The solubility of a gas at 7.0 atm of pressure is 0.52 g/L . How many grams of the gas would be disolved per 1.0 L if the pressure increased 40 percent?
Didn't Devron work this problem for you earlier? I think so.
According to the following reaction and molar enthalpies of formation
SiO2(s)
+ 4HF(aq)
®
SiF4(g)
+ 2H2O
(kJ/mol)
-910.9
-320.8
-1615
-285.8
The heat released (-) or absorbed (+) in the reaction of 25.3 grams of SiO2 (quartz) with excess hydrofluoric acid will be kJ.
The heat released (-) or absorbed (+) in the reaction of 25.3 grams of SiO2 (quartz) with excess hydrochloric acid will be what in KJ?
To calculate the new solubility of the gas when the pressure increases by 40 percent, we first need to determine the new pressure and then use the solubility constant to find the new solubility.
Step 1: Calculate the new pressure
To obtain the new pressure, we need to increase the initial pressure by 40 percent.
Increase = 40% of 7.0 atm
= (40/100) * 7.0 atm
= 2.8 atm
New pressure = Initial pressure + Increase
= 7.0 atm + 2.8 atm
= 9.8 atm
Step 2: Calculate the new solubility
We know that solubility is directly proportional to pressure, meaning that as pressure increases, solubility also increases.
Given: Initial solubility = 0.52 g/L
Initial pressure = 7.0 atm
New pressure = 9.8 atm
Using the concept of proportionality, we can set up a ratio between the initial and new solubilities:
Initial solubility / Initial pressure = New solubility / New pressure
Plugging in the values:
0.52 g/L / 7.0 atm = New solubility / 9.8 atm
Solving for New solubility:
New solubility = (0.52 g/L / 7.0 atm) * 9.8 atm
New solubility = 0.0743 g/L
Therefore, when the pressure is increased by 40 percent, the new solubility of the gas is 0.0743 g/L.