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DiKsa [7]
3 years ago
15

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

f F- must be added to a cylindrical water reservoir having a diameter of 3.36x10^2 m and a depth of 21.80 m? B) How many grams of sodium fluoride, NaF, contain this much fluoride?
Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

a) <u>1.740 g</u> 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 =

\pi  r^{2}h ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

r=\frac{d}{2}

=\frac{3.36x10^{2}}{2}

= 168 m

Volume =

=\frac{22\times 168^{2}\times 21.80}{7}

=1.93\times 10^{6} m^{3}

2.Convert ppm to g/m3 and Solve for mass of F-

1ppm = 1g/m^{3}

0.9ppm = 0.9g/m^{3}

Because both ppm and g/m3 are same quantity .

g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}

0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}

mass\ of\ F- =1.740g

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>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)

2Na+F_{2}\rightarrow 2NaF

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 =

\frac {84}{38}\times 1.74

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

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Answer:

a)  molar composition of this gas on both a wet and a dry basis are

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4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

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0.18           3× 0.18

moles        moles

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