Answer:
g = 1.11x10⁻⁵Ω.
Explanation:
The membrane conductance (g) can be calculated by dividing membrane current (I) through the driving force (Vm - E) as follows:
<em>where Vm: is the membrane potential and </em><em>: is the equilibrium potential for the ion or reversal potential. </em>
The equilibrium potential for the ion can be calculated using the Nernst equation:
<em>where R: is the gas constant = 8.314 J/K*mol, F: is the Faraday constant = 96500 C/mol, T: is the temperature (K), z: is the ion's charge, [ion]out and [ion]ins: is the concentration of the ion outside and inside, respectively. </em>
Now, we can calculate the membrane conductance (g) using equation (1):
Therefore, the membrane conductance is 1.11x10⁻⁵Ω.
I hope it helps you!
For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:
R₁/R₂ = √(M₂/M₁)
Let 1 be N₂ while 2 be Cl₂
255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s
<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
Answer:
the standard cell potential value
Explanation:
For every cell, we can calculate its standard electrode potential from the table of standard electrode potentials listed in many textbooks.
However, from Nernst's equation;
Ecell= E°cell - 0.0592/n log Q
Hence the standard cell potential (E°cell) affects the value of the calculated cell potential Ecell from Nernst's equation as stated above.