The final molarity of HCl is 2.284 M
We'll begin by listing what was given from the question. This is shown below:
Initial volume (V₁) = 5.56 mL
Initial Molarity (M₁) = 4.108 M
Final volume (V₂) = 5.56 + 4.44 = 10 mL
<h3>Final Molarity (M₂) =? </h3>
The final molarity of the HCl solution can be obtained by using the dilution formula as illustrated below:
<h3>M₁V₁ = M₂V₂</h3>
4.108 × 5.56 = M₂ × 10
22.84048 = M₂ × 10
Divide both side by 10
M₂ = 22.84048 / 10
<h3>M₂ = 2.284 M</h3>
Therefore, final molarity of the HCl solution is 2.284 M.
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Answer: 1+
Justification:
The ionization energy is the amount of energy needed to loose electrons and becomes ions.
The first ionization energy is the energy needed to liberate one one electron and form the ion with oxidation state 1+.
The second ionization energy is the energy to release a second electron and form the ion with oxidation state 2+.
The third ionization energy is the energy to leave a third electron free and form the ion with oxidation state 3+.
The relatively low first ionization energy of element 2, shows it it will lose an electron easily to form the ion with oxidations state 1+.
The second and third ionization energies are very high meaning that the ions with oxidation staes 2+ and 3+ will not be formed.
Therefore, the answer is that the expected oxidation state for the most common ion of element 2 is 1+.
The amount of precipitate produced will be proportional to the amount of NH₃ reacted with water to produce NH₄OH.
<h3>What is precipitate?</h3>
Precipitates are the crystal type formation, when the solute is no more dissolving in the solvent.
Imagine mixing 1 tablespoon of Epsom salt with 2 cups of ammonia, the reaction is
2NH₃ + MgSO₄ + 2H₂O → Mg(OH)₂ + (NH₄)₂SO₄
The amount of precipitate produced will be proportional to the amount of NH₃ reacted with water to produce NH₄OH.
Learn more about precipitate.
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Answer:
Explanation:
Hello!
In this case, since the net ionic equations are ionic representations of the molecular equation in which the spectator ions (those at both reactants and products sides) are cancelled out, we first write the complete ionic equation for this reaction, considering that the solid silver chloride is not ionized due to its precipitation:
Whereas the nitrate and sodium ions are cancelled out for the aforementioned reason as they are the spectator ions, to obtain:
Which is the required net ionic equation.
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