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elena55 [62]
3 years ago
10

A sample of solid NH 4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (at

m) of ammonia, assuming that some solid NH 4HS remains.
Chemistry
1 answer:
UNO [17]3 years ago
5 0

Answer:

The answer to the question is

The equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains 0.26 atm.

Explanation:

To solve the question, we write out the chemical equation as follows

NH₄HS (s) ⇄ NH₃ (g) + H₂S (g)

From the above equation, it is observed that only the gaseous products contribute to the partial pressure

Kp =PNH₃·PH₂S where at Kp = 0.070 and PNH₃, PH₂S are the partial pressures of the gases

However since the number of moles of both gases are equal, therefore by Avogadro's law PNH₃ = PH₂S

Then PNH₃  = √(0.07) = PH₂S = 0.2645 atm. ≅ 0.26 atm.

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explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
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The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
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The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

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Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

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Learn more about Graham's law of diffusion:

brainly.com/question/14004529

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2 years ago
Which of the following ions is formed when a base is dissolved in a solution?
Aneli [31]

Answer:

\huge\boxed{\sf OH-}

Explanation:

<u>According to Arrhenius concept of acid and base:</u>

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

<u>For example:</u>

NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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2 years ago
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