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agasfer [191]
2 years ago
7

Calculate the concentration of clo2− at equilibrium if the initial concentration of hclo2 is 2. 25×10^−2 m.

Chemistry
1 answer:
jek_recluse [69]2 years ago
6 0

The concentration of ClO₂⁻ at equilibrium if the initial concentration of  HClO₂ is 0.0654.

<h3>What is concentration?</h3>

The concentration of any substance is the quantity of that substance in per square of the space or container.

The reaction is

HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻

The pH is 0.454 M

Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]

2. 25 × 10⁻² m = [x][x] / 0.454-x]

2 + 0.011 - 0.004994 = 0

solve the quadratic equation

x = 0.0654 = [H3O+] = [ClO2-]

pH = -log (H3O+)

pH = -log(0.0654)

pH = 1.2

equilibrium concentrations of

[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M

[ClO2- ] = x = 0.0654

Thus, the equilibrium concentrations  is 0.0654.

To learn more about concentration, refer to the link:

brainly.com/question/16645766

#SPJ4

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Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm e
Delicious77 [7]

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9  L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= 0.789 g/cm^3=0.789 g/mL

1 cm^3=1 mL

m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g

Mass percent

(w/w)\%=\frac{m}{m+M}\times 100

\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%

Molality :

m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

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