Chemistry: Post your doubts here!

[cute97]

^^^^^ I CHANGED THE LINK PLZZ CHECK ITS NOVEMBER 08

 

[6Astarstudent]

#2
n(CS2):n(O2):n(CO2):n(SO2) = 1:3:1:2
so 10 cm3 of CS2 will react with 30cm3 of O2 leaving 20cm3 of O2
but this will form 10cm3 of CO2 and 20cm3 of SO2 so 20+20+10 = 50

then you add NaOH
NaOH + CO2 -> NaHCO3
so all 10cm3 of CO2 disappears
NaOH + SO2 -> NaHSO3
so all 20cm3 of SO2 disappears
so 50-10-20 = 20
answer is C

#30
first you find theoretical value
n(ethanol) = 30/46=0.652mol
n(ethanoic acid) = 30/60 = 0.5mol
so ethanoic is limitting
they are all in 1 to 1 ratio so 0.5mol of ester will be formed
mass of ester = mol x molar mass = 0.5 x 88 = 44

actual yield is 22
so 22/44 x100% = 50%
which is C

#36
1 is wrong because only temperature alters Kc and Kp

by the way post more carefully next time, I answered all the questions for you in s08 then realized you wanted w08.

 

[cute97]

#2
n(CS2):n(O2):n(CO2):n(SO2) = 1:3:1:2
so 10 cm3 of CS2 will react with 30cm3 of O2 leaving 20cm3 of O2
but this will form 10cm3 of CO2 and 20cm3 of SO2 so 20+20+10 = 50

then you add NaOH
NaOH + CO2 -> NaHCO3
so all 10cm3 of CO2 disappears
NaOH + SO2 -> NaHSO3
so all 20cm3 of SO2 disappears
so 50-10-20 = 20
answer is C

#30
first you find theoretical value
n(ethanol) = 30/46=0.652mol
n(ethanoic acid) = 30/60 = 0.5mol
so ethanoic is limitting
they are all in 1 to 1 ratio so 0.5mol of ester will be formed
mass of ester = mol x molar mass = 0.5 x 88 = 44

actual yield is 22
so 22/44 x100% = 50%
which is C

#36
1 is wrong because only temperature alters Kc and Kp

by the way post more carefully next time, I answered all the questions for you in s08 then realized you wanted w08.

thank you frnd sorry but i was a stressed and confused

 

[lee mee..]

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..

Thanks alot

 

[strangerss]

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

how do I draw isopropyle methanoate? Why in Q 28 A and b are absolutely wrong? Sorry but I didn't get it and what about first question link? thanks a lot for the help

 

[6Astarstudent]

how do I draw isopropyle methanoate? Why in Q 28 A and b are absolutely wrong? Sorry but I didn't get it and what about first question link? thanks a lot for the help

iso propyl methanoate is like
H-C=O
|
O
|
C - CH3
|
CH3


and AB are wrong because molecular formula doesnt match, H is different

 

[cute97]

#2
n(CS2):n(O2):n(CO2):n(SO2) = 1:3:1:2
so 10 cm3 of CS2 will react with 30cm3 of O2 leaving 20cm3 of O2
but this will form 10cm3 of CO2 and 20cm3 of SO2 so 20+20+10 = 50

then you add NaOH
NaOH + CO2 -> NaHCO3
so all 10cm3 of CO2 disappears
NaOH + SO2 -> NaHSO3
so all 20cm3 of SO2 disappears
so 50-10-20 = 20
answer is C

i didnt get the first question first part where you added 20+20+10=50 ?? can u plz explain

 

[6Astarstudent]

i didnt get the first question first part where you added 20+20+10=50 ?? can u plz explain

ok so there are 20cm^3 of O2 left, and 20cm^3 of SO2 formed, and 10cm^3 CO2 formed
so total volume of gas is 20+20+10 = 50

 

[HassounFox]

anyone winter 2012 / qp:12 question: 17?

 

[6Astarstudent]

anyone winter 2012 / qp:12 question: 17?

first you find n(O2) the number of mol of oxygen
we know that 1 mol of gas under standard condition occupies 24dm3
so 300cm3 O2 = 0.3/24 = 0.0125mol of O2 or 0.025mol of O
since its a metal from group 1 or 2 so its either MO or M2O
so mol number = mass/molar mass
molar mass = mass/ mol number = 1.15 / 0.025 = 46
so M = 46 or 2M = 46 M =23
no element is 46, so M must be 23 which is sodium

 

[1357913579]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
34 anyone pleaseeeeeeeeeeeeeee

 

[HassounFox]

first you find n(O2) the number of mol of oxygen
we know that 1 mol of gas under standard condition occupies 24dm3
so 300cm3 O2 = 0.3/24 = 0.0125mol of O2 or 0.025mol of O
since its a metal from group 1 or 2 so its either MO or M2O
so mol number = mass/molar mass
molar mass = mass/ mol number = 1.15 / 0.025 = 46
so M = 46 or 2M = 46 M =23
no element is 46, so M must be 23 which is sodium

thank you

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
34 anyone pleaseeeeeeeeeeeeeee

in order to let the right hand side rise, there needs to have an increase in pressure at R
1 is correct because the reaction is endothermic, so forward reaction is favoured by the increase in temperature
so more 2NF2 is formed, hence more pressure.
2 is wrong because
CH3NC(g) -> CH3CN(g) they are 1:1 pressure will not be changed by equilibrium
3 is wrong because same gas..

so answer is D

 

[1357913579]

in order to let the right hand side rise, there needs to have an increase in pressure at R
1 is correct because the reaction is endothermic, so forward reaction is favoured by the increase in temperature
so more 2NF2 is formed, hence more pressure.
2 is wrong because
CH3NC(g) -> CH3CN(g) they are 1:1 pressure will not be changed by equilibrium
3 is wrong because same gas..

so answer is D

why would increase in production of n2f4 increase pressure? X isnt NF2 is could also be N2H4 isnt it?

 

[strangerss]

iso propyl methanoate is like
H-C=O
|
O
|
C - CH3
|
CH3


and AB are wrong because molecular formula doesnt match, H is different

thanks a lot I AM DONE WITH CHEMISTRY AS

 

[waleed82]

As-salam-mu-alikum,

This is Walid. Can anyone please tell me that , can we appear as a private student for CAMBRIDGE INTERNATIONAL AS / A Levels CHEMISTRY?

Please reply as soon as possible! I'll be waiting for the reply!

Thanks.

JAZA-KALAH-KHAIR!

 

[AbbbbY]

As-salam-mu-alikum,

This is Walid. Can anyone please tell me that , can we appear as a private student for CAMBRIDGE INTERNATIONAL AS / A Levels CHEMISTRY?

Please reply as soon as possible! I'll be waiting for the reply!

Thanks.

JAZA-KALAH-KHAIR!

Yeah, you can.

 

[waleed82]

Yeah, you can.

But what about the practicals?????

 

[AbbbbY]

But what about the practicals?????

You'll probably be allotted a school to give the practical in.

 

[waleed82]

You'll probably be allotted a school to give the practical in.

ahha... well do we have the option of ATP in cambridge chemistry?