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NeTakaya
3 years ago
7

Classify these definitions as that of an Arrhenius acid, an Arrhenius base, or other. Arrhenius acid definition Arrhenius base d

efinition OtherA substance that can donate a hydrogen ion to another substanceA substance that produces an excess of hydroxide ion (-OH) in aqueous solution.A substance that produces an excess of hydrogen ion (H+) in aqueous solution
Chemistry
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

       This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

  • Sodium hydroxide NaOH
  • Potassium hydroxide KOH

A substance that produces an excess of hydrogen ion (H+) in aqueous solution

      This is an arrhenius Acid

An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.

Examples are;

  • Hydrochloric acid HCl
  • Hydroiodic acid HI
  • Hydrobromic acid HBr
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

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The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

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F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

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Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

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C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

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