To find AH°rxn, we use the following equation:
What we're going to do is to sum the enthalpy of the products and then substract with the enthalpy of the reactives:
As you can see, we need to multiply by the coefficients of the reaction.
Now, just replace the values of the table:
So the answer is -822.2kJ/mol.
For b:
Now, just replace the values of the table:
The answer for b is -1036kJ/mol.
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:
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>