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ocr a f325 revision thread

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Reply 2700

otrivine

Original post by DavidH20

Do you mean V₂O₅?

Yes so lets say I DID not know that formula.

how can I work out there are 5 oxygens?

Reply 2701

Dr00n

Can anyone be unbelievable great and post the Jan 2013 paper again, i tried the search bar but i couldn't find it

Reply 2702

kevloui

Original post by otrivine

Yes I was looking for complex ions

my turn

dammnn haha, I knew that.
Explain what observations you will see when [Co(H20)6]2+ is reacted with 4Cl-

Reply 2703

alltimeemilyxo

does anyone have any predications on what will come up? so unprepared and freaking out

Reply 2704

DudeBoy

Original post by Dr00n

Can anyone be unbelievable great and post the Jan 2013 paper again, i tried the search bar but i couldn't find it

Done

F325 Jan 2013 MS.pdf577.9KB

F325 Jan 2013 QP.pdf286.6KB

Reply 2705

bilala2

hey, can sum1 please explain to me how to do the last question on the jan 2013 f325 paper.

Reply 2706

kuku2013

how is a buffer solution formed?

Reply 2707

meg0001

How to you work out the rate determing step? the worked example in the book is really confusing :s-smilie:

Reply 2708

LosMutos

Original post by Dr00n

Can anyone be unbelievable great and post the Jan 2013 paper again, i tried the search bar but i couldn't find it

http://www.thestudentroom.co.uk/showthread.php?p=42553934&highlight=f325%20jan13

Reply 2709

otrivine

Original post by kevloui

dammnn haha, I knew that.
Explain what observations you will see when [Co(H20)6]2+ is reacted with 4Cl-

Pink solution ---> blue solution

Reply 2710

kevloui

Original post by kuku2013

how is a buffer solution formed?

With the acid and the acid salt (conjugate base)

Reply 2711

User944645

Original post by otrivine

Yes so lets say I DID not know that formula.

how can I work out there are 5 oxygens?

You know the oxidation state of vanadium is +5. The oxidation state of oxygen is always -2 (in the examples we'll come across), so to balance out the oxidation numbers in the compound overall 2 vanadium atoms and five oxygen ones are required.

Reply 2712

kevloui

Original post by otrivine

Pink solution ---> blue solution

nicee, my turn, but more energy related

Reply 2713

Pride

Original post by Satta101

Can anyone help me with this please??!

ok, you gave me the answers. You don't get 2.38x10^-3 when you put in the numbers, that's out by a factor of 10. If you want to try, maybe I'm missing something silly. But that screws up the rest of my calculation really.

Reply 2714

nothepreacher

I have got this tricky question i am finding hard to get my head around. Please help!
dichromate ions ad managnese ions question at the end of the paper.
http://www.mediafire.com/view/x3yztd...2003_jun_w.pdf
I'm getting the oxidation state of mn after oxidation to be +6 beacuse
3 X 4H+ +mn2+ +so4 2- ----> mn^n+ + h2o +so2 +2e. To balance charge on both sides oxidation state of Mn must be +6. But in the mark scheme it s +4

Reply 2715

alltimeemilyxo

Original post by kuku2013

how is a buffer solution formed?

mixing a weak acid and its salt with a strong base?

Reply 2716

JimmyA*

Original post by otrivine

After further treatment, the water could be used for drinking. In the drinking water produced, the OH– concentration was 100 times greater than the H+ concentration.
What was the pH of this drinking water?

why is it sometimes we can say Kw= [H+]^2 ?

Kw= [H+] * 100[H+]
This is because the concentration of OH- is 100 times more
kw=100[H+]^2
(1*10^-14)/100= [H+]^2
1*10^-16 = [H+]^2
1*10^-8= [H+]
-log(1*10^-8)= 8

Reply 2717

Dr00n

Original post by DudeBoy

Done

you're unbelievable great, it seems ive repped you too much to do it again, so thank you

Reply 2718

DudeBoy

Original post by Dr00n

you're unbelievable great, it seems ive repped you too much to do it again, so thank you

It's all good mate, If you get stuck message me :smile:

gotta go for about 1 hr

Reply 2719

kuku2013

Original post by kevloui

With the acid and the acid salt (conjugate base)

thanks, do we have to show the dissociation equation for the salt of the weak acid or just the weak acid. e.g. CH3COOH ==> H+ + CH3COO- AND CH3COO-Na+ ==> Na+ + CH3COO- e.