Chemistry 9th Edition

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

Chapter 11 - Properties of Solutions - Exercises: 44

Answer

a) $H_2O$ b) $C_6H_{14}$ c) $H_2O$ d) $C_6H_{14}$ e) $H_2O$ f) $C_6H_{14}$

Work Step by Step

a) $Cu(NO_3)_2$ is an ionic compound due to it being composed of a cation ($Cu^{2+}$) and an anion ($NO^-$) making it most soluble in the polar $H_2O$. b) $CS_2$ is a nonpolar compound due to its linear shape making it most soluble in the nonpolar $C_6H_{14}$. c) $CH_3OH$ is a polar compound due to its polar alcohol functional group making it most soluble in the polar $H_2O$. d) $CH_3(CH_2)_{16}CH_2OH$ is a relatively nonpolar compound because although it has a polar alcohol functional group, its long nonpolar carbon chain makes it predominantly nonpolar making it most soluble in the nonpolar $C_6H_{14}$. e) $HCl$ is an ionic compound due to it being composed of a cation ($H^+$) and an anion ($Cl^-$) making it most soluble in the polar $H_2O$. f) $C_6H_6$ is a nonpolar compound due to it being composed of only nonpolar $C-H$ bonds making it most soluble in the nonpolar $C_6H_{14}$.
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