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2 years ago

What is the pOH of a solution with an OH- ion concentration of 8.4E-7?

Chemistry

1 answer:

Andreyy892 years ago

6 0

Answer:

Explanation:

The pOH of a solution can be found by using the formula

$pOH = - log [ {OH}^{-} ]$

From the question we have

$pOH = - log(8.4 \times {10}^{ - 7} ) \\ = 6.075720...$

We have the final answer as

Hope this helps you

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Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

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Gd(g) → $Gd^{+}(g) + e^{-}$ , $Gd^{+}(g)$ → $Gd^{2+}(g) + e^{-}$ , $Gd^{2+}(g)$ → $Gd^{3+}(g) + e^{-}$

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Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ $Gd^{+}(g) + e^{-}$

The second ionization energy is the energy required to remove next/second electron from $x^{+}$ ion. The second ionization energy is always higher than the first:

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The third ionization energy is the energy required to remove third electron from $x^{2+}$ ion:

$Gd^{2+}$ (g) → $Gd^{3+}(g) + e^{-}$

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