Example 9A copper surface area, A=100 cm?, is exposed to an acid solution. After 24 hour, the lossof copper due to corrosion (Oxidation) is 15*10*(a) The current density in µA/cm2.(b) The corrosion rate in mm/yGiven that: molar mass for copper = 63.54 g/mole, p 8.96 g/cmgram. Calculate:

From the given data Area (A)= 100cm2 time (t)=24 hour= 24×60×60= 86400 seconds weight lost of copper…

A copper surface area, A=100 cm?, is exposed to an acid solution. After 24 hour, the loss of copper due to corrosion (Oxidation) is 15*10 gram. (a) The current density in µA/cm?. (b) The corrosion rate in mm/y Calculate: Given that: molar mass for copper = 63.54 g/mole, p= 8.96 g/cm %3D

A copper surface area, A=100 cm?, is exposed to an acid solution. After 24 hour, the loss of copper due to corrosion (Oxidation) is 15*10 gram. (a) The current density in µA/cm?. (b) The corrosion rate in mm/y Calculate: Given that: molar mass for copper = 63.54 g/mole, p= 8.96 g/cm %3D

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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