Answer:
Kb = 1.77x10⁻⁵
Explanation:
When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
And the dissociation constant, Kb, for this equilibrium is:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
To find Kb you need to find the concentration of each species. The equilibrium concentrations are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
<em>Where X is reaction coordinate.</em>
You can know [OH⁻] and, therefore, X, with pH of the solution, thus:
pH = -log [H⁺] = 11.612
[H⁺] = 2.4434x10⁻¹²
As 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]
4.0926x10⁻³ = [OH⁻] = X
Replacing, concentrations of the species are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
[NH₃] = 0.9459M
[NH₄⁺] = 4.0926x10⁻³M
[OH⁻] = 4.0926x10⁻³M
Replacing in Kb expression:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]
<h3>Kb = 1.77x10⁻⁵</h3>
Answer:
most likely c
Explanation:
the volume is increased by the quickly moving water particles
..............................
Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>
