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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

c. is the answer

by love gaming