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**Please answer quickly!** 5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?

Respuesta :

Answer:

418.352441397 Liters

Explanation:

You start off with your equation of PV=nRt, where P= Pressure, V= Volume, n= The amount of moles, R= the gas constant which is 0.0821, and t= The temperature.

You are solving for volume so you would divide the left side by P so you can isolate the Volume. Now you have V=nRt/P.

Next you would fill in all of the variables. n=5.1169, R=0.0821, t=911K, and P=0.9148atm.

You would get the equation V=(5.1169*0.0821*911)/0.9148

Your final answer would be 418.352441397 Liters, or rounded to 418.35 Liters.

The volume of the container is 418.15 L

Ideal gas law

From the question, we are to determine the volume of the container

From the ideal gas equation, we have that

PV = nRT

Where P is the pressure

V is the volume

n is the number of moles

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

and T is the temperature

From the given information

P = 0.9148 atm

n = 5.1169 mol

T = 911 K

Putting the parameters into the formula, we get

0.9148 × V = 5.1169 × 0.08206 × 911

Then,

[tex]V = \frac{5.1169 \times 0.08206 \times 911}{0.9148}[/tex]

V = 418.15 L

Hence, the volume of the container is 418.15 L

Learn more on Ideal gas law here: https://brainly.com/question/25736513

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