## Chemistry 12th Edition

Published by McGraw-Hill Education

# Chapter 8 - Periodic Relationships Among the Elements - Questions & Problems: 8.57

#### Answer

Please see the work below.

#### Work Step by Step

We know that $E_1=-(2.18\times 10^{-18}J(2)^2)(6.02\times 10^{23})=-5.25\times 10^{3}\frac{KJ}{mol}$

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