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larisa [96]
3 years ago
8

Injecting salt crystals over the ocean to grow cloud droplets has been proposed in efforts to make the clouds __________ thereby

affecting the radiation budget.
Chemistry
1 answer:
anastassius [24]3 years ago
7 0
Injecting salt crystals over the ocean to grow cloud droplets has been proposed in efforts to make the clouds brighter thereby affecting the radiation budget. The light of the sun shines on Earth, some of that light is reflected by the clouds back to space and some of the light reaches the earth and warms our planet. The earth and the hot oceans emit infrared radiation (IR), which we feel as heat. That IR "light"; returns to space through the atmosphere. Most are trapped by greenhouse gases, which keep the earth warm. Soon after, the IR radiation returns to space. Scientists call this "energy budget of the Earth" this cycle of incoming and outgoing energy.
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The functions of the ears are:

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<h3>What is the function of the ear?</h3>

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Read 2 more answers
Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).
Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce  C2H5NH3^+ + Cl^-

The base hydrolysis reaction:  C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11  

Let's solve for x. In this equation,  It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x  is approximately equal to 0.26.  We won't know until we do the calculation.

We get:  x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x  is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0
3 years ago
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