Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH = 
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
We need to see your problem/question you’re talking about to answer it
The two electrons that share an orbital repel each other.
All electrons bear a negative charge. They are held in their orbits by the attractive force of charged protons. The farther away an orbital is to the atomic nucleus the easier it is to expunge an electron from this distant orbital shell.
Explanation:
Because electrons have the same negative charge, they repel each other especially when they occupy the same orbital shell in an atom. To reduce this repulsion, each of the electrons in the orbital shell (remember electrons occupy orbital shells of atoms in 2s) assumes an opposite quantum (M<em>s</em>) spin; one with – ½ while the other + ½ .
Learn More:
For more about electrons check out;
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Answer is: sodium (Na) and iodine (I₂).
<span>
First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
