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Advanced Higher Chemistry 2014/2015

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Original post by Zain-A
lol i'm planning on taking this subject next year and reading just this last page of the thread I can tell its gonna be hard.


Its probably the easiest and most boring AH science dw

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Original post by Pennyarcade
Its probably the easiest and most boring AH science dw

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Defo the most boring aha, sometimes it can be so so dull

organic i'm looking at you
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spfl
6
What's the best revision material? Can't really even do past papers questions. This exam is gonna be fun!
Could someone be a legend and explain 2, 24 and 33 from this paper?
Original post by Pennyarcade
Could someone be a legend and explain 2, 24 and 33 from this paper?


2. So that makes the electron arrangement 1,8,3 then just remember your stuff from higher about full outer shell etc

24. You have K+/K on one side of the cell and Fe3+/Fe on the other side. Which one of these is being reduced and which one is being oxidised. Write out the redox equation and add together the delta E.

33. You should know what each of the things do - bromine adds across C=C bonds, LiAlH4 is a reducing agent, acidified dichromate is an oxidising agent, sodium will be involved in ion ****


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Original post by Asklepios


24. You have K+/K on one side of the cell and Fe3+/Fe on the other side. Which one of these is being reduced and which one is being oxidised. Write out the redox equation and add together the delta E.


What numbers do you use? I did this but I used 2.92 - 0.04

Original post by Asklepios

33. You should know what each of the things do - bromine adds across C=C bonds, LiAlH4 is a reducing agent, acidified dichromate is an oxidising agent, sodium will be involved in ion ****


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Ah ok, I drew the molecule wrong.
(edited 8 years ago)
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.A_C.
14
Original post by Pennyarcade
What numbers do you use? I did this but I used 2.92 - 0.04



Ah ok, I drew the molecule wrong.

Hello :smile: Someone a few pages previously asked PPA predictions, so does anyone have any :/? They seem to ask EDTA and Nickel titration, manganese in steel, propanone/iodine, cyclohexanol to cyclohexene, the ethyl benzoate/benzoic acid and certainly 2,4 dinitrophenylhydrazine derivative/unknown ketone thing quite A LOT. However there are the odd few such as as aspirin etc that pop up. I personally think the thermodynamic prdeiction one(with the gas syringe and oil bath) will come up since it's not appeared since like 2005. Any other suggestions tho? That would be much appreciated , but I know it's impossible to predict exactly... But it's worth a try to figure out the frequency of the PPA's and the possible ones that are more likely to come up than others. Thanks again! :smile:
Can someone please explain the last part of this question to me :frown: ImageUploadedByStudent Room1432460708.751042.jpg


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Original post by jen123456
The answer is B because the Cl has 7 outer electrons - it shares one with carbon and has 6 left so 3 lone pairs on the Cl!


Thank you!!!! :smile:


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ImageUploadedByStudent Room1432462678.574950.jpg

Could someone please help? :/


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Original post by studentt_11
ImageUploadedByStudent Room1432462678.574950.jpg

Could someone please help? :/


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There should be a page in the data books that tells you the wavelength of light each of those metals emit


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Original post by Pennyarcade
What numbers do you use? I did this but I used 2.92 - 0.04



Ah ok, I drew the molecule wrong.


Soz I was wrong before, it's not a voltaic cell but an electrolytic cell. That is I- is being oxidised and Fe3+ is being reduced.


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Original post by SerLorasTyrell
There should be a page in the data books that tells you the wavelength of light each of those metals emit


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Copper - 325 nm
Potassium - 405 nm
Barium - 554 nm
Lithium - 671 nm

So to get the highest frequency, I need to take the lowest wavelength.. This gives me the answer A, but it's supposed to be B??


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Original post by SerLorasTyrell
Can someone please explain the last part of this question to me :frown: ImageUploadedByStudent Room1432460708.751042.jpg


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Initially you should worked out that the average oxidation number of Cu was 2.33 (ie one Cu was in +3 and the other two were +2). So if you reduce the last Cu to +2, then overall the combined oxidation states of the metals becomes less positive. This means to balance this the combined oxidation states of all the O must become less negative.

I've explained the general concept of what's going on here, hopefully you can see how it works and get the numbers right.


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can someone please explain these questions to me? :smile: Screen Shot 2015-05-24 at 12.38.40.png
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Original post by jen123456
can someone please explain these questions to me? :smile: Screen Shot 2015-05-24 at 12.38.40.png
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For question 12, this is just a guess as I'm only doing Higher haha:

So the pH of the diluted solution is 2, which means that its concentration = 1*10-2 mol l-1 (maybe?)
N=cv=1*10-2 * 0.25 = 0.25*10-2 mol

Then c=n/v for the original solution = 0.25*10-2 / 0.005 = 5*10-1 mol l-1 = D

I have no idea if you can do it like this, hopefully someone more knowledgeable will help :smile:


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ah4p
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can someone please explain this question and their working to me?

from 2012 AH Chemistry the ANSWER is B
I usually find these questions easy but i keep getting an answer which isnt one of the options :frown:

thank you :biggrin:

2012 14. mc.PNG
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ah4p
16
could someone please xplain how to work this out?

thank you!

its from 2012 chemistry

2012 16.PNG
Original post by studentt_11
Copper - 325 nm
Potassium - 405 nm
Barium - 554 nm
Lithium - 671 nm

So to get the highest frequency, I need to take the lowest wavelength.. This gives me the answer A, but it's supposed to be B??


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If that's the case then I'm confused, are you sure the answer is absolutely B? I swear I done it by just looking at the data book values once, if it is B then I'm not sure sorry :/


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Original post by ah4p
could someone please xplain how to work this out?

thank you!

its from 2012 chemistry

2012 16.PNG


The latter 3 can all be measured from burning something so that's pretty standard experimental procedure, however the first one will take a few different steps to reach the final answer I believe


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