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ella [17]
4 years ago
10

You are starting with 2 pounds of 30% acetic acid solution. You want to dilute

Chemistry
1 answer:
Tpy6a [65]4 years ago
8 0

Answer:

The mass of water to be added is 2 pounds

Explanation:

The given parameters are;

The mass of the given solution = 2 pounds

The concentration of the given solution = 30%

The desired concentration of the solution = 15%

The mass, m of the acetic acid in the given solution = 30% × 2 pounds

m = 30/100 × 2 pounds = 0.6 pounds

To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;

15% of X = 0.6 pounds

15/100 × X = 3/20 × 0.6 pounds

∴ The mass of the solution required  X = 0.6 × 20/3 = 4 pounds

The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds

Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.

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Answer:

Kc = 3.1x10²

Explanation:

At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:

aA + bB ⇄ cC + dD

Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}

Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).

For the reaction given, let's make an equilibrium chart:

Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)

1.1*10⁻³       8.2*10⁻⁴           0                  <em> Initial</em>

  -x               -x                  +x                  <em>Reacts</em> (stoichiometry is 1:1:1)

1.1*10⁻³ -x   8.2*10⁻⁴ -x       x                 <em>  Equilibrium</em>

x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:

[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M

[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M

[FeSCN⁺²] = 1.8*10⁻⁴ M

Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])

Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)

Kc = 306 = 3.1x10²

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A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution
mariarad [96]

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Explanation:

pH=-log [H+]

3.26 = -log [H+]

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K_a=\frac{[x][x]}{[0.020-x]}

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Answer:

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Explanation:

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