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Chemistry: Post your doubts here!

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19. Divide the equation into half equations, the oxidation half and reduction half:

Oxidation half:
1/2Cl2 + 6OH- -------> ClO3 - + 3H2O + 5e

Reduction Half:
(Cl2 + e ------> Cl- ) * 5

1/2Cl2 + 6OH- ------> ClO3 + 3H20 + 5e
5/2Cl2 + 5e -------> 5Cl-
------------------------------------------------------
3Cl2 + 6OH- -------------> 5Cl- + ClO3 + 3H20

So the answer is C.
how do I know the half equations are that way?:(
 
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:confused::confused:I think u misunderstood the paper; Mj 15 paper 1 variant 2( the answer to my question is C btw)
Ooops :p Sorry.

29. HO2CCH2C(OH)(CO2H)CH2CO2H

1 mole NaOH reacts with 1 mole -CO2H.
So, as citric acid contains 3 moles of CO2H, so 1 mole of ctric acid will require 3 mole of NaOH.
We have 0.005 mole of Citric acid, so they will require 0.005*3 mole of NaOH.

so volume required will be:
n = c * v
v = n/c
v = (0.005*3)/0.4
v = 0.0375dm^3
= 37.5cm^3

Answer is C.
 
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Plz could anyone calculate the enthalpy change for this question using the values already given image.jpg
 
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Please can someone explain what is meant by the carbon atoms lying in the same plane? How do we interpret such mcqs and what are the conditions for this. Must reply plz. Thanks.
 
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questions 1 and 2 please
I won't help with question 1 cuz it's too simple :)

Question 2:

CH4 + 2O2 ---> CO2 + 2H2O

C2H6 + 3.5O2 ----> 2CO2 + 3H2O

So if 1 mole each of methane and ethane are burned, 3 moles of CO2 is produced in total.

So since 10cm3 each was used, we expect 30cm3 total CO2 to be released, which would be absorbed by KOH(aq) due to its acidic nature.
 
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