The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>
Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies
When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles
<span>molar mass of NH3 = 17 g/mole </span>
<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
Answer:
Kb = 6.22x10⁻⁷
Explanation:
Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:
C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)
Kb is defined from concentrations in equilibrium, thus:
Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]
The equilibrium concentration of these compounds could be written as:
[C₆H₁₅O₃N] = 0.486M - X
[C₆H₁₅O₃NH⁺] = X
[OH⁻] = X
pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M
Also, Kw = [OH⁻] ₓ [H⁺];
1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]
1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]
5.495x10⁻⁴M = [OH⁻], that means <em>X = 5.495x10⁻⁴M</em>
Replacing in Kb formula:
Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]
<em>Kb = 6.22x10⁻⁷</em>
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I would say D would be the answer
