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Debora [2.8K]
3 years ago
7

How would you describe the process of a gas condensing into a liquid and then back into a gas? What is happening to the atoms of

the gas?
Biology
1 answer:
Lina20 [59]3 years ago
3 0

During the process of a gas condensing into a liquid and then back into a gas, all the molecules go flying apart and become a gas. This process is called Evaporation.

Explanation:

The process in which the molecules of a gas slow down, come together and form a liquid. This process is known as condensation.

When the gas condenses to liquid, it releases thermal energy absorbed to become a gas. The temperature does not change during this process.

The water vapor condenses to form the liquid water droplets in clouds. When heating the solid it produces energy and this causes the particle to move and the solid to change into liquid ,or melt.

If the liquid is heated they evaporate and turn into a gas. The energy is taken away from the gas may leads to turn into a liquid.

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Which type of gas or gasses are ressponsible for the start of life on earth
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assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
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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 )

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}}

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)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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