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Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 +0.052R 0.041 R=0.138 R 3.37lb mol

Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 +0.052R 0.041 R=0.138 R 3.37lb mol

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Please answer the 2 items. Show how it derived to that answers.
Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 + 0.052R 0.041 R = 0.138 R 3.37lb mol
Transcribed Image Text:Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 + 0.052R 0.041 R = 0.138 R 3.37lb mol
Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 + 0.052R 0.041 R = 0.138 R 3.37lb mol
Transcribed Image Text:Solutions Chapter 12 1.385+0.0931R 0.052 (27.35 + R) = 1.523 + 0.052R 0.041 R = 0.138 R 3.37lb mol
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