Nitrogen dioxide reacts with water and oxygen in the atmosphere to produce acid rain.

4NO2(g) + 2H2O(g) + O2(g) = 4HNO3(g)

Assuming that the temperature and the pressure remain constant, calculate the volume of air at 1 atm, 20 C, necessary to dilute the HNO3(g) to 0.080 ppm (by volume) when 532 g of NO2(g) is allowed to react.

I am wondering what should I do with the 0.080 ppm. I am pretty sure I know how to do everything else but, I am not sure how to use the ppm in to the equation of if it used after the pv=nRT equation.

To calculate the volume of air necessary to dilute the HNO3(g) to 0.080 ppm (by volume), we first need to understand what ppm (parts per million) means in this context.

ppm is a unit of concentration, representing the ratio of the number of parts of a substance (in this case, HNO3) to every one million parts of the total volume of the mixture (in this case, air).

To use ppm in our calculation, we need to convert it into a more commonly used concentration unit, such as molarity (moles per liter). Here's how to proceed:

1. Convert 0.080 ppm of HNO3 to a molar concentration (M):
ppm can be converted to molar concentration by dividing the ppm value by the molar mass of HNO3. The molar mass of HNO3 can be found by adding up the atomic masses of its constituent elements (nitrogen, hydrogen, and oxygen):
HNO3: (1 × 1.01) + (1 × 14.01) + (3 × 16.00) = 63.02 g/mol

0.080 ppm HNO3 can be converted to molarity using the following equation:
M = (mass of HNO3 / molar mass of HNO3) / (volume of air in liters)

Since we want to find the volume of air needed, let's rearrange the equation:
Volume of air in liters = (mass of HNO3 / molar mass of HNO3) / (M x 1,000,000)

2. Calculate the number of moles of NO2:
Given that 532 g of NO2 is allowed to react, we can calculate the number of moles of NO2 using its molar mass. The molar mass of NO2 is calculated as:
NO2: (1 × 14.01) + (2 × 16.00) = 46.01 g/mol

Number of moles of NO2 = mass of NO2 / molar mass of NO2

3. Use the balanced equation to relate the number of moles of NO2 to the number of moles of HNO3:
From the balanced equation, we can see that 4 moles of NO2 react to produce 4 moles of HNO3.

4 moles of NO2 = 4 moles of HNO3

4. Substitute the values into the equation to find the volume of air:
Volume of air in liters = (mass of HNO3 / molar mass of HNO3) / (M x 1,000,000)
= (532 g / 63.02 g/mol) / (4 moles HNO3 / 4 moles NO2)
= (532 g / 63.02 g/mol)

By plugging in the value for the mass of HNO3 and the given molar mass, you can calculate the volume of air in liters needed to dilute the HNO3(g) to 0.080 ppm.