Class 12th

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New answer posted

11 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

11 months ago

0 Follower 90 Views

V
Vishal Baghel

Contributor-Level 10

Birch reduction takes place of carbon connected with electron withdrawing group (-NO2)

New answer posted

11 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

MnF4+4 MnF3+3

E.C = [Ar] 3d3                    [Ar]3d4

MnF2+2 [Ar]3d5 [EMn3+|Mn2+0=1.57VEMn4+|Mn2+0=1.27V]

New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

ω = Δ k . E = 1 2 m [ v t 2 v i 2 ]

= 1 2 m [ b * 4 5 / 2 ] 2 1 2 m [ 0 ] 2

= 1 2 * 0 . 5 [ 0 . 2 5 2 * 4 5 ] = 2 4 = 1 6 J

New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

 C2H5OH+PCl3C2H5Cl+H3PO3

H3PO3+PCl3H4P2O5+HCl

 

New answer posted

11 months ago

0 Follower 80 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

11 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

To free the electron from metal surface minimum energy required, is equal to the work function of that metal.

              So Assertion A, is correct

              have = w0 + K.Emax

                        If have = w0

              Þ K.Emax = 0

              Hence reas

...more

New answer posted

11 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

Both indicated hydrogen atoms repel each other so due steric hinderance the given compound becomes non planar. It is non-aromatic.

New answer posted

11 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

ΔTb=iKbm1

ΔTbΔTf=Kb*1Kf*236=12=Kbkf*12

KbKf=1x=1

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