Suppose that the resistance between the walls of a biological cell is 9.00 x 10^

Question

Suppose that the resistance between the walls of a biological cell is 9.00 x 10^9 olhms. (A) what is the current when the potential difference between the walls is 40 mV? (B) if the current is composed of Na+ ions (q=+e), how many such ions flow in 1.4 s? Suppose that the resistance between the walls of a biological cell is 9.00 x 10^9 olhms. (A) what is the current when the potential difference between the walls is 40 mV? (B) if the current is composed of Na+ ions (q=+e), how many such ions flow in 1.4 s? (A) what is the current when the potential difference between the walls is 40 mV? (B) if the current is composed of Na+ ions (q=+e), how many such ions flow in 1.4 s?

Explanation / Answer

A)

I = V / R

I = (40 x 10^-3 V) / (9 x 10^9 )

I = 4.44 pA

B)

q = I*t

q = (4.44 x 10^-12 A) x (1.4 s) = 6.216 pC

Q = No.(Na) x 1.6 x 10^-19

No. of Na Ions = 6.216 x 10^-12/1.6 x 10^-19

No. of Na Ions = 3.885 x 10^7 ions

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