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Arisa
15 hours ago
14

Two gases, A and B, are at equilibrium in a sealed cylinder. Individually, gas A is colorless, while gas B is dark colored. The

reaction might be written as: A(g) equilibrium reaction arrow 2 B(g). (a) The cylinder should appear (color or colorless) (b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic) Suppose the reaction equation were written as follows: 2 B(g) equilibrium reaction arrow A(g) (c) Which of these statements would then be true? (Select all that apply.) a The value of K would not change. b The ΔH value would have the same magnitude value but opposite sign. c The K expression would be inverted. d The color of the cylinder would be darker.
Chemistry
1 answer:
Alekssandra [968]14 hours ago
7 0

Answer:

(a) colored

(b) exothermic

(c)

b The ΔH will be equal in value but with reversed sign.

c The K expression will be the inverse.

Explanation:

Let us examine the subsequent reaction at equilibrium.

A(g) ⇄ 2 B(g)

colorless dark colored

(a) The appearance of the cylinder will be (color or colorless)

<pwhen equilibrium="" is="" reached="" a="" blend="" of="" and="" b="" exists="" resulting="" in="" colored="" appearance="" the="" cylinder.="">

(b) If the system's temperature decreases, the cylinder will become much lighter in color. Hence, the reaction is (endothermic or exothermic)

According to Le Chatelier's Principle, a system at equilibrium responds to changes to mitigate their effects. Cooling the system prompts it to elevate temperature by releasing heat, indicating the reaction is endothermic, thus favoring the formation of colorless A.

If the reaction were expressed as: 2 B(g) ⇄ A(g)

(c) What can be inferred about the following statements?

a The K value would remain unchanged. FALSE. The new K will be the reciprocal of the original K.

b The ΔH value will retain the same absolute value but change in sign. TRUE. This aligns with Lavoisier-Laplace Law.

c The K expression would be the reversed form. TRUE. The initial products now act as reactants and the reverse is true.

d The cylinder's color would appear darker. FALSE. Modifying the reaction's expression has no bearing on the equilibrium state.

</pwhen>
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A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T
KiRa [971]

Answer:

\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}

(ii) Volume of each component

\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}

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Assuming mixing doesn't alter the total volume.

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c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
1 day ago
What volume of 0.550 M KBr solution can you make from 100.0 mL of 2.50 M KBr?
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M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
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From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
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14 days ago
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Explanation:

Initial moles of ethanoic acid = 0.020 mol

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The sample solution might be tainted.

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If the nichrome wire is contaminated, sodium impurities could be causing the yellow flame. The wire is initially placed in the flame without the sample to check for such impurities.

The testing solution could also be contaminated, causing it to display a color different from the anticipated shade of the test ion.

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I think the state change illustrated in the diagram is deposition. 
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