Answer:
The answer that completes the question are in BOLD:
At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.
Explanation:
In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.
The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.
When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.
N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.
<span>Argon and chlorine are both gases at room temperature because they are non-metals.</span>
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
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Mixture......................
