Chemistry: Post your doubts here!

[♣♠ Magnanimous ♣♠]

ah

Link ?

hhhhh ok
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
ques 9

 

[not.maria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Plz help me with 3 questions from o/n/6
Q9-C
Q11-B
Q25-B
Thanku

q9


q 11 assume the moles used up were x.


so u have equation

4= x^2/(1-x)^2

taking square roots on both sides we have

2=x/(1-x)

2-2x=x

2/3 =x


since volume is not mentioned you can assume it to be 1 dm^3

 

[♣♠ Magnanimous ♣♠]

q9

q 11 assume the moles used up were x.

so u have equation

4= x^2/(1-x)^2

taking square roots on both sides we have

2=x/(1-x)

2-2x=x

2/3 =x


since volume is not mentioned you can assume it to be 1 dm^3

where the square come from in ques 11

 

[Thought blocker]

ah

hhhhh ok
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
ques 9

Okay, here goes:

CO2(g) + H2(g) -> CO(g) + H2O(g)

CO2 is being broken, so instead of -283, we take +283
H2O is being formed, so we use -286
BUT it is being formed in gaseous state, not liquid state, so we also take +44.

283 + 44 - 286 = +41.

So the answer is C

 

[♣♠ Magnanimous ♣♠]

Okay, here goes:

CO2(g) + H2(g) -> CO(g) + H2O(g)

CO2 is being broken, so instead of -283, we take +283
H2O is being formed, so we use -286
BUT it is being formed in gaseous state, not liquid state, so we also take +44.

283 + 44 - 286 = +41.

So the answer is C

sorry but muje kuch samaj me nahi aya

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q1
Q11
Q13 how exactly?
Q14
Q16 doesn't it give white ppt?
Please help

 

[Thought blocker]

sorry but muje kuch samaj me nahi aya

Ask someone else, I have 109 sums left to solve ._.
I'll tell u tomorrow at school.
not.maria help him out, I have 109 sums of physics left to solve.

 

[not.maria]

where the square come from in ques 11

look at the table

 

[♣♠ Magnanimous ♣♠]

which table!

look at the table

 

[Browny]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf

Can anyone explain question number 13 ans B, 28 ans D and 35 ans B.

Questions are;
In 13 why can't A be correct.
In 28 why is D correct, I think D is correct as tertiary alcohols don't oxidise.
In 35 I don't get a point.

 

[not.maria]

which table!

this one

also look at my post again.I edited it.
the pics were not uploading

 

[not.maria]

this one
View attachment 44479
also look at my post again.I edited it.
the pics were not uploading

kc =(Product)/(reactant)
=(x)(x)/(1-x)(1-x)

do you get it ?

 

[sadiaali]

Please Q11,I cant understand a word
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

 

[♣♠ Magnanimous ♣♠]

kc =(Product)/(reactant)
=(x)(x)/(1-x)(1-x)

do you get it ?

okay!

 

[ZaqZainab]

sorry but muje kuch samaj me nahi aya

hope it makes sense

 

[♣♠ Magnanimous ♣♠]

hope it makes sense

Thank you so much

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q1
Q11
Q13 how exactly?
Q14
Q16 doesn't it give white ppt?
Please help

q.1
moles = (V × M)/1000

mole ratio of KOH to CaSO₄ = 2 : 1
moles of KOH = (25 × 0.01)/1000
moles of KOH = 2.5 × 10⁻⁴

hence moles of CaSO₄ = (1/2) × 2.5 × 10⁻⁴ = 1.25 × 10⁻⁴

Concentration of CaSO₄ = (1.25 × 10⁻⁴ × 1000)/50
Concentration of CaSO₄ = 2.5 × 10⁻³ mol/dm³
Answer: A

 

[not.maria]

Please Q11,I cant understand a word
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

the examiner report gave a pretty nice explanation:
Some candidates might have found it easier to recognise A as
correct if they had rewritten the equation using familiar substances, for example, CH4(g) →C(g) + 4H(g).
Dividing ΔH by 4 will give a value for the C-H bond energy.

 

[ZaqZainab]

q.1
moles = (V × M)/1000

mole ratio of KOH to CaSO₄ = 2 : 1
moles of KOH = (25 × 0.01)/1000
moles of KOH = 2.5 × 10⁻⁴

hence moles of CaSO₄ = (1/2) × 2.5 × 10⁻⁴ = 1.25 × 10⁻⁴

Concentration of CaSO₄ = (1.25 × 10⁻⁴ × 1000)/50
Concentration of CaSO₄ = 2.5 × 10⁻³ mol/dm³
Answer: A

how did you know the volume for CaSO4 was 1000 dm^3?

 

[Hadi Murtaza]

how did you know the volume for CaSO4 was 1000 dm^3?

dats not da volume, volume is 50 cm³
M = (moles × 1000)/V