2)
if 5.0 moles of o2 and 3.0 moles of n2 are placed in a 30.0 l tank at a temperature of 25°
c, what will the pressure of the resulting mixture of gases be?

Total pressure = 6.55 atm

Explanation:

Given data:

Number of moles of oxygen = 5.0 mol

Number of moles of nitrogen = 3.0 mol

Volume of tank = 30.0 L

Temperature = 25°C (25+273 = 298 K)

Total pressure of mixture = ?

Solution:

Pressure of oxygen:

PV = nRT

P = nRT/V

P = 5.0 mol ×0.0821 atm.L/mol.K × 298 k /30.0 L

P = 122.33 atm /30.0

P = 4.1 atm

Pressure of nitrogen:

PV = nRT

P = nRT/V

P = 3.0 mol ×0.0821 atm.L/mol.K × 298 k /30.0 L

P = 73.4 atm /30.0

P = 2.45 atm

Total pressure = Partial pressure of oxygen + partial pressure of nitrogen

Total pressure =  4.1 atm +  2.45 atm

Total pressure = 6.55 atm

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the answer to your question is c