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October 2, 2014

October 2, 2014

Posted by **manny** on Tuesday, September 4, 2007 at 4:05pm.

I need help with this problem--- I have no idea how to go about it

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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

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Thanks for all the help you guys can provide,

manny

- chem -
**DrBob222**, Tuesday, September 4, 2007 at 5:04pmHow about this hint.

(7.0 x 10^6 g/m^{3}) x 0.75 x 1/2 = amount of consumable lead. Convert 8.5 x 10^{3}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.

- chem -
**manny**, Tuesday, September 4, 2007 at 5:55pmok! 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

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