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Chemistry: Post your doubts here!

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please pelase make it a bit more specific.
I though enthalpy change = enthalpy change of products - enthalpy change of reactants?
I have learned it in another way and its a whole new concept but if you wanna know it i will explain my way it works everytime
 
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W.salam,
Mr(P) = 31
Mr(P2O5) = 2(31) + 5(16) = 142
In 100 g of fertiliser, there is 30.0% P2O5
That means mass of fertiliser is
= (100/30) * Mr(P2O5)
= (100/30) * 142
= 473.3 g

% of phosphorus in fertiliser is
= (2*31 / 473.3) * 100 [(2*31) because there are 2 atoms of P in one molecule of P2O5 ;) ]
= 13.1 %
So, A is the answer!
:)

how do you know is 100g? Or is it already a standard?
 
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Video on Nov 2011 P11 uploaded

http://www.youtube.com/playlist?list=PL4Jmce8VJnNX-EKGH10S1O8HEK2elI72j

This is likely to be the final paper uploaded before your exams.

I will try to work on the rest of the papers (both A and O levels) for the next few months and upload them eventually.
But I must say its not likely to remain free access for the future batches.

I'll try to help out by answering questions in this thread until Jun 10. After which you guys can enjoy a good break. (y)
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf

Need help on Q3, 2o, 27 ( isnt the tertiary alcohol will be resistant to oxidation? ) and 40 ( why 2 cant be seen after distillation? )

Hi, please include the answers to make it easier for us to reply.

w03qp1

Q3. C6H12 + 9O2 --> 6CO2 + 6H2O
moles of H2O : moles of CO2
6:6

mass of H2O absorbed : mass of CO2 absorbed
6 x 18: 6 x 44
0.41:1

Q27. Butanol has the following structural isomers
Picture 2.png

2-Butanol has a chiral carbon, so it has two optical isomers. Total 5 isomers

Q27. Both are primary alcohols, both can be oxidised.
The first one is resistant to dehydration but not the second one.

Q40. Substance 2 is a ionic compound (high boiling point) , it would be found left in the distilling flask and not the distillate.
 
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf

Can anybody please explain question 35 ans B

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf

Question 34 ans B

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_12.pdf

question 4 ans A and question 31 ans B

Question 4 I get 630 but the answer suggests 655.

THANKS IN ADVANCE!!!:)(y)

For question-4,
Heat energy required to break the bonds-Heat energy released in forming the bonds=Enthalpy Change
496+994-2x=180
2x=1310
x=655
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
q26 the answer is D but how...
Q34 couldn't understand at all....the answer is D
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 how to do it.....the answer is B...
Q19 answer is B
Q21 answer is A
Q33 answer is C
Please help me wid this ASAP thanks a lot...

For May/June 2007:

Q26 the answer is C. For the alcohol to not react with potassium per manganate, the alcohol must be a tertiary alcohol. So, the central chiral carbon atom is attached to a methyl group, an ethyl group, a propyl group and a hydroxyl group. That makes 7 carbon atoms.

Q34 If the mercury level in the right hand limb must rise the gas at R must have an increase in pressure with a rise in the temperature. For the first optio, the no. of moles increases with increase in temperature. So pressure rises in R. For the second option the no. of moles remains unchanged even if the temperature rises. The no. of moles same is same on both sides.

For O/N 2007:

Q9 By calculation, it can be found out that two mole of the salt reacts with 1 mole of sulfite. The half equation for the oxidation of sulfur shows us that 2 electrons are lost in the process. This means that the metal in the salt must be reduced and two electrons must be gained. Since two moles of metal ions reacts , the oxidation number decreses from +3 to +2, so that two metal ions can gain a total of two electrons to balance it out.

Q19 It may be difficult to explain this due to the limitations of drawing the structures here. The first compound is chiral. The chiral carbon is attached to a bromine, and iodine a hydrogen and an ethyl group. The second compound is chiral too. The chiral carbon is attached to a bromine, iodine, hydrogen another carbon atom which has a double bond to the third carbon atom. The third compound's chiral carbon has a hydrogen, an iodine, a methyl group and an iodomethyl group.

Q21 If chlorine free radicles collide with tetramethyl lead(IV) methyl free radicles will be generated.

Q33 Since the pressure has not been changed in any way, but only the temperature, the pressure inside cannot be any different from the outside one. Besides, the plunger is moveable, so theere is no pressure difference.
 
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s07qp1

Q26. I keep reworking the question as I kept getting C as the answer. I think you typed the answer wrongly. :confused:

You need a tertiary alcohol (not oxidised easily)
25.jpg

For the alcohol to be chiral, smallest number of carbons for the 3 R groups would be 1,2 and 3 respectively. So minimum carbon would be 7.

Q34. Question is asking if the pressure of the left flask would increase more than the right flask when temperature of whole setup is increased.
A. Temp increase, left flask eqm shift right as forward reaction is endothermic. Left flask more gases particles than before, higher pressure than right flask.

B. Temp increase, left flask eqm shift left as backward reaction is endothermic. No change in moles of gases. So change in pressure is same for both flasks.

C. Change in pressure is same for both flasks.
OHHHHHHHH I AM SO SORRY......MY BAD THE ANSWER IS C....and i am soooo stupid...i didnt think about the chirality.....THANKS ALOT FOR UR HELP.........
 
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http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
q26 the answer is D but how...
Q34 couldn't understand at all....the answer is D
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 how to do it.....the answer is B...
Q19 answer is B
Q21 answer is A
Q33 answer is C
Please help me wid this ASAP thanks a lot...

Metanoia sorry for botheration can u see if u cud help me wid this.....@kruti u also have a look....THANKS IN ADVANCE
 
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Hi there, please type the answers next to the questions to make it easier to reply.

Q23. Products formed CH3CH2OH, CH3CH2COONa
% of CH3CH2OH = Mr of CH3CH2OH/(Mr of CH3CH2O- + Mr of CH3CH2COONa) x 100% = 46/(46+96) x 100% = 32.3%

Q30.
Statement 2: True
PV=nRT
n/V=P/RT
density =P/RT
Therefore, density is inversely proportional to T

Statement 3: False
Temperature should be in Kelvins not degree Celsius

Q40. Conditions for nucleophilic substitution of halogen (on halogen alkane) with NH2, is to heat the halogenalkane with conc NH3 in sealed tube.
 
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Metanoia sorry for botheration can u see if u cud help me wid this.....@kruti u also have a look....THANKS IN ADVANCE

s07qp1

Q26 and 34, I've already answered in post
#9717

w07qp1

Q9. This is probably what prompted me to do the youtube videos, cause this question I've typed out at least 5 times in the last 2 days. :eek:

moles of SO3 2- : moles of electrons : moles of metal
0.0025 : ? : 0. 005
1 :? : 2
1 : 2 : 2

2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q19. Are you able to draw the structures? Which one are you having difficulty viewing as containing a chiral carbon?

Q21. Have to realise its a free radical reaction, and link it to free radical concepts. We need CH3 radicals in the initiation and propagation steps, which is provided by the (CH3)4Pb

Q33.
Statement 1: (False) The gas is exerting the same pressure as the atmosphere outside, otherwise the plunger will either move left or right.
Statement 2: (True) When plunger moves inwards, it causes the pressure to increase, the eqm will shift to the left to reduce pressure (produce less moles of gas).
Statement 3: (True) For every one mole of PCl5 dissociated, two moles of gases were formed (1 mol of PCl3 and 1 mol of Cl2). So total volume of gases after dissociation is more than original volume of gas without dissociation.
 
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Metanoia , ZaqZainab , Suchal Riaz
Could u PLZ give me a brief explanation for SN1 and SN2 Mechanisms :)
Thnx in advance

I'll limit the discussion to substitution of halogen alkanes, might be a better idea if you google youtube videos to visualise easier.

sN2
usually for primary halogen alkanes
nucleophile forms bond with partial positive carbon at the same time as bond between halogen (X) and carbon is broken
Nu-----C------X
rate depends on both concentration of Nucleophile and halogen alkane

sN1
usually for tertiary halogen alkanes
bond between halogen and carbon is broken first, a carboncation is formed.
C---X --> C+ and X-

then the nucleophile attacks the carbocation
rate depends on only on concentration halogen alkane
 
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