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Nadya [2.5K]
3 years ago
13

When the concentration of a system is changed:

Chemistry
1 answer:
Alex Ar [27]3 years ago
8 0

Answer:

The concentrations change to maintain the original value of K.

Explanation:

The concentration changes to maintain the original value of K when the concentration of the system changes.

This is the effect concentration has on a reaction at equilibrium.

Le Chatelier's principle proposes that "if any conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".

  • The equilibrium constant K is temperature dependent.
  • An increase in concentration of a specie favors the direction that uses it up and vice versa.
  • This does not change the value of the equilibrium constant.
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Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

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A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi
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Answer:

hypochlorite ion

Explanation:

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HCl + ClO⁻ → HClO + Cl⁻.

Where more hypochlorous acid is produced.

That means, the HCl reacts with the hypochlorite ion present in solution

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