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Anit [1.1K]
3 years ago
5

The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea

ctant, pcl5, at equilibrium?
Chemistry
1 answer:
sineoko [7]3 years ago
4 0
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

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Classify unopened root beer as an element, a compound, a homogeneous mixture, or a heterogeneous mixture
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4 0
3 years ago
Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w
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3 0
3 years ago
A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h. The inflow water contains 0.01 mol/m3 of chemical. Chem
Naily [24]

Answer:

C ≈ 1.44 × 10⁻³ mol/m³

Explanation:

The given information are;

The liquid volume the pond can hold = 104 m³

The volume of inflow into the pond = 5 m³/h

The volume of outflow into the pond = 5 m³/h

The concentration of the chemical in the inflow water = 0.01 mol/m³

The concentration of the chemical discharged directly into the water = 0.1 mol/h

The concentration, c_{(inflow)}, of chemical that enters the water through inflow per hour is given as follows;

c_{(inflow)} = 0.01 mol/m³ × 5 m³/h = 0.05 mol/h

The concentration, c_{(discharge)}, of chemical that enters the water through direct discharge per hour is given as follows;

c_{(discharge)} = 0.1 mol/h

The total concentration that enters the pond per hour is given as follows;

c_{(inflow)} + c_{(discharge)} = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h

Whereby the water in the pond properly mixes with the pond, we have;

The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³

C ≈ 1.44 × 10⁻³ mol/m³.

4 0
3 years ago
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