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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
where the square come from in ques 11q9
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
Okay, here goes:
sorry but muje kuch samaj me nahi ayaOkay, 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
Ask someone else, I have 109 sums left to solve ._.sorry but muje kuch samaj me nahi aya
look at the tablewhere the square come from in ques 11
look at the table
kc =(Product)/(reactant)
okay!kc =(Product)/(reactant)
=(x)(x)/(1-x)(1-x)
do you get it ?
hope it makes sensesorry but muje kuch samaj me nahi aya
Thank you so muchhope it makes sense
q.1http://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
the examiner report gave a pretty nice explanation: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
how did you know the volume for CaSO4 was 1000 dm^3?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
dats not da volume, volume is 50 cm³how did you know the volume for CaSO4 was 1000 dm^3?
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