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Arisa
3 months 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 [3K]3 months 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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There are eight isotopes of lawrencium, and each one is radioactive. I will perform calculations for ²⁶²Lr, which has a half-life of 3.6 hours. Let A₀ denote the initial quantity of lawrencium. The remaining amount after one half-life is ½A₀. After two half-lives, the quantity left is ½ × ½A₀ = (½)²A₀. After three half-lives, the remaining amount is ½ × (½)²A₀ = (½)³A₀, and this pattern continues. We can generalize the formula for the amount remaining: A = A₀(½)ⁿ, where n indicates the number of half-lives. Data: A₀ = 5 g,  t = 30 min. Calculations: (a) Convert the half-life to minutes. (b) Compute n. (c) Determine A.
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2 months ago
For each group of compounds listed, which is the strongest acid? I. HIO2, HIO3, HIO4 II. H2Se, H2S, H3As III. HPO2, HClO2, HBrO2
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Answer:

In group I, the most potent acid is HIO_4.

For group II, the strongest acid is H_2Se.

In group III, the strongest acid identified is HClO_2.

Explanation:

Group I:

The strength of an acid is influenced by the stability of the conjugate base formed after H+ ion loss.

From the acids provided, the following conjugate bases result upon H+ release:

HIO_2 \rightarrow H^+ + IO_2^-\\HIO_3 \rightarrow H^+ + IO_3^-\\HIO_4 \rightarrow H^+ + IO_4^-

In IO_4^-, the negative charge is spread across three electronegative oxygen atoms, rendering it more stable compared to the other two conjugate bases.

Thus, HIO_4 becomes the strongest acid in this group.

Group II:

The H-Se bond is less robust than H-S bonds, making it more acidic since breaking the H-Se bond more easily releases H+.

For H_3As, arsenic (As) has lower electronegativity than selenium (Se). Due to the higher electronegativity of Se, it attracts the electrons of the H-Se bond, facilitating the removal of H+.

Consequently, H_2Se is the strongest acid in this grouping.

Group III:

Acidity is also impacted by the distance between bonds and the central atom's electronegativity.

In this scenario, electronegativity takes precedence over bond distance.

Among the elements listed, chlorine (Cl) exhibits the highest electronegativity. This significant electronegativity causes the electrons to be pulled toward Cl, making the release of H+ easier compared to the other two.

Therefore, HClO_2 is the strongest acid in this set.

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3 months ago
Potassium and iodine have formed a bond. Prior to this the potassium gave up an electron. It became a _____. positive ion neutra
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2 months ago
Read 2 more answers
Question 1: Which of the following statement is false about conjugated systems?
Tems11 [2777]

Answer:

A conjugated system can solely consist of two alkenes: Incorrect

B. A cumulated diene is not as stable as an isolated diene.

Incorrect

Explanation:

A. A conjugated system has greater stability compared to an unconjugated system.: Correct

Owing to resonance, a conjugated system exhibits increased stability.

B. The s-trans form is preferred over the s-cis form.: Correct

Because of repulsion in the s-cis form, they are less stable.

C. A conjugated system has multiple resonance forms.Correct

This contributes to their stability.

D. A conjugated system can only comprise two alkenes: Incorrect

A conjugated system may consist of alternating double and triple bonds.

2) A trans alkene liberates less heat during hydrogenation compared to a cis alkene.

Correct

Trans alkenes are more stable than their cis counterparts, resulting in a lower heat of hydrogenation for trans alkenes.

B. A cumulated diene is less stable than an isolated diene.

Incorrect

C. Numerous products can potentially form when a conjugated diene reacts with HBr. Correct

This can result in either 1,2 or 1,4 addition.

D. Conjugated systems occur only between sp2-hybridized atoms.

Correct

Conjugation enables resonance, which is feasible in sp2 systems.

E. Conjugated double bonds rely on unhybridized p-orbitals that align with one another.

Correct

The double bond forms as a pi bond due to the lateral overlap of unhybridized p orbitals.

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