Answered

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The pressure exerted by 1.30 mol of gas in a 13.0 L container at 295 K is

Sagot :

Answer:

The pressure of the gas is 2.42atm.

Explanation:

The given information from the exercise is:

- Number of moles (n): 1.30 moles

- Volume (V): 13.0L

- Temperature (T): 295K

With the Ideal Gases Law formula we can calculate the pressure (P) of the gas, by replacing the values of n, V and T:

[tex]\begin{gathered} P*V=n*R*T \\ P*13.0L=1.30mol*0.082\frac{atm*L}{mol*K}*295K \\ P*13.0L=31.45atm*L \\ P=\frac{31.45atm*L}{13.0L} \\ P=2.42atm \end{gathered}[/tex]

So, the pressure of the gas is 2.42atm.

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