Chemistry: Post your doubts here!

[♣♠ Magnanimous ♣♠]

Oh, thats right, i apologised.

Option A is actually C-F bond broken.
For option C, it is the C-Cl bond broken.

So its is option C, but using the explanation I mentioned in post #9124

thanks

 

[♣♠ Magnanimous ♣♠]

can i know the year please?

2003

 

[♣♠ Magnanimous ♣♠]

2003 what session may or oct?

may june!

 

[♣♠ Magnanimous ♣♠]

it's not there!!!!!!!

ques 27 see it

 

[Metanoia]

D should be the correct answer!
becoz faster the reaction, lower the Ea and lower the hump and since reaction 2 is faster it should have a lower hump!

You are right in that in is option D and not C, but the approach is a bit risky.

Hmm...are you comparing the 2nd hump of option C vs the 2nd hump of option D? That would be a dangerous approach.

At most we should be comparing the
1) 1st hump of option C vs 2nd hump of option C
or
2) 1st hump of option D vs 2nd hump of option D

And in both option C and D, the 1st step is slower than 2nd step, so we can't decide between C or D just by this alone.

 

[♣♠ Magnanimous ♣♠]

SORRY

hahaha

 

[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...