Chemistry: Post your doubts here!

[Metanoia]

http://www.youtube.com/playlist?list=PL4Jmce8VJnNXSmX2z10ACyurQ7mWgY0fq

Video of June 2011 is uploaded.

 

[kanchan dhiman]

Here you need to know two concepts/formulas:
mole fraction = number of moles of molecule A / total number of moles
Partial pressure of A = mole fraction of A x total pressure

If 20% of steam had been converted to hydrogen and oxygen, there must be 80% of steam left. So, if you before had 1 mole of steam, you now have 0.8 moles of steam (80% of 1.). Because of the 2:2:1 ratio in the equation, 2H2O ---> 2H2 + O2 , if 0.2 moles of steam had been converted into its products, then by simple stoichiometry the change for H2 is +0.2 and for O2 +0.1.

So you now have the moles of each substance present at equilibrium. If you now use the formulas from the beginning, to calculate the partial pressure, you would see you need the mole fraction of your gas and the total pressure. So, mole fraction = number of moles of molecule A / total number of moles

For steam, you have 0.80 moles, for H2 0.20 and for O2 0.10. If you add them up this give 1.1 - the total number of moles. Therefore, the mole fraction of steam is 0.80/1.1, for H2 is 0.20/1.1 and for O2 is 0.10/1.1. To get the partial pressure of each gas you now times the mole fraction of each substance by the total pressure at which the experiment is done - 1 atm. Hence for steam the partial pressure is 0.8 x 1 / 1.1 and so on...

Hope it helps.

How do you get mole of H2 and O2 please explain?????

 

[not.maria]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_12.pdf

can anyone kindly explain:
q6 10 11 23 26

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_12.pdf

can anyone kindly explain:
q6 10 11 23 26

6:B because the graph will move to the left with its peak becoming higher (hence A will move up) but to maintain the number of particles, B and C will have to move down because the number if particles having a lower energy will increase and those having higher energies will decrease
10:C draw up ratios. The number of atons of and percentages of O,H and Ca are already given to you. The percentage divided by the number if atoms will always give a co start value. Find that value and equate it to the ratio of the percentages & number of atoms of the other atoms. You'll end up with he same answer for both Si and Al ie 3

 

[sweet.sugar]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q1-c
i got D cz pb(c2h5)4 +o2 -->pbo + co2
2*4=8 c and 5*4=20h on react side so if we balance i got 8 oxy frm co2 and 10 oxygens from h2o n 1 oxy from pbo total 27 :/
Q4-c (explain it)
Q6-a (h2=hr-h1 so -395+297=-98 so y is it a ?)
Q12-d (i wrote the eq n for mg n s it is 1 mol of o2 n for al it is 3 moles (3o2) so y is the curve for s the highest ie ans shud be c ) :/
Q20-b (plz write the isomers )
Plz help

 

[sadiaali]

O__O x.x help
Q40
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf

 

[kitkat <3 :P]

O__O x.x help
Q40
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf

B ?

 

[kingo44]

omg from where now this schoking question came plz guys post some f in solutions

 

[sadiaali]

B ?

Its A

 

[kitkat <3 :P]

Its A

sorry
but i think all the compounds have C:H ratios dont they :/?

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_12.pdf

Q1

 

[ZaqZainab]

O__O x.x help
Q40
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf

1 aldehydes and ketones actually contain the same no of C and H so its defiantly a yes
2 take a carboxylic acid you decide?
for example ethanoic acid C2H4O2
or just CnH2nO2
C:H= n/2n=1/2
now lets take an ester
methyl ethanote C3H6O2
or just CnH2nO2
C:H= n/2n=1/2

3 alkenes CnH2n (n/2n)
ketones CH3COCH3
C3H6O
CnH2nO
C:H= n/2n

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_12.pdf

Q1

The 2 gases produced at the anode are oxygen and carbon dioxide , the equations are
O2-= 2e +O2 oxidation number changes from -2 to 0 ( increases by 2)
C+O2=CO2 oxidation number changes from 0 to -2 ( decreases by 2)

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_12.pdf

Q1

Al2O3
all i knew about was O2 but as they are indicating Carbon electrodes CO2 might also be released as O2 reacts with Carbon electrode
so we have O2- from Al2O3 which become O2 and then CO2
so from -2 to 0 to -2

 

[sweet.sugar]

Am i

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q1-c
i got D cz pb(c2h5)4 +o2 -->pbo + co2
2*4=8 c and 5*4=20h on react side so if we balance i got 8 oxy frm co2 and 10 oxygens from h2o n 1 oxy from pbo total 27 :/
Q4-c (explain it)
Q6-a (h2=hr-h1 so -395+297=-98 so y is it a ?)
Q12-d (i wrote the eq n for mg n s it is 1 mol of o2 n for al it is 3 moles (3o2) so y is the curve for s the highest ie ans shud be c ) :/
Q20-b (plz write the isomers )
Plz help

invisible :/

 

[ZaqZainab]

Am i

invisible :/

Q1
Pb(C2H5)4 +13.5 O2= 8 CO2+10 H2O+PbO
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

 

[sweet.sugar]

Q1
Pb(C2H5)4 +O2= 8 CO2+10 H2O+PnO3
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

Yeah thanks tht helped
Bt a small doubt for q-12 u said for sulphur only 1 mol of o2 is used ryt n thts wt i said too bt in the graph it says sulphur used the most moles o o2 as the curve for S is the highest

 

[ZaqZainab]

Yeah thanks tht helped
Bt a small doubt for q-12 u said for sulphur only 1 mol of o2 is used ryt n thts wt i said too bt in the graph it says sulphur used the most moles o o2 as the curve for S is the highest

we have 1/2 3/4 and 1
which one is the most? 1 right?

 

[sweet.sugar]

Q1
Pb(C2H5)4 +O2= 8 CO2+10 H2O+PnO3
this is the formula what about H2O??
Q4 C as Hydrogen atom is attached to Oxygen atom
Q6
Because of the moles
the enthalpy change is -297 then 1 mole of SO2 is formed but we need 2 moles to be formed and so enthalphy change wil be (-297*2)
and so for the SO3 it will be (-395*2)
now do you your calculations
Q12 with Mg it is 1/2 mole O2 Mg+1/2O2=MgO
with Al it is 3/4 O2 Al+3/2 O2= 1/2 Al2O3
with Sulpur it is 1 mole of O2 S+O2=SO2
Q20
CHCl=CHCl
CH2=CCl2
And a cis trans
for CHCl=CHCl
like in one of them H will be on the top while in another at the bottom for the second carbon

Besides im impressed by the explantion of q1 cz there is no suh thing pno3 n i managed to get the ans by myself we divides 27 by 2 n ans is 13.5 cz o2 molecule has a 2 so tht would be something lile 13.5 * 2 = 27
Thx anyway

 

[ZaqZainab]

Besides im impressed by the explantion of q1 cz there is no suh thing pno3 n i managed to get the ans by myself we divides 27 by 2 n ans is 13.5 cz o2 molecule has a 2 so tht would be something lile 13.5 * 2 = 27
Thx anyway

There such thing as called PbO3 it was typo
and i wanted you to get it yourself that's why i stated just the equation