C. Saturated

When something is Saturated, it means that it is unable to absorb or dissolve any more of a solute at the given temperature and pressure.

6 0

4 years ago

Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. Given the reaction CH3COOH(aq) ↔ CH3COO−(aq) + H+(aq)

likoan [24]

Answer:

The correct answer is: 1.035 x 10⁻³ M

Explanation:

The dissociation equilibrium for acetic acid (CH₃COOH) is the following:

CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵

The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):

$Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}$

Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:

$\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}$

⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M

8 0

3 years ago

A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac

kicyunya [14]

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3 0

3 years ago

2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)

ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

$n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}$

2. Calculate the mass of KCl

$m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}$

3. Prepare the solution

Measure out 224 g of KCl.
Dissolve the KCl in a few hundred millilitres of distilled water.
Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.

8 0

3 years ago