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)
SUGAR = that’s the pro-fix glu in glucose
Answer:
Aquatic life quickly suffers the effects of watershed pollution, while new pollutants introduced into ecosystems alter wildlife habitats. This reduces biodiversity by eliminating some species and introducing new, invasive ones that destroy the native species.
Explanation: