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

Diamond Crafter
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#181
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#181
(Original post by JakeDaw)
But water does exist as both ice and liquid at the same temperature? On a cold day the top of a pond will freeze over but there will still be water underneath. In fact water exists as all three states at the same temperature, water vapour in the air, liquid as the pond, and solid as the frozen top of the pond (its even got a special name and occurs for other molecules as well - the triple point)
I'm not going to argue with that as I don't know enough on the subject of the "triple point". However the reason why water doesn't freeze in a pond is because the water at the top of the pond is in direct contact with the cold air which makes the top freeze however because ice is less dense than water it floats to the top and acts as a barrier of protection to the rest of the water underneath. This is why the top of the pond freezes but the rest of the water underneath is insulated enough (i.e. temperature is high enough) to still exist in it's liquid state.
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Diamond Crafter
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#182
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#182
(Original post by Turtlefushsia)
Sorry to burst in, but is phenylammine the same as aminobenzene?
Yes but aminobenzene is the "old" way of saying it. In exams they still allow it but they'd prefer you to use phenylamine - Quote from my teacher.
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JakeDaw
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#183
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(Original post by Turtlefushsia)
Sorry to burst in, but is phenylammine the same as aminobenzene?
Yes

(Original post by Diamond Crafter)
I'm not going to argue with that as I don't know enough on the subject of the "triple point". However the reason why water doesn't freeze in a pond is because the water at the top of the pond is in direct contact with the cold air which makes the top freeze however because ice is less dense than water it floats to the top and acts as a barrier of protection to the rest of the water underneath. This is why the top of the pond freezes but the rest of the water underneath is insulated enough (i.e. temperature is high enough) to still exist in it's liquid state.
Yes that is true and I probably chose a poor example, but if you imagine when the pond is just freezing over and the layer hasn't formed yet, the top of the pond will be at the same temperature but the ice crystals forming stay on the top of the water opposed to sinking, which is probably the concept the book is trying to get at
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Diamond Crafter
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#184
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(Original post by JakeDaw)
Yes that is true and I probably chose a poor example, but if you imagine when the pond is just freezing over and the layer hasn't formed yet, the top of the pond will be at the same temperature but the ice crystals forming stay on the top of the water opposed to sinking, which is probably the concept the book is trying to get at
Yes, I can see and agree with that. Well you learn something new and perhaps-not-even-relevant-to-your-imminent-exam everyday! Haha.
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JakeDaw
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#185
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#185
Well I may not have learnt anything from my revision today but at least I can now draw a hexagon really well
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LordGaben
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#186
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#186
Everyone ready?
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pinkgorilla
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#187
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This may seem like a silly question but how do you answer questions like these ? Can someone show me an example please ? Name:  ImageUploadedByStudent Room1434320147.862526.jpg
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pinkgorilla
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#188
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#188
(Original post by LordGaben)
Everyone ready?
I hope so
You ? XD


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greencat55
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#189
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(Original post by JakeDaw)
Well I may not have learnt anything from my revision today but at least I can now draw a hexagon really well
Haha! I think I've drawn about 50 hexagons in the last couple days and only about 2 have come out perfect :P I might have to frame them and put them on the wall

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imedico10
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#190
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#190
Can someone help me with question 4 d) (v) on january 2013 f335 please. I don't understand how the answer becomes 2.3? I thought it was 2.12
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LordGaben
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#191
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#191
(Original post by pinkgorilla)
I hope so
You ? XD


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Can only hope for the best. Good luck!
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MrPolish
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#192
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(Original post by greencat55)
Haha! I think I've drawn about 50 hexagons in the last couple days and only about 2 have come out perfect :P I might have to frame them and put them on the wall

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im resitting f335, and i still cant draw a perfect benzene
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JakeDaw
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#193
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(Original post by greencat55)
Haha! I think I've drawn about 50 hexagons in the last couple days and only about 2 have come out perfect :P I might have to frame them and put them on the wall

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I did the January 2012 past paper today and not a single question asked about aromatic synthesises and I know that if it is like that tomorrow I may have to cry... so much wasted time
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Diamond Crafter
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#194
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#194
Good luck everyone. May the odds be ever in our favour.:crossedf:
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Shostakovish
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#195
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#195
No. I am not ready. Nowhere close!!!
Well, good luck everyone.
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JakeDaw
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#196
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(Original post by imedico10)
Can someone help me with question 4 d) (v) on january 2013 f335 please. I don't understand how the answer becomes 2.3? I thought it was 2.12
You forgot to account for the 10cm3 of NaOH you just added which dilutes the solution and reduces the concentration of H+ ions
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hannnahcox
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#197
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#197
Good luck everyone, especially to people sitting 2 exams tomorrow !

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MrPolish
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#198
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#198
when a mark scheme says: (primary) amine
does that mean we HAVE to have the primary to gain the mark ?
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NFFC95
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#199
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(Original post by MrPolish)
when a mark scheme says: (primary) amine
does that mean we HAVE to have the primary to gain the mark ?
No, whenever there's something in brackets in the mark scheme, it means that the phrase/word is something that could be added to the answer, but isn't essential to gain the mark.
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ThePhoenixLament
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(Original post by imedico10)
Can someone help me with question 4 d) (v) on january 2013 f335 please. I don't understand how the answer becomes 2.3? I thought it was 2.12
Even with the mark scheme in front of me it took a good 20 minutes to work out!

Basically, as nitric is a strong acid, it fully dissociates. Therefore you have a 0.015 mol/dm3 concentration of H+ ions. However, the NaOH is a strong alkali so will neutralise the H+, but as you only have 10cm3 of NaOH versus the 20cm3 of HNO3, you can only have 10cm3 of acid left to provide H+ ions .
This means that you now have 10cm3 of 0.015 mol/dm3 solution of H+ ions.

To work out moles (conc x volume):
0.015 x (10/1000) = 1.5x10-4
-log(1.5x10-4) = 2.3

It was a very nasty question
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