• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

HubbaBubba

HubbaBubba

Messages
213
Reaction score
135
Points
53
Raiyan3 said:
Hubbabubba looks like you have done w11 paper12. How did you do question number 4? :/
Click to expand...

Sorry for the late reply, I had to revise a topic in chemistry. BTW I did Summer 2011 not Winter but I'll help you

q4 - It says there's 96kg of Hydrogen leftover, so 120-96 = 24kg of H2 was used in the reaction.

no. of moles of H2 = 24kg/2 = 12 moles.
N2 + 3H2 ---> 2NH3.

Ratio of Hydrogen to Ammonia is 3: 2
that means number of moles of ammonia produced is 8 moles.
Then you calculate the mass of ammonia which is moles * Molar Mass (8 * 17) = 136kg. Answer is C
 
HubbaBubba

HubbaBubba

Messages
213
Reaction score
135
Points
53
21 An alkene has the formula CH3CH=CRCH2CH3 and does not possess cis-trans isomers. What is R?
A) H
B) Cl
C) CH3
D) C2H5

Why can't it be B? Why is the answer D?
 
6

6Astarstudent

Messages
140
Reaction score
106
Points
53
HassounFox said:
winter 2012 qp:13 question 15? can anybody explain?
Click to expand...
#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
 
H

HassounFox

Messages
12
Reaction score
0
Points
1
6Astarstudent said:
#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
Click to expand...
thank you

how about question 14???
 
6

6Astarstudent

Messages
140
Reaction score
106
Points
53
Rabb94 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
question-4,11,15
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
question- 9,13,14
plzzzzzzzzz anyone
Click to expand...

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
 
6

6Astarstudent

Messages
140
Reaction score
106
Points
53
HassounFox said:
thank you

how about question 14???
Click to expand...
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
 
H

HassounFox

Messages
12
Reaction score
0
Points
1
6Astarstudent said:
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
Click to expand...


Thank you very much your effort in helping me is very appreciated :D
 
Raiyan3

Raiyan3

Messages
124
Reaction score
49
Points
38
HubbaBubba said:
21 An alkene has the formula CH3CH=CRCH2CH3 and does not possess cis-trans isomers. What is R?
A) H
B) Cl
C) CH3
D) C2H5

Why can't it be B? Why is the answer D?
Click to expand...

Thankyou so much! I was just leaving for the examination center! Haha! Hope it goes well for you :)
 
Aymen Ezazi

Aymen Ezazi

Messages
32
Reaction score
32
Points
8
Kyusam said:
http://papers.xtremepapers.com/CIE/Cambridge International A and 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 !!
Click to expand...


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
 
You must log in or register to reply here.
Top