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lisov135 [29]
3 years ago
11

Choose the pair of substances that are most likely to form a homogeneous solution.

Chemistry
2 answers:
oksano4ka [1.4K]3 years ago
7 0
Answer is: <span>a. c6h14 and c10h20.
This pair will </span>most likely form a homogeneous solution because they are both nonpolar substances and "li<span>ke 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 C</span>₅H₁₂ is nonpolar).
Masja [62]3 years ago
6 0

Homogeneous solution will be formed by \boxed{{\text{a}}{\text{. }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}{\text{ and }}{{\text{C}}_{{\text{10}}}}{{\text{H}}_{{\text{20}}}}}

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. {{\text{C}}_6}{{\text{H}}_{14}} and {{\text{C}}_{10}}{{\text{H}}_{20}}

Both are non-polar hydrocarbons and completely dissolve in each other. So a homogeneous solution is formed by {{\text{C}}_6}{{\text{H}}_{14}} and {{\text{C}}_{10}}{{\text{H}}_{20}}.

b. LiBr and {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}

LiBr is an ionic compound whereas {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}} is a non-polar hydrocarbon. Both will not dissolve in each other. So homogeneous solution will not be formed by LiBr and {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}.

c. {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} and {\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}

{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} is a symmetric molecule and non-polar in nature whereas {\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}} is a polar molecule. Both will not dissolve in each other. So homogeneous solution will not be formed by {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} and {\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}.

d. {{\text{C}}_6}{{\text{H}}_{14}} and {{\text{H}}_2}{\text{O}}

{{\text{C}}_6}{{\text{H}}_{14}} is a non-polar hydrocarbon whereas {{\text{H}}_2}{\text{O}} is a polar molecule. Both will not dissolve in each other. So homogeneous solution will not be formed by {{\text{C}}_6}{{\text{H}}_{14}} and {{\text{H}}_2}{\text{O}}.

Learn more:

1. Characteristics of a mixture: brainly.com/question/1917079

2. Example of physical change: 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,  

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At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g
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Answer:

Total pressure at equilibrium is 0.2798atm.

Explanation:

For the reaction:

H₂S(g) ⇄ H₂(g) + S(g)

Kp is defined as:

Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834

If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:

H₂S(g): 0.150atm - x

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Replacing in Kp:

\frac{X*X}{0.150atm-X} = 0.834

X² = 0.1251 - 0.834X

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Solving for X:

X = -0.964 → False solution: There is no negative pressures

X = 0.1298

Thus, pressures are:

H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>

H₂(g): <em>0.1298atm</em>

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Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

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(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

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\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

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3 years ago
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mariarad [96]

Answer: The volume of sample at 400 K is 285cm^3

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T     (At constant pressure and number of moles)

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V_2 = final volume of gas  = ?

T_1 = initial temperature of gas  = 360 K

T_2 = final temperature of gas  = 400 K

Putting in the values we get:

{\frac{257}{360}=\frac{V_2}{400}

V_2=285cm^3

Thus volume of sample at 400 K is 285cm^3

6 0
3 years ago
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