Question
6.77510-4 mol of O3(g) (ozone gas) is generated into a 165.0 mL headspace, at 25.0 C, thus increasing the pressure by 0.09902 atm. If we
6.775×10-4 mol of O3(g) (ozone gas) is generated into a 165.0 mL headspace, at 25.0 °C, thus increasing the pressure by 0.09902 atm. If we assume that the ozone gas obeys Boyle's Law and Charle's Law, what would be the expected molar volume at STP?
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Volume of O 3 at STP 14968 mL Solution mole of O 3 6775x10 4 mol First condi...
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Fundamentals Of Biostatistics
Authors: Bernard Rosner
8th Edition
130526892X, 978-1305465510, 1305465512, 978-1305268920
