Class 12th

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New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Egg white will stabilize blue ink easily.

New answer posted

10 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

P? = X? (P? – P? ) + P?
ATQ
550 = 1/4 (P? – P? ) + P?
2200 = P? – P? + 4P?
560 = 1/5 (P? – P? ) + P?
2200 = P? – P? + 5P?
P? + 3P? = 2200
P? + 4P? = 2800
P? = 600
P? = 400 mmHg

New question posted

10 months ago

0 Follower 3 Views

New question posted

10 months ago

0 Follower 2 Views

New answer posted

10 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

Charge (q) = (it/96500) F = (1x15x60)/96500 = 900/96500 = 9/965 F = 0.0093F
No. of moles of Au? = 0.025 & No. of moles of Ag? = 0.025
Species with higher value of SRP will get deposited first at cathode.
(i) Au? (aqs) + e? → Au (s)
0.025 0.0093 mole
So only Au will get deposited

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

In Calcination and roasting CO? and SO? are released which are responsible for Global warning and acid rain.

New answer posted

10 months ago

0 Follower 26 Views

V
Vishal Baghel

Contributor-Level 10

[Co (H? O)? F? ]Co³? = 3d?4s? ⇒ t? g²? ¹? ¹ e? ¹? ¹
CFSE = [−0.4nt? g + 0.6neg]? o + n (P)
= [−0.4 * 4 + 0.6 * 2]? o + 0 = −0.4? o

New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

We know I = 4I? cos² (φ/2) but φ = 2πx/λ
I = 4I? cos² (πx/λ)
(i) when x = λ, I = k
i.e. k = 4I? cos²π
k = 4I?
(ii) when x = λ/6
I' = kcos² (π (λ/6)/λ) = kcos² (π/6) = k (√3/2)²
I' = 3k/4
Correction from image: The provided image has an error in the calculation. I' = k ( (√3/2)² ) = 3k/4. The image shows 9k/12 which simplifies to 3k/4. I will use the value from the OCR which seems to be a typo.
I' = kcos² (π/6) = k (3/4)
I' = 9k/12

New answer posted

10 months ago

0 Follower 18 Views

V
Vishal Baghel

Contributor-Level 10

λ = 6 * 10? M
d = 6 x 10? M
I = I? [ (sin (β)²/β²] ; β = (πdsinθ)/λ
θ = π/2 ; β = πd/λ
= (π (6 * 10? )/ (6 x 10? ) = 100π

So at ∞ also minima will form total number of minima = 2 * 100 = 200

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