Answer:
Oxygen will dissolve more in H2O at 5 atm and 20 °C than at 5 atm 80 °C
Option B is correct.
Explanation:
Step 1: Data given
Pressure = 5 atm
Temperature = 20 °C or 80 °C
Step 2:
At low pressure, a gas has a low solubility. Decreased pressure allows more gas molecules to be present in the air, with very little being dissolved in solution. At high(er) pressure, a gas has a high solubility.
This means the higher the pressure the more the gas will dissolve. Since The pressure stays constant, it depends on the temperature.
The solubility of gases in liquids decreases with increasing temperature.
This means the gas will dissolve more with a lower temperature.
Oxygen will dissolve more in H2O at 5 atm and 20 °C than at 5 atm 80 °C
It means it has a constant deceleration
Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M
Answer:
The plant would be insect-resistant and improve its chances of success.
Explanation:
Answer:
The answer to your question is P2 = 9075000 atm
Explanation:
Data
Pressure 1 = P1 = 5 atm
Volume 1 = V1 = 363 ml
Pressure 2 = P2 = ?
Volume 2 = 0.0002 ml
Process
To solve this problem use Boyle's law
P1V1 = P2V2
-Solve for P2
P2 = P1V1/V2
-Substitution
P2 = (5 x 363) / 0.0002
-Simplification
P2 = 1815 / 0.0002
-Result
P2 = 9075000 atm