You have a 6.75 l container of pure gaseous oxygen at a constant pressure of 1.25 atm. if the temperature is 298 k, calculate the number of grams of oxygen (o2) gas in the container. (r = 0.0821 l∙atm/mol∙k) a. 11.1 g b. 92.8 g c. 2.90 g d. 0.346 g

There is 11.1 grams of O2 in the container ( option A)

Explanation:

Step 1: Data given

Volume of the container = 6.75 L

Pressure = 1.25 atm

Temperature = 298 K

Step 2: Calculate number of mol of oxygen

p*V = n*R*T

⇒ with p = the total pressure of oxygen = 1.25 atm

⇒ with V = the volume of the oxygen gas = 6.75 L

⇒ with n = the number of moles of oxygen = TO BE DETERMINED

⇒ with R = the gas constant = 0.0821 L*atm/K*mol

⇒ with T = the temperature in the container = 298 K

n = (P*V)/(R*T)

n = (1.25*6.75)/ (0.0821*298)

n = 0.345 moles O2

Step 3: Calculate mass of O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.345 moles O2 = 32 g/mol

Mass O2 ≈ 11.1 grams O2

There is 11.1 grams of O2 in the container ( option A)

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.