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

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4 years ago
#121
That is incorrect. It will be scaled down I'm sure. I got 42/45 and would say that's around 81 UMS. I pray I'm wrong please would someone clarify.
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#122
(Original post by Boom Boom Bam)
That is incorrect. It will be scaled down I'm sure. I got 42/45 and would say that's around 81 UMS. I pray I'm wrong please would someone clarify.
My teacher told me that 1 mark was 2 ums but I don't know for sure

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4 years ago
#123

Can I anyone help me with this please? I can do the normal calculations but when it gets more complicated I can never figure it out and the mark scheme doesn't help. Thanksssss it's June 2013

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4 years ago
#124
thst is definitely wrong because last year 38/45 was 72 UMS
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4 years ago
#125
(Original post by izipony)

Can I anyone help me with this please? I can do the normal calculations but when it gets more complicated I can never figure it out and the mark scheme doesn't help. Thanksssss it's June 2013

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The H+ concentration is equal to the acidity constant given in the question, because the salt/acid ratio is 1:1. From there the pH is equal to -log (H+). Hope this is a good explanation.
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4 years ago
#126
(Original post by davvv)
The H+ concentration is equal to the acidity constant given in the question, because the salt/acid ratio is 1:1. From there the pH is equal to -log (H+). Hope this is a good explanation.
And for the second part the concentration of the salt is doubled as a base is added, and the concentration of the acid is halved because half of it has reacted with the base to produce the salt. So the salt/acid ratio is 3:1, so the acidity constant is now 3 x H+.
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4 years ago
#127
(Original post by izipony)

Can I anyone help me with this please? I can do the normal calculations but when it gets more complicated I can never figure it out and the mark scheme doesn't help. Thanksssss it's June 2013

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The way I see it the ratio is 1:1 or 2:2. If half the methanoic acid is reacted with you now have 1:2 in terms of methanoic acid:salt, but that half that reacted became more salt so in reality it's 1:3 for methanoic acid: salt.

I hope that helps
1
4 years ago
#128
Could anyone explain how to do this question to me?
The mark scheme isn't clear and I can't figure out how to get the answer. :/
Thank you

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4 years ago
#129
How would you draw energy level diagram for January 2011 2d?
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4 years ago
#130
(Original post by thelizzister)
holy crap, i've never felt so under pressure, i have geography on the same day and then another exam the day after, if f335 is anything as bad as f334 there is literally no hope for me...
WE CAN DO IT :-) hopefully f335 might be a bit nicer as f334 was so hideous!!

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#131
(Original post by EggsterminateMe)
How would you draw energy level diagram for January 2011 2d?
It's an AS synoptic question, when I did it I drew two diagrams one for copper and one for arsenic where the electrons were falling back to ground state from different energy levels
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4 years ago
#132
When talking about difference in boiling points, sometimes the answer is there a more imb, stronger imb and sometimes both
For example Jan 2012 Q4
Oleic acid and Elaidic acid (they're cis trans isomers btw)
I don't understand why there would be stronger imb in elaidic acid
I'd have thought elaidic acid would just have more imb but that's not right
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4 years ago
#133
(Original post by Diamond Crafter)
X.
Thanks guys, hope we can all do well!
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4 years ago
#134
(Original post by Shantellea)
It's an AS synoptic question, when I did it I drew two diagrams one for copper and one for arsenic where the electrons were falling back to ground state from different energy levels
Ok that's fine. Exactly what i did. Mark scheme just confuses me. Thanks
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4 years ago
#135
(Original post by greencat55)
Could anyone explain how to do this question to me?
The mark scheme isn't clear and I can't figure out how to get the answer. :/
Thank you

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Calculate number of moles of both. Then do moles of acid-moles of alkali to get moles of remaining H+ in 30cm. Divide this by 30 then multiply by 1000.
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4 years ago
#136
(Original post by greencat55)
Could anyone explain how to do this question to me?
The mark scheme isn't clear and I can't figure out how to get the answer. :/
Thank you

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This gives you concentration of H+ ions then do -log of it.
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4 years ago
#137
From experience (Or by talking about lowest grade boundaries) What would you consider the hardest F335 paper? Going to test myself
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4 years ago
#138
(Original post by FudgeMonkey97)
The way I see it the ratio is 1:1 or 2:2. If half the methanoic acid is reacted with you now have 1:2 in terms of methanoic acid:salt, but that half that reacted became more salt so in reality it's 1:3 for methanoic acid: salt.

I hope that helps
(Original post by davvv)
And for the second part the concentration of the salt is doubled as a base is added, and the concentration of the acid is halved because half of it has reacted with the base to produce the salt. So the salt/acid ratio is 3:1, so the acidity constant is now 3 x H+.
thank you both
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4 years ago
#139
(Original post by jonesy123)
Calculate number of moles of both. Then do moles of acid-moles of alkali to get moles of remaining H+ in 30cm. Divide this by 30 then multiply by 1000.
Ohhhh, thank you! I got confused and thought that the number of moles of H+ was the concentration, so I didn't convert it to conc. :P
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4 years ago
#140
Is the value of Kw always 1x10^-14 ? Whilst working out strong bases ?
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