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Bogdan [553]
4 years ago
14

Complete this equation for the dissociation of the electrolyte NaCl(aq). Omit water from the equation because it is understood t

o be present.
And
Complete this equation for the dissociation of Fe(ClO4)3(aq). Omit water from the equation because it is understood to be present.
Chemistry
2 answers:
ahrayia [7]4 years ago
8 0

The equation for the dissociation of NaCl is  \boxed{{\text{NaCl}}\left({aq}\right)\to{\text{N}}{{\text{a}}^ + }\left({aq}\right)+{\text{C}}{{\text{l}}^ - }\left({aq}\right)}

The equation of the dissociation of {\mathbf{Fe}}{\left({{\mathbf{Cl}}{{\mathbf{O}}_{\mathbf{4}}}}\right)_{\mathbf{3}}} is \boxed{{\text{Fe}}{{\left({{\text{Cl}}{{\text{O}}_4}}\right)}_3}\left({aq}\right)\to {\text{F}}{{\text{e}}^{3+}}\left({aq}\right)+3{\text{ClO}}_4^ -\left( {aq}\right)}

Further Explanation:

The attraction between atoms, molecules or ions which results in the formation of chemical compounds is known as a chemical bond. It is formed either due to electrostatic forces or by the sharing of electrons. There are many strong bonds such as ionic bonds, covalent bonds, and metallic bonds while some weak bonds like dipole-dipole interactions, London dispersion forces, and hydrogen bonding.

Ionic compounds are the compounds that are formed from the ions of the respective species. Ions are the species that are formed either due to loss or gain of electrons. A neutral atom forms cation by the loss of electrons and anion by the gain of electrons.

For example, {\text{MgC}}{{\text{l}}_2} is an ionic compound formed from one {\text{M}}{{\text{g}}^{2 + }} and two {\text{C}}{{\text{l}}^ - } ions. Its dissociation occurs as follows:

{\text{MgC}}{{\text{l}}_2}\rightleftharpoons {\text{M}}{{\text{g}}^{2+}}+{\text{2C}}{{\text{l}}^ - }

(A) NaCl is an ionic compound formed by one  {\text{N}}{{\text{a}}^ + } and one {\text{C}}{{\text{l}}^ - } ions. The reaction for  dissociation of NaCl is,

{\text{NaCl}}\to{\text{N}}{{\text{a}}^+}+{\text{C}}{{\text{l}}^ - }

(B) {\mathbf{Fe}}{\left({{\mathbf{Cl}}{{\mathbf{O}}_{\mathbf{4}}}}\right)_{\mathbf{3}}}  is an ionic compound formed by one {\text{F}}{{\text{e}}^{3 + }} and three {\text{ClO}}_4^ - ions. The reaction for dissociation of  {\mathbf{Fe}}{\left({{\mathbf{Cl}}{{\mathbf{O}}_{\mathbf{4}}}}\right)_{\mathbf{3}}} is,

{\text{Fe}}{\left({{\text{Cl}}{{\text{O}}_4}}\right)_3}\left({aq}\right)\to{\text{F}}{{\text{e}}^{3+}}\left({aq}\right)+3{\text{ClO}}_4^ -\left({aq}\right)

Learn more:

1. Identification of ionic bonding: brainly.com/question/1603987

2. Basis of investigation for the scientists: brainly.com/question/158048

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Ionic and covalent compounds

Keywords: ionic compound, cation, anion, ions, dissociation, neutral atom, electrons, dissociation, equation, chemical bond, electrostatic forces, mutual sharing of electrons.

ella [17]4 years ago
5 0

1. Answer;

Equation;

-NaCl(aq) → Na{+}(aq) + Cl{-}(aq)

Explanation;

-The polar property of water allows ionic compounds such as sodium chloride to dissolve in water. (sodium chloride is an ionic compound).

-Sodium chloride dissociates in water into sodium ions and chloride ions as shown by the equation. Water molecules surrounds the negatively charged chloride ions and positive sodium ions and pulls them away into solution.

2. Answer;

Equation;

Fe(ClO4)3(aq) → Fe 3+(aq) + 3ClO4 - (aq)

Explanation;

Iron (iii) perchlorate (Fe(ClO4)3 dissociates in water into iron (iii) ion and perchlorate ion. Iron (iii) ion is positively charged while the perchlorate ion is negatively charged, they are pulled apart by the polar water molecules, a process called dissociation.

-

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It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

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