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:
Empirical formula of ionic compound formed by two ions 
and 
is 
(for 
) of AB (for x = y)
The above empirical formula is in accordance with charge neutrality principle
Here each cation (
and 
) can form two ionic compounds by combining with two given anions (
and 
).
So the four ionic compounds are:
Answer:
Salt water
Explanation:
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Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
