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.
Answer:
Explanation:
density = mass / volume = 85 / 92 = 0.92 g/cm3
The answer is B or number 2
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