Question

The water in a fishery became polluted when toxic waste was dumped into its pond, causing the fish population to substantially decline. The percents of fish that survived are recorded in the table.
Day 1 2 3 4 5
% Survival 79 55 38 31 19


a) What is the order of the reaction live fish ------> dead fish?

b) What is the rate constant?

c) If the fish continue to die at this rate, how many fish will be alive after 10 days?

Answer 1

In the pond, the percentage of fish left after ten days is 4%.

Given that;

N(t) = Noe^-kt

Where;

N(t) = percentage of fish at time t

No = Initial percentage of fish

k = rate constant

t = time taken

Given that we had 100% fish population at the beginning, we can see that this reaction is first order.

The rate constant is obtained from;

When t = 5

19 = 100e^-5k

19/100 =e^-5k

0.19 = e^-5k

ln 0.19 = -5k

k = ln (0.19)/-5

k = 0.33

In ten days;

N(10) = 100e^-0.33(10)

N(10) = 4%

Learn more: https://brainly.com/question/5624100