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How much water must be evaporated from 240 gallons of a 3 percent salt solution to produce a 5 percent solution?
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∙ 14y ago
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∙ 14y ago
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What is the answer to 60kg of an 80 percent acid solution to produce a 50 percent solution?
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).
How many milliliters of a 50 percent acid solution and how many milliliters of a 20 percent acid solution must be mixed to produce 36 mL of a 30 percent acid solution?
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
What quantity of a 80 percent acid solution must be mixed with a 30 percent solution to produce 300 mL of a 40 percent 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.
Do Asian farmers produce fifty percent seventy-five percent or ninety percent of the worlds rice?
they produce about 90%
How much 5 percent acid must be added to 25 percent acid to get 200 ml of 8 percent acid?
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
What quantity x of a 65 percent acid solution must be mixed with a 20 percent solution to produce 300 mL of a 45 percent solution?
133.33
A solution of 30 percent acid and a solution of 60 percent acid are to be mixed to produce 50 liters of 57 percent acid How many liters of each solution should be used?
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.
How many ml should be added to 100ml of a 25 percent solution to get produce a 20 percent solution?
add 25ml more of solution x * 20 = 100 * 25 x = 25
What happens to salt water when it is evaporated?
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.
What is the answer to 60kg of an 80 percent acid solution to produce a 50 percent solution?
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).
How many gallons of hypochlorite has to be added to 300 gallons of water to produce 100 mgl chlorine solution?
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.
How do you produce 1L of 10 percent ammonia solution from 25 percent ammonia solution?
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).
How many gallons of water produce an average car?
About 39,000 gallons
What makes brine?
Salt and water make a basic brine compound
Why do you have floods if water evaporates?
Evaporated water turns into clouds and clouds will produce rain.
How many gallons of ethanol will 183 bushels of corn produce?
About 500 gallons.
How many milliliters of a 50 percent acid solution and how many milliliters of a 20 percent acid solution must be mixed to produce 36 mL of a 30 percent acid solution?
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
