Chemistry: Post your doubts here!

[qwertypoiu]

Isn't it 400 for D ?

But violet isn't in the option

 

[The Sarcastic Retard]

Here you go. It would be a really crowded structure in a space fill model.

Thanks.

 

[holoholo]

But violet isn't in the option

how about this question ?

Value in c(i) is 7 x 10^-7 . in the mark scheme it's written concentration of product if 0.1 but shouldn't it be 0.1 - (7 x 10^-7 )?

 

[The Sarcastic Retard]

how about this question ?

Value in c(i) is 7 x 10^-7 . in the mark scheme it's written concentration of product if 0.1 but shouldn't it be 0.1 - (7 x 10^-7 )?

Why will u do that?

 

[holoholo]

Because at first the concentration is of Ag+ is 0.1 then it is reduced to 7 x 10^-7 . As it is 1:1 mole ratio shouldn't the concentration of the product increase by 0.1 - 7 x 10^-7 ?

 

[The Sarcastic Retard]

how about this question ?

Value in c(i) is 7 x 10^-7 . in the mark scheme it's written concentration of product if 0.1 but shouldn't it be 0.1 - (7 x 10^-7 )?

You made ur equation of Kc.
Just substitute the values and find [NH3(aq)]
o.1 is conc of silver nitrate soln and (7 x 10^-7 ) is conc of Ag.

 

[holoholo]

You made ur equation of Kc.
Just substitute the values and find [NH3(aq)]
o.1 is conc of silver nitrate soln and (7 x 10^-7 ) is conc of Ag.

Yes but we need to use the concentration of Ag(NH3)2 not of silver nitrate

 

[Abdullahassan_99]

Can anyone plz explain me why the answer for chemistry may june 2007 v1 p1 first answer is C?

 

[Awesome12]

View attachment 59644 Can anyone plz explain me why the answer for chemistry may june 2007 v1 p1 first answer is C?

= (Mass no x Relative Abundance) / (Relative Abundance)

= {(10 * 1) + (11 * 4)} / 1 + 4

= 10.8

There is no peak for the mass no. 12, so we don't need to consider it. Note: Even if you consider it, it will be 12 * 0 = 0

 

[holoholo]

Explanation for question (c) please ?

 

[holoholo]

Can someone please sketch this graph ?

 

[The Sarcastic Retard]

View attachment 59647

Explanation for question (c) please ?

From part (a) you can see that there is no effect of Cl- ion on production of O2 where as if u see the reaction of chlorine the equilibrium shifts to form more chlorine and moves in direction of electrons thus the E value will be less positive that is it will decrease.

 

[Muhammad Amer]

I don't know if it has been uploaded here before, but I've been looking for a PDF version of this book for ages, and I found it.

It can be downloaded from here.

You need a torrent down-loader to download it, μtorrent should be fine.

 

[Muhammad Amer]

I don't know if it has been uploaded here before, but I've been looking for a PDF version of this book for ages, and I found it.

It can be downloaded from here.

You need a torrent down-loader to download it, μtorrent should be fine.

I believe the CD has been uploaded here before, so you could find it by using the search bar in xtremepapers. If not, here is a torrent link for the CD.

 

[The Sarcastic Retard]

View attachment 59648

Can someone please sketch this graph ?

qwertypoiu
My try :

 

[Zoha Ali]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_11.pdf

Can anyone please explain question 11 and 12?

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_11.pdf

Can anyone please explain question 11 and 12?

The tricky thing involved in the 11 question is the balancing of the redox reaction equation. You're given this reaction:
Sn+2 + MnO4 -1 + 4H+ ----> Mn+2 + Sn+4 + 2H2O

You can see that this equation is not balanced. So first balance the equation. Split the equation into oxidation and reduction halves.

Oxidation Half:
MnO4 -1 + 4H+ + 5e ----> Mn+2 + 2H2O

Reduction Half:
Sn+2 ---> Sn+4 + 2e

Equal the no. of electrons in both the equations by multiplying the equation for oxidation half with 2, and the reduction half with 5. You'll get:

Oxidation half:
2MnO4 -1 + 8H+ + 10e ----> 2Mn+2 + 4H2O

Reduction half:
5Sn+2 ---> 5Sn+4 + 10e

Now combine these equations:
5Sn+2 + 2MnO4 -1 + 8H+ ----> 5Sn+4 + 2Mn+2 + 4H2O

^ This is the balanced equation for the reaction. Now use the ratios thingy, to calculate the moles.
You're given that there're 0.05 moles of Sn+2.
5 Moles of Sn+2 give rise to 2 moles of Mn+2, so 0.05 moles of Sn+2 will give :

0.05 * 2 /5 = 0.02 moles of Mn+2

So the answer is B.

 

[Eugene99]

Is my enthalpy cycle for Q12 correct?

If it is, how do I calculate the enthalpy change of combustion now?

 

[Rizwan Javed]

View attachment 59660
Is my enthalpy cycle for Q12 correct? View attachment 59661

If it is, how do I calculate the enthalpy change of combustion now?

Your cycle is correct, but you don't need to make any cycles here. Do this question with bond energies. Find the energy absorbed for bond-breaking, and the energy released when bonds were formed. Then simply subtract the energy released from energy absorbed to get the enthalpy change.

 

[Rizwan Javed]

There are 4 C--H bonds and 2 O=O bonds broken. So calculate the energy absorbed for this:

4 * [C--H] + 2 [O=O] = 4 * 410 + 2 * 496 = 2632

There are 2 C=O bonds and 4 O--H bonds formed. So calculate the energy released:
2 * 740 + 4 * 460 = 3320

Now find the enthaply:

energy absorbed - energy released
= 2632 - 3320
= - 688 kJ/mol

So the answer is D.