Answer:
The answer to your question is V2 = 29.6 l
Explanation:
Data
Pressure 1 = P1 = 12 atm
Volume 1 = V1 = 23 l
Temperature 1 = T1 = 200 °K
Pressure 2 = 14 atm
Volume 2 = V2 = =
Temperature 2 = T2 = 300°K
Process
1.- To solve this problem use the Combine gas law.
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (12)(23)(300) / (200)(14)
3.- Simplification
V2 = 82800 / 2800
4.- Result
V2 = 29.6 l
Answer:
44
Explanation:
Given that :
Mass of solute = Mass of urea = 16g
Mass of water = 20g
Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g
Percentage Mass = (mass of solute / mass of solution) * 100%
Percentage Mass = (16 / 36) * 100%
Percentage Mass = 0.4444444 x 100%
Percentage Mass = 44.44%
Percentage Mass = 44%
Tearing paper, Physical Change
Answer: Since this is merely a Physical Change, it is improper to write it as a chemical equation.
The water molecules received enough energy through the heat for them to change to a liquid state by unfixing themselves from their fixed positions.