Answer
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Work Step by Step
From the text:
$IE=h\nu-KE$
$h.c/\lambda=3.40×10^{-18}\ J$
$IE=2.98×10^{-18}\ J$
$2.98×10^{-18}\ J\div 1.602×10^{-19}\ J/eV=18.6\ eV$
$2.98×10^{-18}\ J×6.022×10^{23}\ 1/mol=$
$1794.0\ kJ/mol$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals - 9-3 Molecular Orbital Theory - Applying Chemical Principles - Questions - Page 367: 4
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