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It is recommended that drinking water contain 1.6 ppm fluoride (F−) for prevention of tooth decay. Consider a reservoir with a diameter of 4.70 ✕ 102 m and a depth of 14.2 m. (The volume is πr2h, where r is the radius and h is the height.) How many grams of F− should be added to give 1.6 ppm?

Respuesta :

Answer:

total amount of fluoride 3.6 *10^6 g

Explanation:

we know that

volume of reservoir [tex]= \pi r^2 h[/tex]

                                  =[tex]\pi* [\frac{4.70*10^2}{2}]^2 *14.2[/tex]

                                  = 2.25*10^6 m^3

denso=ity of water = 1000 kg/m3

mass of water = 1000*2.25*10^6

                         = 2.25*10^9 kg = 2.25*10^12 g

fluoride 1.6 ppm

total amount of fluoride [tex]= \frac{1.6}{10^6} * 2.25*10^{12} g[/tex]

                                         = 3.6 *10^6 g