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.
Answer:
45200J
Explanation:
Given parameters:
Heat of vaporization of water = 2260J/g
Mass of steam = 20g
Temperature = 100°C
Unknown:
Energy released during the condensation = ?
Solution:
This change is a phase change and there is no change in temperature
To find the amount of heat released;
H = mL
m is the mass
L is the latent heat of vaporization
Insert the parameters and solve;
H = 20g x 2260J/g
H = 45200J
Both theoretical and actual yields would be needed to calculate percent yield.
The percent yield of a reaction is defined as the ratio of the actual yield to that of the theoretical yield. It can be mathematically expressed as:
percent yield = actual yield/theoretical yield x 100%
The actual yield is the amount of a specific product produced from a reaction while the theoretical yield is the stoichiometric amount of the same product.
More on percent yield can be found here: brainly.com/question/2506978?referrer=searchResults
This is true.
The reaction rates are affected by how often the particles collide.
