Chemistry: Post your doubts here!

[Zsiddiqui]

HELP Please :/

 

[étudiante]

I asked this question already, but just thought I post it again since we're on a new page.

An aqueous solution was prepared containing 1.0mol of AgNO3 and 1.0mol of FeSO4 in 1.00 dm3
of water. When equilibrium was established, there was 0.44mol of Ag+(aq) in the mixture.
Ag+(aq) + Fe2+(aq) --> Ag(s) + Fe3+(aq)

What is Kc?

A. 0.35 B. 0.62 C. 1.62 D. 2.89

I got C as my answer but the correct one is D, can anyone explain how to get D?

I'm not getting any of the options
wat method r u using please?

 

[Zsiddiqui]

I asked this question already, but just thought I post it again since we're on a new page.

An aqueous solution was prepared containing 1.0mol of AgNO3 and 1.0mol of FeSO4 in 1.00 dm3
of water. When equilibrium was established, there was 0.44mol of Ag+(aq) in the mixture.
Ag+(aq) + Fe2+(aq) --> Ag(s) + Fe3+(aq)

What is Kc?

A. 0.35 B. 0.62 C. 1.62 D. 2.89

I got C as my answer but the correct one is D, can anyone explain how to get D?

even i am getting option C. Can you tell me which year paper is it ?

 

[saudha]

I'm not getting any of the options
wat method r u using please?

me neither trying frm long!!

 

[TheMan123]

HELP Please :/
View attachment 8856

2. Total Flask Volume=5+10= 15dm3=15/1000 m3
Next work out the number of moles of helium and neon with pv=nRT
n=pv/RT
Since RT is constant throughout, let's make life easier by letting RT be an unknown, let's say A
For helium: n=[12000*(5/1000)]/A
For Neon: n=[6000*(10/1000)]/A
Total n=120/A

To find total P, use pv=nRT=nA

pV=nA
p=nA/V
p=[(120/A)*A]/(15/1000)
p=8000Pa=8kPa

 

[étudiante]

me neither trying frm long!!

same here
what methods r u using? could u share please?

 

[TheMan123]

even i am getting option C. Can you tell me which year paper is it ?

its winter 2011 variant 13 question 12. No examiners report out yet unfortunately
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

 

[smartangel]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q4..plz explain!

 

[TheMan123]

HELP Please :/
View attachment 8856

I answered your first question, here is the next one:
When you polymerise, you start with many monomers to make one or a few polymers, because you are combining monomers.
so if you start with one mole of monomers, you combine them and you end up with less monomers, therefore the answer could be from anything less than 1 mole to the minimum of 1 molecule, which the inverse of 6.02*10^23, option 3 of the question

 

[MysteRyGiRl]

10 cm3 of hydrocarbon is burned in 70cm3 of oxygen(excess) the funal gaseous mixture contains 30cm3 of ccarbon dioxide n 20 cm3 of unreacrted oxygen.whats the formula of hydrocarbon?
a) C2H5 b)C3H6 c)C3H8 d)C4H10
hows it c? these r supoosed 2 be easy questions bt cant get the answers

 

[TheMan123]

me neither trying frm long!!

No point telling you my method, i got the wrong answer too

 

[Zsiddiqui]

its winter 2011 variant 13 question 12. No examiners report out yet unfortunately
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

In Kc we do not take the concentration of the solid. So in this case we will not include the concentration of Ag(s)
So it will be as follows:
Kc= (0.56) / (0.44^2)
= 2.89
That's option D

 

[TheMan123]

In Kc we do not take the concentration of the solid. So in this case we will not include the concentration of Ag(s)
So it will be as follows:
Kc= (0.56) / (0.44^2)
= 2.89
That's option D

Thanks! That was a tough one

By the way I answered your question too

 

[étudiante]

In Kc we do not take the concentration of the solid. So in this case we will not include the concentration of Ag(s)
So it will be as follows:
Kc= (0.56) / (0.44^2)
= 2.89
That's option D

thanks a lot! i was trying the same thing of excluding Ag(s) but.... i thought we need to claculate its number of moles too ryt? then wat do we do wid da number of moles of Ag(s)?

 

[TheMan123]

thanks a lot! i was trying the same thing of excluding Ag(s) but.... i thought we need to claculate its number of moles too ryt? then wat do we do wid da number of moles of Ag(s)?

You just ignore it the formula would be Kc=[Fe 3+]/[Ag +][Fe 2+]=.56/.44*.44

 

[smartangel]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q4..plz explain!

 

[Muhammad Asif]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q4..plz explain!

the ansewer is C you have to keep in mind that bond making is exothermic and bond breaking is endothermic.......
In P Br2 is broken to 2Br thus energy is given so ∆H=+193
In Q 2cl are converted into Cl2 thus energy is released so ∆H=-244
In R CH3+Cl a bind is made b/w C and Cl so ∆H=-340
In S CH4-----> CH3+H thus a bind b/w C and H is broken ∆H=+410
if we arrange ∆H in increasing order it is:-340,-244,+193,+410
thus R,Q,P,S

 

[étudiante]

You just ignore it the formula would be Kc=[Fe 3+]/[Ag +][Fe 2+]=.56/.44*.44

alright! so we dont need the moles of Ag den... THANKS A LOT

 

[TheMan123]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf
Q4..plz explain!

Remember this: Bond breaking is endothermic(positive) and Bond making is exothermic(negative)
Enthalpy changes:
P=+193 Br-Br bond broken
Q=-244 Cl-Cl made
R=-340 C-Cl made
S=+410 C-H broken
Answer C. RQPS

 

[smartangel]

the ansewer is C you have to keep in mind that bond making is exothermic and bond breaking is endothermic.......
In P Br2 is broken to 2Br thus energy is given so ∆H=+193
In Q 2cl are converted into Cl2 thus energy is released so ∆H=-244
In R CH3+Cl a bind is made b/w C and Cl so ∆H=-340
In S CH4-----> CH3+H thus a bind b/w C and H is broken ∆H=+410
if we arrange ∆H in increasing order it is:-340,-244,+193,+410
thus R,Q,P,S

thanks..could you please solve 21 of the same paper?