PH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg[OH⁻]
pH=14+lg(8.7*10⁻¹²)=14-11.06=2,94
Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
Answer:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. The distinction between physical and chemical change is not clear cut.
