Answer
0.927g Ni(OH)2
Work Step by Step
0.0200 mol KOH* ( 1 mol Ni(OH)_2)/( 2 molKOH) * (92.71 g Ni(OH)_2)/(1 mol Ni(OH)_2 ) = 0.927g Ni(OH)2
Chemistry: The Central Science (13th Edition)
by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.
Published by
Prentice Hall
ISBN 10:
0321910419
ISBN 13:
978-0-32191-041-7
Chapter 4 - Reactions in Aqueous Solution - Exercises - Page 161: 4.87d
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