Copper(II) fluoride contains 37.42% F by mass. Calculate the mass of fluorine (in g) in 55.5 g of copper(II) fluoride.
1 answer:
Answer:
There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride
Explanation:
x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen
37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.
So, 100 g of copper (II) fluoride contains 37.42 g of F
55.5 g of copper (II) fluoride contains g of F or 20.8 g of F
Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.
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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:
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