Answer :
Final answer:
The density of Ar (g) at -11°C and 675 mm Hg is approximately 1.52 g/L.
Explanation:
The density of a gas can be calculated using the ideal gas law equation:
Density (g/L) = (molar mass of the gas * pressure) / (R * temperature)
Given that the temperature is -11°C and the pressure is 675 mm Hg, we need to convert the temperature to Kelvin and the pressure to atm. The molar mass of Ar (Argon) is 39.948 g/mol.
Converting the temperature to Kelvin: -11°C + 273.15 = 262.15 K
Converting the pressure to atm: 675 mm Hg * (1 atm / 760 mm Hg) = 0.8882 atm
Now, we can plug in the values into the equation:
Density (g/L) = (39.948 g/mol * 0.8882 atm) / (0.0821 L atm/mol K * 262.15 K)
Calculating the density gives: Density (g/L) ≈ 1.52 g/L
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