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

the solar system used to have much dust particles option no d

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :