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3 years ago

What occurs to the electromagnetic radiation from the sun enters earths atmosphere

Chemistry

2 answers:

Ilya [14]3 years ago

7 0

Answer:

When electromagnetic radiations enter earth surface , it passes through several layers of our atmosphere in meantime some raditions get reflected back and some reaches to us .

Darina [25.2K]3 years ago

7 0

Answer:

Hey!

The electromagnetic radiation the Earth recieves form the sun is a mixture of Infrared, Visible and Ultraviolet (UV)...

As these different wavelengths of Electromagnetic radiation approaches the Earth, these are absorbed by different parts of the atmosphere or reach the surface (where Humans are present)...

A layer of OZONE absorbs UV rays...

Visible light is of course not absorbed by the Atmosphere and reaches the surface of the Earth (This is what we see as daylight)

The layer of GREENHOUSE GASES (Carbon-Dioxide, Methane) absorbs and reflects the Infrared rays (aka HEAT) and sometime re-emitted back into space...

Explanation:

HOPE THIS HELPS!!

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Consider the decomposition of the compound C5H6O3 as follows: C5H6O3(g)  C2H6(g) + 3CO(g) A 5.63 g sample of pure C5H6O3(g) was

deff fn [24]

Answer:

$K_{eq}=1.02x10^{-4}$

Explanation:

Hello,

In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:

$5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3$

After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:

$n_{total}^{eq}=\frac{P_{total}^{eq}V}{RT}=\frac{1.63atm*2.50L}{0.082\frac{atm*L}{mol*K}*473.15K} =0.105mol$

Then, by knowing the moles at the equilibrium considering the change " $x$ ", which yields to:

$\ \ \ \ \ C_5H_6O_3(g) \leftrightarrow C_2H_6(g) + 3CO(g)\\I\ \ \ \ \ 0.0494mol\ \ \ \ \ \ 0mol \ \ \ \ \ \ \ \ 0mol\\C\ \ \ \ \ \ -x\ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x\\E\ \ 0.0494mol-x\ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ 3x$

The total moles at the equilibrium turn out:

$n_{total}^{eq}=0.0494mol-x+x+3x$

By solving for " $x$ ", we've got:

$3x=0.105mol-0.0494mol\\x=\frac{0.0556mol}{3}\\x=0.0185mol$

Finally, the equilibrium constant is:

$K_{eq}=\frac{(x)(3x)^3}{0.0494-x}=\frac{(0.0185mol)(3*0.0185mol)^3}{0.0494mol-0.0185mol}=1.02x10^{-4}$

Best regards.

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