Chemistry: Post your doubts here!

[Rizwan Javed]

Mj 15 P12, Q 12, 19, 29
can someone pls give me explainations to these?

12. In this question, in the options all the compounds given are the group II compounds. So a general equation for all the decomposition reactions of Nitrates of Group II is:

2X(NO3)2 ----> 2XO + 4NO2 + O2

You are given, that 1.53g is lost, so the mass of XO formed is: 3 - 1.53 = 1.47g

1 mole of X(NO3)2 forms 1 moles of XO, find the moles of XO and X(NO3)2, put them equal and equate like this:
3/(X + 124) = 1.47/(X + 16)
solve it, you'll get
X = 87.76

From the periodic table, Strontium is the metal which has the Mr approximately equal to 87.76, so the answer is D.

 

[Rizwan Javed]

Mj 15 P12, Q 12, 19, 29
can someone pls give me explainations to these?

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

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

Reduction Half:
(1/2Cl2 + e ------> Cl- ) * 5

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

So the answer is C.

 

[Rizwan Javed]

Mj 15 P12, Q 12, 19, 29
can someone pls give me explainations to these?

29. Here you are asked about the RATE of formation of cloudiness. Here RATE depends on the strength of the bond between R-X (where X is the Halogen). R-I is weakest bond, so it will dissociate readily, and the rate of formation of cloudiness will be higher. R-Br is weaker than R-Cl but stronger than R-I, so the rate of formation of cloudiness when R-Br is present will be higher than R-Cl but lower than R-I. Same goes for R-Cl.

Hence the only option satisfying this is A.

 

[mistique_bee]

29. Here you are asked about the RATE of formation of cloudiness. Here RATE depends on the strength of the bond between R-X (where X is the Halogen). R-I is weakest bond, so it will dissociate readily, and the rate of formation of cloudiness will be higher. R-Br is weaker than R-Cl but stronger than R-I, so the rate of formation of cloudiness when R-Br is present will be higher than R-Cl but lower than R-I. Same goes for R-Cl.

Hence the only option satisfying this is A.

nerd

 

[Rizwan Javed]

nerd

 

[slisjunknown]

29. Here you are asked about the RATE of formation of cloudiness. Here RATE depends on the strength of the bond between R-X (where X is the Halogen). R-I is weakest bond, so it will dissociate readily, and the rate of formation of cloudiness will be higher. R-Br is weaker than R-Cl but stronger than R-I, so the rate of formation of cloudiness when R-Br is present will be higher than R-Cl but lower than R-I. Same goes for R-Cl.

Hence the only option satisfying this is A.

I think u misunderstood the paper; Mj 15 paper 1 variant 2( the answer to my question is C btw)

 

[slisjunknown]

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?

 

[Rizwan Javed]

I think u misunderstood the paper; Mj 15 paper 1 variant 2( the answer to my question is C btw)

Ooops 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.

 

[slisjunknown]

So Na reacts with C(OH) and not NaOH?

 

[Rizwan Javed]

So Na reacts with C(OH) and not NaOH?

Yes, NaOH will only react with carboxylic acid group not with alcohol group. If it were Na instead of NaOH then we would have considered the -OH groups in the citric acid.

 

[slisjunknown]

Yes, NaOH will only react with carboxylic acid group not with alcohol group. If it were Na instead of NaOH then we would have considered the -OH groups in the citric acid.

okay

 

[Awesome12]

Question 22 - Why isnt it D?

 

[Mohammed Elatta]

View attachment 60669
Question 22 - Why isnt it D?

k2Cr2O7 is a weak oxidising agent only capable of oxidising the alcohol group not the alkene whereas if KMnO4 was stated it would have been able to oxidise both with no issues.

 

[Hamody]

Plz could anyone calculate the enthalpy change for this question using the values already given

 

[qwertypoiu]

Plz could anyone calculate the enthalpy change for this question using the values already given View attachment 60681

Reaction 1 is adding Mg to Left Hand Side (LHS) of Reaction 3. Reaction 2 is adding Mg to Right Hand Side (RHS) of Reaction 3. Same products are formed.
Reaction 3 = Reaction 1 - Reaction 2 = -444 - -504 = +60

 

[fatimarehman]

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.

 

[Abdullahassan_99]

questions 1 and 2 please

 

[sfhn_128]

pls answer this ques!
thanks in advance!

 

[qwertypoiu]

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.

Chemistry: Post your doubts here!

 

[qwertypoiu]

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.