Respuesta :
Answer is: a. c6h14 and c10h20.
This pair will most likely form a homogeneous solution because they are both nonpolar substances and "like dissolves like".
Other pairs will not form homogeneous solution because nonpolar substances have low solubility in polar or ionic substances (for example LiBr is ionic and C₅H₁₂ is nonpolar).
This pair will most likely form a homogeneous solution because they are both nonpolar substances and "like dissolves like".
Other pairs will not form homogeneous solution because nonpolar substances have low solubility in polar or ionic substances (for example LiBr is ionic and C₅H₁₂ is nonpolar).
Homogeneous solution will be formed by [tex]\boxed{{\text{a}}{\text{. }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}{\text{ and }}{{\text{C}}_{{\text{10}}}}{{\text{H}}_{{\text{20}}}}}[/tex]
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”.
a. [tex]{{\text{C}}_6}{{\text{H}}_{14}}[/tex] and [tex]{{\text{C}}_{10}}{{\text{H}}_{20}}[/tex]
Both are non-polar hydrocarbons and completely dissolve in each other. So a homogeneous solution is formed by [tex]{{\text{C}}_6}{{\text{H}}_{14}}[/tex] and [tex]{{\text{C}}_{10}}{{\text{H}}_{20}}[/tex].
b. LiBr and [tex]{{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}[/tex]
LiBr is an ionic compound whereas [tex]{{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}[/tex] is a non-polar hydrocarbon. Both will not dissolve in each other. So homogeneous solution will not be formed by LiBr and [tex]{{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}[/tex].
c. [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}[/tex] and [tex]{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}[/tex]
[tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}[/tex] is a symmetric molecule and non-polar in nature whereas [tex]{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}[/tex] is a polar molecule. Both will not dissolve in each other. So homogeneous solution will not be formed by [tex]{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}[/tex] and [tex]{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}[/tex].
d. [tex]{{\text{C}}_6}{{\text{H}}_{14}}[/tex] and [tex]{{\text{H}}_2}{\text{O}}[/tex]
[tex]{{\text{C}}_6}{{\text{H}}_{14}}[/tex] is a non-polar hydrocarbon whereas [tex]{{\text{H}}_2}{\text{O}}[/tex] is a polar molecule. Both will not dissolve in each other. So homogeneous solution will not be formed by [tex]{{\text{C}}_6}{{\text{H}}_{14}}[/tex] and [tex]{{\text{H}}_2}{\text{O}}[/tex].
Learn more:
1. Characteristics of a mixture: https://brainly.com/question/1917079
2. Example of physical change: https://brainly.com/question/1119909
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Solutions
Keywords: homogeneous solution, LiBr, H2O, C6H14, N2O4, C10H20, NH4Cl, C5H12, polar, non-polar, symmetric, ionic compound, solubility, like dissolves like, hydrocarbon,
