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Vadim26 [7]
3 years ago
11

Balance the chemical equation. (Blank) N2O3 → (Blank) N2 + (Black) O2

Chemistry
2 answers:
Anna71 [15]3 years ago
6 0

  The balanced equation is as below

 2 N₂O₃ →  2N₂ + 3O₂

<u><em> Explanation</em></u>

According  to the law  of mass conservation the  number of atoms  in reactants side must  be equal  the to the number of atoms in products  side.

 Therefore the  balanced reaction  above is balanced  since the number of atoms  in reactants side  are equal to number of atoms in products side.

For  there 4 atoms  of N   in reactants  side and  4 in products side.


Crank3 years ago
4 0

The balanced equation is  2 N₂O₃ →  2N₂ + 3O₂

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Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

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i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

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According to the reaction represented by the unbalanced equation above, how many moles of so2(g) are required to react completel
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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.

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