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OCR Salters Chemistry F335 12th June 2013 Exam revision thread

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Right I think I've eventually got it, I reckons its due to the fact that we're told 1/3 of the solution is diluted, and the ratio of [salt]/[acid] HAS to be 2 (from reverse engineering the answer).

So my final calculating came to be:

(0.0045*0.1)/((2/3)*0.05*0.027) which works out to be 1/2 exactly. I think the trick of this question is that the concentration of the acid (HA) has been partially diluted. :smile:

Hi, sorry if i'm being really stupid but could you walk me through how you got the answer. I've been trying to figure it out for an hour. I thought to work out the conc of H+ that it was ka * [HA]/[A-]

but in your calculation you have the salt being divided by the acid, and i'm confused why. Sorry if i'm being thick, i've been staring at it too long and my brain has turned to mush :colondollar:

Reply 181

Tikara

Original post by puddinboy

Can someone please explain to me how to work out how many hydrogen environments there are in a compound in NMR?

When explaining, examples would be most appreciated.

Thank you

Posted from TSR Mobile

there are two peaks here so two different Hydrogen environments - you can also tell it's low res cause there are no little peaks directly next to the main ones.

So as theres two different environment this rules out the possibility of it being Compound D cause that has three (it's important to look at the everything the C is bonded to).

You can actually work out which one it is without the i.r. spectrum but I suppose you use that to make sure .

(edited 10 years ago)

Reply 182

totw

Hi, could anyone help me with a four marker on the June 2011 paper? It is question 5d, I really can't work it out :/

Thanks

Reply 183

super121

Yea, I think the range is from 104-109 (or something like that anyway) but they defo accept 109 :smile:

Posted from TSR Mobile

Reply 184

calmpeach

Original post by totw

Hi, could anyone help me with a four marker on the June 2011 paper? It is question 5d, I really can't work it out :/

Thanks

here you go :smile:

hope the step by step helps you :smile:

http://i41.tinypic.com/dcu4ph.jpg

yes

Original post by calmpeach

here you go :smile:

hope the step by step helps you :smile:

http://i41.tinypic.com/dcu4ph.jpg

I didn't even ask this question and it was still helpful to me haha, thanks!

Reply 187

totw

Original post by calmpeach

here you go :smile:

hope the step by step helps you :smile:

http://i41.tinypic.com/dcu4ph.jpg

Thanks a lot, really clear :smile:

Good luck with the exam next week :wink:

This was posted from The Student Room's iPhone/iPad App

Reply 188

desiboy20395

hi

Just wondering, what do I have to get to achive A* overall? Does it mean i need to get 480 UMS with 90% UMS average over the 3 A2 Units or only just 2 of the A2 Units?

Reply 189

hayleyvictoria

Hello all

Does anyone have some condensed chemical storylines notes? I only ask because I remember a really helpful person doing them for F334 :smile:

Good luck
Xxx

Reply 190

username970812

Hey there, I got a question from Jan 2012 F335 It's question 2b that im stuck on.

Its a question on diborane. Asking about its 'three centre bonds'.

To my knowledge this was never covered or taught in any respect in my college.

So are we meant to know about '(three?) centre bonds'? Or is the question just for context and it links to other chemistry knowledge that I'm just not making the link to?

Question:
The structure of diborane can be represented as (picture)

H         H        H
   -    =    =   -
     B         B
   -    =    =   -
H         H        H

the - are straight lines while the = are meant to be dotted lines. showing each bond.

The B---H---B arrangement is unusual and is known as a 'three centre bond'.
(i) How many electrons are present in each three centre bond?
(ii) a Dot Cross diagram can not be used to describe the bonding in diborane. Suggest why chemist continue to use it.

I get part (ii) [ its the 'norm' - accepted and established method to show bonding]
Just part (i)

(edited 10 years ago)

Reply 191

Salmonidae

Original post by Kirity

H         H        H
   -    =    =   -
     B         B
   -    =    =   -
H         H        H

So the answer is 2 electrons per bond and the best way to think about it is purely logically. Between 2 hydrogens each with one electrons and each of the Boron's one last electron that is free to bond there are four electrons available for these three centre bonds. Therefore it has to be 2 each!

I had never come across this before doing that paper the other day either!

S

Reply 192

Tikara

Guysss can I get some help with nmr spectroscopy? The mark scheme says neither of the peaks have splitting :frown:

Can you not have splitting when the H is bonded to an O ? or when there is in O next to the C?

Reply 193

Tikara

Original post by Salmonidae

whaaat that's what I thought but then I thought if there are four of these bonds then there must be 1/4 of all the electrons so 1 arghhhhhh oh wait, haha silly me
a three centre bond is B----H-----B and not just B----H which I was trying to work out xD

Reply 194

super121

Original post by Tikara

Guysss can I get some help with nmr spectroscopy? The mark scheme says neither of the peaks have splitting :frown:

Can you not have splitting when the H is bonded to an O ? or when there is in O next to the C?

There is no splitting of hydrogens in the OH and NH bonds or of the hydrogens on the next door carbons

Reply 195

Salmonidae

Original post by super121

There is no splitting of hydrogens in the OH and NH bonds or of the hydrogens on the next door carbons

Sorry misclicked at derepped you :frown:

This is correct, worth remembering splitting only occurs for hydrogens on adjacent CARBONS, not any other element.

Reply 196

Tikara

Original post by super121

There is no splitting of hydrogens in the OH and NH bonds or of the hydrogens on the next door carbons

thanks very much thats cleared up alost - I couldn't really seem to find any of that in the chemical ideas or rev book

Reply 197

Tikara

Original post by Salmonidae

Sorry misclicked at derepped you :frown:

This is correct, worth remembering splitting only occurs for hydrogens on adjacent CARBONS, not any other element.

lmao- yeah I was just about to ask whether I should assume if it's not a carbon adjacent then there's no splitting so thanks for that :smile:

(edited 10 years ago)

Reply 198

super121

Original post by Salmonidae

Sorry misclicked at derepped you :frown:

This is correct, worth remembering splitting only occurs for hydrogens on adjacent CARBONS, not any other element.

It's ok

Original post by Tikara

thanks very much thats cleared up alost - I couldn't really seem to find any of that in the chemical ideas or rev book

Welcome

Yea, salter's just expect us to know some things :rolleyes:

Reply 199

Tikara

omg I can't ever believe how a solid buffer question gets 1 mark and no indication in the mark scheme whatsoever of how to properly do it - must I be doing them wrong or something?

can anyone help on Jan 12 question 5 fii ? It's another annoying proportion thing but I got the previous bit right (somehow) but can't get 2.73 for the answer