Class 11th

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New answer posted

a year ago

0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

 ΔH=−165kJ/mole

T =?

ΔS=−550JK−1

At equilibrium ; ΔG=0

∴T=ΔHΔS=−165*1000−550K

=3*100K=300K

New answer posted

a year ago

0 Follower 60 Views

P
Payal Gupta

Contributor-Level 10

E1H=−2.2*10−18J

Li→Li+2+2e−, n=2

ELi+2=E1H*Z2n2=−2.2*10−18*3222

λ=4*10−8 m

New answer posted

a year 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

a year 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

a year 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

a year ago

0 Follower 2 Views

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

a year ago

0 Follower 73 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

a year 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

a year ago

0 Follower 15 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

a year 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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