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2 years ago

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

Chemistry

2 answers:

Ilya [14]2 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]2 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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On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

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Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

Calculating the % ionization:

$= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\$

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