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Annette [7]
3 years ago
7

A reaction begins with 1.00 mol of HCl in a 4.00 liter container at a certain temperature.

Chemistry
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

[HCl]_{eq}=0.05M

Explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action at equilibrium results:

Kc=\frac{[H_2]_{eq}[Cl_2]_{eq}}{[HCl]^2_{eq}}

Next, in terms of the change x due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:

4.00=\frac{(x)(x)}{(0.25-2x)^2}

Thus, solving for x via quadratic equation or solver, the following results are obtained:

x_1=0.1M\\x_2=0.17M

Clearly, the solution is x=0.1M as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:

[HCl]_{eq}=0.25M-2*0.1M

[HCl]_{eq}=0.05M

Best regards.

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The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

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1 year ago
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