for each solute, click the button under the better solvent. | solute | which is the better solvent? | | ---- | ---- | | (chemical structure of a hydrocarbon) | $\ce{h_{2}o}$ (radio button)
$\ce{ch_{3}(ch_{2})_{4}ch_{3}}$ (selected radio button) | | $\ce{cacl_{2}}$ | $\ce{ccl_{4}}$ (radio button)
$\ce{ch_{3}oh}$ (selected radio button) | | $\ce{nh_{4}br}$ | (chemical structure of a hydrocarbon derivative) (radio button)
$\ce{ch_{3}ch_{2}oh}$ (radio button) |

Question

Accepted Answer

# Brief Explanations:
- For the first solute (benzene - a non - polar molecule), the "like dissolves like" principle is applied. Benzene ($C_6H_6$) is non - polar. $H_2O$ is polar, while $CH_3(CH_2)_4CH_3$ (a non - polar hydrocarbon) is a better solvent for benzene.
- For $CaCl_2$ (an ionic compound), $CH_3OH$ (a polar solvent with a hydroxyl group that can interact with ions) is a better solvent than $CCl_4$ (a non - polar solvent). Ionic compounds tend to dissolve in polar solvents.
- For $NH_4Br$ (an ionic compound), although $CH_3CH_2OH$ is polar, the first option (the non - polar benzene - like structure) is incorrect. $CH_3CH_2OH$ (a polar solvent) is a better choice for $NH_4Br$ compared to the non - polar structure shown.

# Answer:
- For the first solute (benzene - like structure): $CH_3(CH_2)_4CH_3$ is the better solvent.
- For $CaCl_2$: $CH_3OH$ is the better solvent.
- For $NH_4Br$: $CH_3CH_2OH$ is the better solvent.

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QUESTION IMAGE

Question

Explanation:

Answer:

solute	which is the better solvent?
$\ce{cacl_{2}}$	$\ce{ccl_{4}}$ (radio button) <br> $\ce{ch_{3}oh}$ (selected radio button)
$\ce{nh_{4}br}$	(chemical structure of a hydrocarbon derivative) (radio button) <br> $\ce{ch_{3}ch_{2}oh}$ (radio button)