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Chemistry: Post your doubts here!

E

Es Jay

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for enthalpy change of atomisation/ dissociation, do the elements initially have to be in there gaseous states?
 
D

darknessinme

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Rahma Abdelrahman said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_11.pdf
Please ... for Q22, why the answer is B? Ok I know that for CH3 to be an electrophile it should have less than 8 electrons, so answer is not D, And for nucleophile it shd have lone pair of electrons.. but how do we apply this here?
And also Q34, Q36,
Click to expand...

22. Draw the structure of CH3:
Nucleophile has a lone pair it can donate. So outer shell has 8e-, 6e- for bonding and 1 lone pair.
Electrophile can accept a pair of electrons. 6e- in outer shell needed to bond, and so it has an incomplete octet. Outer shell can accept a lone pair.
Free Radical has an unpaired electron, so 6 used for bonding and one lone e- by itself, so 7e- in total.
Answer is A.

34.
1. True, Learn it.
2. True, Learn it.
3. False. General knowledge that table salt in water is neutral.
Answer B

36.
1. True,Bond energy of Hal-Hal decreases down group, because bond length increases. So Astatine molecules should dissociate more easily than chlorine.
2. False, Reactions between halogen and hydrogen get less vigorous down group 7. Becomes an equilibria with I2, so conclude that it reacts slowly with Astatine.
3. False, Halogens get weaker as oxidising agents down the group.
Answer D
 
S

strangerss

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233
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O
Warrior66 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_33.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_ms_33.pdf
I guess you're talking about 1 c part iii, you have to multiply the answer you got in part ii with 10 because if 25 cm3 of Fa1 have the amount of moles you found out for ii, then 250 cm3 of FA1 has 10 times as much amount of moles..
see, i think you're confused because you probably think that the II mark on the mark scheme is for part ii. But it's not. It is one whole mark consisting of what calculation you did in part ii and iii.
Then for iv you multiply the Mr of FA1 with the moles from iii and then divide it by the mass of FA1 you found in 1a. 'weighing out the salt' part and times it with 100.
Click to expand...
Okay I got it , thanks a lot for the help , jazak alallah 5ayran.
 
E

Es Jay

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strangerss said:
the smallest reading in a stop watch would be 1 second so , 1/2 = 0.5 seconds , any one do please correct me if I'm wrong
Click to expand...
the smallest reading is 1 millisecond, of this i'm sure. But i was thinking, don't we consider the reaction time when we write time uncertainties?
 
S

strangerss

Messages
233
Reaction score
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Points
38
Es Jay said:
the smallest reading is 1 millisecond, of this i'm sure. But i was thinking, don't we consider the reaction time when we write time uncertainties?
Click to expand...
No we don't , but if they tell you to write the minimum reaction time then you subtract the uncertainty fro the reaction time and vice versa for the maximum reaction time , hope you got it:)
 
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