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Galina-37 [17]
3 years ago
10

using the water cycle diagram, match the number with the following; condensation runoff [precipitation and evaporation. once you

've identified the process for each number, explain how water moves and changes form in each step (you only need to do this for the 4 basic steps)

Chemistry
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

1. Evaporation

2. Condensation

3. Precipitation

4. Runoff

Explanation:

The water cycle shows how water moves on the earth as it passes through different phases.

Water on land is stored in oceans, rivers and other water bodies. This water coupled with those in plants and animals are turned into vapor through EVAPORATION. Evaporation facilitates the movement of water on the surface into the atmosphere.

In the atmosphere, the vapor condenses on dust particles found up there. The vapors forms a nuclei around the dust particles and water condenses at the saturated vapor pressure. This forms cloud.

As the water collects more and more, gravity forces the water to fall in form of PRECIPITATION. The precipitation can be in form of snow or rainfall.

When precipitation occurs, they move on the surface as SURFACE RUNOFFS. Some of the runoff goes back into oceans and rivers. Others infiltrates into the ground and collects in the ground water pool under the subsurface. Subsurface water can also get into into other water bodies when the water table coincides with the steam level.

The water in these bodies can then go into the cycle again. The sun is the source of energy for this process.

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The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

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