C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) =
0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) =
2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch
equation:
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>
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Answer:

Explanation:
The pressure at the bottom of the tank is:


The force exerted on the circular bottom is:
![F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}](https://tex.z-dn.net/?f=F%3D%2873581.921%5C%2CPa%29%5Ccdot%20%28%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%5Ccdot%20%5B%2812%5C%2Cft%29%5Ccdot%20%28%5Cfrac%7B0.305%5C%2Cm%7D%7B1%5C%2Cft%7D%20%29%5D%5E%7B2%7D)

I hope this answers helps it’s D