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Chemistry: Post your doubts here!

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Please please can somebody help me with this question of bond angles!
Please explain this because bond angles are so frustrating! !Screenshot_2015-10-27-19-37-59.png
 
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View attachment 57308plz explain this!! thnx in advance!

This takes a bit of visualization. Imagine viewing the molecule from the side, the blue parts are the 3 front groups, the red parts are the 3 back group.

Fix your front COOH at 12 o'clock position, fix your back COOH at 6o'clock position.
Change the positions of H and OH groups to get unique structures.

Screen Shot 2015-10-28 at 9.41.36 AM.png
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_43.pdf

I have a few problems with question 6.
First one is 6 ei) How many different dipeptides is it possible to synthesise, each containing two of the three amino acids alanine, serine and lysine?
I first thought 6 (which is correct) but then i looked at the structure of lysine. it has two NH2 groups so if peptide bonds can form at either NH2 wouldn't that make it 12?? Wouldn't joining the NH2 joined to (CH2)4 would form a diff molecule if joined to the NH2 that is from the CH???

Second one is f ii) Which of the structures G, H or J is identical to structure F?
I thought it was G. But its J. I have no idea why. No matter how you flip J it doesn't look like F.

EDIT: i dont get f iii) either, why cant i just switch the positions of NH2 and COOH. that looks different

Please help and thanks in advance

A pretty interesting analysis. I would say a peptide bond is when the N-terminus forms a bond with the C-terminus of the amino acid (the N-terminus and C-terminus should be both connected to the same carbon).

800px-Peptidformationball.svg.png


The side chain of (CH2)4 NH2 is not considered the N-terminus.

fii) Imagine looking at the molecules from the right. The black groups are the groups at the end nearer to you and the red groups are at the end further away from you.

Screen Shot 2015-10-28 at 12.49.01 PM.png

fiii) Not too sure what your structure looks like. Perhaps you can draw them like I did in the picture above, then we can guide you from there.
 
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upload_2015-10-28_14-21-23.png
How to do part b?
I understand that the values are 0=not dependant 1=proportional and 2=sqaured etc...
But I don't understand how to determine this when there are 3 different steps for 1 reaction... please explain

The answers are
1 1 0
1 1 1
1 2 2
 
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View attachment 57322
How to do part b?
I understand that the values are 0=not dependant 1=proportional and 2=sqaured etc...
But I don't understand how to determine this when there are 3 different steps for 1 reaction... please explain

The answers are
1 1 0
1 1 1
1 2 2

If reaction 1 is slowest
Rate determining step is based on reaction 1: H2O2 + I- --> IO- + H2O
The powers of the rate equation is based on the stoichiometric ratio of the reactants , so its [H2O2]^1 and [I-]^1

If reaction 2 is slowest, we combine reaction 2 with the reactions that occur before it (i.e. combine with reaction 1), resulting in
H2O2 + I- + H+ --> H2O + HOI
Again, the powers of the rate equation is based on the stoichiometric ratio of the reactants , so its [H2O2]^1 , [I-]^1 and [H+]^1

If reaction 3 is slowest, we combine reaction 3 with the reactions that occur before it (i.e. combine with reactions 1 and 2), resulting in
H2O2 + 2I- + 2H+ --> I2 + 2H2O
Again, the powers of the rate equation is based on the stoichiometric ratio of the reactants , so its [H2O2]^1 , [I-]^2 and [H+]^2
 
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upload_2015-10-28_19-17-24.png
Why part iii no change?
Wouldn't the equlibrium shift to the left so the potential will decrease so wouldn't the Ecell increase?
 
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upload_2015-10-28_19-23-0.png
mark scheme says concentration of fe3+ is 0.2 but question is showing 0.1... how is it 0.2??
Also for Ka don't you multiply Ka by the concentration of the solution and not just Fe3+ ions??
 
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If anyon . ANYONE has solved w/14 33 could you please please please upload it here? I want to see how the graph was drawn in q1.
 
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