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,
If we assume a system with an anodic overpotential, the variation of n as a function
of current density:
1. at low fields is linear 2. at higher fields, it follows Tafel's law
Obtain the range of current densities for which the overpotential has the same value
when calculated for 1 and 2 cases (maximum relative difference of 5% compared to
the behavior for higher fields).
To which overpotential range does this correspond?
Data: i = 1.5 mA cm², T = 300°C, B = 0.64, R = 8.314 J K1 mol-1 and F = 96485 C mol-1.
Answer by equation please
Some of the theories used to describe interface structure can be distinguished by:1. the measured potential difference.2. the distribution of ions in solution.3. the calculation of charge density.4. the external Helmoltz plane.
Chapter 12 Solutions
OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
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