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Chemistry: Post your doubts here!

S

shaikh

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The equation below represents the combination of gaseous atoms of non-metal X and of
hydrogen to form gaseous X2
H6 molecules.
2X(g) + 6H(g) → X2
H6(g) ∆H = –2775kJmol–1
The bond energy of an X–H bond is 395kJmol–1
.
What is the bond energy of an X–X bond?
A –405.0kJmol–1
B –202.5kJmol–1
C +202.5kJmol–1
D +405.0kJmol–1


Can someone plz answer this ?
 
Rahma Abdelrahman

Rahma Abdelrahman

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Nov 09 P12, can someone explain Q 18 , Q20 , Q22_Why not B? And Q40 ... :)
And in June 2010 P11 Q35?!!! How is it C?!! How is Carbon wrong while Sulfur is right?!!! I really don't get it!
Please reply asap ...
 
Fma 07

Fma 07

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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf

Q6. Answer is D
Q9. Answer is C (im getting 2.952 grams)
Q11. Answer is A
Q20. Answer is C
Q21. Answer is C (im getting 5? )
Q22 Answer is B
Q28 Answer is D
Q29 Answer is B
Q30 Answer is D ( which one is the chiral centre? )
Q32 Answer is D which one is reduction?
Q35 Answer is C.

( I realise theyre alot of questions..sorryy :p )
 
soso247

soso247

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Fma 07 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf

Q6. Answer is D
Q9. Answer is C (im getting 2.952 grams)
Q11. Answer is A
Q20. Answer is C
Q21. Answer is C (im getting 5? )
Q22 Answer is B
Q28 Answer is D
Q29 Answer is B
Q30 Answer is D ( which one is the chiral centre? )
Q32 Answer is D which one is reduction?
Q35 Answer is C.

( I realise theyre alot of questions..sorryy :p )
Click to expand...


hey
so Question 6
NH4NO3→ N2O + 2H2O
treat NH4+ and NO3- as separate and get the oxidation state on each ..... N in NH4+ has the oxidation state of -3 and in NO3- it has an oxidation state of +5 and since in N2O the oxidation state is 1 you get from -3 to 1 which is +4 and from +5 to +1 is -4

question 9
i am getting 2.995 and hence it is the only one closest to 3 so it is right and so it is C

question 11
when you add acid the H+ react with OH- to form water and hence the equilibrium of the equation shift to the right to create more HOCl and so reducing OCl- and hence it cannot for cl- under uv light

question 20
anything flammable burns in oxygen and because oxygen cannot burn in oxygen (obviously) then oxygen is non-flammable

question 21
i will post a picture for explanation later

question 22
later

question 28
only tertiary halogenoalkane are unaffected by the concentration of OH- and D is the only tertiary halogenoalkane

Q29
but-1-ene, cis-but-2-ene, trans-but-2-ene are all isomers of butene formed from but-2-ol

Q30
the 3rd Carbon is the chiral centre
 
daredevil

daredevil

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biba

biba

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Dr.MMM said:
Oh and ive got another question:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf Q:40th
Click to expand...

only 1 and 2 are correct ... in 1 u need to use ur info related to esters... the NaOH wil break the compound from wher -CO2 is present henca an alcohol and corboxylic salt wld be formed... in 1 they have shown the alcohol and in 2 they have shown u the salt! 3 is incorrect because if u look at the stucture in 3 they have totally eliminate the double bond, that is incorrect
 
Rahma Abdelrahman

Rahma Abdelrahman

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soso247 said:
here is another one can you help me with it please :)
Click to expand...

LOL! I actually have the same doubts.. I posted them yesterday but no one answered yet!
:LOL:
daredevil Can u help here?
soso247 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
please help
question 21, 23, 28, 29, 31
Click to expand...
Rahma Abdelrahman said:
Nov 09 P12, can someone explain Q 18 , Q20 , Q22_Why not B? And Q40 ... :)
And in June 2010 P11 Q35?!!! How is it C?!! How is Carbon wrong while Sulfur is right?!!! I really don't get it!
Please reply asap ...
Click to expand...

:D Thanks in advance :)
 
ahmed abdulla

ahmed abdulla

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biba said:
only 1 and 2 are correct ... in 1 u need to use ur info related to esters... the NaOH wil break the compound from wher -CO2 is present henca an alcohol and corboxylic salt wld be formed... in 1 they have shown the alcohol and in 2 they have shown u the salt! 3 is incorrect because if u look at the stucture in 3 they have totally eliminate the double bond, that is incorrect
Click to expand...

can u help plz ;)
 

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S

strangerss

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cute97 said:
Noo :( Q19 (B) and Q28 (A) according to the mark scheme
Click to expand...
answer is B
amount of sulphite use = conc x vol
= 0.1 x (25/1000)
=2.5x10^-3mol
2 electrons are lost
amount of electrons lost = 2 x 2.5 x 10^-3 ==== 5 x 10^-3 mol
amount of electorns gained by metallic salt = amount of electrons lost by sulphite
= 5 x 10^-3 mol
amount of metallic salt used = 0.1 x (50/1000) ==== 5 x10^-3
amount of electrons gained PER MOLE of salt = (5 x 10^-3)/(5x10^-3) = 1 unit
hence oxidation state of metallic salt decreases by 1 unit meaning the oxidation satate becomes +2 from +3!!! this is for the question 9 , some other person posted it :D
 
Rahma Abdelrahman

Rahma Abdelrahman

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soso247 said:
here is another one can you help me with it please :)
Click to expand...

LOL! I actually have the same doubts.. I posted them yesterday but no one answered yet!
:LOL:

soso247 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
please help
question 21, 23, 28, 29, 31
Click to expand...
Rahma Abdelrahman said:
Nov 09 P12, can someone explain Q 18 , Q20 , Q22_Why not B? And Q40 ... :)
And in June 2010 P11 Q35?!!! How is it C?!! How is Carbon wrong while Sulfur is right?!!! I really don't get it!
Please reply asap ...
Click to expand...

:D Thanks in advance :)
strangerss Can u try to help here? :)
 
S

strangerss

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Rahma Abdelrahman said:
LOL! I actually have the same doubts.. I posted them yesterday but no one answered yet!
:LOL:




:D Thanks in advance :)
strangerss Can u try to help here? :)
Click to expand...
SOSO247 OKay , in question 21 the only way to solve it is by drawing the displayed formula of the the compound. the first thing I did is I drew the cyclohexene ring you know how right? in the cyclo hexene there is 12 hydrogen atoms and 6 carbon atoms , then from the sixth carbon atom I drew a bond and counted till I reached the 11th carbon atom where I put a double bond because in the question the cis isomer is between the 11th and 12th carbon atom right? the I drew the remaing carbon atoms after the 11th one which are 9 at the last carbon atom I put the aldehyde group because it's pretty obvious it's always at the last. now the last and probably most important step is the number of double bonds between the carbon atoms , now we know that there are 12 hydrogen in the cyclohexene and two for the cis isomer and the total number of hydrogen atoms is 28(from the question) so 28-14=14 carbons atoms are left to be bonded to the carbon chain , so if you give each carbon atom a hydrogen till you used 14 , you add a double bond between the carbon atoms with only three bonds and you will get six double bonds ..I will put the pic. I drew if you didn't get it ok?...question 23 it's C because we get a polychroalkene when we polymerise the monomer given right? then by reacting with water a nucleaphilic substitution reaction takes place in which the hydroxide ions which acts as a nucleaphile displaces the weaker nucleaphile the chlorine atom , it's exactly like when reacting a halognoalkane with sodium hydroxide and we get an alcohol , in A we cannot hydrate an alkane because there are no double bonds , and in B and D it's the same case we cannot oxidise an alkane , so C is correct. question 28 I really got no clue :/ . question 29 it's B because we know that a halogenoalkane plus CN will give a nitrile and when we hydrolyse a nitrile we get a carboxylic acid , when CN displaces Br in B , the final product formed will have a carbon atom attached to an ethyl group which is shown in the diagram given , notice how all the other products formed will not have an ethy group that is the C2CH3 being attached to the carbon atom that was attached to the br , did you get it ?and question 31 I don't know :/ ..hope I helped :)
 
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