Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
I would say soil would be your best option. This is because out of all these, soil collects a lot of different substances and could have easily absorbed something that then killed the organism.
I think the answer is B. static charge. I hope this helps! :)
Answer:
the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>
Explanation:
Carbonate-bicarbonate is:
HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25
Using Henderson-Hasselbalach formula:
pH = pka + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
7,2 = 10,25 + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
8,91x10⁻⁴ =
<em>(1)</em>
Also:
0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>
Replacing (2) in 1:
<em>[HCO₃⁻] = 0,03996 M</em>
Thus:
<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>
I hope it helps.