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galben [10]
3 years ago
6

All factors that cause an increase in the rate of dissolving have what in common?

Chemistry
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

They increase the attractive forces between the solute and solvent particles.

Explanation:

The dissolution of a solute in a solvent depends on interaction between the solute and the solvent. The more the attractive force and interaction between solute and solvent, the greater the greater the rate of dissolution of the solute in the solvent.

The absence of interaction between solute and solvent molecules means that the substance can not dissolve in that particular solvent. Hence, any factor that enhances solute-solvent interaction will enhance dissolution of a solute in a particular solvent.

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The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air
IceJOKER [234]

Answer:

1.75\cdot 10^{-4} M

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

S = k_H p^o

Where k_H is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

S_1 = 2.67\cdot 10^{-4} M

Also, at sea level, we have an atmospheric pressure of:

p = 1.00 atm

Given mole fraction:

\chi_{O_2} = 0.209

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

p^o = \chi_{O_2} p

Then the equation becomes:

S_1 = k_H \chi_{O_2} p

Solve for k_H:

k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

p = 0.657 atm

Apply Henry's law using the constant we found:

S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M

8 0
3 years ago
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