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>
Higher probability of loss. Chorionic villus sampling (CVS) and Amniocentesis (AC). The prenatal diagnosis technique can be done earlier in fetal development CVS (first trimester --> 10-13 weeks). AC (second trimester --> 16-20 weeks)
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Answer:
option D= Gold (I) nitride
Explanation:
The name of the given compound is gold(I) nitride.
Molar mass can be determine by following way:
molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)
molar mass Au3N = (196.97 × 3 ) + ( 14 )
molar mass of Au3N = 590.91 g/mol + 14 g/mol
molar mass of Au3N = 604.91 g/mol
The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.
Au3N
3(1+) + (-3) = 0
+3 - 3 = 0
0 = 0
The overall charge is 0, the compound will be neutral.
Answer:
The correct answer is c) 134L
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.
1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K
V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm
V = 133, 86828 l