Soil that is damaged and loses its nutrients is quick to regain nutrients.
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This reaction is called as <span>Chemiluminescence. A type of reaction in which light is produced due to chemical reaction. While in case of Phosphorescence and Fluorescence light is emitted due to absorption of photons.
When dianion of luminol is reacted with Oxygen, a unstable peroxide is formed, and elimination of Nitrogen gas takes place with the de-excitation of electrons from higher state to ground state and the energy is released in the form of Blue Light. </span>
When dianion of luminol is reacted with Oxygen, a unstable peroxide is formed, and elimination of Nitrogen gas takes place with the de-excitation of electrons from higher state to ground state and the energy is released in the form of Blue Light. </span>
In this problem we need to identify the relative molar amounts of:
NH₄⁺ , ClO₄⁻ , NH₃ , HClO₄ , H₃O⁺ , H₂O, OH⁻
- The salt is completely ionized through this equation:
NH₄ClO₄ → NH₄⁺ + ClO₄⁻
0.01 M 0.01 M 0.01 M
So:
[ClO₄⁻] = 0.01 M (conjugate base of strong acid)
[HClO₄] = 0 M
Note that: NH₄⁺ is weak acid ionized as follow:
NH₄⁺ ⇄ NH₃ + H⁺
Ka =![\frac{[ H^{+}][ NH_{3}]}{ [NH_{4} ^{+}] }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B%20H%5E%7B%2B%7D%5D%5B%20NH_%7B3%7D%5D%7D%7B%20%5BNH_%7B4%7D%20%5E%7B%2B%7D%5D%20%20%7D%20)
= 5.6 x 10⁻¹⁰
[H⁺] = [NH₃] = x and [NH₄⁺] = 0.01 - x (because Ka is very small so x will be neglected compared to 0.01)
Ka = 5.6 x 10⁻¹⁰ =
[H⁺] = [NH₃] = x = 2.37 x 10⁻⁶ M
[NH₄⁺] = 0.01 M
[OH⁻] =![\frac{1 X 10^{-14} }{[H^{+} ]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%20X%20%2010%5E%7B-14%7D%20%7D%7B%5BH%5E%7B%2B%7D%20%5D%7D%20)
= 4.22 X 10⁻⁹ M
because water is the solvent & this is very dilute solution:
[H₂O] =
= 55.6 M H₂O
NH₄⁺ , ClO₄⁻ , NH₃ , HClO₄ , H₃O⁺ , H₂O, OH⁻
- The salt is completely ionized through this equation:
NH₄ClO₄ → NH₄⁺ + ClO₄⁻
0.01 M 0.01 M 0.01 M
So:
[ClO₄⁻] = 0.01 M (conjugate base of strong acid)
[HClO₄] = 0 M
Note that: NH₄⁺ is weak acid ionized as follow:
NH₄⁺ ⇄ NH₃ + H⁺
Ka =
[H⁺] = [NH₃] = x and [NH₄⁺] = 0.01 - x (because Ka is very small so x will be neglected compared to 0.01)
Ka = 5.6 x 10⁻¹⁰ =
[H⁺] = [NH₃] = x = 2.37 x 10⁻⁶ M
[NH₄⁺] = 0.01 M
[OH⁻] =
because water is the solvent & this is very dilute solution:
[H₂O] =