Under certain water conditions, the free chlorine(hypochlorous acid, hocl) in a swimming pool decomposes according to the law of uninhibited decay. After shocking a pool, the poolboy, geoff, tested the water and found the amount of free chlorine to be 2.7 parts per million (ppm). Twenty-four hours later, geoff tested the water again and found the amount of free chlorine to be 2.5 ppm. What will be the reading after 4 days (that is, 96hours)? When the chlorine level reaches 1.0 ppm, geoff must shock the pool again. How long can geoff go before he must shock the pool again? After 4 days, or 96 hours, the amount of free chlorine in the pool will be ___ ppm.
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The function represents the amount A of a material present at a time t where k<0 represents the rate of decay.
substitute 2.5ppm from and 2.2ppm for A(t) and 24 hours or 1 day for t in the equation and find k
Divide both sides of the equation
Take the natural logarithm of both sides of the equation.
ln(0.88)=ln
ln0.88=k
-0.1278=k
so the value of k is -0.1278
substitute 2.5ppm for and t=3 and k=-0.1278
=
=2.5(0.6811)
=1.7
so the amount of chlorine after 3 days will be 1.7ppm
divide both sides of the equation by 2.5
Take the natural logarithm of both sides of the equation
ln0.4=
ln0.4=-0.1278t
divide both sides of the equation by -0.1278
t=7.17
so the pool must be shocked after 7.17days or 172hours
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