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Chemistry: Post your doubts here!

XPFMember

XPFMember

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CaptainDanger said:
Answer is B. Explain?

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Assalamoalaikum wr wb!
yup..
enthalpy change of formation is the formation of 1 mole of product, with the reactants and products in the standard state...!!

iodine is solid under standard conditions...so total enthalpy change of formation of 2 moles of ICl3 is 38 + -214 = -176

standard is for one mole so -176/2 = -88
 
G

geek101

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gary221 said:
Cud sum1 tell me hw v calculate the no. of moles required?? Ans is C, btw
View attachment 7849
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It says that one side chain is converted to a compound with 2 double bonds instead of 3, which means one bond was hydrogenated, so one mole of H2 was used. Then it says that 2 chains were converted to compounds with only one double bond each, which means that 2 bonds were hydrogenated per chain, so 2 moles of H2 used per chain. Now if you add em up its 1 mole for the first chain, 2 moles for the second, and 2 moles for the third. So 5 moles of H2 were used. :)
 
mominzahid

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i have a problem in these type of questions where the examiner asks about the possible products of oxidation or reduction.
Please help
 
hmlahori

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Student12 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Q40 - the second point can someone explain ?
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Statement 2 says: It can be esterified both by ethanoic acid and by ethanol, in the presence of H+ ions
this means that the compound shown (mevalonic acid) can form an ester both by reacting with ethanol and ethanoic acid. This statement is correct because the compound contains both an acidic(CO2H) group(on its extreme left) and alcoholic(OH) groups (on the top and extreme right). This means it can react with both an alcohol(ethanol) to from an ester in which its acidic group will react and with an acid(ethanoic acid) in which one of its alcoholic groups will react to form the ester. In the presence oh H+ ions(or an acid) is the condition necessary for ester formation.
Hope it makes sense now!
 
smzimran

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hmlahori said:
Can someone explain these questions from paper 1

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_er.pdf
Q11, Q29 and Q33.
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AoA,
Q11:
Straightforward simple question.
Ammonia is more basic than water.
B says it is completely ionised, it cannot do that as it is a weak acid.
C says it is strongly acidic, false!
D says ammonia is more polar, wrong because H2O is highly polar, it can form hydrogen bonds!

Q29:
You should know that the reagent for this reaction is conc. H2SO4
So, A & D are incorrect!
To purify means to remove or neutralise any excess acid present!
So to neutralise acid, we have to add alkali...
That means solution Z is NaOH
Thus, C is correct!

Q33:
This one is difficult!
The forward reaction is endothermic means graphite is more stable or simply, C-C bond is stronger in graphite.
This is why 2 is correct!

Stronger C-C bond in graphite means it requires more heat to break and enthalpy change of atomisation of graphite is greater.
This is why option 1 is correct!

In combustion of both allotropes the compounds formed will be same.(As they both are carbon allotropes and have chemical properties of carbon)!
So, energy released in bond formation in combustion reaction is the same in both cases
Diamond has to be given lesser heat because its C-C bond energy is lower.
So the gap or difference in bond breaking energy and bond formation energy is larger in diamond!
That is why its enthalpy change is greater!

So, the answer is A
 
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