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Chemistry: Post your doubts here!

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oh goog... look.. we have 3 products.. ethene methane and propene...

if i mole is produced by cracking, then o.5% will be of ethene(as mentioned in the q) then 0.25 of propene and .25 of methane is produced... .25+.25+.5=1...

to convert .25 into proper fraction, we need to multiply it by 4 to bring it to a whole number... by doing this.. we will get 1 mole of propene, 1 mole of methane and 2 moles of ethene(.5*4).. no of Carbons now we have are 8 all together... so B is the answer...
Hello, why are propene and methane .25% respectively? How do you know that they are equally .25%? It could be .15 and .35 also too, right?
 
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4)
Untitled.jpg

5)
Shape of BF3 is trigonal planner so option B and D are ruled off.
Due to presence of lone pair at nitrogen atom will repel B - F bond pairs hence its shape will not be trigonal planner its bond angle will decrease so A cannot be answer hence its C.

38)
Only 1 is correct.
First step will be Nucleophilic substitution reaction. Cl will be replaced by CN, on hydrolysis of CN with NaOH(aq) it will give COO-Na+ :)
2 is incorrect, I guess there wont be any such reaction. :)
3 is incorrect as Cl will be replaced by OH.
 
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Well 39 it is B coz

Catalytic converter converts oxides of nitrogen to nitrogen gas :D

and CO and unburnt hydrocarbons to CO2 which means it is oxidation. :D

No idea why 24 is wrong :')

Qn 24, upon dehydration, we have either

2-methylbut-2-ene

Screen Shot 2015-10-31 at 6.45.00 AM.png

or

3-methylbut-1-ene

Screen Shot 2015-10-31 at 6.46.26 AM.png

Neither versions have cis-trans isomerism, so there are only a total of 2 possible alkenes.
 
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View attachment 57326
Why part iii no change?
Wouldn't the equlibrium shift to the left so the potential will decrease so wouldn't the Ecell increase?

When you say equilibrium shifts to the left, which equilibrium equation are you referring to? And how do you conclude it would shift to the left?

View attachment 57327
mark scheme says concentration of fe3+ is 0.2 but question is showing 0.1... how is it 0.2??
Also for Ka don't you multiply Ka by the concentration of the solution and not just Fe3+ ions??

For weak acid,

[H+] = square root (Ka*[acid])

The concentration of Fe2(SO4)3 is o.1 mol/dm3. But since we are using concentration of Fe3+ , we do need to multiply it by 2 as there are two Fe3+ ions for each Fe2(SO4)3.

[H+] = square root (8.9 x 10^-4 *(0.2)) = 0.0133

pH = - log [H+] = - log (0.0133) = 1.87

Note: Answer edited due to spotted error.
 
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4)
View attachment 57351

5)
Shape of BF3 is trigonal planner so option B and D are ruled off.
Due to presence of lone pair at nitrogen atom will repel B - F bond pairs hence its shape will not be trigonal planner its bond angle will decrease so A cannot be answer hence its C.

38)
Only 1 is correct.
First step will be Nucleophilic substitution reaction. Cl will be replaced by CN, on hydrolysis of CN with NaOH(aq) it will give COO-Na+ :)
2 is incorrect, I guess there wont be any such reaction. :)
3 is incorrect as Cl will be replaced by OH.

i got it(y) , thx for taking ur time for solving.:)
by the way which variant of papper 1 are u giving?
i also have doubt in q2 and q15 same papper:(
 
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i got it(y) , thx for taking ur time for solving.:)
by the way which variant of papper 1 are u giving?
i also have doubt in q2 and q15 same papper:(
2) Use of N from NPK will be 15/60
Moles of N = (15/60)/14
Concentration of N in fertilizer : (15/60)/(14*5) = 0.03

15) CaCO3 --------> CaO + CO2
Mr of CaCO3 = 100.1
Total mass of CaCO3 = 1200g
n of CaCO3 used in total = 1200/100.1 = 11.9
1 : 1 ratio of CaCO3 and CO2
Mr of CO2 = 44
m of CO2 = 44 * 11.9 = round up to 527g
 
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  1. The following half reactions occur when potassium iodate(V), KIO3, in hydrochloric acid solution oxidises iodine to ICl –.

    IO– +2Cl– +6H+ +4e– →ICl– +3HO 322

    I +4Cl– →2ICl– +2e– 22

    What is the ratio of IO – to I in the balanced chemical equation for the overall reaction? 32

    A1:1 B1:2 C1:4 D2:1
 
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9 Nitrogen reacts with hydrogen to produce ammonia.
N2(g) + 3H2(g) 2NH3(g)

A mixture of 2.00mol of nitrogen, 6.00mol of hydrogen, and 2.40mol of ammonia is allowed to reach equilibrium in a sealed vessel of volume 1 dm3 under certain conditions. It was found that 2.32 mol of nitrogen were present in the equilibrium mixture.

What is the value of Kc under these conditions?

(1.76)2 (2.32)(6.96)3

(1.76)2 (2.32)(6.32)3

(2.08)2 (2.32)(6.32)3

(2.40)2 (2.32)(6.00)3
 
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26 How many moles of hydrogen, H2, are evolved when an excess of sodium metal is added to one mole of citric acid?

CO2H
HO
page10image21736
C
page10image21968
CH2CO2H

CH2CO2H citric acid

A1B2C3D4
 
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  1. The following half reactions occur when potassium iodate(V), KIO3, in hydrochloric acid solution oxidises iodine to ICl –.

    IO– +2Cl– +6H+ +4e– →ICl– +3HO 322

    I +4Cl– →2ICl– +2e– 22

    What is the ratio of IO – to I in the balanced chemical equation for the overall reaction? 32

    A1:1 B1:2 C1:4 D2:1
To balance the number of electron, you need to multiply second equation with 2.
So 2I will be there and hence electrons will be canceled out.. :)
LEaving us with the ratio of 1 : 2
 
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9 Nitrogen reacts with hydrogen to produce ammonia.
N2(g) + 3H2(g) 2NH3(g)

A mixture of 2.00mol of nitrogen, 6.00mol of hydrogen, and 2.40mol of ammonia is allowed to reach equilibrium in a sealed vessel of volume 1 dm3 under certain conditions. It was found that 2.32 mol of nitrogen were present in the equilibrium mixture.

What is the value of Kc under these conditions?

(1.76)2 (2.32)(6.96)3

(1.76)2 (2.32)(6.32)3

(2.08)2 (2.32)(6.32)3

(2.40)2 (2.32)(6.00)3
Initial Moles = IM and eq mole = EM
IM of N2 = 2 mol
IM of H2 = 6 mol
IM of NH3 = 2.4 mol
EM of N2 = 2.32
We have eqantion, N2 + 3H2 <====> 2NH3
So,
EM of H2 is 6 - 3(x)
EM of NH3 is 2.4 + 2(x)
we know x is -0.32
Keep in those equation :
EM of H2 = 6.96mol
EM of NH3 = 1.76mol
molar Concentration = no of moles at eq. (As V = 1dm^3)
Kc = [NH3]^2 / [N2][H2]^3
Kc = (1.76)^2 / (6.96)^3 x (2.32)
 
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26 How many moles of hydrogen, H2, are evolved when an excess of sodium metal is added to one mole of citric acid?

CO2H
HO
page10image21736
C
page10image21968
CH2CO2H

CH2CO2H citric acid

A1B2C3D4
Na reacts with both alcohols and acids as well. So 4 moles of Na is there.
We know hydrogen molecule exist as H2 so 2 moles of H2 needed to balance the reaction.
 
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2) Use of N from NPK will be 15/60
Moles of N = (15/60)/14
Concentration of N in fertilizer : (15/60)/(14*5) = 0.03

15) CaCO3 --------> CaO + CO2
Mr of CaCO3 = 100.1
Total mass of CaCO3 = 1200g
n of CaCO3 used in total = 1200/100.1 = 11.9
1 : 1 ratio of CaCO3 and CO2
Mr of CO2 = 44
m of CO2 = 44 * 11.9 = round up to 527g
Q15 how u find total mass?
 
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How do you draw the apparatus to collect water and then N2O?
1c:
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s12_qp_52.pdf
 
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