Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook).

Part A. 0.270 M

Part B. 7.84×10-2 M

Part C. 1.97×10-2 M

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

=

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  

percentage% (∝) =

= 0.704%

For Part B; where Concentration of B = M

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C=

percentage% (∝) = 0.02608  × 100%

= 2.60%

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2