Chemistry: Post your doubts here!

[Studydayandnight]

is it B

Yes it is, how did you figure it out? :/

 

[Boss201]

Yes it is, how did you figure it out? :/

The book has the answer sheet.

its an old paper J90/1/5

 

[The Sarcastic Retard]

when 20cm (cube) of a gaseous hydrocarbon were completely burnt in excess of oxygen, 60cm(cube) of carbon dioxide and 40cm(cube) of water vapour were formed, all volumes being measured at same temperature and pressure.

what is the formula of the hydrocarbon ?

A.C2H6
B.C3H4
C.C3H6
D.C3H8

ANS : B

how to work out this question ?

Thanks

Vol of oxygen : 80cm^3
Hydrocarbon + O2 --------> CO2 + H2O
Set ratio.
1Hydrocaron + 4O2 ---> 3CO2 + 2H2O
You can see there are 3 carbon and 4 hydrogen hence C3H4

 

[princess Anu]

there is no intermediate for this reaction. bcs it proceeds by SN 2 mechanism.. and a hump on the enthalpy change diagram is seen only when there is a intermediate

Yes it is, how did you figure it out? :/

 

[Boss201]

Vol of oxygen : 80cm^3
Hydrocarbon + O2 --------> CO2 + H2O
Set ratio.
1Hydrocaron + 4O2 ---> 3CO2 + 2H2O
You can see there are 3 carbon and 4 hydrogen hence C3H4

HOw you got volume of oxygen 80cm(cube)

is there any kind of formula you used ?

 

[The Sarcastic Retard]

HOw you got volume of oxygen 80cm(cube)

is there any kind of formula you used ?

All volumes being measured at same temperature and pressure.
100cm^3 on RHS
20cm^3 on LHS
so there should be 80cm^3 more on LHS to balance the volume on both side.

 

[princess Anu]

help please

 

[The Sarcastic Retard]

help please

Initial moles H2O = 1
Initial moles CO = 1
Eq moles H2O = 1-x
Eq moles CO = 1-x
Eq moles H2 = x
Eq moles CO2 = x

Total moles = (1-x) + (1-x) + x + x = 2
if hydrogen occupies 33% = 2/3 moles hence CO2 is also 2/3 moles and CO and H2O have 1/3 moles

Now put the values into the equilibrium law.

 

[Mahnoorfatima]

http://www.sheir.org/a-level-chemistry-11-june2011.pdf
Question 8?
And in 11 why's D wrong?
13 My answer's D. What did I do wrong in this calculation:
Al2O3: 6HCL
0.02x6= 2xvolume/1000
AND I DONT GET 15 AND 16 AT ALL!

 

[fantastic girl]

http://www.sheir.org/a-level-chemistry-11-june2011.pdf
Question 8?
And in 11 why's D wrong?
13 My answer's D. What did I do wrong in this calculation:
Al2O3: 6HCL
0.02x6= 2xvolume/1000
AND I DONT GET 15 AND 16 AT ALL!

FOR 11 - In D see the forces are getting stronger as it goes frm liquid to solid but in case of C we need to break the forces in solid to form liquid...So in C we need to overcome the forces.

 

[Mahnoorfatima]

Why cant we use NaOH(aq) and K2Cr2O7/H2SO4 to convert BrCH2CH2Br to X and to HO2CCH2CH2CO2H??

 

[fantastic girl]

http://www.sheir.org/a-level-chemistry-11-june2011.pdf
Question 8?
And in 11 why's D wrong?
13 My answer's D. What did I do wrong in this calculation:
Al2O3: 6HCL
0.02x6= 2xvolume/1000
AND I DONT GET 15 AND 16 AT ALL!

Q13- 4Al + 3O2 ---> 2AlO3
Al2O3 + 6HCl ----> 2AlCl3 + 3H2O

Ratio of Al : Al2O3
4 = 2
0.02 = ?
=0.01 moles of Al2O3
now ratio of Al2O3 : HCl
1 = 6
0.01 = ?
=0.06
use the formula no.of moles = vol x conc. to find the volume
that will be 0.06/2 = 0.03 dm3

 

[Mahnoorfatima]

Q13- 4Al + 3O2 ---> 2AlO3
Al2O3 + 6HCl ----> 2AlCl3 + 3H2O

Ratio of Al : Al2O3
4 = 2
0.02 = ?
=0.01 moles of Al2O3
now ratio of Al2O3 : HCl
1 = 6
0.01 = ?
=0.06
use the formula no.of moles = vol x conc. to find the volume
that will be 0.06/2 = 0.03 dm3

But why don't you take the moles of Al2O3 to that of HCl directly?

 

[fantastic girl]

But why don't you take the moles of Al2O3 to that of HCl directly?

is the moles given in the question ?
they hve given us Al moles only

 

[fantastic girl]

http://www.sheir.org/a-level-chemistry-11-june2011.pdf
Question 8?
And in 11 why's D wrong?
13 My answer's D. What did I do wrong in this calculation:
Al2O3: 6HCL
0.02x6= 2xvolume/1000
AND I DONT GET 15 AND 16 AT ALL!

check that video for Q8

 

[Abdul Hanan]

So we have an upcoming chemistry MCQ exam and i keep getting MCQs for organic chemistry wrong
any advice?
Thanks

 

[princess Anu]

Initial moles H2O = 1
Initial moles CO = 1
Eq moles H2O = 1-x
Eq moles CO = 1-x
Eq moles H2 = x
Eq moles CO2 = x

Total moles = (1-x) + (1-x) + x + x = 2
if hydrogen occupies 33% = 2/3 moles hence CO2 is also 2/3 moles and CO and H2O have 1/3 moles

Now put the values into the equilibrium law.

but my answer comes wrong if I do it this way
I have the solution for this question with the correct ans but
I don't get why we have to multiply moles of H2 by 2

 

[Metanoia]

but my answer comes wrong if I do it this way
I have the solution for this question with the correct ans but
I don't get why we have to multiply moles of H2 by 2

You can post the answer scheme here for checking.

This question is familiar, I believe it was answered before.

View attachment 51471

 

[princess Anu]

You can post the answer scheme here for checking.

This question is familiar, I believe it was answered before.

Yeah it was

 

[princess Anu]

..