OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

I guess because Metal + Acid makes a Metal Salt of the Acid, you would have to reduce the amount of acid used as the metal is reacting with it?

Question 6a for the June 15 paper http://stsmith.co.uk/data/documents/F325-01-QP-Jun15.pdf

How the F325 am I supposed to know which is more negative?

The answer is ECD I think.

Look at the first cell. The right electrode is the cathode, so that means E is more negative than C.

Look at the second electrode. The left electrode is the cathode, so that means C is more negative than D.

Therefore, E is most negative, followed by C, followed by D.

If it's in the blue boxes, knowing OCR it's bound to be a 6 marker. We're screwed.

Can I get some PogChamps in the tsr chat for another U grade student 4Head.

I guess because Metal + Acid makes a Metal Salt of the Acid, you would have to reduce the amount of acid used as the metal is reacting with it?

Question 6a for the June 15 paper http://stsmith.co.uk/data/documents/F325-01-QP-Jun15.pdf

How the F325 am I supposed to know which is more negative?

lol I got like 2/3 marks out of what was it 5? That Q is tough cause I think you also got to x2 the 0.25.

But it's a metal and should be added to RHS. I think cause it's a ion or something. I just took it away from both sides haha. **** this exam

I am retaking this exam too, what did you get last year?

Oh so whatever is the side with the negative is the most negative, wow when I put it that way it looks really obvious haha thanks.

I got ABBB for AS CCD for A2. I had to go hopsital last year for medical reasosn and didnt study for the last month before exam(came back 1 week before). so I had good reason to do **** in my subjects but this year I aced F324 which I had U in last year but I spent too much time revising that cause of how bad the last year exam frightened me. I gave up at that point and got U for F325. I've done the same thing for bio A in all the units expect the last one U and this year i've left it too late haha.

Last year I had an excuse this year I don't. I think I will get a C in this or higher though. about 50 marks are easy enough as long as you get the questions. getting to a B won't be too hard too.

Wbu?

I got CDD, last year for A2, not really sure why, lack of ability probably and erm yh retaking my other exams however Chemistry I did pretty horifically in both units but a U in F324, I think I did well in it this year but the problem is I still think I'd need like a really high B or most likely an A but I think it's not looking likely at this point.

We are both in the same situation then haha . Exactly the same last year in C and similar grades CCD is not much better haha. So what do you need for uni overall?

I think I got an A for AS chem(retook unit 1) and at lowest B for unit 4 and A in both course works and probably can get a C/D for an overall B.

Hell I got D last year with 2 Us in A2. So I got loads of marks but my *****y planning made me not revise this for like a month and I forgot this all aswell as bio haha. This is okay though not as hard to recall due to quick practise.

lol I got like 2/3 marks out of what was it 5? That Q is tough cause I think you also got to x2 the 0.25.

But it's a metal and should be added to RHS. I think cause it's a ion or something. I just took it away from both sides haha. **** this exam

Iron starts as Iron (III) Oxide Therefore its formula is likely Fe2O3 and when reacted with a solution containing hydroxide ions forms FeO4 2- Where the oxidation number of the iron is +6.
So a change from +3 to +6 means it is being oxidised.

To start with you have to balance atoms

Fe2O3 + Cl2 + OH- ---> FeO4 2- + 2Cl-
You need 2 iron atoms on the right hand side making it now
Fe2O3 +OH- + Cl2 --> 2FeO4 2- + 2Cl-
Now there are too many oxygens on the right hand side so you have to increase the number of OH-
Fe2O3 + 5OH- +Cl2 --> 2FeO4 2- +2Cl-
As there are more hydrogens on the left than right you add water molecules to the right hand side. However each water molecule is H2O so you have to double the amount of OH-
Fe2O3 +10OH- +Cl2 --> 2FeO4 2- +2Cl- + 5H2O

Now all the atoms are balanced you have to balance charges.
The charge of the left hand side is 10- and of the right hand side is 6-
If you increase the number of Cl- to 6Cl- you will have balanced charges. Then change the number of Cl2 so it balances. Overall you will get
Fe2O3 + 10OH- + 3Cl2 --> 2FeO4 2- + 6Cl- +5H2O

Sup, can anyone solve this? The corrects answer is 4.22 but I have no idea how. Thank you!

Not too sure if you guys need the help... Just think of it as a buffer solution. The magnesium reacts with the acid to form the salt of the acid, thereby increasing the concentration of the conjugate base (which is provided for by the salt). The molar ratio is 2:1 so we have to add the twice the number of moles of magnesium to the conjugate base and minus twice the number of mg moles from the acid. Then you can work out the concentration.

Good practice of this is with buffers and how to calculate changing ph upon addition of naoh/hcl.