## Chemistry 9th Edition

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