Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 2 - Atoms, Molecules, and Ions - Exercises: 79

Answer

a) Copper (I) Iodide b) Copper (II) Iodide c) Cobalt (II) Iodide d) Sodium Carbonate e) Sodium Hydrogen Carbonate f) Tetrasulfur Tetranitride g) Selenium Tetrachloride h) Sodium Hypochlorate i) Barium Chromate j) Ammonium Nitrate

Work Step by Step

a) $Cu$ = Copper $I$ = Iodide Since the compound is ionic, Copper (I) Iodide b) $Cu$ = Copper $I_{2}$ = Iodide Since the compound is ionic, it is Copper (II) Iodide c) $Co$ = Cobalt $I_{2}$ = Iodide Therefore since the compound is ionic, $CoI_{2}$ is Cobalt (II) Iodide d) $Na_{2}$ = Sodium $CO_{3}$ = Carbonate Therefore, $Na_{2}CO_{3}$ is Sodium Carbonate e) $Na$ = Sodium $H$ = Hydrogen $Co_{3}$ = Carbonate Therefore, $NaHCO_{3}$ is Sodium Hydrogen Carbonate f) $S_{4}$ = Tetrasulfur $N_{4}$ = Tetranitride Since $S_{4}$$N_{4}$ is a covalent compound, it's Tetrasulfer Tetranitride g) $Se$ = Selenium $Cl_{4}$ = Tetrachloride Since $SeCl_{4}$ is a covalent compound, it is Selenium Tetrachloride h) $Na$ = Sodium $OCl$ = $ClO$ $ClO$ = Hypochlorite Therefore, $NaOCl$ is Sodium Hypochlorite i) $Ba$ = Barium $CrO_{4}$ = Chromate Therefore, $BaCrO_{4}$ is Barium Chromate j) $NH_{4}$ = Ammonium $NO_{3}$ = Nitrate Therefore $NH_{4}NO_{3}$ is Ammonium Nitrate
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