Answer:
[N₂] = 0.032 M
[O₂] = 0.0086 M
Explanation:
Ideal Gas Law → P . V = n . R . T
We assume that the mixture of air occupies a volume of 1 L
78% N₂ → Mole fraction of N₂ = 0.78
21% O₂ → Mole fraction of O₂ = 0.21
1% another gases → Mole fraction of another gases = 0.01
In a mixture, the total pressure of the system refers to total moles of the mixture
1 atm . 1L = n . 0.082L.atm/mol.K . 298K
n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles
We apply the mole fraction to determine the moles
N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂
O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂
T₁ = 40°C + 273.15 = 313.15 Kelvin T₂ = 30°C + 273.15 = 303.15 Kelvin
Solving Gay-Lussac's Law for P₁ we get:
P₁ = P₂ • T₁ ÷ T₂ P₁ = 760 torr • 313.15 K ÷ 303.15 K P₁ = 785.07 torr
Using the calculator, we click on the P1 button.
We then enter the 3 numbers 760 313.15 and 303.15 into the correct boxes then click "CALCULATE" and get our answer of 785.07 torr.
Answer:
a) Volume of vial= 9.626cm3
b) Mass of vial with water = 62.92 g
Explanation:
a) Mass of empty vial = 55.32 g
Mass of Vial + Hg = 185.56 g
Therefore,
Density of Hg = 13.53 g/cm3
b) Volume of water = volume of vial = 9.626 cm3
Density of water = 0.997 g/cm3
Answer:
141g of CCl₄
Explanation:
First, we have to write the balanced equation.
CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)
We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.
V = 14.9 dm³ = 14.9 L
T = 21°C + 273.15 = 294.15 K
P = 1.48 atm
R = 0.08206 atm.L/mol.K
We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.
Answer:
Explanation:
a strongly acidic solution of the gas hydrogen chloride in water
