Answer:
Amount of reactant after four hours = 4,26 grams
Explanation:
Suppose y denotes the amount of reactant at the time (t)
The given function:
[tex]\dfrac{dy}{dt} = -0.7 y[/tex]
[tex]\dfrac{dy}{y} = -0.7 dt[/tex]
Taking integral on both sides
㏑(y) = -0.7t + c¹
[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]
[tex]y(t) = Ce ^{-0.7t}[/tex]
At t = 0 ; y (t) = 70
∴
[tex]70 = Ce^{-0.7(0)}[/tex]
C = 70
As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]
After four hours, the amount of the reactant is:
[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]
[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]
[tex]\mathtt{y(t) = 4.26}[/tex]
Amount of reactant after four hours = 4,26 grams
