Answer:
Explanation:
Hello,
In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:
![\% ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
![[H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B%5C%25%20ionization%2A%5BHA%5D%7D%7B100%5C%25%7D%20%3D%5Cfrac%7B4%5C%25%2A0.30M%7D%7B100%5C%25%7D%3D0.012M)
Therefore the Ka is computed by using the equilibrium expression:
![Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%5Cfrac%7B0.012M%2A0.012M%7D%7B0.30M-0.012M%7D%5C%5C%20%5C%5CKa%3D5x10%5E%7B-4%7D)
And the pH:
![pH=-log([H^+])=-log(0.012)\\\\pH=1.92](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29%3D-log%280.012%29%5C%5C%5C%5CpH%3D1.92)
Regards.
Explanation:
it's true that Lysosomes break down old and worn out cell parts....
Explanation:
science can answer any and all questions
a. 30 moles of H₂O
b. 2.33 moles of N₂
<h3>Further explanation</h3>
Given
a. 20 moles of NH₃
b. 3.5 moles of O₂
Required
a. moles of H₂O
b. moles of N₂
Solution
Reaction
4NH₃+3O₂⇒2N₂+6H₂O
a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :
=6/4 x mol NH₃
= 6/4 x 20 moles
= 30 moles
b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :
=2/3 x mol O₂
= 2/3 x 3.5 moles
= 2.33 moles
Answer:
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