Answer:
Incomplete question
Complete question:
8. 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.
Answer: D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
Explanation:
Diffusion is the movement of molecules from region of higher concentration to lower concentration through a semipermeable membrane.
Since the k+ is the permeable membrane, the k+ ion in the KCl would move in equal magnitude and direction in the solution.
Answer:
VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Explanation:
Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.
=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Review Graham's Law => Effusion Rate ∝ 1/√formula mass.
B. 0.9 <span>
</span>D.Light intensity has no effect on whether electrons are emitted or not.
and
A. X=1.9eV,Y=0.2eV
I already took the gizmo so I know these are right. The first one I got wrong b/c there was no graph and the last one I didn't understand. Basically for the last one you calculate the work function for the metals and find their difference.
Hope this helps.
