Class 12th

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New answer posted

10 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Using the Ideal Gas Law, PV = nRT:

P = 1 bar

V = 20 mL = 0.020 L

R = 0.083 L·bar·mol? ¹·K? ¹

T = 273 K

n = PV / RT = (1 * 0.020) / (0.0831 * 273) = 8.8 * 10? mol of Cl?

Number of Cl? molecules (N) = n * N_A = (8.8 * 10? ) * (6.022 * 10²³) = 5.3 * 10²? molecules.

Number of Cl atoms = 2 * (5.3 * 10²? ) = 1.06 * 10²¹.

The answer, rounded off to the nearest integer for the power of 10²¹, is 1.

New answer posted

10 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

For the dissociation of K? [Fe (CN)? ]? 4K? + [Fe (CN)? ]? , the number of ions produced (n) is 5.

The degree of dissociation (α) is related to the van't Hoff factor (i) by α = (i-1)/ (n-1).

Given α = 0.4: 0.4 = (i - 1) / (5 - 1) => 1.6 = I - 1 => I = 2.6

Using the depression in freezing point formula (ΔTf = i·Kf·m), and equating the ΔTf for two different solutions:
ΔTf (K? [Fe (CN)? ]) = ΔTf (A)
i? ·Kf·m? = i? ·Kf·m?
(2.6) * [ (18.1 / M) / (100-18.1)/1000] = (1) * [ (w? /M? ) / (W? /1000)]
The problem simplifies to finding the molar mass M of solute A:
2.6 * [18.1 / (M * 81.9)] * 1000 = [w? /M? ] * [1000/W? ]
Assuming the secon

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New answer posted

10 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

f (x) = (cos (sin x) - cos x) / x? We need lim (x→0) f (x) = 1/k.
Using cos C - cos D = -2 sin (C+D)/2) sin (C-D)/2).
f (x) = -2 sin (sin x + x)/2) sin (sin x - x)/2) / x?
For small x, sin x ≈ x.
lim (x→0) f (x) = lim -2 * ( (sin x + x)/2 ) * ( (sin x - x)/2 ) / x?
Using series expansion: sin x = x - x³/3! + x? /5! - .
sin x + x = 2x - x³/6 + .
sin x - x = -x³/6 + x? /120 - .
f (x) ≈ -2 * ( (2x)/2 ) * ( (-x³/6)/2 ) / x?
≈ -2 * (x) * (-x³/12) / x?
≈ (2x? /12) / x? = 2/12 = 1/6.
So, 1/k = 1/6 ⇒ k = 6.

New answer posted

10 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

The equation of the plane passing through the line of intersection is:
(2x - 7y + 4z - 3) + λ (3x - 5y + 4z + 11) = 0.
The plane passes through the point (-2, 1, 3).
(2 (-2) - 7 (1) + 4 (3) - 3) + λ (3 (-2) - 5 (1) + 4 (3) + 11) = 0
(-4 - 7 + 12 - 3) + λ (-6 - 5 + 12 + 11) = 0
(-2) + λ (12) = 0 ⇒ 12λ = 2 ⇒ λ = 1/6.
Substitute λ back into the equation:
(2x - 7y + 4z - 3) + (1/6) (3x - 5y + 4z + 11) = 0
Multiply by 6:
6 (2x - 7y + 4z - 3) + (3x - 5y + 4z + 11) = 0
12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0
15x - 47y + 28z - 7 = 0.
This is the equation ax + by + cz - 7 = 0.
So, a=15, b=-47, c=28.
We need to find the value of 2a + b

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New answer posted

10 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

10 months ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

10 months ago

0 Follower 55 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

10 months ago

0 Follower 13 Views

R
Raj Pandey

Contributor-Level 9

R_eq = 10 + (50 * 20) / (50 + 20) = 170/7 Ω
⇒ I = 170 / (170/7) = 7A
⇒ x = 10 * 7 = 70V ⇒ Voltage across 10Ω resistor

New answer posted

10 months ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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