Posted by manny on .
I need help with this problem--- I have no idea how to go about it
An average adult breathes about 8.5 x 10^3 L of air per day. The concentration of lead in higly polluted urban air is 7.0 x 10^-6 g of lead per one m3 of air. Assume that 75% of the lead is present as particles less than 1.0 x 10^-6 m in diameter, and that 50% of the particles below that size are retained in the lungs. Calculate the mass of lead absorbed in this manner in one year by an average adult living in this environment.
***Note answer has two sigfigs
Thanks for all the help you guys can provide,
How about this hint.
(7.0 x 10^6 g/m3) x 0.75 x 1/2 = amount of consumable lead. Convert 8.5 x 103 L to cubic meters (or the other way around--it doesn't matter which one you convert) and calculate the amount consumed per day. Then multiply by 365 to convert to amount per year.
ok! thanks... Dr. Bob your the best!!!
I assumed you meant (7.0 x 10^-6 g/m3)
I got 8.1 x 10^-3 g Pb / year