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# ocr a f325 revision thread watch

1. (Original post by patterson)
when is the exam? morning or afternoon?
Afternoon
2. (Original post by DudeBoy)
Ok so say you have 0.2 moles of HCl and 0.2 moles of NaOH
The equation is HCl+NaOH--->H2O + NaCl
So 0.2 moles of water is formed, you need to scale it up to form 1 mole of water.

So do 1/0.2 = 5 so you have to scale up the Q=mcT by 5 to form 1 mole
you're a lifesaver! thankyou so when don't you scale? because i dont normally i dont think!? sorry for all the questions!

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3. (Original post by rival_)
Hi guys my answer is 1:1 but I don't know if anyone can calculate anything different?
Ignore the my scribbles I had no idea at the time of what to do.
So first do 10^-pH to find [H+]
=2.2387*10^-5
Rearrange the Ka equation to make Ka/[H+]=[A-]/[AH]
So [A-]/[AH]=2.78/1
So you need 2.78 moles of [A-] for every 1 mole of [AH]

I think anyway
4. (Original post by georgiaaaxo)
you're a lifesaver! thankyou so when don't you scale? because i dont normally i dont think!? sorry for all the questions!

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It's ok This helps me revise too.

You always need to scale because by definition "The enthalphy change of neutralisation is the enthalpy change that occurs when an acid and a base react fully to form 1 mol of water"
So always scale to 1 mole
5. Does anyone find themselves overthinking/ over-complicating questions? I was doing a q just now and I couldn't figure out whether to use K or cetigrade in a Q=mct question, even though I've done this type of q a million times before :/ I always think ocr are trying to catch us out
6. (Original post by Myocardium)
Does anyone find themselves overthinking/ over-complicating questions? I was doing a q just now and I couldn't figure out whether to use K or cetigrade in a Q=mct question, even though I've done this type of q a million times before :/ I always think ocr are trying to catch us out
Remember, Q=mcT is degrees
^G=^H-T^S is in kelvin
7. (Original post by DudeBoy)
It's ok This helps me revise too.

You always need to scale because by definition "The enthalphy change of neutralisation is the enthalpy change that occurs when an acid and a base react fully to form 1 mol of water"
So always scale to 1 mole
I think when the mean scale, your meant to use the 1:1 ratio. so if 0.2 mols of HCL is reacted and 1:1 ratio of acid to water, the moles of water you must divide Q by is 0.2

Look atq2b of Jan 2013 for example
8. (Original post by DudeBoy)
First work out the moles of KMnO4.
cv/1000=n
So 0.02*23.45/1000 = 4.69*10^-3
Look at the overall all equation, 5 moles of H2O2 reacts with 2 moles of MnO4-
So (5/2)*(4.69*10^-3)
=0.011725 moles
Now this is the mole of H2O2 in the DILUTED solution, it asks for undiluted so
You have to times moles by a factor of 10.
=0.11725 mole of H2O2

Now it wants it in gdm-3 not, molcm-3
So the mr of H2O2 is 34.
Now I am really not sure where they get the 40 from the markscheme from tbh.... maybe some one could help?
You're nearly there just one more step..c=n/v, so divide ur number of moles by 25/1000 as that's the volume of h2o2..multiplying by 40 and dividing by 25/1000 is the same thing

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9. (Original post by Myocardium)
Does anyone find themselves overthinking/ over-complicating questions? I was doing a q just now and I couldn't figure out whether to use K or cetigrade in a Q=mct question, even though I've done this type of q a million times before :/ I always think ocr are trying to catch us out
it doesnt really matter, the temperature change of K is the same value as C, all there is, is a 273 difference
10. (Original post by needtosucceed=))
You're nearly there just one more step..c=n/v, so divide ur number of moles by 25/1000 as that's the volume of h2o2..multiplying by 40 and dividing by 25/1000 is the same thing

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See that's what I don't get, we work out no. of moles for 250cm3 so wouldn't be 4?
11. (Original post by DudeBoy)
See that's what I don't get, we work out no. of moles for 250cm3 so wouldn't be 4?
I think the original amount was 25 wasnt it? it was diluted to 250, then a sample of 25 was taken
12. Why is standard enthalpy change of formation, endothermic ?
13. (Original post by DudeBoy)
Remember, Q=mcT is degrees
^G=^H-T^S is in kelvin
Thanks! I'm going to stay calm and try not to confuse myself during the exam -_-

Btw what is the unit for G?
14. (Original post by Joey952)
when working out buffers do the units have to be in moles or conc
bump
15. hey guys what do we need to know about hexadentate ligand (EDTA)
16. In a neutral solution does [H+]=[OH-] ?
17. (Original post by Joey952)
bump
Use concentration
18. (Original post by otrivine)
Because if equal volumes = 1+1, double of the existing volume being added would be 1 (existing) +2 (added volume)

Idk, can someone refute or verify this? I could just be complicating things, I probably am.

is this true? for the dilute thing that the question was asked
equal volumes of water are added but conc stays the same so the overall volume doubles and if they're asking for conc in the overall volume you'll half the conc if that makes sense?
19. (Original post by Myocardium)
Thanks! I'm going to stay calm and try not to confuse myself during the exam -_-

Btw what is the unit for G?
kJmol^-1

If you can't remember work it out

^H = kJmole^-1
T=K
^S=(kJmole^-1)K^-1
T^S therefore cancels out the K
= kJmole^-1 - kJmole^-1
So ^G is kJmole^-1
20. (Original post by Myocardium)
In a neutral solution does [H+]=[OH-] ?
Yes

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