Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
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The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Global warming, the hole inside the atmosphere in Oklahoma , california .
Answer:
Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase
Explanation:
The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion
I hope you undestand me
4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
In which region is the substance in both the solid phase and the liquid phase ?
2
