Class 12th

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New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

Consider the following image

New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

It does not contains 2nd most abundant element by weight in earth crust because that is Si Calgon

W a t e r s o l u b l e 2 N a + [ N a 4 ( P O 3 ) 6 ] 2

C a 2 + 2 N a + [ N a 2 C a ( P O 3 ) 6 ] 2

New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

Ceric ammonium nitrate is used to test alcohol while CHCl3/alc. KOH is used to test 1° amine

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

Due to equal and similar charge particle will repel each other, hence will never precipitate.

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6 months ago

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alok kumar singh

Contributor-Level 10

Due to common ion effect solubility of AgCl will decreases in KCl, AgCl and AgNO3 but in deionized water, no common ion effect will takes place so maximum solubility.

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6 months ago

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alok kumar singh

Contributor-Level 10

Covalent character is L i C l > N a C l > K C l > C s C l .  As the cationic size increases polarization decreases.

New answer posted

6 months ago

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alok kumar singh

Contributor-Level 10

(A) Cr = [Ar]3d54s1

              (B) m =   l t o + l

              (C) According to Aufbau principle, orbital are filled in order of their increasing energies.

              (D) Total nodes = n – 1

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6 months ago

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alok kumar singh

Contributor-Level 10

Moles of Fe3O4 =   4 . 6 4 * 1 0 3 2 3 2 = 2 0

              Moles of CO =   2 . 5 2 * 1 0 3 2 8 = 9 0

              So limiting reagent = Fe3O4

              So moles of Fe formed = 60

              Weight of Fe = 60 * 56 = 3360

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

Draw y = cos2x and y = 2x + 2

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6 months ago

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A
alok kumar singh

Contributor-Level 10

3cos22θ + 6cos2θ -   1 0 ( 1 + c o s 2 θ ) 2 + 5 = 0

c o s 2 θ ( 3 c o s 2 θ + 1 ) = 0

 Þ cos2θ = 0,     1 3

Draw y = cos2θ, y = 0 and y =  1 3 ,  find the pt. of intersection.

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