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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
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?
t = number of liters of 30% acid solution s = number of liters of 60% acid solution t+s=57 .30t+.60s=.50*57 t=57-s .30(57-s)+.60s=.50*57 .30*57 -.30s +.60s = .50*57 .30s = .50*57 - .30*57 = .20*57 s = .20*57/.30 = 38 liters of 30% solution t = 57 - s = 57-38 = 19 liters of 60% solution
How many gallons of hypochlorite has to be added to 300 gallons of water to produce 100 mgl chlorine solution?
1.2 gallons
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
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
How much pure acid must be added to 12 ounces of a 40 percent acid solution in order to produce 60 percent acid solution?
6 ounces
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 a rock can form by evaporation?
Evaporated salt water will produce rock salt.
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
