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Roman55 [17]
3 years ago
6

The most abundant gases in the atmosphere are _____. trace gases variable gases greenhouse gases primary gases

Chemistry
2 answers:
elena-s [515]3 years ago
4 0

Answer:

greenhouse gases

Explanation:

White raven [17]3 years ago
3 0
The correct answer is greenhouse gases. It is the most abundant gases among the choices in the atmosphere. These gases are water vapor, methane, nitrous oxide, ozone and carbon dioxide. Without these gases, the temperature of Earth will be about -18 degrees Celsius.
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Which of the following situations best shows deposition?
KATRIN_1 [288]

Answer:

chemistry, deposition occurs when molecules settle out of a solution. Deposition can be viewed as a reverse process to dissolution or particle re-entrainment. It is a phase change from the gaseous state to a solid, without passing through the liquid state, also called re-sublimation.

Explanation:

Examples include beaches, deltas, glacial moraines, sand dunes and salt domes. In severely cold temperatures frost will form on windows because the water vapor in the air comes into contact with a window and immediately forms ice without ever forming liquid water.

8 0
3 years ago
What's the answer to this question?
Lena [83]
The answer is B- 124 degrees
4 0
3 years ago
Read 2 more answers
What process occurs during the corrosion of iron?
Butoxors [25]

Answer:

A

Explanation:

The iron corrodes so it oxidized

8 0
3 years ago
I need help fast pls someone
Neko [114]

Answer:

I would say A. I'm no expert, but it can't be C obviously, and I think wind would hit all of it, wearing off the top as well like the great pyramids. B would be my next choice, but A i think would be best.

3 0
2 years ago
600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c
Ludmilka [50]

Answer:

k=1.20x10^{-3} s^{-1}

Explanation:

For a first order reaction the rate law is:

v=\frac{-d[A]}{[A]}=k[A]

Integranting both sides of the equation we get:

\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

ln\frac{[A]}{[A]_{0} } =-kt

[A]=[A]_{0}e^{-kt}

ln[A]=ln[A]_{0} -kt

k=\frac{ln[A]_{0}-ln[A]}{t}

therefore k is

k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}

8 0
3 years ago
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