The Table 17 − 1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated. Concept introduction: The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air. To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
The Table 17 − 1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated. Concept introduction: The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air. To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
Solution Summary: The author analyzes the relationship between cell potential and Gibbs free energy change by analyzing the Table 17-1 showing reduction potential values of various elements.
The Table
17−1 showing reduction potential values of various elements is given. The reason behind the corrosion of most of the metals and the difficulty in corrosion of noble metals like, gold, platinum, and silver is to be stated.
Concept introduction:
The metal undergoes oxidation and the oxygen undergoes reduction during corrosion. So, the metal with reduction potential values lower than oxygen easily gets corroded in the air.
To determine: The reason of corrosion of most of the metals in air and the reason of difficulty in the corrosion of noble metals like gold, platinum, and silver.
Epoxides can be opened in aqueous acid or aqueous base to produce diols (molecules with two OH groups). In this question, you'll explore the
mechanism of epoxide opening in aqueous acid.
2nd attempt
Be sure to show all four bonds at stereocenters using hash and wedge lines.
0
0
Draw curved arrows to show how the epoxide reacts with hydronium ion.
100 +1:
1st attempt
Feedback
Be sure to show all four bonds at stereocenters using hash and wedge lines.
See Periodic Table
See Hint
H
A
5
F
F
Hr
See Periodic Table See Hint
03 Question (1 point)
For the reaction below, draw both of the major organic products. Be sure to consider stereochemistry.
>
1. CH₂CH₂MgBr
2. H₂O
3rd attempt
Draw all four bonds at chiral centers. Draw all stereoisomers formed.
Draw the structures here.
e
130
AN
H
See Periodic Table See Hint
P
C
Br
You may wish to address the following issues in your response if they are pertinent to the reaction(s) you propose to employ:1) Chemoselectivity (why this functional group and not another?)
2) Regioselectivity (why here and not there?)
3) Stereoselectivity (why this stereoisomer?)
4) Changes in oxidation state.
Please make it in detail and draw it out too in what step what happens. Thank you for helping me!
Chapter 17 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
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