Federal regulations set an upper limit of 50 parts per

Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown

Problem 115IE Chapter 4

Chemistry: The Central Science | 12th Edition

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Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown

Chemistry: The Central Science | 12th Edition

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Problem 115IE

Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00 × 102 mL of 0.0105 MHCl. The NH3 reacts with HCl according to:

NH3(aq) + HCl(aq)→NH4Cl(aq)

After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point.

(a) How many grams of NH3 were drawn into the acid solution?

(b) How many ppm of NH3 were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)

(c) Is this manufacturer in compliance with regulations?

Step-by-Step Solution:
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GPH210AirQuality SpringSEMESTER2016 Professor:Dr.ElizabethLarson EliteNotetaker:Phoebe(phoebe@studysoup.com) 1. The Killer Smog ○ Donora, Pennsylvania (1948) ■ Deadly air pollution ■ 50 died ■ thousands ill ■ Zinc­smelting factory ■ Airborne...

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Chapter 4, Problem 115IE is Solved
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Textbook: Chemistry: The Central Science
Edition: 12
Author: Theodore E. Brown
ISBN: 9780321696724

The answer to “Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00 × 102 mL of 0.0105 MHCl. The NH3 reacts with HCl according to:NH3(aq) + HCl(aq)?NH4Cl(aq)After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH3 were drawn into the acid solution? (b) How many ppm of NH3 were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?” is broken down into a number of easy to follow steps, and 146 words. Since the solution to 115IE from 4 chapter was answered, more than 268 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Chemistry: The Central Science, edition: 12. The full step-by-step solution to problem: 115IE from chapter: 4 was answered by , our top Chemistry solution expert on 04/03/17, 07:58AM. This full solution covers the following key subjects: air, acid, HCL, solution, regulations. This expansive textbook survival guide covers 49 chapters, and 5471 solutions. Chemistry: The Central Science was written by and is associated to the ISBN: 9780321696724.

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