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weqwewe [10]
3 years ago
9

Which process in the atmosphere is responsible for moving heat from warm equatorial regions to higher latitudes

Chemistry
1 answer:
Anni [7]3 years ago
8 0

Answer:

Heat transfer in the atmosphere from the equatorial regions to higher latitudes occurs through the process of Convection

Explanation:

The equatorial regions of the earth's surface receive the most heat from the sun than any other region of the earth. This is because the are most directly in line with the direct heat from the sun.

Due to this heat from the sun, atmospheric air around the equatorial regions are hot and less dense than air in regions of higher latitudes, and thus, rises above the equator. The rising air at the equator is replaced by colder and denser air from higher latitudes north and south of the equator. As the rising air of the equatorial regions are being replaced by colder and denser air from higher latitudes, the hot and humid air moves away from the equator, toward regions of higher latitude, north and south thereby setting up a convection current of heat flow.

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Answer:

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In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign
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Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

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