David J. answered 03/11/21
Tutor
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PhD chemist with extensive teaching experience.
8.314 m3⋅Pa⋅K−1⋅mol−1 ------> 0.008314 m3-kPa/K-mol
Or, in liters.......8.314 L-kPa/K-mol
Taylor B.
asked 03/11/21
David J. answered 03/11/21
PhD chemist with extensive teaching experience.
8.314 m3⋅Pa⋅K−1⋅mol−1 ------> 0.008314 m3-kPa/K-mol
Or, in liters.......8.314 L-kPa/K-mol
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