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lora16 [44]
4 years ago
14

Write balanced complete ionic equation for hcl(aq)+lioh(aq)→h2o(l)+licl(aq)

Chemistry
2 answers:
DedPeter [7]4 years ago
8 0

The balanced complete ionic equation is \boxed{{{\text{H}}^ + }\left( {aq} \right) + {\text{C}}{{\text{l}}^ - }\left( {aq} \right) + {\text{L}}{{\text{i}}^ + }\left( {aq} \right) + {\text{O}}{{\text{H}}^ - }\left( {aq} \right) \to {{\text{H}}_2}{\text{O}}\left( l \right) + {\text{L}}{{\text{i}}^ + }\left( {aq} \right) + {\text{C}}{{\text{l}}^ - }\left( {aq} \right)}.

Further Explanation:

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The reactants and products remain in <em>undissociated</em> form in the <em>molecular equation</em>. In the case of <em>total ionic equation</em>, all the ions that are <em>dissociated</em> and present in the reaction mixture are represented while in the case of net or overall ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the net ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, HCl reacts with LiOH to form LiCl and {{\text{H}}_{\text{2}}}{\text{O}}. The balanced molecular equation of the reaction is as follows:

{\text{HCl}}\left({aq}\right)+{\text{LiOH}}\left({aq}\right)\to{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)+{\text{LiCl}}\left({aq}\right)

Step 2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phases remain same. The total ionic equation is as follows:

{{\text{H}}^+}\left({aq}\right)+{\text{C}}{{\text{l}}^-}\left({aq}\right)+{\text{L}}{{\text{i}}^+}\left({aq}\right)+{\text{O}}{{\text{H}}^-}\left({aq}\right)\to{{\text{H}}_2}{\text{O}}\left(l\right)+{\text{L}}{{\text{i}}^+}\left({aq}\right)+{\text{C}}{{\text{l}}^-}\left({aq}\right)

Therefore, the complete ionic equation obtained is as follows:

{{\text{H}}^+}\left({aq}\right)+{\text{C}}{{\text{l}}^-}\left({aq}\right)+{\text{L}}{{\text{i}}^+}\left({aq}\right)+{\text{O}}{{\text{H}}^-}\left({aq}\right)\to{{\text{H}}_2}{\text{O}}\left(l\right)+{\text{L}}{{\text{i}}^+}\left({aq}\right)+{\text{C}}{{\text{l}}^-}\left({aq}\right)

Learn more:

1. Balanced chemical equation: brainly.com/question/1405182

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: net ionic equation, complete ionic equation, Li+, OH-, Cl-, LiCl, HCl, H2O, LiOH, solid phase, liquid phase, aqueous phase.

olga2289 [7]4 years ago
4 0
<span>H and Li have a +1 charge. Cl and OH have a -1 charge. When written out it should look like this: H(+1) + Cl(-1) + Li(+1) + OH(-1) --> H2O + Li(+1) + Cl(-1) if you wanted the net ionic equation it would be: H(+1) + OH(-1) --> H2O</span>
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Se ha añadido un evaporador para una alimentación de 11500 kg/dia de zumo de pomelo de forma que evapore 3000 kg/dia de agua por
mamaluj [8]

Answer:

37 %.

Explanation:

¡Hola!

En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:

F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia

Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:

x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia

Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:

x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37

Que es igual al 37%.

¡Saludos!

6 0
4 years ago
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geniusboy [140]

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Explanation:

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8 0
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The reaction 2 ClO2(g) + F2(g) → 2 FClO2(g) is first-order in both ClO2 and F2. When the initial concentrations of ClO2 and F2 a
kupik [55]

Answer:

3. 75.0%

Explanation:

2 ClO2(g) + F2(g) → 2 FClO2(g)

First order with respect to ClO2 and F2.

This means the rate equation is given as;

Rate = k [ClO2][F2]

When the initial concentrations of ClO2 and F2 are equal?

Let's assume an initial value of 1 for both reactants, so rate equation is given as;

Rate = k * 1 * 1 = k

The rate after 25% of the F2 has reacted is what percent of the initial rate?

The concentration left of F2 is 75% ( 100% - 25%) = 0.75

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So rate equation is given as;

Rate = k * 1 * 0.75 = 0.75 k

Comparing 0.75k and k.

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