Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
The answer is b i just took the test
The flotation process used in metallurgy involves the separation of gangue from ore.
<h3>How does the flotation process work?</h3>
The various wettability qualities of a material's surface are the foundation of flotation operations. The basic principles of flotation are quite similar to those of a sink and float process, where the materials' relative densities to the medium in which they are deposited determine the basis of the separation.
<h3>What is the process of separating minerals from gangue known as?</h3>
Mineral processing, mineral dressing, or ore dressing are all terms for the process of separating minerals from gangue. It is an important and frequently necessary part of mining. Depending on the type of minerals used, the process may be difficult.
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The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
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