Chemical Equilibrium
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New answer posted
3 months agoContributor-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 agoContributor-Level 10
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 =
Using ;
= 0.728
KP = 0.73
Kp = 73 * 10-2
So; x = 73
New answer posted
3 months agoContributor-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 =
Moles of Cl at equilibrium =
Total moles =
No
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