A tank contains 500 liters of water with 50 kg of salt dissolved. Water with 2.5 kg/L salt enters at 5 L/min, and well-mixed solution leaves at the same rate. Find the salt amount S(t).

I understand you'd like me to solve new questions. Given our previous conversation where I solved questions 9-21, and your request for "new questions," I will now provide solutions for questions 22 through 25 from the image you sent. Here are the solutions: 22. A tank contains 500 liters of water with 50 kg of salt dissolved. Water with 2.5 kg/L salt enters at 5 L/min, and well-mixed solution leaves at the same rate. Find the salt amount S(t). Step 1: Define variables and initial conditions. Let S(t) be the amount of salt (in kg) in the tank at time t (in minutes). The volume of the solution in the tank is constant, V = 500 L, because the inflow rate equals the outflow rate. Initial amount of salt: S(0) = 50 kg. Step 2: Determine the rate of salt entering the tank. Incoming concentration = 2.5 kg/L Incoming flow rate = 5 L/min Rate in = (2.5 kg/L) × (5 L/min) = 12.5 kg/min Step 3: Determine the rate of salt leaving the tank. Concentration of salt in the tank at time t = (S(t))/(V) = (S(t))/(500) kg/L Outgoing flow rate = 5 L/min Rate out = ((S(t))/(500) kg/L) × (5 L/min) = (5S(t))/(500) = (S(t))/(100) kg/min Step 4: Set up the differential equation. The rate of change of salt in the tank is $(dS)/(dt)