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Tresset [83]
3 years ago
12

How does global evaporation compare to global precipitation​

Chemistry
1 answer:
GenaCL600 [577]3 years ago
3 0

Answer:

On a global scale, the amount of water evaporating is about the same as the amount of water delivered to the Earth as precipitation. There is no net difference.

Explanation:

Evaporation is the process by which water changes from a liquid to a gas or vapor. Evaporation is the primary pathway that water moves from the liquid state back into the water cycle as atmospheric water vapor. Studies have shown that the oceans, seas, lakes, and rivers provide nearly 90 percent of the moisture in the atmosphere via evaporation, with the remaining 10 percent being contributed by plant transpiration.

Evaporation from the oceans is the primary mechanism supporting the surface-to-atmosphere portion of the water cycle. After all, the large surface area of the oceans (over 70 percent of the Earth's surface is covered by the oceans) provides the opportunity for large-scale evaporation to occur. On a global scale, the amount of water evaporating is about the same as the amount of water delivered to the Earth as precipitation. This does vary geographically, though. Evaporation is more prevalent over the oceans than precipitation, while over the land, precipitation routinely exceeds evaporation. Most of the water that evaporates from the oceans falls back into the oceans as precipitation. Only about 10 percent of the water evaporated from the oceans is transported over land and falls as precipitation. Once evaporated, a water molecule spends about 10 days in the air. The process of evaporation is so great that without precipitation runoff, and groundwater discharge from aquifers, oceans would become nearly empty.

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You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when ex
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a. 1.78x10⁻³ = Ka

2.75 = pKa

b. It is irrelevant.

Explanation:

a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.

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HA ⇄ H⁺ + A⁻

And Ka is defined as:

Ka = [H⁺] [A⁻] / [HA]

The HA reacts with the base, XOH, thus:

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It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:

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Replacing:

2.75 = pKa + log₁₀ [A⁻] / [HA]

As [HA] = [A⁻]

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<h3>2.75 = pKa</h3>

Knowing pKa = -log Ka

2.75 = -log Ka

10^-2.75 = Ka

<h3>1.78x10⁻³ = Ka</h3>

b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.

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