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How many moles of gaseous boron trifluoride, $BF_{3}$, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of $BF_{3} ?$
5.553 $\mathrm{g}$
01:21
Aadit S.
Chemistry 101
Chapter 9
Gases
Carleton College
Rice University
University of Kentucky
University of Toronto
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so we were taking a look at the ideal cast equation. PV equals n r. T. So what we're gonna be doing is re arranging for moles. So moles is equal to P V over r T. We're gonna be plugging in some values. So we have miles is equal to 1.2 to 0 atmosphere multiply that are pressure multiplied by our volleying that is 4.3417 liters and then that is all divided by no point No. 8 to 06 Atmosphere Polyta the multi minus one Calvin's minus one multiply by 788 Calvin So our malls is equal to not 0.98190 models. Therefore the moles of BF three is equal to not point no. 819 several moles. So from this we can calculate a mass. So we have mass. BF three is equal toothy moles not 30.98190 Moles well supplied by the molecular weight which is 67.8062 grams per mall. Our mass is equal to 5.553 g
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