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. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

i just learnt this recently but i need to look through my notes again