Chemical Equilibrium

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3 months ago

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V
Vishal Baghel

Contributor-Level 10

Let total mole = 1

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

X Y 2 X + 2 + 2 y - K   sp   = X + 2 [ X ] 2 = 10 - 3 2 * 10 - 3 2 = 4 * 10 - 9

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

500 m L has H C l = 25 * 10 - 3 m o l = 25 m m o l

120 m L has = 1 m m o l

and 500 m L has C H 3 C O O H = 1 20 * 10 3 m m o l

so 20 m L has 10 3 20 * 20 500 = 2 m m o l

N a O H added in 20 m L is 5 * 1 2 = 2.5 m m o l

So, N a O H (left) + C H 3 C O O H ? C H 3 C O O H + H 2 O
1.5
2
1.5

p H = p K a + l o g ?   salt     acid   = 4.75 + l o g ? 0.4771 = 5.2271

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Ba (OH)? ? Ba? ² + 2 (OH)?
Initial: 0.005M, 0, 0
Final: 0, 0.005, 0.010
[OH? ] = 0.01 M = 10? ² M
Kw = [H? ] [OH? ]
[H? ] or [H? O? ] = Kw/ [OH? ] = 10? ¹? /10? ² = 10? ¹² M
∴ [H? O? ] = 1 * 10? ¹² M

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

AB2 (g)? A (g)+2B (g)

Initial1 mole -

At equilirbium 1-x mole x mole2x mole

V = 25 L and

T = 300 K.

At equilibrium, P = 1.9 atm

Total moles at equilibrium, n = 1 + 2x

V = 25 L

T = 300 K

Using, PV = nRT

1.9 * 25 = (1 + 2x) * 0.08206 * 300

x = 0.465

Now;Partial pressure of AB2 at equilibrium = 0.5351.93*1.9atm

Using ; Kp=PA.PB2PAB2

Kp= (0.465*1.91.93) (2*0.4651.93*1.9)20.5351.93*1.9 = 0.728

KP = 0.73

Kp = 73 * 10-2

So; x = 73

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Initially ->              1 mol                   -

At eq.                      1-x mol               2x mol

Here; molecules of Cl2 = atoms of Cl

i.e moles of Cl2 = moles of Cl

So : 1 – x = 2x        x = 1/3

Moles of Cl2 at equibrium =   2 3

Moles of Cl at equilibrium =   2 3

Total moles =   4 3

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