Class 12th

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S
Satyendra Dhyani

Beginner-Level 5

Shiksha focuses on completely fulfilling the needs of CBSE class 12 board students. We provide NCERT Solutions for classes 11th and 12th in class for physics, chemistry and mathematics.

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New answer posted

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A
alok kumar singh

Contributor-Level 10

Na+ C + N + S ®NaSCN

Fe3+ + SCN [ F e ( S C N ) ] 2 + B l o o d r e d

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alok kumar singh

Contributor-Level 10

Potassium permanganate in alkaline medium oxidise lodide to lodate.

2 M n O 4 + H 2 O + I Θ 2 M n O 2 + 2 O H Θ + I O 3 Θ ( A )            

Compound A is  . Therefore, oxidation state of 1is + 5.

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a month ago

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A
alok kumar singh

Contributor-Level 10

From structure it is clear [CO2 (CO)2] has bridging carbonyl ligand.

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A
alok kumar singh

Contributor-Level 10

In presence of light allylic substitution occur.

In presence of CCl4, addition reaction will occur.

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alok kumar singh

Contributor-Level 10

KMnO4 decomposes upon heating at 513 K and forms K2MnO4 and MnO2.

2KMnO4  K2MnO4 + O2 + MnO2

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a month ago

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A
alok kumar singh

Contributor-Level 10

  a + 5 b  is collinear with c  

  a + 5 b = c           …(1)

b + 6 c is collinear with a  

⇒   b + 6 c = μ a               …(2)

From (1) and (2)

  b + 6 c = μ ( λ c 5 b )          

-> ( 1 + 5 μ ) b + ( 6 λ μ ) c = 0

? b and c  are non-collinear

-> 1 + 5m = 0 μ = 1 5  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

b = 6 c = 1 5 a

5 b + 3 0 c = a

a + 5 b + 3 0 c = 0 a + α b + β c = 0 ]

On comparing

α = 5, β = 30  α + β = 35

New answer posted

a month ago

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alok kumar singh

Contributor-Level 10

y (x) = 2x – x2

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5

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a month ago

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alok kumar singh

Contributor-Level 10

( s i n x c o s x ) s i n 2 x t a n x ( s i n 3 x + c o s 3 x ) d x

( s i n x c o s x ) s i n x c o s x s i n 3 x + c o s 3 x d x , put sin3x + cos3x = t(3 sin2x*cosx – 3cos2xsinx) dx = dt

-> 1 3 d t t

= l n t 3 + c

= l n | s i n 3 x + c o s 3 x | 3 + c

             

           

New answer posted

a month ago

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alok kumar singh

Contributor-Level 10

f ( x ) = ( 2 x + 2 x ) t a n x t a n 1 ( 2 x 2 3 x + 1 ) ( 7 x 2 3 x + 1 ) 3

f ( x ) = ( 2 x + 2 x ) . t a n x . t a n 1 ( 2 x 2 3 x + 1 ) . ( 7 x 2 3 x + 1 ) 3

f ' ( x ) = ( 2 x + 2 x ) . s e c 2 x . t a n 1 ( 2 x 2 3 x + 1 ) . ( 7 x 2 3 x + 1 ) 3 + t a n x . ( Q ( x ) )

f ' ( 0 ) = 2 . 1 π 4 . 1

= π

 

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