• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

Messages
98
Reaction score
80
Points
28
Well, lets look upon (D) first, shall we?
Now when the tap B is open, what will happen is pressure decreases as well as N2 concerntration decreases cos Li+N2 will occur. Position of equilibrium shifts to the left, causing % of NH3 to decrease. (D) is wrong.
Looking upon (C)
When A is open, pressure decreases, yield of ammonia decreases as position of equilibrium shifts to the left.
Looking upon (B)
When both are open, oh my god, the pressure will be even lower,....
Looking upon (A)
Now this is the answer. This is because pressure is higher so position of equilibrium is shifted to RHS so % of ammonia increases
 
Messages
98
Reaction score
80
Points
28
Messages
150
Reaction score
222
Points
53
BTW
volume of M = V
volume of N = 3 V.
Total volume = 4V.
P1V1=P2V2.
(1 x 10^5)(V)=P2(4V)
P2 = 25000 Pa

Change in temp ; 20 to 100 degree Celsius.
New pressure P2/T1=P3/T2.
25000/293=P3/373
Hence, P3 is 3.18 x 10^4 Pa.
BTW there is an easier way.

USE this : [P1V1]/T1 = [P2V2]/T2 and put in the values you put in. Much quicker and less hassle!

hope this helped you:)
 
Messages
98
Reaction score
80
Points
28
The answer should be C.
This is because of the Markonokov addition principle of alkenes. It states that when a H-X (where X can be any halogen) reacts with an alkene, the X shoould go and bond with the carbon containg more alkyl groups attatched to it. This can be seen in the mechanism. During the mechanism, a carbo cation is formed which makes it relatively unstable, so to make it stable, the H should go and bond with the carbon containg lower alkyl groups to make more alkyl groups bonded to the carbocation which will reduce the charge density of it and hence, finally, making it stabiluzed.
 
Messages
363
Reaction score
194
Points
53
The answer should be C.
This is because of the Markonokov addition principle of alkenes. It states that when a H-X (where X can be any halogen) reacts with an alkene, the X shoould go and bond with the carbon containg more alkyl groups attatched to it. This can be seen in the mechanism. During the mechanism, a carbo cation is formed which makes it relatively unstable, so to make it stable, the H should go and bond with the carbon containg lower alkyl groups to make more alkyl groups bonded to the carbocation which will reduce the charge density of it and hence, finally, making it stabiluzed.
THANKYOUUU VERY MUCH! !!!!
 
Messages
17
Reaction score
23
Points
13
Your answer is correct, but if you look in the Data Booklet, In the Bond Energies section in polyatomic molecules section: there are different values for C double bond C and C Double Bond C in CO2.
It depends on booklet to booklet. old booklet which is used for papers 2016 below have only one bond energy value for carbon=carbon bond which is 740 and the new booklet which is used for papers above 2016 has two bond values 740 and 805

and ur question is from 2013 paper so they assumed the bond value as 740
thats y they got -688 kj per mole
hope this helped you:D:D
 
Messages
150
Reaction score
222
Points
53
It depends on booklet to booklet. old booklet which is used for papers 2016 below have only one bond energy value for carbon=carbon bond which is 740 and the new booklet which is used for papers above 2016 has two bond values 740 and 805

and ur question is from 2013 paper so they assumed the bond value as 740
thats y they got -688 kj per mole
hope this helped you:D:D
OH, I didnt know about this, thanks for informing me about it!!:D:D
 
Messages
150
Reaction score
222
Points
53
Messages
438
Reaction score
3,645
Points
253
volume of M = V
volume of N = 3 V.
Total volume = 4V.
P1V1=P2V2.
(1 x 10^5)(V)=P2(4V)
P2 = 25000 Pa

Change in temp ; 20 to 100 degree Celsius.
New pressure P2/T1=P3/T2.
25000/293=P3/373
Hence, P3 is 3.18 x 10^4 Pa.

Thanks .
By the way, what is this formula P2/T1=P3/T2.:confused:
Can you tell me. :unsure:
Thanks again
 
Messages
665
Reaction score
13,617
Points
503
may 02-NO-2,6,8,21,30,34,39 do help me please THANKYOU
q2
the eq will be CH2O -> H2O + CO2
first find the number of moles of carbs digested
Mr of CH2O = 30
moles = 1.8/30 = 0.06
using mole ratios the number of moles of CO2 r also 0.06
to calculate mass of CO2 multiply the number of moles by CO2 molar mass
thatll be 44 x 0.06
the answer will be 2.64 i.e D

q6 is a bit tough for me too

q8
for enthalpy change of formation of CO2 the equation is
C + O2 -> CO2
n for enthalpy change of combustion of carbon it is the same equation so both the enthalpy changes r for same reaction n must be equal
so its A

q21
1- C2H5Cl
2- C2H4Cl2 (these should be 2 cuz it may be dichloromethane or 1,2- dichloromethane)
3- C3H3Cl3 (same goes for this...all Cl may be attached to one C only or there may be 2 atached to 1 C n 1Cl to the other C)
4- C2H4Cl2 (same is the case with this)
5- C2H5Cl
6- CCl6
so as 3 compounds have two structural isomers the total should be 9.
answer is C.

q30
its an ester so the COO bond will be broken to give one alcohol n one acid
the side linked to C=O will break to form COOH so the acid formed will be HCOOH
while the side linked with O-C will become an alcohol
so the answer is D
 
Messages
363
Reaction score
194
Points
53
Thanks .
By the way, what is this formula P2/T1=P3/T2.:confused:
Can you tell me. :unsure:
Thanks again
P3 is the final pressure and P2 is the answer I got before. I just set up a variables.

The real formula is p1V1/T1 = p2V2/T2 Mstudent says that this gives a correct answer too but I couldn't get one so, I did P1V1 = P2V2 1st and then P2/T1 = P3/T2 (T2 being the new temperature) It just made more sense to me.
 
Top