First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)
Let's find the molecular mass of each element.




Now, let's find the mass of each compound.


We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.


Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).


Turn that into a percentage and you get 45.431%.
Hope this helps! :)
Answer:
28.9%
Explanation:
Let's consider the following balanced equation.
2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂
We can establish the following relations:
- The molar mass of Fe₂O₃ is 159.6 g/mol
- 1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
- 1 mole of Fe is in 1 mole of FeS₂
- The molar mass of Fe is 55.84 g/mol
The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

The percent of Fe in 1.25 g of the ore is:

The answer to your question is,
Metalloids. They are a mix of elements that are both metals and non-metals in one.
-Mabel <3
You’ll need to be sure to count all the atoms in each side of the chemical equation.