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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Answer: The coefficient we should change first is for .
Explanation:
The given reaction equation is as follows.
Here, number of atoms present on reactant side are as follows.
- Al = 1
- Ni = 1
= 2
Number of atoms present on the product side are as follows.
- Al = 1
- Ni = 1
To balance this equation, multiply Al by 2 and by 3 on reactant side. Also, multiply
by 2 and Ni by 3 on the product side.
Hence, the equation can be rewritten as follows.
Now, number of atoms present on reactant side are as follows.
- Al = 2
- Ni = 3
Number of atoms present on product side are as follows.
- Al = 2
- Ni = 3
Since, the atoms on both reactant and product side are same. Hence, it is now a balanced chemical equation.
Thus, we can conclude that the coefficient we should change first is for .