http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_13.pdf Q29 AnsD Q35 ansC Q36 ansA
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Chemistry: Post your doubts here!
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http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q3(C), 4(D), 21(C) and 25(B)
Need help!
Q3(C), 4(D), 21(C) and 25(B)
Need help!
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Which compounds can be obtained from ethene in a
single
reaction?
1 CH3CH3
2 -(- CH2CH2-)-n
3 HOCH2CH2OH
single
reaction?
1 CH3CH3
2 -(- CH2CH2-)-n
3 HOCH2CH2OH
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Metanoia , ZaqZainab , @ Every One, Could u Plz help me with this
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
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
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!!!!
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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is it A?
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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.
THANKS!!!
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Q3 answer is C buts isn't it suppose to be A?
Q20 answer is A
Q23 isn't moles of H2 suppose to be 3?? answer is A
Q28 answer is B
Q33 isnt 3 suppose to be correct too if the activation energy is higher than 500 the reaction wouldnt have even started
Q40 i have no idea answer is C
Q3 answer is C buts isn't it suppose to be A?
Q20 answer is A
Q23 isn't moles of H2 suppose to be 3?? answer is A
Q28 answer is B
Q33 isnt 3 suppose to be correct too if the activation energy is higher than 500 the reaction wouldnt have even started
Q40 i have no idea answer is C
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I hope it helped My pleasure
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A
Ethene,C2H2
1. C2H2--->C2H6
using H2 and Ni
2. Polymerization
3. using cold acidified KMO4
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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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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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3: x is 104 y is 109 z is 107 so C is correct.
y has no lone pair so greatest angle x has two so smallest.
20:
23: H2 does not break CO bond.
28: Reactant should have 1 less carbon as one is added by reaction with sodium cyanide. So C is wrong.
CN will be attached in place of Br and then it will form carboxylic acid. Only B will form the correct carboxylic acid.
33:The activation energy remains same at different temperatures.
40: A would mean polymer has same number of molecules as monomer and would be identical to it so not possible.
If all 6.02x10^23 molecules react, it will form one molecule but 1/6.02x10^23 moles.
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w06qp1
Q3.
P is option A
Ca is option B
Kr is option D
So that leaves element X as option C.
Q4.
Use any of the alkane as an example and construct a balance equation for complete combustion, for here, I'll use CH4
I prefer to use table, but not sure if the alignment shows when typed out.
CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (l)
10 .............70 ................0 .................. - .......... initial
-10 ............-20 ............ +10 ............. - .......... change
0 ............ 50 .............. 10 ............ - ............ final
Total gases at the end = 50 + 10 = 60 cm3, this fits option D.
Q21
This is a bit tedious, you need to draw and figure out the number of structural isomers for the following halogenalkanes.
C2H5Cl (1 isomer)
C2H4Cl2 (2 isomers)
C2H3Cl3 (2 isomers)
C2H2Cl4 (2 isomers)
Q25. Divide by n throughout for easy balancing.
C3H6 + O2 --> CO2 + H2O
C3H6 + 4.5 O2 --> 3CO2 +3H2O
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I'll just add on to Q40 since Ahmed Aqdam has addressed the other questions.
Q40. Requires a bit of reasoning, if 1 mole of monomer join together, we will get less than 1 mole of polymer.
If we end up with 1 mole of polymer in the end, it means none of the monomer has joined to any other!
Its like 1000 bricks (monomer) will build one house (polymer) , the number of monomer is expected to be more than polymers.
Hope it makes sense.
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Can some one plz plz plz explain the question to me... I would really be grateful
There is a lone pair of electrons on the N, so there are 2 bond pairs and 1 lone pair. The closest angle would be 120 degrees.
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Q6. need to make sure units are correct
pV= nRT
R = 8.31
T= 303K
n= 0.56/28= 0.02 mol
p= 120 000 Pa
You can solve for v (in terms of m^3)
Q7.
Formation of propanone
3C (s) + 3H2 (g) + 1/2 O2 (g)--> C3H6O (g)
heat of reaction = heat of combustion of reactants- heat of combustion of products
= 3(-394)+3(-286)-(-1786)
=-254
Q13. ? What answer?
Q14.
I'm guess the hurdle is getting the balanced equation?
10Al + 3Ba(NO3)2 --> 5Al2O3 + 3BaO + 3N2
moles of barium nitrate = 0.783 / 261 = 0.003
moles of N2 = 0.003
vol of N2 at RTP = 0.003 x 24000 = 72 cm^3
Q38.
Mr of alcohol = 74
Moles of alcohol used = 70/74 = 0.946
Moles of alcohol converted into products = 0.946 x 62% = 0.586
1. If 0.586 mol of butanone was obtained, it would weight 0.586 x Mr of butanone = 0.586 x 72 = 42.2 g (statement 1 true)
2. If 0.586 mol of butanoic acid was obtained, it would weight 0.586 x Mr of butanoic acid = 0.586 x 88 = 51.6 g (statement 2 true)
3. If 0.586 mol of 2-methylpropanoic acid was obtained, it would weight 0.586 x 2-methylpropanoic acid = 0.586 x 88 = 51.6 g (statement 3 true)