Given the solubility of an unknown gas at partial pressure and temperature, its solubility at partial pressure 24.6 atm has to be calculated. Concept Introduction: Solubility of a gas in liquid is explained by Henry’s law which states – “At a constant temperature, the amount of a gas dissolved in a given volume of liquid is directly proportional to the partial pressure of the gas that is in equilibrium with the liquid.” It is expressed as, S α P Where, S = Solubility P = Partial pressure Introducing proportionality constant, S = k H P Where, k H is Henry’s constant . Solubility of a same gas at two different pressures can be calculated as, S 2 S 1 = k H P 2 k H P 1 = P 2 P 1
Given the solubility of an unknown gas at partial pressure and temperature, its solubility at partial pressure 24.6 atm has to be calculated. Concept Introduction: Solubility of a gas in liquid is explained by Henry’s law which states – “At a constant temperature, the amount of a gas dissolved in a given volume of liquid is directly proportional to the partial pressure of the gas that is in equilibrium with the liquid.” It is expressed as, S α P Where, S = Solubility P = Partial pressure Introducing proportionality constant, S = k H P Where, k H is Henry’s constant . Solubility of a same gas at two different pressures can be calculated as, S 2 S 1 = k H P 2 k H P 1 = P 2 P 1
Solution Summary: The author explains how the solubility of an unknown gas at partial pressure and temperature is calculated by Henry's law.
Given the solubility of an unknown gas at partial pressure and temperature, its solubility at partial pressure 24.6 atm has to be calculated.
Concept Introduction:
Solubility of a gas in liquid is explained by Henry’s law which states –
“At a constant temperature, the amount of a gas dissolved in a given volume of liquid is directly proportional to the partial pressure of the gas that is in equilibrium with the liquid.”
It is expressed as,
SαP
Where,
S = SolubilityP = Partial pressure
Introducing proportionality constant,
S=kHP
Where,
kH is Henry’s constant.
Solubility of a same gas at two different pressures can be calculated as,
14. Calculate the concentrations of Ag+, Ag(S2O3), and Ag(S2O3)23- in a solution prepared by mixing
150.0 mL of 1.00×10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3
Ag+ + S20
Ag(S203)¯
K₁ = 7.4 × 108
Ag(S203)¯ + S20¯ = Ag(S203)
K₂ = 3.9 x 104
ΗΝ,
cyclohexanone
pH 4-5
Draw Enamine
I
I
CH3CH2Br
THF, reflux
H3O+
I
Drawing
Draw Iminium Ion
:0: :0:
Select to Add Arrows
:0:
(CH3)2NH
:0:
■ Select to Add Arrows
:0:
:0:
(CH3)2NH
■ Select to Add Arrows
Chapter 12 Solutions
Student Solutions Manual for Ebbing/Gammon's General Chemistry, 11th
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