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calculate the density of gaseous sf6 at a temperature of 15C and a pressure of 0.942 atm

User ProDec
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To calculate the density of gaseous sulfur hexafluoride (SF6) at a temperature of 15°C and a pressure of 0.942 atm, we can use the ideal gas law:

PV = nRT

where:

P is the pressure (in atm)

V is the volume (in liters)

n is the number of moles of gas

R is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹)

T is the temperature (in K)

We can rearrange this equation to solve for the density (ρ):

ρ = n/V = P/(RT) * M

where M is the molar mass of SF6 (146.06 g mol⁻¹).

First, we need to convert the temperature to Kelvin:

T = 15°C + 273.15 = 288.15 K

Now, we can plug in all of the known values to calculate the density of SF6:

ρ = (0.942 atm) / (0.08206 L atm mol⁻¹ K⁻¹ * 288.15 K) * 146.06 g mol⁻¹

ρ = 589.588 g/L

Therefore, the density of gaseous SF6 at a temperature of 15°C and a pressure of 0.942 atm is 589.588 grams per liter.

User Sfussenegger
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