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Vilka [71]
3 years ago
6

Scientists are studying cold and dry environments on earth that are like the environment on Mars. What kind of prokaryotes do yo

u think they might find in these environments on earth? Explain.
Chemistry
2 answers:
mote1985 [20]3 years ago
8 0

Answer:

The correct answer will be- Archean bacteria (extremophiles)

Explanation:

The life on Earth began in the conditions of the deep oceans which moved to the extreme areas and evolved into the prokaryotic group of bacteria called Archaebacteria.

The Archaebacteria are modified themselves to live in the very harsh conditions form humans like the very high temperature in which the Archae called the thermophiles could be found and the very cold conditions in which the cryophiles group of bacteria could be found. These Archaebacteria are also known as the extremophiles.

Thus, Archean bacteria (extremophiles) is the correct answer.

FinnZ [79.3K]3 years ago
7 0

The answer is; extremophiles

These organisms are found in places on earth that would be damaging to most life on earth. There are different categories of extremophiles depending on their environment. For example, those that live in the extreme cold (even below  150C ) are called Psychrophiles while Xerophiles live extreme desiccation.


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percent ionization = 50.01%; pH = 4.75

Explanation:

To solve this question we must write the acetic acid equilibrium (Where HX will be acetic acid and X⁻ the sodium acetate):

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Where equilibrium constant, Ka, is defined as:

Ka = 1.76x10⁻⁵ = [H⁺] [X⁻] / [HX]

<em>Where the concentration of each ion is:</em>

[H⁺] = X

[X⁻] = 0.10M + X

[HX] = 0.10M - X

Replacing in Ka expression:

1.76x10⁻⁵ = [X] [0.10-X] / [0.10+X]

1.76x10⁻⁶ + 1.76x10⁻⁵X = 0.10X - X²

X² - 0.0999824 X + 1.76×10⁻⁶ = 0

X ≈ 0.1M → False solution. Decreases a lot the concentration of HX

X = 0.0000176M → Right solution.

The concentration of each ion is:

[H⁺] = 0.0000176062M

[X⁻] = 0.10M + 0.0000176M = 0.1000176M

[HX] = 0.10M - 0.0000176M = 0.0999824M

Percent ionization:

[X-] / [X-] + [HX] * 100 =

0.1000176M / 0.2M =

<h3>50.01%</h3><h3 />

And pH = -log [H+]

<h3>pH = 4.75</h3><h3 />

As you can see, [H+]≈ Ka

3 0
3 years ago
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