Class 12th

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New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Let the charge of on each drop be q and radius of each drop be r.

k q r = 2

When all drops are joined, radius,

r ' = ( 5 1 2 ) 1 3 r = 8 r

Potential of the new drop,

V = k . 5 1 2 q 8 r = 6 4 k q r = 1 2 8 V

New answer posted

10 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Acidic order

Simple concept of conjugate base stability

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10 months ago

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V
Vishal Baghel

Contributor-Level 10

2 π f 1 L C L = λ 2 4 π 2 C c 2 = 9 6 0 2 4 * 3 . 1 4 2 * 2 . 5 6 * 1 0 6 * 9 * 1 0 1 6 =10-7 = 10 * 10-8 H

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10 months ago

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P
Payal Gupta

Contributor-Level 10

Cu, Zn, Ni

Composition of german silver [Cu-50%, Zn-20%, Zi-30%]

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10 months ago

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V
Vishal Baghel

Contributor-Level 10

E 1 E 2 = 3 8 0 7 6 0 = 1 2                                                                                                      

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

Primary amine, whether aliphatic or aromatic when warmed with CHCl3, alcoholic (KOH) it forms isocyanide

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10 months ago

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V
Vishal Baghel

Contributor-Level 10

v = p m v p : v d : v α = 1 : 1 2 : 1 4 = 4 : 2 : 1

F m a g = q v B F P : F d : F α = 1 * 4 : 1 * 2 : 2 * 1 = 2 : 1 : 1

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

Please find the below image

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

[h] = ML2T-1

[E] = ML2T-2

[V] = ML2T-2C-1

[P] = MLT-1

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

I > III > IV > II

[FeF6]3Fe3+3d5 all are unpaired (n = 5)

[Co (NH3)6]3+Co3+3d6 all are paired (n = 0)

[NiCl4]2Ni2+3d8 (n=2)

[Cu (NH3)4]2+Cu++3d9 (n=1)

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