Chemistry: Post your doubts here!

[ashcull14]

 

[Suchal Riaz]

Can you help me with Q1 D,
22D,
30A,
33D,
37B of this paper
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_13.pdf

37:
dichromate will oxidise CHO to CO2H
tollens and fehling reagent will oxidise CHO to CO2H
aqueous bromine will react with C=C
so first two are correct
third is wrong because there is no -CO2H to react with Na2CO3

 

[ashcull14]

 

[Metanoia]

View attachment 45178

11qp12

short answer, cross the Al3+ and C4- charges, though it might not always work.

Long answer
Understand that whatever moles of C we have at the beginning will eventually be the same number of moles of C in CO2 at the end.

Let x and y be the ratio of Al and C respectively.
AlxCy --->???---> CO2

Working backwards from CO2,
moles of CO2 = 72/24 000 = 0.003 moles = moles of C in AlxCy

Going through the options
A Al 2C3, moles of Al2C3 = 0.144/90 = 0.0016, moles of C = 0.0016 x 3 = 0.0048 (incorrect)
B Al 3C4 , moles of Al3C4 = 0.144/129 = 0.0016, moles of C = 0.0016 x 4 = 0.0045 (incorrect)
C Al 4C3 , moles of Al4C3 = 0.144/144 = 0.001, moles of C = 0.001 x 3 = 0.003 (BINGO!)
D Al 5C3

 

[Suchal Riaz]

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i am working on it Metanoia
i had to erase some answer because you had already answered them. please tell me which one you are doing so i do other queries instead.

 

[ashcull14]

11qp12

short answer, cross the Al3+ and C4- charges, though it might not always work.

Long answer
Understand that whatever moles of C we have at the beginning will eventually be the same number of moles of C in CO2 at the end.

Let x and y be the ratio of Al and C respectively.
AlxCy --->???---> CO2

Working backwards from CO2,
moles of CO2 = 72/24 000 = 0.003 moles = moles of C in AlxCy

Going through the options
A Al 2C3, moles of Al2C3 = 0.144/90 = 0.0016, moles of C = 0.0016 x 3 = 0.0048 (incorrect)
B Al 3C4 , moles of Al3C4 = 0.144/129 = 0.0016, moles of C = 0.0016 x 4 = 0.0045 (incorrect)
C Al 4C3 , moles of Al4C3 = 0.144/144 = 0.001, moles of C = 0.001 x 3 = 0.003 (BINGO!)
D Al 5C3

THNKS N nice explanation

 

[ZaqZainab]

question33:
the questions says that methanoic acid IS stronger acid because it has a LOWER pH. it means that obviously it disassociates more. and if the rest two were true then it would have been weaker acid. tha

dumb me thanks again

w13qp13

Q1.
N in NH4+ is -3, increases by 4 units to from +1 in N2O
N in NO3- is +5, decreases by 4 units to from +1 in N2O

Q22.
A. Is not good as we might get multi substitutions
B. Is not good as we get dichloropropane
C. Cl- will not replace OH
D. Cl will replace at the OH position

Q30.
A. CH2-CH2-CCl2-CCl2-CH2-
Coloured are the repeating units that come from the monomers

Q33. Lower pH means higher conc of H+, due to more H+ released. Which statements are you confused about?


Q37. Which statement are you unsure of?
Statement 3: No COOH group to react with sodium carbonate

Thanks and for
Q37 the first one how can it be oxidized when no alcohol is present

 

[ZaqZainab]

i am working on it Metanoia
i had to erase some answer because you had already answered them. please tell me which one you are doing so i do other queries instead.

lol happens with me so many times but i don't erase it might help the person understand more

 

[GCE As and a level]

dumb me thanks again

Thanks and for
Q37 the first one how can it be oxidized when no alcohol is present

Aldehyde is present which get oxidised to -COOH

 

[Snowysangel]

ZaqZainab this is how to solve

Won't it be x2= 4 (1-x)^2 -> x= 2(1-x) -> x= 2-2x -> 3x=2 -> x=2/3?

 

[Suchal Riaz]

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calcium has 2+ charge so there must be two CO2H
in case (1) there will be two ethanoic acids: 2CH3CO2- = C4H6O4(2-)
in case two it is butandioc means there are di=two CO2H groups in one molecule: O2CCH2CH2CO2=C4H6O4(2-)
in case three: there is again same butanedioc acid(it's isomer) so same formula: C4H6O4(2-)
so (2) and (3) are correct

 

[Suchal Riaz]

Won't it be x2= 4 (1-x)^2 -> x= 2(1-x) -> x= 2-2x -> 3x=2 -> x=2/3?

yes i made a mistake. it had to be x²

 

[Suchal Riaz]

dumb me thanks again

Thanks and for
Q37 the first one how can it be oxidized when no alcohol is present

aldehyde will be oxidised to CO2H

 

[Ahmed Aqdam]

but the answer to question 23 is c tht is 4 esters

Yes, two with methanoic acid, one with ethanoic and one with propanoic. So total are 4.

 

[♣♠ Magnanimous ♣♠]

Won't it be x2= 4 (1-x)^2 -> x= 2(1-x) -> x= 2-2x -> 3x=2 -> x=2/3?

wah... thanks

 

[Suchal Riaz]

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question 23:
it means that the number of moles of ethene are 1/2 of total moles of products. it means that there are two moles of ethene, one mole of methane and one mole of propane.
total number of carbon=2 from each ethene, one from methane and three from propane/ 4+1+3=8

 

[Metanoia]

i am working on it Metanoia
i had to erase some answer because you had already answered them. please tell me which one you are doing so i do other queries instead.

Maybe we can quote and "click reply" immediately.

And then take your time to add the answers by editing the posts.

If it turns out we are repeating answers even after that, just leave them posted. Sometimes, it might be good to have differing approaches to the same questions.

Quite a few questions are repeated, so I just copied and paste what I type just a day or two ago.

Anyway, I will be here for just only a while longer .

 

[Suchal Riaz]

thats they way i am doing. adding each question one by one. but instead on new post because the answer won't get notification when i add new answer to the same post.

Maybe we can quote and "click reply" immediately.

And then take your time to add the answers by editing the posts.

If it turns out we are repeating answers even after that, just leave them posted. Sometimes, it might be good to have differing approaches to the same questions.

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Q28

 

[Suchal Riaz]

Maybe we can quote and "click reply" immediately.

And then take your time to add the answers by editing the posts.

If it turns out we are repeating answers even after that, just leave them posted. Sometimes, it might be good to have differing approaches to the same questions.

Quite a few questions are repeated, so I just copied and paste what I type just a day or two ago.

Anyway, I will be here for just only a while longer .

you can take rest. i will handle these childish questions