Answer:
166 torr
Explanation:
Let’s call ethane Component 1 and propane Component 2.
According to Raoult’s Law,
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Data:
p₁° = 304 torr
p₂° = 27 torr
n₁ = n₂
1. Calculate the mole fraction of each component
χ₁ = n₁/(n₁ + n₂)
χ₁ = n₁/n₁ + n₁)
χ₁ = n₁/(2n₁)
χ₁ = ½
χ₁ = 0.0.5
χ₂ = 1- χ₁ = 1- 0.5 = 0.5
2. Calculate the vapour pressure of the mixture
Answer:
Explanation:
On the reactant side, XeO3 is a strong oxidizing agent. It can be a solid or gas. Most likely it is in a solid state.
H+ and Br- combine to form HBr which is a weak acid. So H+ and Br- are in an aqueous state.
On the product side, both Br2 and Xe are not very soluble in water. They will be in gaseous state and H2O is water i.e. liquid state.
The formula for density is density =
. Substituting the given, the density is 162.69 g/cm3.
Answer:
The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up.
Explanation:
D) reacts with oxygen
All other are physical properties.