2 years ago

Question 7 (2 points)

Chemistry

1 answer:

serg [7]2 years ago

4 0

Answer:

C. decrease the velocity.

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c

malfutka [58]

Answer:

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

$Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)$

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

$[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\$

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Best regards.

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2 years ago