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alexira [117]
2 years ago
7

Which one of the following equations represents the net ionic equation for the reaction between aqueous potassium chloride and a

queous silver nitrate? A. Cl- (aq) + Ag+ (aq) AgCl (s) B. K+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) AgCl (s) + K+ (aq) + NO3- (aq) C. KCl (aq) + AgNO3 (aq) AgCl (s) + KNO3 (aq) D. K+ (aq) + NO3- (aq) KNO3 (aq)
Chemistry
1 answer:
Nana76 [90]2 years ago
4 0
<h3>Answer:</h3>

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

<h3>Explanation:</h3>

The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.

<h3>Step 1: Writing a balanced equation for the reaction.</h3>
  • The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;

KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)

  • AgCl is the precipitate formed by the reaction.
<h3>Step 2: Write the complete ionic equation.</h3>
  • The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.

K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)

  • Only ionic compounds are split into ions.
<h3>Step 3: Write the net ionic equation for the reaction.</h3>
  • The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
  • The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

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Answer: 2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

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The balanced combustion reaction for butane is,:

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Given the following thermodynamic data, calculate the lattice energy of LiCl:
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Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

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