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What is the volume of \( 3.0 \) moles of hydrogen at \( 800.00 \) mmHg and \( 25^\circ \text{C} \)?

A. \( 25.1 \, \text{L} \)
B. \( 37.6 \, \text{L} \)
C. \( 50.1 \, \text{L} \)
D. \( 62.7 \, \text{L} \)

Answer :

Final answer:

The ideal gas law (PV=nRT) can be used to find the volume of a gas. For 3.0 moles of hydrogen at 800 mmHg and 25°C, the ideal gas law predicts a volume of about 69.7 L.

Option d.

Explanation:

You can use the ideal gas law to solve this problem. The ideal gas law relates the pressure $(P)$, volume $(V)$, temperature $(T)$, and quantity $(n)$ of a gas. The formula is:

PV = nRT

where:

R is the gas constant, which is equal to $0.082057 [tex]\dfrac{L \cdot atm}{mol \cdot K}$[/tex]

We can rearrange the equation to solve for the volume $(V)$ :

V = [tex]\dfrac{L \cdot atm}{mol \cdot K}$[/tex]

We are given the following values:

n (amount of hydrogen gas) = 3.0 moles

T (temperature) = 25°C + 273.15 K = 298.15 K (convert Celsius to Kelvin)

P (pressure) = 800.00 mmHg (convert to atm)

Pressure conversion:

1 atm = 760 mmHg

P = 800.00 mmHg[tex]\times \dfrac{1 atm}{760 mmHg}[/tex]= 1.0526 atm$

Now we can plug the values into the equation and solve for $V$ :

V = [tex]\dfrac{(3.0 mol) \left( 0.082057 \dfrac{L \cdot atm}{mol \cdot K} \right) (298.15 K)}{(1.0526 atm)}$[/tex]

V = 69.73 L$

Therefore, the volume of 3.0 moles of hydrogen at 800.00 mmHg and 25°C is approximately 69.73 L. The closest answer choice is d) (62.7 , L).However, it's important to note that the ideal gas law is a simplification and may not perfectly represent the behavior of real gases at high pressures or low temperatures.

Option d.

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