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vlada-n [284]
2 years ago
6

A reaction vessel contains nh3, n2, and h2 at equilibrium at a certain temperature. the equilibrium concentrations are [n2] = 0.

31 m, [h2] = 1.51 m, and [nh3] = 0.75 m. calculate the equilibrium constant, kc, if the reaction is represented as
Chemistry
1 answer:
Nat2105 [25]2 years ago
5 0
The balanced equation for the reaction is,
          N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Since the given concentrations are taken at the equilibrium state, we can use those directly for the calculation.

Kc = [NH₃(g)]² / ([N₂(g)] x [H₂(g)]³)
Kc = (0.75 M)² / ((0.31 M) x (<span>1.51 M</span>)³)
Kc = 0.527 M⁻²
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A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
2 years ago
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 276 mL .
garri49 [273]
Mass = ?

Density = 2.70 g/mL

Volume = 276 mL

Therefore:

D = m / V

2.70 = m / 276

m = 2.70 x 276

m = 745.2 g

5 0
2 years ago
Calculate the mass of MNO2 needed to produce 25.0 g of Cl2
Vedmedyk [2.9K]
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O. 
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>
7 0
2 years ago
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Which statement is true of a reversible reaction at equilibrium?
Vedmedyk [2.9K]

Answer:

D.

The concentration of reactants and the concentration of products are constant.

Explanation:

pls mark as brainliest

7 0
2 years ago
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Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46

<u>Explanation:</u>

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

                    4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)

<u>Initial:</u>               2.2          2.4

<u>At eqllm:</u>        2.2-4x      2.4-3x         2x        6x

Evaluating the value of 'x':  

\Rightarrow 2x=0.99\\\\x=0.495

So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm

Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm

Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of K_p for above equation follows:

K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}  

Putting values in above equation, we get:

K_p=\frac{(0.99)^2\times (1.98)^6}{(0.22)^4\times (0.915)^3}\\\\K_p=32908.46

Hence, the pressure equilibrium constant for the reaction is 32908.46

5 0
2 years ago
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