Answer:
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Explanation:
<u>Step 1:</u> Data given
A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K
Moles of CO2 = 1.27 moles
Moles of CO = 3.04 moles
Moles of Ar = 1.50 moles
<u>Step 2:</u> Calculate total number of moles
Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles
<u>Step 3:</u> Calculate mol fraction Ar
Mol fraction Ar = 1.50 mol/5.81 mol = 0.258
<u>Step 4</u>: Calculate partial pressure
1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Answer:
Gas at room temperature
Strong, unpleasant odor
Dissolves easily in water
Explanation:
Just did the assignment
Answer:
The true statement is option A.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 1 atm
V = Volume of gas = ?
n = number of moles of gas = 1 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273.15 K

V = 22.42 L
This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.
So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.