Respuesta :
Answer:
32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.
Explanation:
Volume of cylindrical reservoir = V
Radius of the cylindrical reservoir = r = d/2
d = diameter of the cylindrical reservoir = d =[tex]4.50\times 10^1 m=45 m[/tex]
r = d/2 = 22.5 m
Depth of the reservoir = h = 10.0 m
[tex]V=\pi r^2 h[/tex]
[tex]=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L[/tex]
[tex]1 m^3=1000 l[/tex]
Volume of water cylindrical reservoir : V
Density of water,d = 1 kg/L
Mass of water cylindrical reservoir = m
[tex]m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg[/tex]
1.6 kilogram of fluorine per million kilograms of water. (Given)
Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water
In 1000,000 kg of water = 1.6 kg of fluorine
Then [tex]15,896,250 kg[/tex] of water have x mass of fluorine:
[tex]\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}[/tex]
[tex]x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg[/tex]
15,896,250 kg water of contains mass 25.434 kg of fluorine.
25.434 kg = 25434 g
25,434 grams of fluorine should be added to give 1.6 ppm.
Percentage of fluorine in hydrogen hexafluorosilicate :
Molar mass hydrogen hexafluorosilicate = 144 g/mol
[tex]F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%[/tex]
Total mass of hydrogen hexafluorosilicate = m'
[tex]79.16\%=\frac{25,434 g}{m'}\times 100[/tex]
m' = 32,127.02 g
32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.
