check out !

1. For the cell ( at 298 K )
Ag(s)| AgCl(s) | Cl- (aq)||AgNo3(aq)|Ag(s)

Which of the following is/are correct-
(A) The cell emf will be zero when [Ag+] in anodic compartment = [Ag+] in cathodic compartment .
(B) The amount of AgCl(s) precipitate in anodic compartment will increase with the working of the cell.
(C) The concentration of [Ag+] remain constant, in anodic compartment during working of cell.
(D) o= E°Ag+|Ag -E° cl- |Agcl| Ag - 0.059/1 log 1/Ksp(Ag c l )

8 Answers

4
UTTARA ·

2) Which of the following carbocations are formed during the formation of given products.

CH3----CH CH----CHCH2 --HCL---->

CH3--CH2---CHCH--CH2---cL + CH3---CH(cL)----CHCH----CH3

(A) CH3---CH2----CH+---CHCH2

(B) CH3 --- CH2--CHCH---CH2+

(C) CH3---CHCH---CH+---CH3

(D) CH3----CHCH---CH2---CH3+

4
UTTARA ·

Compound (x) C3H6O decolourises Br water , with dil H2So4it undergoes hydrolysis to give (y) and (z). The correct statements are :

(A) (x) is an ether (b) (X) is an alkene

(C) (X) is cyclic compound (d) Both y and z are carbonyl compounds

4
UTTARA ·

ALL THE ABV R MULTI ANS TYPE

11
Tush Watts ·

Ans 1) b, c (NOT SURE) ????????

13
Avik ·

1)Electrochem. no idea
2)AC
3)AB

1
souravkundu ...... ·

3)A

4
UTTARA ·

ANS GIVEN

1) A B D

( But I got B D)

2) A B C

( I want the mechanism)

3) a b

(But how??? I want explanation)

13
Avik ·

3) Hydrolysis gives 2 products, matlab it is an ether, tht too non-cyclic.
As, there is 1 degree unsturation, & it is declourising Br2 water, a double bond is present.

So, X will be.... CH2=CH-O-CH3.
Y & Z are- CH3-CHO & CH3-OH.

Hence, only A & B are rite.....

Your Answer

Close [X]