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Chemistry: Post your doubts here!

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help mee plzz
 

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
Q6 c (iii)
The amine group will obviously react with the acyl chloride, but the phenol part shouldn't react, right? Because it needs to be reacted with NaOH first. However, the MS shows that the phenol part is also reacting and forming ester. how is this possible?

Acyl chlorides are used because of their high reactivity, meaning it will also react the same way as a carboxylic acid does.
 
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help plzz

9701.mj.06 q, 25
9701.on.06 q,37,3
9701.mj.01 q 34,26,
9701.on.07 q 33,37,4
9701.mj.08 q 27 ,
9701.on.09 ( varient 11) 28,24,21
9701.on.10(varient 12) 39,27,13
9791.0n.10 (varient 11) q 38 ( what product will form if tollen reagent react with it ),35,29
9701.mj.10 (varient 11) q 12,
9701.mj.11(varient 12) q 38,29,28,24,11,
*9701.mj.11(varient 11) q 36,27,34,26,16
9701.on.11.(varient 12) q 9
9701.on.11 (varient 11) q 27,29,21,
9701.mj.12, (varient 12) q 24,23
9701.mj.12 (varient 11) q 29,23,22,9,
9701.on.12 (varient 11) q 26,
 
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Q1. the mass that is given is the mass of Ag so use it to find out the moles of Ag
n of Ag = 0.216/108 = 2 x 10^-3
now multiply this with avagadro's constant to find the no. of atoms in this much mass (or moles) of Ag
no. of atoms of Ag present = [2 x 10^-3] X [6.02 x 10^23] = 1.204 x 10^21
now they asked us no.of atoms per unit area (on 1 cm2) and have already given us the area 150 cm2
divide the total no. of atoms by the given area and you will have atoms per unit area
no. of atoms per unit area = [1.204 x 10^21]/150 = 8.03 x 10^18, thus A is your answer

Q10. moles of O2 is given use this to find out the moles of NO and moles of NO2 is also given
use cross multiplication by seeing the ratio of moles from the equation provided
2 : 1
x : 0.8

x = 2 x 0.8 = 1.6
so we have moles of each as
O2 = 0.8
NO = 1.6
NO2 = 4 - 1.6 = 2.4 (because 1.6 moles of NO2 have been converted to NO)

Kc = ([NO]^2 x [O2])/[NO2]^2
Kc = ([1.6]^2 x [0.8])/ [2.4]^2
thus our answer is D

Q19. I was stuck at it so i opened the Examiner's Report
and this was the answer in it

In Question 19 candidates were asked to recognise any chiral centre in three compunds for which only the
molecular formula was given. Many candidates failed to notice that C3H6I2 could have the structure
CH3CHICH2I.

so our answer is B

Q37. no idea sorry :(
Thanks a lot :)
 
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I am not sure whether my explanation is exactly correct, but here it is:
C is (Cu(NH3)4)2+ (or (Cu(NH3)4)2+ SO4 2-)
when NaOH is added precipitation occurs. CuOH forms alongside (Na)2SO4.
when you heat it CuOH-----> CuO + H2O. (CuO is the black solid.)
upon addition of water (Na)2SO4 dissolves, as it is a soluble salt and the black solid CuO is left behind.
so D is CuO and E is Na2SO4.
I hope its correct. Please correct me if I went wrong somewhere.
 
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help plzz

9701.mj.06 q, 25
9701.on.06 q,37,3
9701.mj.01 q 34,26,
9701.on.07 q 33,37,4
9701.mj.08 q 27 ,
9701.on.09 ( varient 11) 28,24,21
9701.on.10(varient 12) 39,27,13
9791.0n.10 (varient 11) q 38 ( what product will form if tollen reagent react with it ),35,29
9701.mj.10 (varient 11) q 12,
9701.mj.11(varient 12) q 38,29,28,24,11,
*9701.mj.11(varient 11) q 36,27,34,26,16
9701.on.11.(varient 12) q 9
9701.on.11 (varient 11) q 27,29,21,
9701.mj.12, (varient 12) q 24,23
9701.mj.12 (varient 11) q 29,23,22,9,
9701.on.12 (varient 11) q 26,

O/N 2006 Q3. C
they gave us a total of 4 elements and 4 choices....the rest of the three choices belong to Ca P and Kr and the last one belongs to X so after you have figured out the configuration fro the rest the remainder is your answer!

O/N 2007
the ans is D because the X axis says "the no. of electrons removed" and since electrons are removed from the outermost shell, and the point where there is a big jump, that is the point for the change in shell and that means that there were only 2 electrons in the outermost shell and thus it belongs to group 2
i got confused because there is no way there can be 3 electrons in the inner most shell! but that should not matter because for knowing that which shell the electron belongs to we only need to see the ionization energies of the first few electrons!
M/J 2008
the answer is D because in this the C atom is surrounded by 3 carbon atoms and it can not be easily oxidized
M/J 2009
Q24. the answer is A!
for each double bound 2 atoms of hydrogen and bromine are required!
we have hydrogen and bromine in the form of H2 and Br2
thus our answer is 2 moles of each
Q28. the answer is C the explenation is in the pics
 

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The intermediate formed in free radial substitution is known as carbocation and the intermediate formed in SN2 is not a carbocation that's why the ans is A rather then B as u can see that in A the mechanism is free radical and in B the mechanism is SN2.
Owhh Godd.... thanku hassan .... my bad i thought B was SN1.. :p
Thankuu soo much for the helpp...
 
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Owhh Godd.... thanku hassan .... my bad i thought B was SN1.. :p
Thankuu soo much for the helpp...
np ..
P.s there's a mistake in the course book ..in SN 2 there's a transition state rather then the formation of intermediate so correct that mistake and also the -ve sign is missing so add that too in it.
 
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