Okay so just add all of them up the fine the x and the y and d the times that by the size of yiurn four head
Answer:
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
Explanation:
First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:
- Ca: 40 g/mole
- F: 19 g/mole
So the molar mass of CaF₂ is:
CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

moles=2.05*10⁻⁵
<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>
Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

mass of CaF₂= 1000 g
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

moles=12.82
<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>
Answer:
The physical properties of a solution are different from those of the pure solvent. ... Colligative properties are those physical properties of solutions of nonvolatile solutes that depend only on the number of particles present in a given amount of solution, not on the nature of those particles.
Answer: C. He didn't measure the weight of the log before it was burned.