Chemistry, 10.07.2019 17:00 kylieweeks052704

Part a choose the pair of substances that are most likely to form a homogeneous solution. choose the pair of substances that are most likely to form a homogeneous solution. co and c6h14 nh2ch3 and ch4 cbr4 and sf2 nf3 and so2 none of the pairs above will form a homogeneous solution.

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A solution on which the solvent and solute are present in a single phase after dissolution is called homogeneous phase. Such solution remains uniform throughout.

The given pairs of substances are:

a. $CO and C_{6}H_{14}$ : CO is a polar gas and $C_{6}H_{14}$ is a non polar liquid. Therefore, this solution will not be homogeneous.

b. $NH_{2}CH_{3}and CH_{4}$ : $NH_{2}CH_{3}$ is a polar liquid while methane is a non polar gas. So, this solution will not be homogeneous.

c. $CBr_{4} and SF_{2}$ : $CBr_{4}$ is a non polar solid while $SF_{2}$ is a polar gas. So, they cannot form homogeneous solution.

d. $NF_{3} and SO_{2}$ : $NF_{3}$ and $SO_{2}$ are both gases. So, they form a uniform homogeneous solution. Both can uniformly interact as they both are polar compounds.

Therefore the correct answer will be $NF_{3} and SO_{2}$

Answer from: valeriegarcia12

$\boxed{{\text{N}}{{\text{F}}_{\text{3}}}{\text{ and S}}{{\text{O}}_{\text{2}}}}$ is the pair that is most likely to form a homogeneous solution.

Further Explanation:

Homogeneous solution consists of only one phase. It is formed when the substances are completely dissolved in each other. Solubility is the property of any substance that makes it soluble in other substances. It is governed by the principle “like dissolves like”.

CO and ${{\text{C}}_6}{{\text{H}}_{14}}$

CO is a polar ionic compound whereas ${{\text{C}}_6}{{\text{H}}_{14}}$ is a non-polar hydrocarbon. Both will not dissolve in each other. So a homogeneous solution will not be formed by CO and ${{\text{C}}_6}{{\text{H}}_{14}}$ .

${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}$ and ${\text{C}}{{\text{H}}_{\text{4}}}$

${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}$ is a polar compound due to the high electronegativity of nitrogen. But ${\text{C}}{{\text{H}}_{\text{4}}}$ is a non-polar covalent compound. Both will not dissolve in each other. So a homogeneous solution will not be formed by ${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}$ and ${\text{C}}{{\text{H}}_{\text{4}}}$ .

${\text{CB}}{{\text{r}}_{\text{4}}}$ and ${\text{S}}{{\text{F}}_2}$

${\text{CB}}{{\text{r}}_{\text{4}}}$ is a non-polar covalent compound whereas ${\text{S}}{{\text{F}}_2}$ is a polar covalent compound. Both will not dissolve in each other. So a homogeneous solution will not be formed by ${\text{CB}}{{\text{r}}_{\text{4}}}$ and ${\text{S}}{{\text{F}}_2}$ .

${\text{N}}{{\text{F}}_{\text{3}}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$

${\text{N}}{{\text{F}}_{\text{3}}}$ is slightly polar in nature and ${\text{S}}{{\text{O}}_{\text{2}}}$ is also polar in nature. Also, both are gases. So both will dissolve in each other and therefore a homogeneous solution is formed by ${\text{N}}{{\text{F}}_{\text{3}}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$ .

Learn more:

1. Characteristics of a mixture:

2. Example of physical change:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: homogeneous solution, NF3, SO2, CH3NH2, CBr4, SF2, CH4, CO, C6H14, polar, non-polar, ionic compound, solubility, like dissolves like, hydrocarbon.

Answer from: Quest

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.

Answer from: Quest

Um i think answer is a

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