Solutions of hydrogen in palladium may be formed by exposing Pd metal to H 2 gas. The concentration of hydrogen in the palladium depends on the pressure of H 2 gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density = 10.8 g cm 3 ). (a) Determine the molarity of this solution. (b) Determine the molality of this solution. (c) Determine the percent by mass of hydrogen atoms in this solution.
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H 2 gas. The concentration of hydrogen in the palladium depends on the pressure of H 2 gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density = 10.8 g cm 3 ). (a) Determine the molarity of this solution. (b) Determine the molality of this solution. (c) Determine the percent by mass of hydrogen atoms in this solution.
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density = 10.8 g cm3).
(a) Determine the molarity of this solution.
(b) Determine the molality of this solution.
(c) Determine the percent by mass of hydrogen atoms in this solution.
In one electrode: Pt, H2(1 atm) | H+(a=1), the interchange current density at 298K is 0.79 mA·cm-2. If the voltage difference of the interface is +5 mV. What will be the correct intensity at pH = 2?. Maximum transfer voltage and beta = 0.5.
In a Pt electrode, H2(1 atm) | H+(a=1), the interchange current density of an electrode is 0.79 mA cm-2. ¿Qué corriente flow across the electrode of área 5 cm2 when the difference in potential of the interface is +5 mV?.
If the current voltage is n = 0.14 V, indicate which of the 2 voltage
formulas of the ley of Tafel must be applied
i
a
a) == exp (1-B).
xp[(1 - ß³):
Fn
Fn
a
b) == exp B
RT
RT
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