Class 12th

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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

Δ T f = K f * m  

( 1 5 6 − 1 5 5 . 1 ) = 2 * 1 . 8 M W * 1 0 0 0 ( 6 2 . 5 * 0 . 8 )                

  0.9 =   2 * 1 . 8 * 1 0 0 0 M W * 5 0


∴ M W = 8 0 g m / m o l      

New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 30 Views

V
Vishal Baghel

Contributor-Level 10

No. of moles =  5 9 2 ( t o l u e n e )

Moles of benzaldehyde = 5 9 2 * 9 2 1 0 0 = 5 * 1 0 − 2  

Mass of benzaldehyde = 5 * 10-2 *106 = 5.3 gm

= 530 * 10-2 gm

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Fehling's solution is a complex of   C u 2 + , C u 2 + = 3 d 9

No. of unpaired e- = 1

MM =  1 ( 1 + 2 ) = 3 = 1 . 7 3 B M

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

  2 K M n O 4 + 3 H 2 O 2 → B a s i c     m e d i u m 2 M n O 2 + 3 O 2 + 2 H 2 O + 2 K O H

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Bauxite = AlOx (OH)3-2x (where O < x < 1)

Siderite = FeCO3

Cuprite = Cu2O

Calamine = ZnCO3

Haemetite = Fe2O3

Kaolinite = Al2 (OH)4Si2O5

Malachite = CuCO3 Cu (OH)2

Magneite = Fe3O4

Sphalerite = ZnS

Limonite = Fe2O3.3H2O

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

C = 500   μ F , V = 100 v, L = 50 mH                                                        

In this LC – oscillation

q = q0 cos   ω t

i = − d q d t = q 0 ω s i n ω t     ω = 1 2 c = 1 5 0 * 1 0 − 3 * 5 * 1 0 − 4

=  1 0 0 0 5 = 2 0 0  

So,     imax =   q 0 ω = 5 0 0 * 1 0 6 * 1 0 0 * 2 0 0

= 10A

 

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

K = 0 . 6 9 3 t 1 / 2 = 0 . 6 9 3 7 0 * 6 0 = 6 9 3 0 7 * 6 * 1 0 − 6

= 165 * 10-6 s-1

New answer posted

a year ago

0 Follower 54 Views

V
Vishal Baghel

Contributor-Level 10

P 0 − P S P 0 = n s o l u t e n s o l v e n t

P 0 − P 0 2 P 0 = n s o l u t e n s o l v e n t

n s o l u t e = n s o l v e n t 2 = 1 0 0 1 8 * 2 = 2 . 7 8     m o l

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  λ A = 2 5 λ , λ B = 1 6 λ      

At t = 0 Þ NA = NB = N0

after t =  1 a λ : − N B N A = N 0 e − 1 6 λ t N 0 e − 2 5 λ t = e ( 2 5 λ − 1 6 λ ) t  

->e =  e ( 9 λ 1 a λ )  


⇒ 9 a = 1 ⇒ a = 9      

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