Chemistry: Post your doubts here!

[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.

 

[Eugene99]

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.

thanks a lot
but wait...why didn't we use the value of 'C=O in CO2' which is 805 as per the data booklet?

 

[Rizwan Javed]

thanks a lot
but wait...why didn't we use the value of 'C=O in CO2' which is 805 as per the data booklet?

Actually this addition of Bond energy value has been made for 2016 examinations. Previously in the old data booklet, there was no discrimination between these two values of C=O. As this question from one of the previous years, so i used the previous data booklet value, because had i used the new one for 2016 syllabus, the answer would have been wrong. But we're to use the new value in all the questions relating to C=O of CO2, in our examinations of 2016 onwards.

 

[Eugene99]

Actually this addition of Bond energy value has been made for 2016 examinations. Previously in the old data booklet, there was no discrimination between these two values of C=O. As this question from one of the previous years, so i used the previous data booklet value, because had i used the new one for 2016 syllabus, the answer would have been wrong. But we're to use the new value in all the questions relating to C=O of CO2, in our examinations of 2016 onwards.

oooh right! that clears it up! thank you

 

[Eugene99]

I'm having great trouble in enthalpy cycles...is this one correct now? and also why did we multiply the enthalpy change of combustion of hydrogen and carbon by 2?
I did the working wrong because I multiplied the combustion of ethene by 2 instead of carbon and hydrogen

 

[holoholo]

qwertypoiu
My try :
View attachment 59658

Is there a specific reason why it started to curve at 10 and ended at 20 ?

 

[qwertypoiu]

I'm having great trouble in enthalpy cycles...is this one correct now? and also why did we multiply the enthalpy change of combustion of hydrogen and carbon by 2?
I did the working wrong because I multiplied the combustion of ethene by 2 instead of carbon and hydrogen
View attachment 59664

There are two moles of C. So you multiply its enthalpy change by 2.
There are two moles of H2. So you multiply its enthalpy change by 2.
There is only one mole of C2H4. So you leave it as it is (or multiply 1 if you wish)

Your cycle is correct. Some may be picky about the box you have, where you only have 1 mole of CO2 and H2O, but in reality you should have 2 each. But that doesn't affect anything, it's just a small technicality.

 

[Mohammed Elatta]

I'm having great trouble in enthalpy cycles...is this one correct now? and also why did we multiply the enthalpy change of combustion of hydrogen and carbon by 2?
I did the working wrong because I multiplied the combustion of ethene by 2 instead of carbon and hydrogen
View attachment 59664

Here , thought its too much to write...

 

[qwertypoiu]

Is there a specific reason why it started to curve at 10 and ended at 20 ?

No the graph drawn here is not correct.
Yes the pH starts at 11.3 and ends at 1.6.

However this titration will have two end points. This is because they told us the compound undergoes two acid base reactions. We even write their equations.

So it will gradually decrease as usual and then at the 10cm3 mark there will be a steep part to show one end point, then it will decrease gradually again, then at 20cm3 mark it will have a steep decrease again to show the second end point. Then gradual decrease again.

Have a read through this page for better understanding this topic:
http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

 

[Eugene99]

There are two moles of C. So you multiply its enthalpy change by 2.
There are two moles of H2. So you multiply its enthalpy change by 2.
There is only one mole of C2H4. So you leave it as it is (or multiply 1 if you wish)

Your cycle is correct. Some may be picky about the box you have, where you only have 1 mole of CO2 and H2O, but in reality you should have 2 each. But that doesn't affect anything, it's just a small technicality.

Here , thought its too much to write...

thank you so much...but if mine is correct why did Mohammed Elatta draw it in another way?

 

[holoholo]

Can someone please explain the second part please ? Why can it only be CH3+ why cant it be H2 or CH2?

 

[holoholo]

No the graph drawn here is not correct.
Yes the pH starts at 11.3 and ends at 1.6.

However this titration will have two end points. This is because they told us the compound undergoes two acid base reactions. We even write their equations.

So it will gradually decrease as usual and then at the 10cm3 mark there will be a steep part to show one end point, then it will decrease gradually again, then at 20cm3 mark it will have a steep decrease again to show the second end point. Then gradual decrease again.

Have a read through this page for better understanding this topic:
http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

Can you please show me a sketch ?

 

[Eugene99]

CaCO3 >> CaO + CO2
total mass 1200 million tonns 1 mol of carbonate will decompose into 1 mol of CO2...
what next?

 

[Saad the Paki]

View attachment 59681
CaCO3 >> CaO + CO2
total mass 1200 million tonns 1 mol of carbonate will decompose into 1 mol of CO2...
what next?

You need to find the moles of CaCO3 first. 1200 million tonnes is 1200x10^12. Let's just ignore the 10^12 as the answer will also be in million tonnes. So moles of carbonate = 1200/100.1=11.988
CO2 will have an equal number of moles too so to find the mass of co2 = moles×molar mass = 11.988×44=527. So B is the answer

 

[Eugene99]

You need to find the moles of CaCO3 first. 1200 million tonnes is 1200x10^12. Let's just ignore the 10^12 as the answer will also be in million tonnes. So moles of carbonate = 1200/100.1=11.988
CO2 will have an equal number of moles too so to find the mass of co2 = moles×molar mass = 11.988×44=527. So B is the answer

oh! that was simple, (maybe I just had a dizzy mind then ) Thank you!..