A substance above its critical pressure is <span>supercritical.
I hope this helps you! Good luck :)</span>
Acquired because you don't inherit a "cut off" finger. You do it yourself.....if your clumsy enough
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
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<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
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<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
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<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Kc= (nh4)
--------
(nh3) + (h2o)
