Humans breathe about 500 mL of air per breath and take about 12 breaths per minute during normal activities. If a person is exposed to an atmosphere containing benzene at a concentration of 10 ppm (by volume), how many grams of benzene will be deposited in the lungs during an 8-hour shift if all the benzene that enters remains in the lungs

mass of benzene in the lungs = 28.8 g of benzene

Explanation:

Volume of air breathed per hour = 500 mL * 12 * 60 = 360000 ml = 360 Litres of air

In 8 hours, volume of breathed air = 360 L * 8 = 2880 Litres of air

Concentration of benzene in air = 10 ppm

Note:  1 ppm = 1 mg per liter (mg/L)

10 ppm = 10 mg/L

Therefore, mass of benzene in the lungs in an 8-hour shift = 10 mg/L * 2880 L

mass of benzene in the lungs = 28800 mg

Converting to grams = 28800/1000

mass of benzene in the lungs = 28.8 g of benzene

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.