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Mariulka [41]
3 years ago
12

What is the [OH-] in a solution with a POH of 7.86? [OH-] = [?] x 10!?)

Chemistry
1 answer:
gogolik [260]3 years ago
5 0
<h3>Answer:</h3>

1.380 × 10^-8 M

<h3>Explanation:</h3>

The pH is the measure of alkalinity or acidity of a substance.

It is the negative logarithm of H+ ions concentration [H+].

pH = -log[H+]

pOH on the other hand is negative logarith of OH- concentration [OH-].

pOH = -log[OH-]

In this case;

pOH = 7.86

We can calculate the [OH-]

pOH = -log[OH-]

But, pOH = 7.86

pOH = -log[OH-]

-log[OH-] = 7.86

[OH-] = Antilog -7.86

       = 1.380 × 10^-8 M

Therefore, [OH-] is 1.380 × 10^-8 M

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For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--&gt; N2(g) + 3H2(g) K, 0.83 1. When
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Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

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Where K is defined as:

K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

Q = \frac{1atm*{1atm}^3}{1atm^2} = 1

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

<em />

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What effect does rapid exhalation of CO2 during exercise have on the concentration of H2CO3 in the blood?
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Hope it helped,


BioTeacher101

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