In thermodynamic of the dissolution of Borax lab, the student's graph is below. Calculate AS? (R= 8.314 x 10-3 kJ/mole K) LN of Ksp vs 1/Temperature in Kelvin LN Ksp 0 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -2 4 -6 -8 y=-9104.5x + 26.39 R2 = 0.9713 Temperature in K-1 Series1 ......... Linear (Series1) A) 0.008075 kJ/mol K B) -75.69 kJ/mol K OC) -0.008075 kJ/mol K OD) -0.2194 kJ/mol K E) 75.69 kJ/mol K OF) 0.2194 kJ/mol K

In thermodynamic of the dissolution of Borax lab, the student's graph is below. Calculate AS? (R= 8.314 x 10-3 kJ/mole K) LN of Ksp vs 1/Temperature in Kelvin LN Ksp 0 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -2 4 -6 -8 y=-9104.5x + 26.39 R2 = 0.9713 Temperature in K-1 Series1 ......... Linear (Series1) A) 0.008075 kJ/mol K B) -75.69 kJ/mol K OC) -0.008075 kJ/mol K OD) -0.2194 kJ/mol K E) 75.69 kJ/mol K OF) 0.2194 kJ/mol K

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter16: Thermodynamics

Section: Chapter Questions

Problem 2E: What is a nonspontaneous reaction?

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In thermodynamic of the dissolution of Borax lab, the student's graph is below. Calculate AS? (R= 8.314 x 10-3 kJ/mole K) LN of Ksp vs 1/Temperature in Kelvin LN Ksp 0 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -2 -4 -6 y=-9104.5x + 26.39 R2 = 0.9713 Temperature in K-1 Series1 .........Linear (Series1) OA) 0.008075 kJ/mol K B) -75.69 kJ/mol K OC) -0.008075 kJ/mol K OD) -0.2194 kJ/mol K OE) 75.69 kJ/mol K OF) 0.2194 kJ/mol K
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