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?

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.

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.

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'