Which method would increase the solubility of a gas? increasing the pressure stirring the solvent quickly increasing the temperature decreasing the amount of solvent


Answer 1

Answer: Option (a) is the correct answer.


Solubility of a gas can be increased by increasing the pressure on a gas. As on increasing the pressure, the molecules come closer to each other and the number of gas molecules will decrease as the molecules are dissolving due to increase in pressure.

On increasing the temperature, the kinetic energy of molecules will increase. Therefore, the gas molecules will move randomly rather and will not be soluble.

Thus, we can conclude that on increasing the pressure we can increase the solubility of a gas.

Answer 2
Answer: Increasing pressure would be the answer to this question

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Describe a procedure to separate a mixture of salt sand and water


Filtration is used to separate salt and water. 

17. Compare the physical and chemical prop- erties of salt and sugar. What properties do
they share? Which properties could you use
to distinguish between salt and sugar?



my answer is freezing it's is a true


If 425.0 grams of butene react completely in excess oxygen, how many grams of each product are produced? How many grams would be produced if the percent yield is 67? At STP what would be the volume of the gas produced for this yield?



A. 1333  g CO₂; 546.0 g H₂O

B. 2200 g CO₂; 370    g H₂O

C.   460 L

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:  56.11                  44.01    18.02

         C₄H₈ + 6O₂ ⟶ 4CO₂ + 4H₂O

m/g:  425.0

A. Theoretical yield of each product

(1) Calculate the moles of C₄H₈

n = 425.0 ×1/56.11

n = 7.574 mol C₄H₈

(2) Calculate the moles of CO₂ and H₂O

In each case, the molar ratio is 4 mol product/1 mol C₄H₈.

CO₂ and H₂O:

n = 7.574× 4/1

n = 30.30 mol

(3) Calculate the masses of CO₂ and H₂O

Mass of CO₂ = 30.30 mol CO₂ × (44.01 g CO₂/1 mol CO₂)

Mass of CO₂ = 1333 g CO₂

Mass of H₂O = 30.30 mol H₂O × (44.01 g H₂O/1 mol H₂O)

Mass of H₂O = 546.0 g H₂O

B. Actual yield of each product

Mass of CO₂ = 3333 g × 67/100

Mass of CO₂ = 2200 g CO₂

Mass of H₂O = 546.0 g × 67/100

Mass of H₂O = 370 g H₂O

C. Volume of gas

At STP, CO₂ is a gas, but H₂O is a liquid.

Moles of gas = moles of CO₂

If the actual yield is 67 %

Moles of CO₂ = 30.30 mol × 67/100

Moles of CO₂ = 20 mol

STP is 1 bar and 0 °C.

The molar volume at STP is 22.71 L.

∴ V = 201 mol × 22.71 L/1 mol

   V = 460 L


What is the oxidation number of Cl in chlorate ion ClO3?


The correct answer to this question is "5." the oxidation number of cl in ClO3 will be a positive 5 because oxygen is naturally a -2 charge. times that by three and then account for the negative charge of the CLO3- ion.

Answer is: the oxidation number of Cl (chlorine) in chlorate ion is +5.

Chlorate ion (ClO₃⁻) has negative charge.

One oxygen atom in this anion (negatively charged ion) has the oxidation number -2. There are three oxygen atoms in chlorate anion.

x + 3 · (-2) = -1.

x = -1 + 6.

x = +5; the oxidation number of chlorine.

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