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60.1 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300. m. P is destroyed by sunlight, and once in the pond it has a half-life of 6.6 h.Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits. m

60.1 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300. m. P is destroyed by sunlight, and once in the pond it has a half-life of 6.6 h.Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits. m

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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O KINETICS AND EQUILIBRIUM Solving applied problems with first-order kinetics Har 60.1 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300. m°. P is destroyed by sunlight, and 3 once in the pond it has a half-life of 6.6 h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits. x10 3 Explation Check Privacy Accessit O2021 McGraw-Hill Education. All Rights Reserved. Terms of Use )国具a 20 ty II
Transcribed Image Text:O KINETICS AND EQUILIBRIUM Solving applied problems with first-order kinetics Har 60.1 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300. m°. P is destroyed by sunlight, and 3 once in the pond it has a half-life of 6.6 h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits. x10 3 Explation Check Privacy Accessit O2021 McGraw-Hill Education. All Rights Reserved. Terms of Use )国具a 20 ty II
Expert Solution
Step 1: Given and Formula

Given:

Mass of industrial waste = 60.1 g

mass per hours;(we know that 1 day = 24 hr)

Mass in hour = 60.1 g/ 24 hr

= 2.50 g/hr

Volume of pond = 300 m3

Half life period = 6.6 h

Mass out = 0

Equilibrium concentration CA =?

Mass out due to sunlight  = k.CA.V -----(1)

Where K = rate constant

CA = Equilibrium concentration

V = Volume

 

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