96
In order to reduce the percentage of acid from 80% to 50%, you would need to add another 36 kg of diluent (e.g. water).
Answer:12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solutionLet x = 50% acid solutiony = 20% acid solution Equations:x + y = 36mL ----equation (1)0.5x + 0.2y = 0.3 * 360.5x + 0.2y = 10.8multiplying by 105x + 2y = 108 ----equation (2)eliminating equations (1) and (2)-2(x + y = 36)-2-2x -2y = -725x +2y = 108=========3x = 36x=12substitute x=12 to equation (1)12 + y = 36y = 36 - 12y = 24thus 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution
Let Q be the quantity of 80% solution required then (300 - Q) is the quantity of 30% solution as together the two solutions must equal 300. Then, [80 x Q] + [30 x (300 - Q)] = [40 x 300] 80Q + 9000 - 30Q = 12000 50Q = 3000 Q = 60...........which means (300 - Q) = 240 60ml of 80% acid solution + 240ml of 30% solution produces 300ml of 40% solution.
they produce about 90%
Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
133.33
Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
add 25ml more of solution x * 20 = 100 * 25 x = 25
When salt water is evaporated, the water portion of the solution evaporates leaving behind the salt. This process is known as evaporation and can be used to produce salt from seawater in a process called solar evaporation.
In order to reduce the percentage of acid from 80% to 50%, you would need to add another 36 kg of diluent (e.g. water).
To determine the amount of hypochlorite needed to produce a 100 mg/L chlorine solution in 300 gallons of water, you need to calculate the total chlorine required. The total amount of chlorine needed is 100 mg/L x 300 gallons = 30,000 mg of chlorine. Knowing that hypochlorite contains about 12-15% available chlorine, you can divide 30,000 mg by 0.15 to get the total amount of hypochlorite required.
To produce 1L of 10% ammonia solution from 25% ammonia solution, you need to dilute the 25% solution by adding a calculated amount of water. To do this, you can calculate the volume of the 25% solution needed and the volume of water needed using the formula: C1V1 = C2V2, where C1 is the initial concentration (25%), V1 is the initial volume, C2 is the final concentration (10%), and V2 is the final volume (1L).
About 39,000 gallons
About 500 gallons.
Salt and water make a basic brine compound
Evaporated water turns into clouds and clouds will produce rain.
Bases produce hydroxide ions (OH-) in solution.