OCR B Salter's Chemistry by Design F335 - 15th June 2015 Watch

LordGaben
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#421
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#421
(Original post by chococup123)
Why is the alkyne 180? there is a triple bond between the two carbon atoms isn't there?
A triple bond is 1 area of electron density
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ioal
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#422
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#422
(Original post by Beasttty)
same 11.3 im 100% sure its right
yes i got this too
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JakeDaw
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#423
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#423
(Original post by Remedium)
Did you take the dilution into account? Dividing it by 0.055? I don't see how it can't be 10...


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Check my working out above
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MasterOfTheSwag
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#424
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#424
(Original post by Minecraft27)
Nvm. I give up.

I was confident I'd done well but after a visit here I appear to have got absolutely everything wrong. Looking forward to getting a D and not getting a place in university.
Sorry mate, I wasn't trying to dishearten you, just stating what I thought about the answer.
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username1082804
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#425
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#425
(Original post by Minecraft27)
The triple bond counts as 1 area of electron density. The hydrogen is the second. If there are 2 area of HED, it's 180°.
omg, yeah! I completely forgot about that :'(
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LordGaben
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#426
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#426
For the NMR structure question was Cl opposite the OH? (Or whatever halogen it was)
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kingcobra007
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#427
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#427
(Original post by JakeDaw)
Attachment 428939
Ah here is my working out for the last pH question, finally lets me upload it
nahh you use Kw/OH = H+ which is then -log
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JakeDaw
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#428
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#428
(Original post by kingcobra007)
nahh you use Kw/OH = H+ which is then -log
That's what I did. The 1x10-4 / 55x10-3 is the concentration of OH- . The Kw is then divided by it which gives 5.5x10-12


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Beasttty
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#429
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#429
(Original post by JakeDaw)
That's what I did. The 1x10-4 / 55x10-3 is the concentration of OH- . The Kw is then divided by it which gives 5.5x10-12


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yup same. people are mad that they're wrong
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jackkilbeyy
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#430
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#430
(Original post by JakeDaw)
That's what I did. The 1x10-4 / 55x10-3 is the concentration of OH- . The Kw is then divided by it which gives 5.5x10-12



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yup
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Millie0118
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#431
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#431
(Original post by king cobra)
2 hydroxy nitrile


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2-cyano-2-hyroxypropane?
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Solarburst
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#432
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#432
(Original post by MasterOfTheSwag)
For the name I put 2-hydroxy-2-methylpropanenitrile.

I thought that question was very unfair though. It's just not on the course.
Think I put this as well, deffo a methyl group in there
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The fish inspect
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#433
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(Original post by JakeDaw)
Attachment 428939
Ah here is my working out for the last pH question, finally lets me upload it
I'm pretty sure you don't have to work out the moles. These [ ] are the concentration. You were meant to use the question before it. Your supposed to do the Kw/conc of NaOH that will give you [H+] then your use ph=-log10(ans] and that gives you the ph. Then the answer minus the answer for the question before

That's 3 steps for 3 marks
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Millie0118
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#434
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#434
(Original post by kingcobra007)
nahh you use Kw/OH = H+ which is then -log
Name:  image.jpg
Views: 252
Size:  471.5 KBI used this with the different moles of butanoic acid and NaOH (salt) they gave us?
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hn123
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#435
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#435
(Original post by Millie0118)
2-cyano-2-hyroxypropane?
oh no I put 2-hydroxy-2-cyanopropane (((((
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ioal
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#436
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#436
(Original post by Millie0118)
Name:  image.jpg
Views: 252
Size:  471.5 KBI used this with the different moles of butanoic acid and NaOH (salt) they gave us?
Kw=[OH-][H+]
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BrokenS0ulz
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#437
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#437
I put 2-cyanopropan-2-ol. Does anybody know what the correct name is?
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JakeDaw
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#438
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#438
(Original post by The fish inspect)
I'm pretty sure you don't have to work out the moles. These [ ] are the concentration. You were meant to use the question before it. Your supposed to do the Kw/conc of NaOH that will give you [H+] then your use ph=-log10(ans] and that gives you the ph. Then the answer minus the answer for the question before

That's 3 steps for 3 marks
Concentration of NaOH is not equal to the concentration of OH- though because some of them react with the acid and are neutralised. Also you cannot minus pH results because it is a logarithmic scale


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nats927
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#439
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#439
Was there a question where u had to name something and amide was part of the answer?? Or am i bugging out
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WaiKon1337
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#440
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#440
I misread the labelling of the NMR question, so I labelled the graph with the group responsible for each peak (eg: I labelled the furthest right peak "OH" and put the no. protons in the environment). Would I get the marks?
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