Chemistry: Post your doubts here!

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf. Q 9 , 32 and 37 please

Qn 9 :

moles of SO3 2- : moles of electrons : moles of metal
0.0025 : : 0. 005
1 : : 2
1 : 2 : 2

2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q32. Info is implying that CO2 is more stable than CO
A. Is correct as CO2 (+4) and CO (+2)
B. CO2 is more stable than CO, so formation of CO2 is more exothermic than CO.
C. Eqm lies strongly to left as we expect a high amt of CO to be converted to CO2. Kc is thus high.

 

[IGCSE13]

Qn 9 :

moles of SO3 2- : moles of electrons : moles of metal
0.0025 : : 0. 005
1 : : 2
1 : 2 : 2

2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q32. Info is implying that CO2 is more stable than CO
A. Is correct as CO2 (+4) and CO (+2)
B. CO2 is more stable than CO, so formation of CO2 is more exothermic than CO.
C. Eqm lies strongly to left as we expect a high amt of CO to be converted to CO2. Kc is thus high.

But 1 mole of so3 2- gives 2 electrons so 2 moles will give 4 ?

 

[omarjaved619]

Answer is C. Why not D?

 

[Metanoia]

But 1 mole of so3 2- gives 2 electrons so 2 moles will give 4 ?

Yes, based on the half equation, your statement that 2 mol of SO3 2- loses 4 mol of e is true, just that it's not relevant to this particular question.

If my previous presentation is not clear, I'll put the reasoning in step by step statement form.

0.025 mol of sulfite reacts with 0.05 mol of metal
1 mol of sulfite reacts with 2 mol of metal
1 mol of sulfite loses 2 mol of electrons, which are gained by 2 mol of metal
Therefore, 1 mol of electron is gained by 1 mol of metal

The key to solving such redox mol calculations is usually to find out the ratio of
Mol of oxidized substance: mol of electron transfered: mol of reduced substance.

Hope it makes sense.

 

[Metanoia]

View attachment 43875

Answer is C. Why not D?

For elimination, the Cl is removed together with H from a "next door" carbon.

Note that there are no H on the "next door" carbon for option D.

 

[omarjaved619]

For elimination, the Cl is removed together with H from a "next door" carbon.

Note that there are no H on the "next door" carbon for option D.

Thanks a lot! The thing that got me confused is that chemguide.com mentions that Tertiary Halogenoalkanes mainly undergo Elimination, whereas Primary undergo substitution.

 

[Metanoia]

Thanks a lot! The thing that got me confused is that chemguide.com mentions that Tertiary Halogenoalkanes mainly undergo Elimination, whereas Primary undergo substitution.

Yes, that is a valid general guideline.

Just be careful that in this question, C and D ( including A and B) are both primary. Though its a common mistake among students to view D as tertiary due to the skeletal structure used.

 

[IGCSE13]

Yes, based on the half equation, your statement that 2 mol of SO3 2- loses 4 mol of e is true, just that it's not relevant to this particular question.

If my previous presentation is not clear, I'll put the reasoning in step by step statement form.

0.025 mol of sulfite reacts with 0.05 mol of metal
1 mol of sulfite reacts with 2 mol of metal
1 mol of sulfite loses 2 mol of electrons, which are gained by 2 mol of metal
Therefore, 1 mol of electron is gained by 1 mol of metal

The key to solving such redox mol calculations is usually to find out the ratio of
Mol of oxidized substance: mol of electron transfered: mol of reduced substance.

Hope it makes sense.

Thanks , what about q37 in the same paper do u know how to solve it ?

 

[kingo44]

The Kc expression for reaction I is
Kc(1)= [X2Y]^2 / [X2]^2 + [Y2]

And for reaction two

Kc(2)= [X2] + [Y2]^1/2 / [X2Y]

Now we should rearrange the first expression in order to get a form similar to second expression.

( [X2]^2 + [Y2]) Kc(1) = [X2Y]^2
( [X2]^2 + [Y2]) Kc(1) / [X2Y]^2 = 1
[X2]^2 + [Y2] / [X2Y]^2 = 1/Kc(1)
now square root all terms to get
[X2] + [Y2]^1/2 / [X2Y] = 1/(Kc(1))^1/2 (see the power of half is a square root)

we can see that

Kc(2) = 1/[Kc(1)]^1/2
= 1/(2)^1/2

but y kc is half how we square root 1 to get 0.5 i didnt ge t this part

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf q6. ,19 and 28 can someone please help me with them

 

[Gehad Mohamed]

Hello! Can someone please explain how to solve Q39 http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_11.pdf

 

[mehria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf q6. ,19 and 28 can someone please help me with them

Q6... first of all we will find the moles of ice => mass=density x volume = 1x1= 1g
the mole of ice = 1/18 =0.06
assumng ice as steam:- at room temp the volume will be 0.06 x 24 = 1.32
and then by the ratio we will find the volume of steam produced at 596 K
V1/T1 = V2/T2
1.32/298 = V2/ 596
V2 = (1.32 x 596) / 298 = 2.64 cm3

Q19:- when Ca(OH)2 reacts with SO2 so initially CaSO3 is formed and then CaSO4 will form... so the ans is C

 

[IGCSE13]

Q6... first of all we will find the moles of ice => mass=density x volume = 1x1= 1g
the mole of ice = 1/18 =0.06
assumng ice as steam:- at room temp the volume will be 0.06 x 24 = 1.32
and then by the ratio we will find the volume of steam produced at 596 K
V1/T1 = V2/T2
1.32/298 = V2/ 596
V2 = (1.32 x 596) / 298 = 2.64 cm3

Q19:- when Ca(OH)2 reacts with SO2 so initially CaSO3 is formed and then CaSO4 will form... so the ans is C

Q28 ?

 

[mehria]

Q28 ?

Q28:- H- is nucleophile so it will attack on partial positive charge of Carbon atom of the Carbonyl group...

 

[Gehad Mohamed]

Anyone plz Help me in Q1 Q 20 Q27 Q28 Q29 Q35 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf

 

[immie.rose]

someone pls explain the following!!



Answer: A


Answer: D



Answer:C

 

[AbbbbY]

someone pls explain the following!!


View attachment 43906
Answer: B

View attachment 43907
Answer: B


View attachment 43908
Answer:C

Are you sure you didn't read the answers wrong? I've gotten to A with the first one, I'm damn sure the second one is D. Could you please recheck?

For the last one,

One mol of alcohol reacts with one mol of the acid to form one mol of ester.

Ethanol: 46
Ethanoic Acid: 60
Ethyl Ethanoate: 88
Since ethanoic acid has the greater Mr and same mass is used, it's the limiting reagentt. 30/60 = 0.5 mol so 0.5 mol of the ester = 44g should be formed.

22/44 = 0.5 so, 50% yield.

 

[immie.rose]

Are you sure you didn't read the answers wrong? I've gotten to A with the first one, I'm damn sure the second one is D. Could you please recheck?

For the last one,

One mol of alcohol reacts with one mol of the acid to form one mol of ester.

Ethanol: 46
Ethanoic Acid: 60
Ethyl Ethanoate: 88
Since ethanoic acid has the greater Mr and same mass is used, it's the limiting reagentt. 30/60 = 0.5 mol so 0.5 mol of the ester = 44g should be formed.

22/44 = 0.5 so, 50% yield.

OMG sorry!! Yes the first one is A, and next is D. M all mess atm, sry agn.
Thanks!!

 

[Abdel Moniem]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_11.pdf
Q20

 

[AbbbbY]

Anyone plz Help me in Q1 Q 20 Q27 Q28 Q29 Q35 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf

1:
D
__
CO2 = 0.8 mol
H2O = 0.8 mol

0.2 : 0.8 : 0.8
= 1 : 4 : 4
So, the Hydrocarbon is giving 4 mols of CO2 and H2O on combustion of 1 mol of the HC. The only options that can give 4 mols of CO2 are C and D. Now, I know by looking the an alkene cannot give me 1:1 CO2:H2O, but even if you don't know that, just combust the hydrocarbon and pick.
C4H4 + O2 -> 4CO2 + 2H2O
C4H8 + O2 -> 4CO2 + 4H2O

20:
A
__

Aah. I've answered this so many times here. Should save the answer in notepad or sth

Anyway.
You need to understand that everytime a ring is formed, or a double bond is formed the compound loses TWO hydrogens.

C20H28O.
So if it were a fully saturated HYDROCARBON, it'd have 42 Hydrogens. C20H42.
We have 28, so 14 are missing.
You have an Aldehyde, a ring, and a double bond in the ring. Thats 6 carbons hydrogens.
14-6 = 8. 8/2 = 4 further double bonds.

Since you're asked total number of double bonds, 4+1 = 5 double bonds.
I hope this makes sense

27:
C?

You basically need a central C with OH, CH2CO2CH3, H and R attached.
A is out. CO2CH3, not CH2CO2CH3
B is out. CH3CO2CH2CH, not CH2CO2CH3
D is out. While CH2CO2CH3 is correct, it can't have (CH3)2 as the Carbonyl is an aldehyde.

28:
B

I often find it much easier to solve these questions the other way around.

Since ethanolic NaCN is used, the starting material has to be a haloalkane giving you a nitrile. The nitrile undergoing acid hydrolysis will give you a carboxylic acid.

Haloalkane -> Nitrile -> Carboxylic acid.

Start with 2-ethyl-3-methylbutanoic acid. Remove the -OOH and replace in with N. Careful not to add an additional C here. A lot of people make this mistake. The addition C is already present since you're backsolving!
Now, just remove the entire CN terminal and replace it with Br. Re-arrange to get B.

BTW, this is a very common approach in SATII. Idk if you've given that, but if you have, use the SAT tricks in MCQs. They help a lot.
If this doesn't make sense, let me know and I'll draw it out for you.

29:
D

It's important to understand what's going on. It's taking an ester and using an acid to hydrolyse it. Your products will be an Alchol and an Acid. The Acid half always has the C=O linkage, so HCOOH + RCH(CH3)OH

35:
D

Beautifully made question I must say. Tricked me. Tricked most of my friends too. I got this wrong back when I did it. Class test or sth iirc.

Yes, the FINAL reaction will give you Br2 + SO2. However, it's important to note that 2 and 3 are incorrect regardless.

1- Correct. HBr is an intermediate which later changes to Br2.
2- WRONG! Sulfuric Acid is an oxidising agent so it is itself REDUCED. Besides, it's going from +6 to +4. That's clearly reduction, NOT oxidation.
3- Wrong again! Br- -> Br2. -1 -> 0. That's oxidation!