Bundle: Organic Chemistry, 9th, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
Bundle: Organic Chemistry, 9th, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
9th Edition
ISBN: 9781305701021
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 31.SE, Problem 23AP

Show the structure of the polymer that results from heating the following diepoxide and diamine:

Chapter 31.SE, Problem 23AP, Show the structure of the polymer that results from heating the following diepoxide and diamine:

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In the electrode Pt, H2(1 atm) | H+(a=1), if the electrode balance potential is -0.118 V and the interface potential difference is +5 mV. The current voltage will be 0.005 - (-0.118) = 0.123 V ¿Correcto?
In the electrode Pt, H2(1 atm) | H+(a=1) at 298K is 0.79 mA cm-2. If the balance potential of the electrode is -0.118 V and the potential difference of the interface is +5 mV. Determine its potential.
In one electrode: Pt, H2(1 atm) | H+(a=1), the interchange current density at 298K is 0.79 mA·cm-2. If the voltage difference of the interface is +5 mV. What will be the correct intensity at pH = 2?. Maximum transfer voltage and beta = 0.5.
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CBSE Class 12 Chemistry || Polymers || Full Chapter || By Shiksha House; Author: Best for NEET;https://www.youtube.com/watch?v=OxdJlS0xZ0Y;License: Standard YouTube License, CC-BY