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Serggg [28]
3 years ago
8

A fog over a lake. what two changes of state must occur to produce the fog? do the water molecules absorb or release energy duri

ng these changes of state? what happens to the motion of the water molecules as a result?
Chemistry
2 answers:
LenKa [72]3 years ago
5 0
<span>Evaporation and condensation occur due to daytime heating of the water molecules in the lake. They absorb heat and transform, water molecules rise into the air forming the fog cloud. </span>
Zolol [24]3 years ago
3 0
The water from the lake must first evaporate from the liquid state to the gaseous state and then condense in the air to form vapors.
The water molecules absorb energy while evaporating and release it when condensing. Their motion gets faster when they are evaporating and slows back down upon condensation.
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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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