Answer:

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Explanation:

1) To determine the volume of gas occupied by 2.34 L of CO2 gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law and the molar volume at STP.

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At STP, the pressure is 1 atmosphere (atm) and the temperature is 273.15 K. The ideal gas constant (R) is 0.0821 L·atm/(mol·K).

Rearranging the ideal gas law equation to solve for volume (V), we have:

V = nRT / P

Substituting the given values:

V = (n * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

Now, we can solve for the number of moles (n) using the volume given:

2.34 L = (n * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

Simplifying the equation:

2.34 L = 22.414 L·mol^(-1) * n

Dividing both sides by 22.414 L·mol^(-1):

n = 2.34 L / 22.414 L·mol^(-1)

n ≈ 0.1045 moles

Therefore, 2.34 L of CO2 gas at STP contains approximately 0.1045 moles of CO2.

2) To determine the number of moles of argon in a sample occupying 56.2 liters at STP, we can use the same approach as in the previous question.

Using the ideal gas law equation, we have:

V = nRT / P

Substituting the given values:

56.2 L = (n * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

Simplifying the equation:

56.2 L = 22.414 L·mol^(-1) * n

Dividing both sides by 22.414 L·mol^(-1):

n = 56.2 L / 22.414 L·mol^(-1)

n ≈ 2.51 moles

Therefore, the sample of argon at STP, occupying 56.2 liters, contains approximately 2.51 moles of argon.