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1 <br /> 100µg/1 <br /> • Mass= (1,005 ft2*40 feet) *(100µg/1)* (28 32 L/ft3) *(35%)* (1g/1,000,000 µg) <br /> =39 85gT ams <br /> =0 04 kilogram(0 088 pound) <br /> 1,000µg/1 <br /> Mass=(650 ft2*40 feet)* (1,000µg/I)* (28 32 L/ft3) *(35%) *(1g/1,000,000µg) <br />' =257 7$rams <br />' =0 258 kilogram(0 569 pound) <br />' 10,000 fg/L <br /> Mass=(154 ft2*40 feet)* (10,000 µg/1)*(28 32 L/ft3)*(35%)*(1g/1,000,000µg) <br />' =610 6 grams <br /> =0 611 kilogram (1 35 pounds) <br />' Total Mass in Ground Water <br /> Total mass =0 04 kg+0 258 kg+ 0 611 kg=0 91 kg(2 01 pounds) <br />' The calculation indicates that approximately 0 91 kilogram (2 01 pounds) <br /> of TPH-g remains in the ground water at the site This mass would be <br />' equivalent to approximately 2 to 3 pints of gasoline <br /> ERM B-3 9343 50/UPRR/11/06/02 <br />