Calculate the percent dissociation of benzoic acid C6H5CO2H in a 1.3M aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits.

the percent dissociation is  0.69 %

Explanation:

Given the data in the question;

benzoic acid C₆H₅CO₂H

C₆H₅COOH  ⇔ C₆H₅COO + H

Ka =  [C₆H₅COO- ][ H+ ] / [ C₆H₅COOH ] = 6.28 × 10⁻⁵

given that it dissociated in a 1.3 M  aqueous solution.

so Initial concentration is;  

[ C₆H₅COOH ] = 1.3

[C₆H₅COO- ] = 0

[ H+ ] = 0

Change in concentration

[ C₆H₅COOH ] = -x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Concentration equilibrium

[ C₆H₅COOH ] = 1.3 - x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Hence,

x² / ( 1.3 - x ) = 6.28 × 10⁻⁵

6.28 × 10⁻⁵( 1.3 - x ) = x²

8.164 × 10⁻⁵ - 6.28 × 10⁻⁵x = x²

x² + 6.28 × 10⁻⁵x - 8.164 × 10⁻⁵ = 0

solve for x

ax² + bx - c = 0

x = [ -b ± √( b² - 4ac ) ] / [ 2a ]

we substitute  

x = [ -6.28 × 10⁻⁵ ± √( (6.28 × 10⁻⁵)² - (4 × 1 × -8.164 × 10⁻⁵ ) ) ] / [ 2 × 1 ]  

x = [ -6.28 × 10⁻⁵ ± 0.01807 ] / [ 2]

x = [ -6.28 × 10⁻⁵ - 0.01807 ] / [ 2]  or [ -6.28 × 10⁻⁵ + 0.01807 ] / [ 2]

x = -0.0090664 or 0.0090036

so x = 0.0090036

hence

[ H+ ] = +x = 0.0090036 M

[C₆H₅COO- ] = +x = 0.0090036 M

Initial concentration of [ C₆H₅COOH ] = 1.3 M

concentration of C₆H₅COOH dissociated = 0.0090036 M

percent dissociation of C₆H₅COOH will be;

⇒ ( 0.0090036 M / 1.3 M ) × 100 = 0.69 %

Therefore, the percent dissociation is  0.69 %

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.

since electrons are negative the answer is neagtaive