D. Carbon moved in a cyclic fashion through the atmosphere and living organisms
<u>Condensation</u> is the next step in the water cycle that is most likely to happen next as water molecules evaporate from a lake near the equator and rise into the atmosphere, as shown in the diagram below. and as the molecules travel away from the equator.
Explanation:
As the warmer air mass rises into the upper atmosphere, at the equator, it begins to cool because temperatures drop with altitude. The relative humidity of the air mass increases until it reachesdew point. The water vapor begins to condense into water droplets.
This air mass begins moving towards the poles (away from the equator) bringing with it clouds (condensed water vapor) that precipitate later on. This air mass begins to sinks because it has become denser with a decrease in temperature. As it sinks it replaces the air mass moving to the equator. The air rushing to the equator is replacing the rising warm air mass at the equator. This cycle continues.
Learn More:
For more on the water cycle check out;
brainly.com/question/11437446?section=related_q
brainly.com/question/978359
#LearnWithBrainly
C. Because mutations are varieties
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)
Answer:
the answer is c just did it on edg and got it right
Explanation:
