Answer:
Explanation:
From the information given:
The cell potential on mars E = + 100 mV
By using Goldman's equation:
![E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}} \Big )](https://tex.z-dn.net/?f=E_m%20%3D%20%5Cdfrac%7BRT%7D%7BzF%7DIn%20%5CBig%20%28%5Cdfrac%7BP_K%5BK%5E%2B%5D_%7Bout%7D%2BP_%7BNa%7D%5BNa%5E%2B%5D_%7Bout%7D%2BP_%7BCl%7D%5BCl%5E-%5D_%7Bout%7D%20%7D%7BP_K%5BK%5E%2B%5D_%7Bin%7D%2BP_%7BNa%7D%5BNa%5E%2B%5D_%7Bin%7D%2B%20P_%7BCl%7D%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%20%20%20%5CBig%20%29)
Let's take a look at the impermeable cell with respect to two species;
and the two species be Na⁺ and Cl⁻
![E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}](https://tex.z-dn.net/?f=E_m%20%3D%20%5Cdfrac%7BRT%7D%7BzF%7D%20In%20%5Cdfrac%7B%5BK%5E%2B%5D_%7Bout%7D%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D)
where;
z = ionic charge on the species = + 1
F = faraday constant
∴
![100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20%5CBig%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B1%5Ctimes%2096485%7D%20%5CBig%29%20%5Cmathtt%7BIn%7D%20%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%200.0257%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=3.981%3D%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\ 53.57 = \dfrac{4}{[K^+]_{in}}](https://tex.z-dn.net/?f=exp%20%28%203.981%29%20%3D%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%5C%5C%20%5C%5C%20%2053.57%20%3D%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D)
![[K^+]_{in} = \dfrac{4}{53.57}](https://tex.z-dn.net/?f=%5BK%5E%2B%5D_%7Bin%7D%20%3D%20%5Cdfrac%7B4%7D%7B53.57%7D)
![[K^+]_{in} = 0.0476](https://tex.z-dn.net/?f=%5BK%5E%2B%5D_%7Bin%7D%20%20%3D%200.0476)
For [Cl⁻]:
![100 \times 10^{-3} = -0.0257 \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20-0.0257%20%5C%20%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![-3.981 = \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)](https://tex.z-dn.net/?f=-3.981%20%3D%20%20%5C%20%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![0.01867 = \dfrac{120}{[Cl^-]_{in}}](https://tex.z-dn.net/?f=0.01867%20%3D%20%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D)
![[Cl^-]_{in} = \dfrac{120}{0.01867}](https://tex.z-dn.net/?f=%5BCl%5E-%5D_%7Bin%7D%20%3D%20%5Cdfrac%7B120%7D%7B0.01867%7D)
![[Cl^-]_{in} =6427.4](https://tex.z-dn.net/?f=%5BCl%5E-%5D_%7Bin%7D%20%3D6427.4)
For [Na⁺]:
![100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%200.0257%20%5CBig%20%28%20%5Cdfrac%7B145%7D%7B%5BNa%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![53.57= \Big ( \dfrac{145}{[Na^+]_{in}} \Big)](https://tex.z-dn.net/?f=53.57%3D%20%5CBig%20%28%20%5Cdfrac%7B145%7D%7B%5BNa%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![[Na^+]_{in}= 2.70](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D_%7Bin%7D%3D%202.70)
I believe your pointing to number 16, I believe the answer is B
Biological augmentation, bioremediation , or reforestation are names for the method.
Semi permeable membrane will allow something to pass through them and something not and it will allow on the basis of size particles
while differentialy permeable will include other factors also other than size and will depend on the need of the cell
this is the main difference between them
hope it helps<span />