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:
6.1×10^8
Explanation:
The reaction is;
Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)
E°cell = E°cathode - E°anode
E°cathode= -0.14 V
E°anode= -0.40 V
E°cell = -0.14-(-0.40)
E°cell= -0.14+0.40
E°cell= 0.26 V
But
E°cell= 0.0592/n log K
E°cell= 0.0592/2 log K
0.26= 0.0296log K
log K = 0.26/0.0296
log K= 8.7838
K= Antilog (8.7838)
K= 6.1×10^8
Answer:
Subtracting the number of protons from the atomic mass.
Answer:

Explanation:
Hello!
In this case, when balancing chemical reactions, we must make sure that the atoms of each element are the same at both reactants and products; thus, for the given reaction, we need two iron and aluminum atoms at each side based on their subscripts in the given oxides as shown below:

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