**Problem Statement:**36. How many *grams* of gaseous boron trifluoride, BF₃, are contained in a 4.341-L bulb at 788.0 K if the pressure is 1.220 atm?**Answer Box:**[ ] g BF₃**Explanation:**To solve this problem, you would typically use the Ideal Gas Law, which is given by:\[ PV = nRT \]Where:- \( P \) is the pressure (in atm)- \( V \) is the volume (in liters)- \( n \) is the number of moles of the gas- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K)- \( T \) is the temperature (in Kelvin)To find the mass in grams, calculate \( n \) using the above equation and then convert moles to grams using the molar mass of boron trifluoride (BF₃).**Steps:**1. Substitute the given values into the Ideal Gas Law to find \( n \).2. Calculate the molar mass of BF₃.3. Multiply the number of moles by the molar mass to get the mass in grams.

Assuming the ideal gas behaviour of BF3 molecules, you can say PV = nRT You have given all other…

Answered: 36. How many grams of gaseous boron…

36. How many grams of gaseous boron trifluoride, BF3, are contained in a 4.341-L bulb at 788.0 K if the pressure is 1.220 atm? g BF3

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter5: Gases

Section: Chapter Questions

Problem 5.108PAE

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