- #1
knottlena
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The Answer is "Yes", but I need to show HOW. Plz Help.
A researcher breaks a thermometer & spills most of the mercury in it on the floor of a laboratory that measures 15.2 m Long, 6.6 m Wide, & 2.4 m High. Does the concetration of mercury exceed the EPA air quality regulation of .050 mg Hg/m3 of air? (Temp of the rm is 20 degrees celcius; the vapor pressure of mercury at this temp is 1.7 x 10 to -6th atm.)
PV=nRT & LxWxH
LxWxH=240.77 m3 x 100 to the 3rd power=2.4x10 to the 8th cm3 x 1mL/1cm3=2.4x10 to the 8th/1000 L = 2.4 x 10 to the 5th L
Hg = 200.59g/mol
n=PV/RT (1.7 x10 to -6th)(2.4 x 10 to the 5th)/ (.0821 Latm/mol K) (293K) = 1.4 x 10 to the 3rd mol
I just don't know where to go from here. I know the answer is 14.1 mg/m3, but I can't figure out how the teacher got there. HELP PLEase. Thank you in advance.
Homework Statement
A researcher breaks a thermometer & spills most of the mercury in it on the floor of a laboratory that measures 15.2 m Long, 6.6 m Wide, & 2.4 m High. Does the concetration of mercury exceed the EPA air quality regulation of .050 mg Hg/m3 of air? (Temp of the rm is 20 degrees celcius; the vapor pressure of mercury at this temp is 1.7 x 10 to -6th atm.)
Homework Equations
PV=nRT & LxWxH
The Attempt at a Solution
LxWxH=240.77 m3 x 100 to the 3rd power=2.4x10 to the 8th cm3 x 1mL/1cm3=2.4x10 to the 8th/1000 L = 2.4 x 10 to the 5th L
Hg = 200.59g/mol
n=PV/RT (1.7 x10 to -6th)(2.4 x 10 to the 5th)/ (.0821 Latm/mol K) (293K) = 1.4 x 10 to the 3rd mol
I just don't know where to go from here. I know the answer is 14.1 mg/m3, but I can't figure out how the teacher got there. HELP PLEase. Thank you in advance.