Answer:
See explanation
Explanation:
The electrolysis of nickel(II) iodide yields nickel solid and iodine gas.
In the oxidation half equation;
2I^-(aq)-------> I2(g) + 2e
In the reduction half equation;
Ni^2+(aq) + 2e -----> Ni(s)
These are the half equations of the process;
The overall equation is;
Ni^2+(aq) + 2I^-(aq)-------> NiI2
Answer:
the nucleus is made up of protons and neutrons
Answer:
When the transfer of electrons occurs, an electrostatic attraction between the two ions of opposite charge takes place and an ionic bond is formed. A salt such as sodium chloride (NaCl) is a good example of a molecule with ionic bonding
The amount of barium ions that must be present in order for the salt to precipitate is 0.245 M.
A solution's solubility product is the result of raising each ion's concentration to the power of its stoichiometric ratio. It is portrayed as
A combination of 1 barium ion and 2 fluoride ions results in the ionic compound known as barium fluoride.
The following equation describes the equilibrium reaction for barium fluoride ionization:
BaF₂ → Ba²⁺ + 2F⁻
Ksp = [Ba²⁺] · [F⁻]²
2.45*
= [Ba²⁺] * ![[1. 00*10^{-2} ]^{2}](https://tex.z-dn.net/?f=%5B1.%2000%2A10%5E%7B-2%7D%20%5D%5E%7B2%7D)
[Ba²⁺]=0.245 M
As a result, 0.245 M of barium ions must be present in order for the salt to precipitate.
<h3>Solubility </h3>
Solubility in chemistry refers to a chemical's capacity to dissolve in another substance, the solvent, to produce a solution. Inability of the solute to create such a solution is the opposite quality, or insolubility. A substance's degree of solubility in a given solvent is often determined by the amount of the solute present in a saturated solution, which is a solution in which no additional solute can be dissolved. The solubility equilibrium between the two compounds is considered to have been reached at this time. If there is no such restriction for a given solute and solvent, the two are referred to as being "miscible in any amounts."
What concentration of the barium ion, ba2 , must be exceeded to precipitate baf2 from a solution that is 1. 00×10−2 m in the fluoride ion, f−? ksp for barium fluoride is 2. 45×10−5
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