k = 5.97 WHAT.
ln(No/N) = kt
ln(100/5) = 5.97*t
Solve for t.
I don't know the units (seconds, min, hours, days, years) because I can't tell from your value of 5.97 1/h.
The first-order rate constant for photodissociation of A is 5.97 1/h. Calculate the time needed for the concentration of A to decrease to the following quantities.
a. 5% of its original value h
b. one-third of its initial concentration h
6 answers
5.97 1/hour
Then the time is in hours.
b is worked the same way.
b is worked the same way.
could yo explain part b?
Same equation.
You can make up any number you want for No, then N is 1/3 that. I would pick a convenient number for No such as 300, then N = 100, and solve for t.
No could be 1 and N = 1/3 etc.
You can make up any number you want for No, then N is 1/3 that. I would pick a convenient number for No such as 300, then N = 100, and solve for t.
No could be 1 and N = 1/3 etc.
Thank You Very !
1 answer