Hardest Chemistry MCQs ever.

[A*(a*)]

I solved whole past papers book from 01 to 12 and found these questions seriously hard!
If anyone good in chemistry help me solving them out, so that my past papers could come to an end.

 

[A*(a*)]

 

[A*(a*)]

 

[A*(a*)]

 

[A*(a*)]

 

[iKhaled]

View attachment 24355

is the answer here 1 and 2?

 

[iKhaled]

the answer of question 37 is A right ?

 

[A star]

39 and 18 are C

 

[A star]

the answer of question 37 is A right ?

yup its A acording to me but the first option they are in solid state and only thats confusing

 

[aalmuhannadi]

Q9: from the question it says 50cm^3 of 0.1moldm^-3 of a metallic salt react exactly with 25cm^3 of 0.1moldm^-3 sodium sulfite

The moles of each is therefore: n = cv = 0.05 * 0.1 = 0.005 mol of the salt and 0.025 * 0.1 = 0.0025 mol of sulfite ions, therefore from this we can now say that 1 mol of the sulfite ions react 'exactly' with 2 mol of the metallic salt. Therefore, if, from the given equation, 2 electrons are lost form the oxidation of the sulfite ions, then it is only logical that half the amount of electrons is gained by the metal (due to the molar ratio I just talked about), so if sulfur is oxidised from +4 to +6, then the metal should be reduced by half the amount, which is from +3 to +2. Hope that helps and please correct me if I'm wrong, or ask me to explain it if you don't get it

 

[iKhaled]

yup its A acording to me but the first option they are in solid state and only thats confusing

it doesn't matter. if you look at the equations of these 3 they all produce DH3..idk why CIE is confusing us..whats the point of that "D" they can just make it look normal with the N -___-

 

[A star]

it doesn't matter. if you look at the equations of these 3 they all produce DH3..idk why CIE is confusing us..whats the point of that "D" they can just make it look normal with the N -___-

just to confuse the weak students thaese small yet wierd tricks are how cie diff from toppers in A levels . you need toa answer most of the trick questions write , write quality answers and provide clear diagrams for a top

 

[aniketjain]

i m sure about 1 & 2 of question 37 (they r ryt ) can anyone plzzz explain the 3 one ?
. 1 &2 produces ND3 not sure about the 3 one

 

[A star]

i m sure about 1 & 2 of question 37 (they r ryt ) can anyone plzzz explain the 3 one ?
. 1 &2 produces ND3 not sure about the 3 one

its the reaction between alkali and a ammonium salt where a salt is formed and ammonia gas is relased

 

[aniketjain]

so i think A is the answer to the question 37

 

[iKhaled]

so i think A is the answer to the question 37

i already answered that above

 

[aniketjain]

explanation is important

 

[aalmuhannadi]

For q37, I think 3 is incorrect because, look at this this way if you just replace the D with H:

NH4Cl + NaOH > H2O + NaCl + NH3, that would be the balanced equation. However, since you have NDH3Cl + NaOD, and you already know the molar rations are all 1 to 1 for everything, how can you get the gas ND3 if you have only 2 D atoms in the reactants?

 

[A*(a*)]

39 and 18 are C

Can you please explain how is 39 C?

 

[A*(a*)]

Q9: from the question it says 50cm^3 of 0.1moldm^-3 of a metallic salt react exactly with 25cm^3 of 0.1moldm^-3 sodium sulfite

The moles of each is therefore: n = cv = 0.05 * 0.1 = 0.005 mol of the salt and 0.025 * 0.1 = 0.0025 mol of sulfite ions, therefore from this we can now say that 1 mol of the sulfite ions react 'exactly' with 2 mol of the metallic salt. Therefore, if, from the given equation, 2 electrons are lost form the oxidation of the sulfite ions, then it is only logical that half the amount of electrons is gained by the metal (due to the molar ratio I just talked about), so if sulfur is oxidised from +4 to +6, then the metal should be reduced by half the amount, which is from +3 to +2. Hope that helps and please correct me if I'm wrong, or ask me to explain it if you don't get it

wow! you are seriously genious!