Compartments A and B are separated by a membrane that is permeable to K+ but not to Na+ or Cl-. At time zero, a solution of KCl is poured into compartment A and an equally concentrated solution of NaCl is poured into compartment B. Which would be true once equilibrium is reached?

A. The concentration of Na+ in A will be higher than it was at time zero.
B. Diffusion of K+ from A to B will be greater than the diffusion of K+ from B to A.
C. There will be a potential difference across the membrane, with side B negative relative to side A.
D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
E. The concentration of Cl- will be higher in B than it was at time zero.

The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.

Explanation:

The cell potential may be defined as the potential difference between the inside of the cell and outside the cell. This difference is important for the transmission of nerve impulse.

The equilibrium condition may be defined as the condition at which  the exit and the entrance of the ions will be equal. As the information in the question, Only potassium ions will move inward and outward of the cell. The potassium ions will move in opposite direction since the equilibrium has been reached.

Thus, the correct answer is option (D).

c. heat from the poles back down towards the equator.

c is the answer. hope this