding transition metals etc in the friking book now :sigh:. Issue for me cus' i have the worst memory when it comes to things like this. If they throw one which is not in the book i'll have a fit. How is everyone else revising right now?
Reply 1133
Avatar for tripodd
tripodd
Can anyone help me with this Q?

Calculate the pH of the buffer solution formed at 298K when 0.125mol of sodium ethanoate is dissolved in 250cm of a 1.00 mol dm solution of ethanoic acid. Ka = 1.70x10^-5

nvm problem solved
(edited 12 years ago)
Original post by student777
I did this question the other day: the answer is Fe2Cl6 :smile: the MS suggested using mole ratios.


Yeah I got the answer but I just thought FeCl3 is a molecule I know, add it's Mr up and what do you know it's half of the given mr so I doubled it. I'm sure there must be a way of doing it properly for 3 marks?
Original post by BA1

Original post by BA1
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!


If its over 48 than as long as you get more than 138 + however many marks you were away from 54 in the practical to get an A*
Reply 1136
Avatar for raj16
raj16
Original post by Medifield
I don't understand how to work out the question at the end of the OCR unit 5 exam June 2010. 7b :frown: can someone please help?



i ave worked it out in the above comment
Original post by haydyb123
Hey I have a question but don't have the mark scheme for it I have the correct answer but I have no idea how to get there using maths rather than just working it out trial and error:

a 167 mg sample of iron reacts with a stream of dry chlorine to form 487mg of solid X the molar mass of Solid X was determned to be 324.6 g mol-1

Calculate the molecular formula of X

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
Original post by tripodd
Can anyone help me with this Q?

Calculate the pH of the buffer solution formed at 298K when 0.125mol of sodium ethanoate is dissolved in 250cm of a 1.00 mol dm solution of ethanoic acid. Ka = 1.70x10^-5


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
Original post by haydyb123
Yeah I got the answer but I just thought FeCl3 is a molecule I know, add it's Mr up and what do you know it's half of the given mr so I doubled it. I'm sure there must be a way of doing it properly for 3 marks?


I think I did the same thing, just trying out different numbers on the calc. As long as you got the answer you got 2 of the 3 marks. And it's old spec so doesn't really matter.

Oh, and thanks for your help on the charges thing :biggrin:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you