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Chemistry: Post your doubts here!

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Q14.
Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)

Mass of XO = 1.47 g
Moles of XO = 1.47/ (Mr of X + 16)

Mass of X(NO3)2 = 3 g
Moles of X(NO3)2 = 3/(Mr of X + 124)

since moles of X(NO3)2 = moles of XO
3/(Mr of X + 124) = 1.47/ (Mr of X + 16)

Solve for Mr of X.

Q10
Heat of neutralization for strong acids and strong alkali is estimated to be -57kJ PER MOLE of H2O formed.

For both experiments. (Involving strong acid and bases) , it reflects twos moles of water formed .

So heat of reaction is -57 x 2 for both equations.
 
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603
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153
Question 20 please...
Q2o. IF all carbons are single bonds, CnH2n+2O, C20H42O (20 carbons , 42 hydrogen, 1 oxygen)

Since 1 aldehyde group is present, subtract 2 H, C20H40O (20 carbons , 40 hydrogen, 1 oxygen)

Since it contains a ring, subtract another 2H, C20H38O (20 carbons , 38 hydrogen, 1 oxygen)

Since retinal contains only 28 H, the missing 1o H are due to 5 double bonds.
 
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Which compounds can be obtained from ethene in a
single
reaction?
1 CH3CH3
2 -(- CH2CH2-)-n
3 HOCH2CH2OH
 
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Metanoia , ZaqZainab , @ Every One :) , Could u Plz help me with this :D
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf

Q2..Answer B
Q13.Answer B>>>Why it is not Mg :confused: ??Mg is just beside Al , so i thought that it will have a similar Electronegativity as Mg.
Q15.Answer C>>>When H2SO4 acts as an acid and when it acts as an oxidizing agent?
Q31.Answer A>>>I know that this is simple, but i am always confused with this type,So is there any hint?? :)
Q32.Answer D
Q33.Answer B
Q40.Answer B

THNX in advance :D

Q15. When reacted with conc H2SO4
NaCl --> HCl
NaBr --> HBr --> Br2
NaI --> HI --> I2
so H2SO4 oxidises Br- and I-.

Q31. Answer seems to be B instead of A?
Al in AlCl3 has 3 bond pairs, 0 lone pairs. Trigonal planar
C in CH+ has 3 bond pairs, 0 lone pairs. Trigonal planar
P in PH3 has 3 bond pairs and 1 lone pair. Trigonal pyramidal.

Q40. Hydrolyse the ester bond
Left fragment forms structure 1 (carboxylate salt)
Right fragment forms structure 2. (alcohol)

Q32. When temperature keeps increasing, all the mass of solid would eventually become gas. So the mass of gas at the end is the same as mass of solid.
Pressure will keep increasing if temperature increases at a constant volume.

Q33. It depends on which experiment could be carried out in the lab. If not feasible, have to use Hess Law to calculate enthalpy change indirectly.
A. Not feasible as hard to tell if all CuSO4 is hydrated by 5 mols of H2O
B. Not feasible as C(s) + 2H2(g) --> CH4 (g) doesn't occur in lab
C. Feasible, was we can burn a know quantity of glucose and measure temperature rise in calorimeter.
 
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For May June 2012 paper variant 11

Please explain
--> q 7 the answer is c
Q 6 the answer is B
Question 13 answer is 13

Q 14 answer is B
Question 38 answer is A

Dunno but found this paper real tough! Please help!!!!
 
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In q32, does the gas partially turn into a liquid too?

Possible. Or they could also remain as gas. The idea is they have attraction forces which causes them to occupy a lower volume than expected. This goes against the assumption that idea gases have no attraction forces among each other.
 
Messages
603
Reaction score
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153

Q16. Barium carbonate does not dissolve in water, but reacts with acid to form a soluble Ba2+.
BaCl2 is soluble in both water and acid.
Ba(OH)2 soluble in both
BaSO4 insoluble in both

Q24. Steamy fumes with PCl5 means there COULD be OH group, either as OH alcohol or COOH acid. We are not sure which.
Since it does not react with 2,4-DNPH. It must not be ketone or aldehyde.
So the only sure conclusion we can make is its not an aldehyde.

For next 2 questions, I copied from the examiners report, if need be will draw out the structures of Q27

"Q27. The question should be approached by counting the OH groups in the product X. Warm dilute H2SO4 will hydrolyse the
ester link in santonin, producing two OH groups. Cold, acidified KMnO4 will oxidise the two C=C bonds in santonin, producing four more OH groups. This makes six in total."

"Q34
The key to ruling out both A and C is statement 3, ‘X and Y must both be in the same Group of the Periodic Table’. This statement is incorrect, X and Y could be in group 2, or one or both of them could be transition metals."
 
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603
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w12qp13

Q29. Since methanoic acid is used, the alcohol would be butanol.
HCOOC4H9

There are 4 isomers of butanol, so depending on which isomer of butanol we used, we have 4 possible structural isomers of esters.
If we use butan-2-ol, the ester also have a chiral carbon and so an extra opital isomer.
Total: 5 possible isomers. It would be best if you can try to draw the 4 structural isomers and see for yourself which has a chiral carbon.

Q35. W is MgO
X is CaCO3
Y is CaO
Z is Ca(OH)2
Try to see which of the statements makes sense from here, check back here if still unclear.

Q36. J is sulfur, K is SO2, L is SO3

Again, try to see which of the statements makes sense from here, check back here if still unclear.


 
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