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1. Calculate the pressure exerted by 0.28 g of sulfur hexafluoride (SF6), a colorless, odorless, non-flammable, extremely potent greenhouse gas, in a 5.43 L steel vessel at 69.5°C.

a. 0.0098 atm

b. 0.0198 atm

c. 0.0398 atm

d. 0.0298 atm

Sagot :

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} & V = \text{5.43 L} \\ & \text{mass} = \text{0.28 g} \\ & T = 69.5^{\circ}\text{C} = \text{342.65 K} \end{aligned}[/tex]

Step 2: Calculate the number of moles of the gas.

The molar mass of SF₆ is 146.06 g/mol.

[tex]\begin{aligned} n & = \frac{\text{mass}}{\text{molar mass}} \\ & = \frac{\text{0.28 g}}{\text{146.06 g/mol}} \\ & = \text{0.0019 mol} \end{aligned}[/tex]

Step 3: Solve for the pressure of the gas by using ideal gas equation.

[tex]\begin{aligned} P & = \frac{nRT}{V} \\ & = \frac{(\text{0.0019 mol})\left(0.082057 \: \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(\text{342.65 K})}{\text{5.43 L}} \\ & = \boxed{\text{0.0098 atm}} \end{aligned}[/tex]

Hence, the answer is a. 0.0098 atm.

[tex]\\[/tex]

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