What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 19,600 atm(d) 84(e) 8.2 atm?

Let's assume that the F₂ gas has ideal gas behavior. 

Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

Moles = mass / molar mass

Molar mass of F₂ = 38 g/mol

Mass of F₂  = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

P = ?
V = 1.5 L = 1.5 x 10⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻¹
T = -37 °C = 236 K

By substitution,

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

                         p = 2616138.67 Pa

                         p = 25.8 atm = 26 atm

Hence, the pressure of the gas is 26 atm.

Answer is "a".

76 g of fluorine gas in a 1.50 L vessel at -37°C exert a pressure of (a) 26 atm.

We have 76 g of fluorine gas in a closed vessel. First, we will convert 76 g to moles using its molar mass (38.00 g/mol).

Then, we will convert -37 °C to K using the following expression.

Finally, we can calculate the pressure exerted by the gas using the ideal gas equation.

76 g of fluorine gas in a 1.50 L vessel at -37°C exert a pressure of (a) 26 atm.

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the answer is true.