Hey there, The answer is A. Free electrons.
Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.
Explanation:
The equation for the reaction will be as follows:

By Stoichiometry,
1 mole of
gives 2 moles of
and 1 mole of 
Thus if solubility of
is s moles/liter, solubility of
is s moles/liter and solubility of
is 2s moles/liter
Therefore,
![K_sp=[Ba^{2+}][F^{-}]^2](https://tex.z-dn.net/?f=K_sp%3D%5BBa%5E%7B2%2B%7D%5D%5BF%5E%7B-%7D%5D%5E2)
![2.45\times 10^{-5}=[s][2s]^2](https://tex.z-dn.net/?f=2.45%5Ctimes%2010%5E%7B-5%7D%3D%5Bs%5D%5B2s%5D%5E2)



Thus the molar solubility of barium fluoride is 0.0183 moles/liter.
When y equals 5, x is 104.3
When y equals 3 then x is 108.3
<em><u>Solution:</u></em>
<em><u>Given expression is:</u></em>

<h3><u>If y equals 5 what is x ?</u></h3>
Substitute y = 5 in given expression
5 = 57.15 - 0.5(x)
5 = 57.15 - 0.5x
0.5x = 57.15 - 5
0.5x = 52.15
Divide both sides by 0.5
x = 104.3
Thus when y equals 5, x is 104.3
<h3><u>If y = 3 what is x ?</u></h3>
Substitute y = 3 in given expression
3 = 57.15 - 0.5(x)
3 = 57.15 - 0.5x
0.5x = 57.15 - 3
0.5x = 54.15
Divide both sides by 0.5
x = 108.3
Thus when y equals 3 then x is 108.3
The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be
V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³
Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire