A 3.00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as BaCO3. The excess base was back-titrated to a phenolphthalein endpoint with 23.6 mL of 0.0108 M HCl. Calculate the parts-per-million of CO2 in the air (i.e. mL CO2/10^6 mL air); use 1.98 g/L as the density of CO2.1A]3.00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116M Ba(OH)2, which caused the CO2 in the sample to precipitate as BaCO3. The excess basewas back-titrated…The post 00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as… appeared first on Nursing Essay Tutors.00 L sample of urban air was bubbled through a solution containing 50.0 mL of 0.0116 M Ba(OH)2, which caused the CO2 in the sample to precipitate as… was first posted on November 26, 2020 at 1:20 pm.©2019 “nursingassignmenttutor”. Use of this feed is for personal non-commercial use only. If you are not reading this article in your feed reader, then the site is guilty of copyright infringement. Please contact me at [email protected]  “Are you looking for this answer? We can Help click Order Now”

 

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