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Given that 1 ppb = 1 part per billion = \(10^{-9}\) parts, the concentration of ozone can be written as 84 parts per billion (84x\(10^{-9}\)). Now, we'll use the ideal gas law, \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We need to find the concentration in g/m³, so we have to rearrange the equation to solve for the mass density of ozone, m/V: \(m/V = \frac{n}{V} * M = \frac{P}{RT} * M\) Assuming standard atmospheric pressure (P = 1 atm) and temperature (T = 298 K), we can use R = 0.0821 L atm/mol K and the molecular weight of ozone (M = 48 g/mol) to calculate the mass density: \(m/V = \frac{1 \text{ atm}}{0.0821 \text{ L atm/mol K}\times 298 \text{ K}} \times 48 \text{ g/mol} \times 84 \times 10^{-9} \) \(m/V = 1.40 \times 10^{-4}\) g/m³

Los Angeles County has an area of approximately 4000 square miles. First, we will convert square miles into square meters using the conversion factor, 1 square mile = \(2.59 \times 10^{6}\) square meters: Area = 4000 square miles * \(2.59 \times 10^{6}\) m²/square mile = \(1.04 \times 10^{10}\) m² Assuming a height of 10 m above the ground, we can calculate the volume of air above Los Angeles County as follows: Volume = Area * Height = \(1.04 \times 10^{10}\) m² * 10 m = \(1.04 \times 10^{11}\) m³

Now we will calculate the moles of ozone using the mass density (in g/m³) and molecular weight of ozone (48 g/mol): Moles of ozone = \(\frac{Mass\ Density\ of\ Ozone × Volume\ of\ Air }{Molecular\ Weight\ of\ Ozone}\) Moles of ozone = \(\frac{1.40 \times 10^{-4} \text{ g/m³} × 1.04 \times 10^{11} \text{ m³}}{48 \text{ g/mol}}\) Moles of ozone = 3.03 x 10^5 mol Solution: There would be approximately 3.03 x 10^5 moles of ozone in the air above Los Angeles County if the ozone concentration was at the 84 ppb EPA standard.