Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 44c

Answer

lead (II) Chromate - $PbCrO_{4}$

Work Step by Step

1. To begin, write the name of the first element with its charge. (We are told that Lead has a charge of 2). Pb$^{2+}$ 2. Then write the name of the second element with its charge. (Note - Chromate is a poly-atomic ion so, you have to remember its charge.) CrO$_{4}^{2-}$ 3. Cross down the charges on both of the elements (+2 and -2) to get: $$PbCrO_{4}$$
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