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Chemistry: Post your doubts here!

salvatore

salvatore

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Ahmedraza73 said:
Guyz please help me with question number 25 and 34.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Please help someone?:(
Click to expand...
The answer for qn. 25 is C.
To convert a halogenoalkane to a nitrile, we need to react it with cyanide ions (Usually potassium cyanide, KCN, in ehtanol). So X is solved.
Hydrolysis is used to convert a nitrile into a carboxylic acid. So, by reacting the butanitrile with dilute HCL, the carbon-nitrogen triple bond is broken down by water to form butanoic acid.

Hope qn 34 is made clear by the other person :)
 
Muhammad Bin Anis

Muhammad Bin Anis

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Ahmad Zaeem said:
M/J 2011 P 12 ... Qs no (11,17,22,29,31,38,40)..Only!!........Concept plsssss!!!!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Click to expand...

11 - should be C
17 - should be B because Magnesium is less reactive than group 1 metals which react vigourously and dissolve in water. Mg reacts to a small extent
22 - A ... C and D are cancelled at first place because we do not itend to increase the carbon chain by adding CN
29 - C .. Na reacts with acids and alcohols... neither of them present here
31 - A ... all bonds to the max possible distance b/w them (C in CH3 has 3 electrons)
38 - A.... for an isomer , empirical formula ought to be same,functional groups same in stereomerism and this is an optical isomer so no change in structure
40 - B ... aqeous bromine reacts with double bonded carbons as in ethene , 2,4DNPH gives pos. result in case of ketone/ aldehyde presence (carbonyl group). Fehling solution gives pos. result only in the presence of aldehyde which isnt present
 
Oliveme

Oliveme

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salvatore said:
I'm re-posting this.. anyone?
Click to expand...
C + O2 → CO2 -395 } this reaction produces C for CS2
S + O2 → SO2 -298 } this reaction produces S for CS2 but since there are 2 S, we multiply the ∆Hc by 2.
CS2 + 3O2 → CO2 + 2SO2 -1110 -> combustion of CS2

for the formation, you add the top two reaction enthalpies together and minus from the overall combustion enthalpy.
C + 2S → CS2 ∆H = -395 + 2(-298) -(-1110) = +119 kJ mol-1
Hope that's clear.
 
salvatore

salvatore

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Oliveme said:
C + O2 → CO2 -395 } this reaction produces C for CS2
S + O2 → SO2 -298 } this reaction produces S for CS2 but since there are 2 S, we multiply the ∆Hc by 2.
CS2 + 3O2 → CO2 + 2SO2 -1110 -> combustion of CS2

for the formation, you add the top two reaction enthalpies together and minus from the overall combustion enthalpy.
C + 2S → CS2 ∆H = -395 + 2(-298) -(-1110) = +119 kJ mol-1
Hope that's clear.
Click to expand...
Thank you for your reply..

The first part is clear. I don't understand the second part where we are supposed to add the two reaction enthalpies and subtract from the overall combustion enthalpy. Could you please explain to me why is it supposed to be that way? I don't get the concept behind doing this..

I'll really appreciate your help in making me understand the concept.

Thanks
 
Oliveme

Oliveme

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salvatore said:
Thank you for your reply..

The first part is clear. I don't understand the second part where we are supposed to add the two reaction enthalpies and subtract from the overall combustion enthalpy. Could you please explain to me why is it supposed to be that way? I don't get the concept behind doing this..

I'll really appreciate your help in making me understand the concept.

Thanks
Click to expand...
I'm not sure how I can clarify it but have a look at the Hess's Law diagram.
And not a problem. untitled.JPG
So in this case, the reactants are C and S and product is CS2. that's why in the end, the subtracting takes place.
 
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