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2.0 L of oxygen gas and 8.0 L of nitrogen gas at STP are mixed together. The gaseous mixture is compressed to occupy 2.0 L at 298 K. What is the pressure exerted by this mixture?

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Dsdrajlinidn

Answer:

5.5 atm

Explanation:

Step 1: Calculate the moles in 2.0 L of oxygen at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

2.0 L × 1 mol/22.4 L = 0.089 mol

Step 2: Calculate the moles in 8.0 L of nitrogen at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

8.0 L × 1 mol/22.4 L = 0.36 mol

Step 3: Calculate the total number of moles of the mixture

n = 0.089 mol + 0.36 mol = 0.45 mol

Step 4: Calculate the pressure exerted by the mixture

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm

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