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bija089 [108]
3 years ago
15

What are the conditions for evaporation?

Chemistry
2 answers:
bixtya [17]3 years ago
5 0

Answer:

Explanation:

1 Evaporation is the phase change of a liquid to a gas. There are three very important requirements for evaporation to take place, 1) available energy, 2) available water, and 3) a vertical moisture gradient. Approximately 600 calories of heat must be added to a gram of water for it to evaporate into the air.

2. Evaporation is the process of a substance in a liquid state changing to a gaseous state due to an increase in temperature and/or pressure. It is a fundamental part of the water cycle and is constantly occurring throughout nature.

3.Water evaporates faster if the temperature is higher, the air is dry, and if there's wind. The same is true outside in the natural environment. Evaporation rates are generally higher in hot, dry and windy climates.

                                thank u

SashulF [63]3 years ago
4 0

Answer:

For evaporation you need a warm, preferably humid temperature. The rate of evaporation increases with an increase in temperature. A windy climate is best, as wind helps to remove the evaporated water vapour, and therefore creating a better scope for evaporation to continue. The speed of wind is important for evaporation because the wind pulls in dry air, increasing the rate of evaporation.

Short Answer- Hot and humid temperature, lots of wind speed.

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nexus9112 [7]

Answer: V = 33.9 L

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V2 = V1 T2 / T1

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What is the abbreviation for the element with atomic number 11?
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Answer:

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The Galapagos Islands formed over a hot spot. The majority of the volcanoes here are ________________ volcanoes. A) shield B) ci
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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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