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Zanzabum
3 years ago
8

Organic molecules are sometimes found within meteorites that crash into Earth. Which meteorite condition in space is similar to

what most likely was present in Earth's early atmosphere?
Abundance of oxygen
Very cold temperatures
Presence of inorganic molecules
Low levels of ultraviolet radiation
Biology
1 answer:
nevsk [136]3 years ago
6 0

The presence of inorganic molecules is the meteorite condition in space similar to what was likely present in Earth's early atmosphere.  

The earth initial atmosphere in its early form was too reducing and it was made up entirely of gases such as carbon monoxide, ammonia, methane, nitrogen and other gases.

This atmosphere did not have free oxygen, since it was the result of the mixture of metals, hydrogen and other chemical elements.

The high temperatures, ultraviolet rays and electrical discharges in the early atmosphere would have caused chemical reactions of the elements to form primitive organic molecules.

Meteorites are fragments of solid matter that contain a record of the chemical substances that existed in the early Solar System.

The most predominant gases still found in the space are hydrogen and helium.

Therefore, we can conclude that hydrogen and the presence of inorganic molecules are the initial elements that make up the Earth's early atmosphere and that they are conditions that are present in the space and so in the formation of meteorites.

Learn more about Earth's early atmosphere here: brainly.com/question/18289168

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Answer:

\boxed {\boxed {\sf C. \ 2.5 \ grams }}

Explanation:

We are asked to find how much of a 40 gram sample remains after 12 years.

Iron-55 has a half-life of 3 years. Therefore, after 12 years, 4 half-lives have been completed.

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Every time a half-life is completed, half of the sample's mass decays. Remember we start with a 40 gram sample.

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There is also a formula that can be used to solve this problem.

A= A_o(\frac {1}{2})^{\frac{t}{hl}

Where A₀ is the initial amount, t is the time, and hl is the half-life.

We know 40 grams is the inital amount, 12 years is the time, and 3 years is the halflife.

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A= 40 \ g (\frac{1}{2})^\frac{12}{3}

A= 40 \ g (\frac{1}{2})^4

A= 40 \ g * 0.0625

\bold {A= 2.5 \ g}

After 12 years, <u>2.5 grams </u> of Iron-55 will remain.

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