Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. typically, the temperature and partial pressure of ozone in the stratosphere are 250 k and 0.001 atm, respectively. how many grams of ozone are present in 1000 liters of air under these conditions? (assume ideal-gas behavior)

2.35 g of ozone are present

Explanation:

From Ideal gas equation,

PV = nRT equation 1

P = pressure of the ozone, n = number of mole of the ozone molecule, R = molar gas constant, V = volume of the ozone, T = Temperature.

making n the subject of the

n = PV/RT equation 2

Where V = 1000 dm³, P = 0.001 atm, T = 250 K, R = 0.082 atm/(dm³Kmol)

Substituting these values into equation 2

n = (1000×0.001)/(0.082×250)

n = 1/20.5

n = 0.049 mole.

Number of mole  = Reacting mass/ molar mass .

Molar mass of ozone gas (O₃) = 48 g/mol

Mass of Ozone = 48 × 0.049

Mass of ozone = 2.35 g.

Therefore, 2.35 g of ozone are present.

c is the answer because when something is left after something evaporates that shows the chemical change.

the answer is c. nacl