What is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

The temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm is 176.6K

HOW TO CALCULATE TEMPERATURE:

PV = nRT

Where;

4.68 × 2 = 0.645 × 0.0821 × T

9.36 = 0.053T

T = 9.36 ÷ 0.053

T = 176.6K

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—96.03°C

Explanation:

We'll begin by writing out the information provided by the question. This includes:

Number of mole (n) = 0.645 mole

Volume (V) = 2.00 L

Pressure (P) = 4.68 atm

Temperature (T) =?

Recall: that the gas constant = 0.082atm.L/Kmol

With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

PV = nRT

4.68 x 2 = 0.645 x 0.082 x T

Divide both side 0.645 x 0.082

T = (4.68 x 2) /(0.645 x 0.082)

T = 176.97 K

Now, We can also express the temperature obtained in celsius as shown below:

Temperature (celsius) = temperature (Kelvin) - 273

Temperature (celsius) = 176.97 - 273

Temperature (celsius) = —96.03°C

The temperature of the Neon gas is

—96.03°C

no, the red giant is the hottest with 2.00.000.000 degrees celsius. while the white dwarf only 25,000 degrees celsius.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.