A sample of 2.00 mol of gas in a 10.00 L container is at 45.0 °C. What is the pressure (in atm) of the gas?

5.22 atm

Explanation:

The following data were obtained from the question:

Number of mole (n) = 2 moles

Volume (V) = 10 L

Temperature (T) = 45 °C

Pressure (P) =?

Next, we shall convert 45 °C to Kelvin temperature. This can be obtained as follow:

Temperature (K) = Temperature (°C) + 273

T (K) = T (°C) + 273

T (°C) = 45 °C

T(K) = 45 °C + 273

T (K) = 318 K

Finally, we shall determine the pressure of the gas by using the ideal gas equation as shown below:

Number of mole (n) = 2 moles

Volume (V) = 10 L

Temperature (T) = 318 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =.?

PV = nRT

P x 10 = 2 x 0.0821 x 318

Divide both side by 10

P = (2 x 0.0821 x 318) /10

P = 5.22 atm

Therefore, the pressure of the gas is 5.22 atm

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :