Respuesta :
Answer:
a) 1.740 g of F- must be added to a cylindrical water reservoir
b) Grams of sodium fluoride, NaF, that contain this much fluoride:
3.84 g
Explanation:
Step 1. calculate the volume of the tank:
Volume of cylinder =
[tex]\pi r^{2}h[/tex] ,
Here r = radius of the cylinder = d/2
h = depth = 21.80m
[tex]r=\frac{d}{2}[/tex]
[tex]=\frac{3.36x10^{2}}{2}[/tex]
= 168 m
Volume =
[tex]=\frac{22\times 168^{2}\times 21.80}{7}[/tex]
[tex]=1.93\times 10^{6} m^{3}[/tex]
2.Convert ppm to g/m3 and Solve for mass of F-
[tex]1ppm = 1g/m^{3}[/tex]
[tex]0.9ppm = 0.9g/m^{3}[/tex]
Because both ppm and g/m3 are same quantity .
[tex]g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}[/tex]
[tex]0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}[/tex]
[tex]mass\ of\ F- =1.740g[/tex]
mass of F- required = 1.740 g
3. Apply mole concept to calculate grams of sodium fluoride produced
mass of 1 mole of F2 = 38 g
mass of 1 mole of NaF = 42 g
(from periodic table calculate molar mass)
[tex]2Na+F_{2}\rightarrow 2NaF[/tex]
Here 1 mole of F2 produce = 2 mole of NaF
So,
38 g of F2 produce = 2 x 42 g of NaF
38 g of F2 produce = 84 g of NaF
1 g of F2 produce = 84/38 g of NaF
1.74 g F2 produce =
[tex]\frac {84}{38}\times 1.74[/tex]
1.74 g F2 produce = 3.84 g of NaF
3.84 g of NaF is produced
