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Scilla [17]
3 years ago
9

Which statement about atmospheric pressure is false?

Chemistry
2 answers:
Tcecarenko [31]3 years ago
7 0

Answer:

The false statement is "With an increase in altitude, atmospheric pressure increases as well" because "With an increase in altitude, atmospheric pressure decreases."

Explanation:

Altitude is the vertical distance that exists between any given point on Earth and sea level.

The weight of the air that makes up our atmosphere puts pressure on the earth's surface. This pressure is known as atmospheric pressure. Generally, the more air there is over an area, the higher the atmospheric pressure. This means that atmospheric pressure changes with altitude: the higher the height of the Earth's surface with respect to sea level (altitude), the lower the air pressure.

So, <u><em>the false statement is "With an increase in altitude, atmospheric pressure increases as well" because "With an increase in altitude, atmospheric pressure decreases."</em></u>

Vinil7 [7]3 years ago
6 0

Answer:

D) With an increase in altitude, atmospheric pressure increases as well.

Explanation:

Generally when altitude increases, the value of pressure decreases. This shows that pressure is inversely proportional to altitude. For example, the higher the altitude, the lower the pressure and vice versa. At very high altitude, the number of molecules of air are smaller than the number of moles of air at very low altitude. Thus, the higher the altitude, the lower the atmospheric pressure and the lower the altitude, the higher the atmospheric pressure. Therefore, option (D) is false.

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Brrunno [24]
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3 years ago
The half-life of radon-222 is 3.82 days. If a sample of gas contains 4.38 μg of radon-222, how much will remain in the ample aft
LekaFEV [45]

Answer:

0.28 \mu g

Explanation:

We can write the amount of mass of radon-222 left after a time t by using the equation:

m(t) = m_0 (\frac{1}{2})^{t/t_{1/2}}

where

m_0 = 4.38 \mu g is the initial mass

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Substituting t = 15.2 d in the formula, we find

m(15.2 d) = (4.38 \mu g) (\frac{1}{2})^{15.2d/3.82 d}=0.28 \mu g

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Answer:

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