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winter 2012 qp:13 question 15? can anybody explain?
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Chemistry: Post your doubts here!
- Thread starter XPFMember
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
question-4,11,15
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
question- 9,13,14
plzzzzzzzzz anyone
question-4,11,15
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
question- 9,13,14
plzzzzzzzzz anyone
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#15
answer is D, -8 in oxidation number
5H2SO4 + 8KI → 4K2SO4 + 4I2 + H2S + 4H2O
there are 2 ways to approach this problem, the obvious one is by by finding the change in oxidation number of the element oxidised which is I
8I- -> 4I2 oxidised from -1 -> 0 and there are 8 I- so the reduction must be 8 as well
the other method is by finding oxidation number of S
H2SO4, H is +1, O is -2, so S is -6 because 2(-1) + S + 4(-2) = 0 so S = +6
next we find oxidation number of S in H2S. H is +1 so S must be -2
6 - (-2) = 8
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w12 qp 11
#4:
first remember bond breaking is + (endothermic) and bond forming is - (exothermic)
N2 + O2 → 2NO
so 1x NN triple bond and 1x O=O is broken
and 2 N->O dative bond is formed
the overall enthalpy change is +180 so
+944+496 -2(N->O) = +180
2(N->O) = +1260
N->O = around 630 answer is A
#11
Answer is A
Think of X as a Carbon atom and Y as a Hydrogen atom
CH4-> C + 4H
4 C-H bond is broken, so enthalpy of C-H bond can be found by dividing ΔH by n
#15
group 2 nitrate decomposition
2 M(NO3)2 -> 4NO2 + O2 + 2 MO
first we find the mass of MO solid formed as NO2 and O2 are both gases
m(MO) = 2-1.32 = 0.68g
and we know m(M(NO3)2) = 2g
so 0.68/2 = m(MO)/m(M(NO3)2) = 16+M / M + (14 + 16x3) x2 = 16 + M /M + 124
solve the equation M = 39.6
so must be calcium answer is B
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group 2 nitrate decomposition
2 M(NO3)2 -> 4NO2 + O2 + 2 MO
first we find the mass of MO solid formed as NO2 and O2 are both gases
m(MO) = 3-1.53 = 1.47g
and we know m(M(NO3)2) = 3g
so 1.47/3 = m(MO)/m(M(NO3)2) = 16+M / M + (14 + 16x3) x2 = 16 + M /M + 124
solve the equation M = 87.7
so must be Strontium answer is D
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Thankyou so much! I was just leaving for the examination center! Haha! Hope it goes well for you
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
Why in question 28 is it C not B.....plzzz help me out I'll be really really thankful
Please please please !!
Why in question 28 is it C not B.....plzzz help me out I'll be really really thankful
Please please please !!
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
questions 9 and 14 plzzz....
questions 9 and 14 plzzz....
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its not B bcoz as they have given in the example that –CH2CO2CH3 forms from CH3CO2CH3 and we will keep that as it is as in C and D.. now an aldehyde reacts which is formed from primary alcohal so D is wrong bcoz its was a teritary alcohol , and C was an aldehyde which is changed to primary alcohal for reaction with CH3CO2CH3
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in this enthalpy change of reation= enthalpy change of formation of Fe2O3 .. and there are 2 moles.. so divide it by 2 u get -824 which is B..
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hey please I have p1 today and mAny doubts can you please clear them?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf q 12,17
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf Q 5, 31
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf Q 21 how isn't 4 the answer?
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf Q24 , I tried a lot but only got 3 isomers
http://papers.xtremepapers.com/CIE/... Level/Chemistry (9701)/9701_s12_qp_12.pdfQ28
http://papers.xtremepapers.com/CIE/...AS Level/Chemistry (9701)/9701_w12_qp_13.pdfq 34 AND THANK SOOO MUCH in adavnce
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#9 is bit hard to explain but
aP -> bQ + cR
initial mol 2 0 0
equilibrium mol 2-x(a/c)-(a/b)Y Y x
we know that at equilibrium total mol is 2+x
so 2 - x(a/c) - (a/b)Y + Y + x = 2 + x
so Y must cancel out so a/b = 1 a:b = 1:1
2 - x (a/c) = 2+x
so a/c = 2 a:c = 2: 1
so a:b:c = 2:2:1
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w05
#12
this is oxidation reactions forming oxides
eg. 1 Mg + 0.5O2 -> 1MgO
1 Al + 0.75O2 -> 0.5Al2O3
1 S + 1O2 -> 1 SO2 because its in excess oxygen 1SO2 + 0.5 O2 -> 1SO3 so overall equation is 1S + 1.5O2 -> 1SO3
the oxidation ratio is 0.5:0.75: 1.5 = 2:3:6 = 1:1.5:3 so answer is D
#17
[Ag(NH3)2]+ Cl- complex ion forms so NH3 is a ligand
answer is B
I'll continue answering each paper 1 post per paper
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w09
#5
answer is B
A is wrong because all the Cl balances out, nonpolar
C is wrong because although molecule is not symmetrical, Cl and O have similar eletronegativity so small overall dipole
D is wrong because its symmetrical nonpolar and O cancels out
B is correct because not symmetrical and H and O has big electronegativty so largest overall dipole
#31
Answer is C = 2,3 correct
think logically, polymer is made of lots and lots of monomers
so 1 mol of monomer will form less than 1 mol of polymer
so 1 is wrong, rest are correct
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s10
#21
answer is 3
remove H from 1st carbon on ethyl group
remove H from 2nd carbon from ethyl group
remove H from 1 of methyl group
s12 qp11
#24
Answer is C, 4 esters
methyl propanoate
ethyl ethanoate
propyl methanoate
isopropyl methanoate
s12 qp12
#28
question asking for name of C8H16Br2
so A and B is absolutely wrong
notice how amine group and propyl group are substituted at 1,5 position so answer is D
w12 qp 13
#34
X:H2 = 1:1 YZ:H2Z = 1:1
and it says same amount of mass (1g) of X and YZ forms same mol of gas
and we know that mol number = mass/molar mass they have same mol number, same mass so molar mass must be same. so 1 is correct
2 is correct because X and Y both form 2+ cation must be metal
3 is wrong because transition metals also form 2+ cations
answer is B
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ANYONE PLS
Q2 plzz ,Q30 when i do it using ethanol i get 38% while using ethanoic acid i get a different answer which is the correct one 50%??? , Q36 the firts point is correct No. of moles more on reactant side so increasing the pressure will increase the product and kp will increase ???
Thanks in advance and please answer