The pressure exerted by 1.5 mol of gas in a 13 L flask at 22 °C is kPa

Pressure is defined as the force exerted by the molecules colliding with the surface of the object and other surfaces. The ideal gas follows the general gas law equation, PV = nRT.

The pressure exerted by the gas is 282.7 KPa.

Given:

Pressure (P) = ?

Gas constant (R) = 0.082atm.L/Kmol  

Temperature (T) = 22°C = 22 + 273°C = 295K

Volume (V) = 13L

Number of mole of (n) = 1.5 mole

Now, using the ideal gas equation:

PV = nRT

Substituting the values, we get:

P x 13 = 1.5 x 0.082 x 295

P x 13 = 36.285

P =

Pressure = 2.79 atm.

Also, 1 atm = 101.32 KPa

2.79 atm = 2.79 x 101.32 = 282.7 KPa.

Thus, the pressure exerted by the gas molecules is 282.7 KPa.

To know more about ideal gas law, refer to the following link:

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282.7KPa

Explanation:

Step 1:

Data obtained from the question.

Number of mole of (n) = 1.5 mole

Volume (V) = 13L

Temperature (T) = 22°C = 22 + 273°C = 295K

Pressure (P) =..?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the pressure exerted by the gas.

This can be obtained by using the ideal gas equation as follow:

PV = nRT

P = nRT /V

P = 1.5 x 0.082 x 295 / 13

P = 2.79atm.

Step 3:

Conversion of 2.79atm to KPa.

This is illustrated below:

1 atm = 101.325KPa

Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa

Therefore, the pressure exerted by the gas in KPa is 282.7KPa

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%