Answer:
Explanation:
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In this case, according to the Henderson-Hasselbach equation, it is possible to write:
![pH=pKa+log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Next, since we are given the pH and the [A–]/[HA] ratio, we can solve for the pKa as shown below:
![pKa=pH-log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pKa%3DpH-log%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Now, we plug in the values to obtain:
Next, Ka is:
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Warm is less heaver then cold air so warm air rise and cold air sinks
The sand on the beach cools down at night then the ocean is because the sand is on dry land.
Answer:
2. Inorganic
Explanation:
All man-made and most carbon-based compounds are inorganic
When NH3 is dissolved in water, it dissociates partially producing NH4+ ions and OH- ions. It has an equation:
NH3 + H2O → NH4+ + OH-
<span>We use the Kb expression to determine the [OH-] concentration,
</span>
<span>Kb = [NH4+] [OH-] /* [NH3] </span>
We can write NH4+ as OH- since they are of equal ratio.
<span>(1.76*10^-5) = [OH-]² / 0.188
</span><span>[OH-]² = 3.3088*10^-6 </span>
<span>[OH-] = 1.819*10^-3 </span>
We calculate for H+ concentration as follows:
<span>[H+] [OH-] = 10^-14 </span>
<span>[H+] = 10^-14 / [OH-] </span>
<span>[H+] = 10^-14 / (1.819*10^-3) </span>
<span>[H+] = 5.50*10^-12 </span>
<span>pH = -log [H+] </span>
<span>pH = -log (5.5*10^-12) </span>
<span>pH = 11.26</span>
