Chemistry: Post your doubts here!

[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?

 

[smartangel]

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

thamks. plz see if u can explain 21 of the same paper as well!

 

[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

can any1 help em out here??

 

[TheMan123]

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

The Cl radical can bond to any CH3 or CH2 carbons by substituting a hydrogen. There are 5 such carbons here, however there are 3 CH3 groups bonded directly to the alpha central carbon. If the Cl radical bonds to any three of this, you get the same molecule. The 3 CH3 groups form only one type of molecule. There are another two carbons which Cl can bond to, which makes 3 Chloroalkanes in total.
Answer: C. 3 (I checked)
Sorry for the messy explanation, not very good at this, ask me to clarify if needed
EDIT: Fixed typo

 

[Muhammad Asif]

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

the answer is C as the 2 CH3 groups are the side chain on same parent chain carbon thus makes no difference but if Cl substitutes any other H in parent chain it gives rise to 2 pther forms thus in total 3

 

[Muhammad Asif]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_1.pdf
what would be the equation in question no 1???????

 

[TheMan123]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
what would be the equation in question no 1???????

The full equation isnt really too important, you just need to know this:
FeTiO3-->TiO2 + other stuff
What you need to know is that 1 mole of FeTiO3 gives one mole of TiO2

 

[MysteRyGiRl]

somebodyyyyy???

 

[TheMan123]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
what would be the equation in question no 1???????

Anyway this is how you do it:
FeTiO3=151.7grams per mol
TiO2=79.9 grams per mol

Therefore the same moles of TiO2 as FeTiO3 will weigh 79.9/151.7 times as FeTiO3

so 79.9/151.7 * 19 tonnes= 10 tonnes

A

 

[Muhammad Asif]

can any1 help em out here??

the general equation for combustion of hydro carbon is CxHy+x(y/4)O2----->XCO2+(y/2)H2O
in such question volume ratio is molar ratio .........1 mol of hydrcarbon produces 3 mol of CO2 thus the carbon atoms are three......... 10cm3of hydrocarbon:50cm3 oxygen(70-20)
thus it is 1:5 if we put 5 in the general equation we get..... x+(y/4)=5 futhur solve t it is y=8.....
thus CxHy=C3H8

 

[TheMan123]

somebodyyyyy???

I'm on it! Give me a minute

 

[TheMan123]

I'm on it! Give me a minute

Nvm seems that Muhammad Asif is on a roll!