CHEM DOUBT

[cooldude100]

Hey, can anyone tell me the solution for question 5 part c and d for the following paper:
http://www.xtremepapers.me/CIE/Cambridg ... 2_qp_3.pdf

 

[haochen]

Hey, can anyone tell me the solution for question 5 part c and d for the following paper:
http://www.xtremepapers.me/CIE/Cambridg ... 2_qp_3.pdf

c)i) we use the given and mole so dont take the 150 coz it is in excess
so (20 * 11)/2 = 110
ii)same method as the first one (20 * 8) /2
iii) 80 + 40 = 120 (we got the 40 from the excess so 150 - 110 = 40)

d)9 / 54 = 0.166
3 * 0.166 = 0.5(we use 3 because the question mentioned that it is 3 mole of water)
0.5 * 18 = 9


hope u got it

 

[cooldude100]

thanx!!!!!!!!!!

 

[JmCullen]

Hey, can anyone tell me the solution for question 5 part c and d for the following paper:
http://www.xtremepapers.me/CIE/Cambridg ... 2_qp_3.pdf

5(c)(i) 110cm3 (11 times 10) because 11 mol.. since 2 mol of butyne.
(ii) 80cm3 (same. times 10)
(iii) 150cm3 - 110cm3 + 80cm3 = 120cm3

(d) number of moles of butyne reacted = 9 (divide) 54 = 0.17mol
number of moles of water formed 0.17 (times) 3 = 0.5 mol
mass of water formed = 0.5 (times) 18 = 9g

I don't know is this correct or not, but base on what i calculated, these are the answers.
hope it helps

 

[haochen]

thanx!!!!!!!!!!

ur welcm

 

[Ramosk95]

haoechen can u asnswer the whole qsn plz bcz the answers aren't in the ms?

 

[Ramosk95]

btw i think for c)i) we should caculate no. of moles of ch4h6 by cross multiplication 1molCH4H6 GIVES 24 Dm^3
???? CH4H6 gives .02dm^3
then use the answer in ci 2mol CH4H6 gives 11 mol of o2
.000833... ???
so the no. of moles of o2 would be .00458..... then multiply this answer by 24dm^3 which is 0.11 dm^3 :unknown:

 

[haochen]

haoechen can u asnswer the whole qsn plz bcz the answers aren't in the ms?

ok

5) a) liquid molecule are still close together but they slide past each other so it take the shape of the container.
while gas molecule are far apart so they are scattered and fill the space of the container.

b) i) Volume of the gas and time
ii) CO2 because it hv a lower Mr than S04.
CO2=12+(16*2)=34----------SO4=32+(16*4)=96

and the rest are in the previous post


hope u got it

 

[haochen]

btw i think for c)i) we should caculate no. of moles of ch4h6 by cross multiplication 1molCH4H6 GIVES 24 Dm^3
???? CH4H6 gives .02dm^3
then use the answer in ci 2mol CH4H6 gives 11 mol of o2
.000833... ???
so the no. of moles of o2 would be .00458..... then multiply this answer by 24dm^3 which is 0.11 dm^3 :unknown:

that is what i did but the way u did it is wrong

 

[Ramosk95]

btw i think for c)i) we should caculate no. of moles of ch4h6 by cross multiplication 1molCH4H6 GIVES 24 Dm^3
???? CH4H6 gives .02dm^3
then use the answer in ci 2mol CH4H6 gives 11 mol of o2
.000833... ???
so the no. of moles of o2 would be .00458..... then multiply this answer by 24dm^3 which is 0.11 dm^3 :unknown:

that is what i did but the way u did it is wrong

so u're sure about that! r8?

 

[Ramosk95]

btw i think for c)i) we should caculate no. of moles of ch4h6 by cross multiplication 1molCH4H6 GIVES 24 Dm^3
???? CH4H6 gives .02dm^3
then use the answer in ci 2mol CH4H6 gives 11 mol of o2
.000833... ???
so the no. of moles of o2 would be .00458..... then multiply this answer by 24dm^3 which is 0.11 dm^3 :unknown:

that is what i did but the way u did it is wrong

no its right man bcz .11 dm^3 =110 cm^3

 

[haochen]

from where u got the CH4H6??
and hw did u know 1 mol of it give 24 dm^3???
coz the uestion said 20 ??

 

[xIshtar]

One mole of gas takes up 24dm^3 at room temperature.

I linked the question on the other thread.

 

[Ramosk95]

from where u got the CH4H6??
and hw did u know 1 mol of it give 24 dm^3???
coz the uestion said 20 ??

c4h6 is written down in the equation
and 1 mol gives 24 dm^3 is written in the periodic table

 

[Ramosk95]

another qsn !
do these react or not
cu+h2so4??
probably they dont bcz h is above cu in the reactivity series but sulphuric acid acts as an oxidising agent so whats the solution here??