Class 12th

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New answer posted

11 months ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

Here,

A = [ 2 1 1 1 0 1 1 1 0 ]            

we get A2 = A and similarly, for

B = A I = [ 1 1 1 1 1 1 1 1 1 ]            

We get B2 = -B  B3 = B

A n + ( ω B ) n = A + ( ω B ) n f o r n N

For ω n  to be unity n shall be multiple of 3 and for Bn to be B. n shell be 3, 5, 7, ….9

n = { 3 , 9 , 1 5 , . . . . . 9 9 }     

Number of elements = 17.

New answer posted

11 months ago

0 Follower 88 Views

V
Vishal Baghel

Contributor-Level 10

2 (only 2 carbon will chiral in product form)

New answer posted

11 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

  C o 3 +  will not get oxidized, so will not liberate H2 gas upon reaction with acidic solution.

No. of unpaired electron = 4

Spin only magnetic moment = 4 ( 4 + 2 ) B M = 2 4 B M 5 B M

New answer posted

11 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Complex absorbing minimum wavelength of light means the value of Δ  for it is maximum.

That complex will be [ C o ( C N ) 6 ] 3

C o 3 + 3 d 6 4 s 0 t 2 g 6 e g 0

No. of unpaired electron = 0

New answer posted

11 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

( t 1 / 2 ) l ( t 1 / 2 ) I I = ( P I P I I ) 1 n n = 1

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

  A n o d e : ( Z n Z n 2 + + 2 e ) * x

C a t h o d e : ( S n x + + x e S n ) * 2 _      

Net call reaction :  x Z n + 2 S n x + x Z n 2 + + 2 S n ( n = 2 x )

Using Nernst equation

No. of electron involved = 4

New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

d = 2 R h          

d = 2 * 6 4 0 0 * h * 1 0 3 ( h i n m )

Area =  π d 2

= ( π * 2 * 6 4 0 0 * h * 1 0 3 ) km2

h = 150m

New answer posted

11 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

E 9 = h c λ = 1 2 4 2 λ ( n m ) = 1 2 4 2 4 0 0 = 3 . 1 0 5

New answer posted

11 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Each wile has resistance = ρ 4 l π d 2 = r

Eight wire in parallel, then equivalent resistance is

r 8 = ρ l 2 π d 2

Single copper wire of length 2 l  has resistance

R = ρ 2 l * 4 π d 1 2 = ρ l 2 π d 2

d1 = 4d

New answer posted

11 months ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

K  V  L Loop 1               

   

6 - l * 1   -l1 * 1 - 4 = 0

2 = l + l1                             - (i)

K  V  L  Loop 2

4 + l1 * 1 - (l – l1) * 1 – 2 = 0

2 + l 1 l = 0                        

l = 2 + 2l1                     &nb

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