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 equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky
Explanation:
we have the following balanced reaction:
CaC2 + 2H2O = C2H2 + Ca(OH)2
the value of K for this reaction will be equal to:
K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)
if we multiply the reaction by the value of 6, we have:
6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2
Again, the value of K for this reaction will be equal to:
K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6
For the second reaction:
6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H
The value of K for this reaction:
K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)
we also have:
K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)
Thus:
K1 = K^6 * Ky
It would form a negative ion... it lacks 1 e in its valence shell., it is easier for F to accept an e than to shed all existing 7.
e= electron
Answer:
The answer is North
Explanation:
The direction of the field is taken to be the direction of the force it would exert on a positive test charge.
D) light is reflecting in the direction indicated by T.