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Chemistry: Post your doubts here!

ousamah112

ousamah112

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JD REBORN said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q11,33 and 34
Click to expand...
for question 11 look at the first equation..in order to increase the concentration of HOCL we need to make the reaction forward bias. So,by acidifying the water reaction will proceed in forward direction according to Le chatelier principle. As acidity increase more OH ion will be produced to lower the acidity,hence reaction will occur in forward direction and conc of HOCL increases..
 
Jaf

Jaf

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D

DPratik

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JD REBORN said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q32 how can statements 2 and 3 be correct.
Q9
Click to expand...


for question 32) as Carbon is more stable in +4 oxidation state as compared to +2 oxidation state.Now being stable in terms of enthalpy can be define as being exothermic,which means the enthalpy change is negative
As the question has said that carbon monoxide readily forms carbon dioxide which means the amount of carbon dioxide will be higher in equilbm and as equilbm constant directly proportional to the amount of product so from this we can say that the equilbm constant of this reaction would be high !!
 
D

DPratik

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Student12 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q1, 4, 8, 13 ,20,28
Q36 - Doesnt the bond length decrease ? and what does the 3rd point mean ?
Q38- doesnt KMnO4 react with secondary alcohol ?
Can you specify the sub topic these msqs are based on ?
Click to expand...



for question 1) moles of carbon:29.7/12 = 2.47
moles of hydrogen:6.19/1=6.19
so carbon and hydrogen are in the ratio: 2.47:6.19
(dividing both by the lowest number)
or the ratio of carbon and hydrogen is 2.47/2.47: 6.19/2.47 and that is equal to 1: 2.5 ( it is based on stoichimetry)

Now as the question has given the emperical formuale C8Hx,so mutlipy both the ratio by 8 you would get C8H20
(the reason for multiplying by 8 was because according to the question we need C8)


for question 4) you need to draw the p-orbital in box form,you can see that when atomic no 13,px has one electron,when atomic no 14,py has one electron,when atomic no 15,py has one electron,when atomic no 16,px has paired electron,when atomic no 17,py has paired electron and when atomic no 18,pz has paired electron....so from these you can come to the conclusion that the number of upaired p-electron increases and again decreases !! (based on electronic configuration)

for question 8) 2*(-187.8) +delta Hd = 2*(-285.8).....you can verify this by drawing Hess cycle and adding the enthalpy the vector rule of addition !! (enthalpy change)


for question 13) the question has stated using calcium hydroxide which is a base...and now check out the following ions mentioned in the question..you can see that there is an acid
HCO3-..so when calcium hydroxide reacts with this ion...calcium carbonate will be formed !!! (group 2)


for question 38) option 1 and option 2 are correct because of the presence of double bond but option 3 isn't because alcohol(OH) is not a stronger acid to react with that NaHCO3..but if it would had been carboxylic acid instead of alcohol then option 3 would also be correct !!

for question 36) as we move from hydrogen chloride, hydrogen bromide and hydrogen iodide..their bond length increases,the ease of oxidation increases(usually halogens are reduced but as we move from chlorine to iodine their reduction property decrease and oxidation property increaaes) but thermal stability decreases because of increasing bond length !!!





fofr question
 
D

DPratik

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iKhaled said:
guys guys guysssss...is Cambridge acting crazy or what??

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf

question 17 in nov/05 P1 is an A2 level..how come they ask us this and we r not even suppose to study it for an AS exam?

can somone explain plssss.. :/
Click to expand...

we have to study this topic in Group 7...there is nothing to know much about A2 for this question...just identify which is the ligand and which is the complex ion ??
 
D

DPratik

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saudha

saudha

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Use of the Data Booklet is relevant to this question.
Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate
nitrogen gas, which inflates the bag.
2NaN3  3N2 + 2Na
A typical driver’s airbag contains 50g of sodium azide.
Calculate the volume of nitrogen this will produce at room temperature.
A 9.2 dm

B 13.9 dm

C 27.7 dm

D 72.0 dm
 
Nikesh

Nikesh

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Most_UniQue said:
Can anyone hep me how to find bond angles? I dont get it!
Click to expand...
See the number of bond pairs and lone pairs in the central atom...
Expected bond angles according to the number of electron pairs available:
1) 2 ------> 180 (linear shape)
2) 3-------> 120 (trigonal planar)
3) 4-------->109 (tetrahedral)
Above are most commonly asked bond angles
But presence of lone pairs in central atom decreases the bond angle due to electron pairs repulsion........
Bond pairs do not have such high deflection to change the bond angle
Remember the Repulsive force of electron pairs which has the trend as follows:
Lone pair - Lone pair > Lone pair - Bond Pair > Bond pair - Bond pair
Hope u got it clear
 
hmlahori

hmlahori

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Jaf said:
Dude don't mislead people.

We ARE supposed to know what ligands are [complex ion formation]. I've seen this term used in the book.
Click to expand...

I have no intention whatsoever of misleading people. This term is used in the books i use too but this is because the AS and A2 books are combined and some information is for AS and some for A2. But it is nowhere mentioned in our syllabus even 'complex ion formation' is not mentioned.. You can check each and every point in the AS part of the syllabus for exams in May/June 2012. I dont know about syllabuses before or after that session it may be in one of them but not in this one. And as far as i know Cambridge RARELY gives thinks that are not in or related to our syllabus.
 
Nikesh

Nikesh

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saudha said:
Use of the Data Booklet is relevant to this question.
Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate
nitrogen gas, which inflates the bag.
2NaN3  3N2 + 2Na
A typical driver’s airbag contains 50g of sodium azide.
Calculate the volume of nitrogen this will produce at room temperature.
A 9.2 dm

B 13.9 dm

C 27.7 dm

D 72.0 dm
Click to expand...

given,
mass of NaN3= 50g
i.e number of moles = 50 / 130 = 0.385 moles
now from the given equation,
2 moles of NaN3 gives 3 moles of N2
0.385 moles of NaN3 gives 3/2 * 0.385 = 0.577 moles of N2

At RTP,
Molar volume of any gas - 24 dm3
n = produced volume / Molar volume
--> 0.577 = v/ 24
i.e. v = 13.9 dm3
So answer is B
 
hmlahori

hmlahori

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Fellows need help with these paper 1 questions
others have explained some of them but im not entirely satisfied.​

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
http://www.xtremepapers.com/papers/...and AS Level/Chemistry (9701)/9701_w08_er.pdf
Q3 how do we work out the answer of this question?(which is D)
Q31 Why is the answer not A?
Q38 Why is the answer not B?(someone in a previous post said that flourine has to be present in order for a hydrocarbon to be a flame retardant. Can someone confirm that and tell me why?)

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf
Q9 Why is the answer not A?
Q31 Why cant the answer be B? The second statement seems correct.​
 
Nikesh

Nikesh

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Student12 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Q1, 4, 8, 13 ,20,28
Q36 - Doesnt the bond length decrease ? and what does the 3rd point mean ?
Q38- doesnt KMnO4 react with secondary alcohol ?
Can you specify the sub topic these msqs are based on ?
Click to expand...
1---> D because use the formula to calculate the mole ratio
to do this follow these steps
i) change % composition to no. of moles i.e. % composition / molar mass
ii) divide number of moles by lowest no. of moles
iii) if mole ratio is in fraction then multiply it by suitable no. to make it whole number
4-----> D because from 13 to 15 electrons start to be filled up in p orbital, and when it reaches 16 it starts to be paired up
8---> B simply use Hess law by forming the intermediate
 
J

JD REBORN

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