Chemistry: Post your doubts here!

[Metanoia]

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.

 

[Metanoia]

Please please can somebody help me with this question of bond angles!
Please explain this because bond angles are so frustrating! !View attachment 57319

1) 4 bond pairs, 0 lone pairs. Tetrahedral
2) 3 bond pairs, 0 lone pairs. Trigonal planar
3) 2 bond pairs, 0 lone pairs. Bent.

 

[Metanoia]

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).

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.



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.

 

[darks]

any idea what ions coming in p33.... which is on 29th?

 

[nehaoscar]

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

 

[Metanoia]

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

 

[nehaoscar]

Why part iii no change?
Wouldn't the equlibrium shift to the left so the potential will decrease so wouldn't the Ecell increase?

 

[nehaoscar]

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??

 

[Muhammad Bilal Shaikh]

Any guess or predicted paper for chemistry p33 which is on 29th?????

 

[darks]

Anyone please any guess for p33 ? 29th oct

 

[awesomaholic101]

when they ask us to add 1 cm^3 of starch indicator for titration, we add approximately how many drops?

 

[Nushad Nahue]

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.

 

[darks]

when they ask us to add 1 cm^3 of starch indicator for titration, we add approximately how many drops?

0.05 cm3 is one drop. so u put it in estimation. usually a dropper will have a mark on it.

 

[awesomaholic101]

0.05 cm3 is one drop. so u put it in estimation. usually a dropper will have a mark on it.

that would come aroung like 20 drops

 

[darks]

that would come aroung like 20 drops

the dropper will have markings of 1-3 cm3.
in case it does not... 15-20 drops then.

 

[awesomaholic101]

the dropper will have markings of 1-3 cm3.
in case it does not... 15-20 drops then.

owkay

 

[lizlovespiano]

oh goog... look.. we have 3 products.. ethene methane and propene...

if i mole is produced by cracking, then o.5% will be of ethene(as mentioned in the q) then 0.25 of propene and .25 of methane is produced... .25+.25+.5=1...

to convert .25 into proper fraction, we need to multiply it by 4 to bring it to a whole number... by doing this.. we will get 1 mole of propene, 1 mole of methane and 2 moles of ethene(.5*4).. no of Carbons now we have are 8 all together... so B is the answer...

Hello, why are propene and methane .25% respectively? How do you know that they are equally .25%? It could be .15 and .35 also too, right?

 

[lemonrsiow]

Does anyone know how to solve the molecular mass question? 9701/52/MJ/15 Q2di

 

[ahmed butt]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q38, Q4,Q5,Q2,Q15? Doubt can anyone, i will be thankfull to him

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q38, Q4,Q5? Doubt can anyone, i will be thankfull to him

4)


5)
Shape of BF3 is trigonal planner so option B and D are ruled off.
Due to presence of lone pair at nitrogen atom will repel B - F bond pairs hence its shape will not be trigonal planner its bond angle will decrease so A cannot be answer hence its C.

38)
Only 1 is correct.
First step will be Nucleophilic substitution reaction. Cl will be replaced by CN, on hydrolysis of CN with NaOH(aq) it will give COO-Na+
2 is incorrect, I guess there wont be any such reaction.
3 is incorrect as Cl will be replaced by OH.