## Chemistry 9th Edition

a) $CCl_4$ b) $H_2O$ c) $H_2O$ d) $CCl_4$ e) $H_2O$ f) $H_2O$ g) $CCl_4$
a) $KrF_2$ is nonpolar due to its tetrahedral shape; therefore, it dissolves best in the nonpolar $CCl_4$. b) $SF_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$. c) $SO_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$. d) $CO_2$ is nonpolar due to its linear shape; therefore, it dissolves best in the nonpolar $CCl_4$. e) $MgF_2$ is polar due to its bent shape; therefore, it dissolves best in the polar $H_2O$. f) $CH_2O$ is polar due to its asymetric polar $C-O$ bond; therefore, it dissolves best in the polar $H_2O$. g) $CH_2=CH_2$ is nonpolar due to it containing only nonpolar $C-H$ bonds; therefore, it dissolves best in the nonpolar $CCl_4$.