Chemistry: Post your doubts here!

[Ahmed Aqdam]

The answer would be C we can have CH3●
we can have CH3CH2●
or just CH●

How CH●? If you remove from the methyl groups they will be same but not CH.

 

[princessnoor]

In statement 1, 2 moles of ethanoic acid will react with 1 mole of calcium ion as charge on calcium is +2. So compound will be Ca(CH3CO2)2 which has empirical formula CaC4H6O4.
In 2 and 3, both H from CO2H will be removed so empirical formula will be CaC4H4O4 and not CaC4H6O4.

hey for how long u will b available today cuz i might hve some more questions nd u know tomorow is ppr

 

[ZaqZainab]

How CH●? If you remove from the methyl groups they will be same but not CH.

C● ehhh sorry

 

[Metanoia]

Also i dont get Qs 11 and 23 in this ppr http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_13.pdf

Hi, please include answers to the questions so its easier to reply.

s12qp13

Q11. Unless you can do them mentally, it might be advisable to adopt a trial and error approach to this question.
Go from options A to D and see which one gives you the correct total moles of gases in the end

I will use B as example as its the answer.



Q23. Draw all possible structural isomers of C4H8O2 that has ester bonds.

E.g CH3CH2COOCH3, CH3COOCH2CH2CH3, etc...

 

[umair1161]

1 mole=24 dm^-1
x moles= 0.3 dm^-1 (volume of O2)

This way we will get the moles of oxygen: 0.0125 mol

Then using the mass of each element we find the moles of that particular element using the formula moles=mass÷molecular mass. Then we form equations of the reaction of each element with oxygen:

Ca + 0.5 O2 -> CaO
Mg +0.5 O2 -> MgO
2K + 0.5 O2 -> K2O
2Na + 0.5 O2 -> Na2O

Since we already have the moles of each element, we can now find the moles of oxygen each element requires and that should be equal to 0.0125.

In this case 0.05 moles of sodium gives us 0.0125 moles of O2. Hence the answer D.

Ahh! I get it now! Thanks alot!

 

[Ahmed Aqdam]

Also i dont get Qs 11 and 23 in this ppr http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_13.pdf

11: A: x mol of R, 2 mol of Q so total 3x of products. P is initially x so as 1 mole of P it will be 2-x. Total are 2+2x
B: x mol of R, 2 mol of Q so 2x. P is 2-2x as 2 moles of P. Total are 2+x
C: x mol of both R and Q and P is again 2-2x so total 2 moles.
D: x mol of R, 0.5 mol of Q, P is 2-x as x will be halved here as R is 2 moles. Total 2+0.5x

23: Two with acid Methanoic, one with ethanoic, one with propanoic.
Two with methanoic acid as one straight chain and the other with methyl group.

 

[ahmed faraz]

THE ANS IS B FOR Q11! BUT u did The Correct Method !

Hi, please include answers to the questions so its easier to reply.

s12qp13

Q11. Unless you can do them mentally, it might be advisable to adopt a trial and error approach to this question.
Go from options A to D and see which one gives you the correct total moles of gases in the end

I will use D as example as its the answer.

View attachment 45167

Q23. Draw all possible structural isomers of C4H8O2 that has ester bonds.

E.g CH3CH2COOCH3, CH3COOCH2CH2CH3, etc...

 

[ZaqZainab]

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

 

[Metanoia]

THE ANS IS B FOR Q11! BUT u did The Correct Method !

oh, I think drew the columns wrongly . I'll edit the post.

 

[ahmed faraz]

oh, I think drew the columns wrongly . I'll edit the post.

its Ok !

 

[Ahmed Aqdam]

hey for how long u will b available today cuz i might hve some more questions nd u know tomorow is ppr

Not much as I have to revise some things myself. There are others here to help, or you can just google the link of the paper as most of the questions are already asked.

 

[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

question number 1:

NH4+: each Hydrogen has +1 oxidation number.total +ve oxidation number by H =4. overall charge is +1.
4-1= 3
N must have -3 oxidation number.
NO3-:each oxygen:-2, all oxygen:-6 overall:-1, -1-(-6)=+5
N2O: oxygen:-2, overall=0; 0-2=+1
change of N of NH4= =+1--3=4
change of N of NO3=+1-5=-4​

22:

B and C not possible. `A` will give many chloropronanes so yeild of 2 chloropropane will be low.​

 

[princessnoor]

Not much as I have to revise some things myself. There are others here to help, or you can just google the link of the paper as most of the questions are already asked.

ok

 

[Young Stunner]

How can we do Q9?

 

[_Ahmad]

w11qp11

Q4. There is 1 lone pair on the N atom. So we have 2 bond pairs and 1 lone pair, which is sightly below 120 degrees.

Q17. Volatility has to do with intermolecular attraction. As electrons/protons increases, the instantaneous dipole- induced dipole gets stronger.

Q20. Cold dilute Mno4 adds 2 OH across the double bond. Hot MnO4 cleaves the double bond, and there are 2 6-member rings left.

Q27. CH3CH2CH2CH3
Replacing any the 6 red H gives us 1-chloropropane
Replacing any of the 4 white H gives us 2-chloropropane.
So by probability, we have 6:4 which is 3:2

Q35:
Cl- --> HCl (no redox)
Br- --> HBr --> Br2 (oxidised)
I- --> HI --> I2 (oxidised)

Q36.
X is N2 (alkaline hydride is NH3)
Y is NO (diatomic)
Z is NO2 (polar)

Sir can you please elaborate Q20

 

[ashcull14]

 

[ZaqZainab]

question number 1:

NH4+: each Hydrogen has +1 oxidation number.total +ve oxidation number by H =4. overall charge is +1.
4-1= 3
N must have -3 oxidation number.
NO3-:each oxygen:-2, all oxygen:-6 overall:-1, -1-(-6)=+5
N2O: oxygen:-2, overall=0; 0-2=+1
change of N of NH4= =+1--3=4
change of N of NO3=+1-5=-4​

22:

B and C not possible. `A` will give many chloropronanes so yeild of 2 chloropropane will be low.​

so we take them separately i didnt know that and i am going for the exam tomorrow thanks

 

[princessnoor]

11: A: x mol of R, 2 mol of Q so total 3x of products. P is initially x so as 1 mole of P it will be 2-x. Total are 2+2x
B: x mol of R, 2 mol of Q so 2x. P is 2-2x as 2 moles of P. Total are 2+x
C: x mol of both R and Q and P is again 2-2x so total 2 moles.
D: x mol of R, 0.5 mol of Q, P is 2-x as x will be halved here as R is 2 moles. Total 2+0.5x

23: Two with acid Methanoic, one with ethanoic, one with propanoic.
Two with methanoic acid as one straight chain and the other with methyl group.

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

 

[Metanoia]

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

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

 

[princessnoor]

f

View attachment 45173 View attachment 45174

for q 40
In statement 1, 2 moles of ethanoic acid will react with 1 mole of calcium ion as charge on calcium is +2. So compound will be Ca(CH3CO2)2 which has empirical formula CaC4H6O4.
In 2 and 3, both H from CO2H will be removed so empirical formula will be CaC4H4O4 and not CaC4H6O4.