Class 11th

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New answer posted

11 months ago

0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

 ΔH=165kJ/mole

T =?

ΔS=550JK1

At equilibrium ; ΔG=0

T=ΔHΔS=165*1000550K

=3*100K=300K

New answer posted

11 months ago

0 Follower 58 Views

P
Payal Gupta

Contributor-Level 10

E1H=2.2*1018J

LiLi+2+2e, n=2

ELi+2=E1H*Z2n2=2.2*1018*3222

λ=4*108 m

New answer posted

11 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

n 1 = 1 m o n o t o m i c

n 2 = 3 d i a t o m i c C v = α 2 R 4 J / m o l k

As, Cv (mix) = 1 * 3 2 R + 3 * 5 2 R 4 = 9 R 4 = α 2 R 4

α = 3

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

For adiabatic process – PVY = const

T 1 V 1 Y 1 = T 2 V 2 Y 1

T 2 T 1 = ( v 1 v 2 ) Y 1 = ( d 2 d 1 ) Y 1 = ( 3 2 ) ( 7 5 1 ) = ( 3 2 ) 2 / 5

= ( 2 ) 2 = 4

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

g ( a t h ) = g ( a t d e p t h α h ) h << R

g ( 1 2 h R ) = g ( 1 α h R )

1 2 h R = 1 α h R 2 h R = α h R α = 2

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

R = u 2 s i n ( 2 * 4 5 ° ) g = u 2 g

R 2 = u 2 2 g = u 2 s i n 2 0 g

s i n 2 θ = 1 2

2 θ = 3 0 ° θ = 1 5 °

New answer posted

11 months ago

0 Follower 71 Views

V
Vishal Baghel

Contributor-Level 10

x 0 1 = 3 ( 1 ( 1 2 ) ) 2 0 1 1 2 = 6 ( 1 1 2 2 0 )

= i = 1 2 0 ( x i ) 2 + ( i ) 2 2 x i i

Now, i = 1 2 0 ( x i ) 2 = 9 ( 1 ( 1 4 ) ) 2 0 ( 1 1 4 ) = 1 2 ( 1 1 2 4 0 )

x ¯ = 2 8 5 8 2 0 + ( 1 2 2 4 0 + 2 2 2 2 0 ) * 1 2 0

[ x ¯ ] = 1 4 2

New answer posted

11 months ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

e E = 1 b 2 a 2 , e H = 2

I f e E = 1 e H a 2 b 2 a 2 = 1 2

k 2 = a 2 * 5 2 + b 2 = 3 2

6 b 2 = 3 2 b 2 = 1 4 a n d a 2 = 1 2

4 ( a 2 + b 2 ) = 3

New answer posted

11 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Put 1 + x 2 = t 2 2 x d x = 2 t d t  

1 2 1 5 ( t 2 1 ) t d t t 2 + t 3 d t Put (1 + t) = u2

3 0 2 3 ( u 4 2 u 2 ) d u dt = 2u du

= 6 3 + 1 6 2 = α 2 + β 3

α = 1 6 , β = 6 α + β = 1 0      

New answer posted

11 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Required number = Total – no character from {1, 2, 3, 4, 5}

= ( 1 0 6 5 6 ) + ( 1 0 7 5 7 ) + ( 1 0 8 5 8 )

= 5 6 ( 2 6 * 1 1 1 3 1 ) = 5 6 * 7 0 7 3 α

= 7073

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