Chapter 3and so the mass of salt in the tank aftertminutes isx(t)=5−4.5e−2t/25.If the concentration of salt in the tank is 0.02 kg/L, then the mass of salt is 0.02×100 = 2 kg,and, to Fnd this moment, we solve5−4.5e−2t/25=2⇒e−2t/25=23⇒t=25 ln(3/2)2≈5.07 (min).3.Letx(t) be the volume of nitric acid in the tank at timet. The tank initially held 200 Lof a 0.5% nitric acid solution; therefore,x(0) = 200×0.005 = 1. Since 6 L of 20% nitricacid solution are ﬂowing into the tank per minute, the rate at which nitric acid is enteringis 6×0.2=1.2 L/min. Because the rate of ﬂow out of the tank is 8 L/min and the rateof ﬂow in is only 6 L/min, there is a net loss in the tank of 2 L of solution every minute.Thus, at any timet, the tank will be holding 200−2tliters of solution. Combining this
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