0
What quantity of 75 per cent acid solution must be mixed with a 30 per cent solution to produce 945 mL of a 50 per cent solution?
Wiki User
∙ 11y ago
A
nswer:525mL of 75% solution and 420mL of 30% solution must be mixed to produce 945mL of a 50% acid solution
Let x = 75% acid solution
y = 30% acid solutionEquations:
x + y = 945mL ----equation (1)
0.75x + 0.3y = 0.5 * 945
0.75x + 0.3y = 472.5
multiplying by 100
75x + 30y = 4725 ----equation (2)
eliminating equations (1) and (2)
-30(x + y = 945)-30
-30x -30y = -28350
75x +30y = 4725
============
45x = 23625
x=525
substitute x=525 to equation (1)
525 + y = 945
y = 945 - 525
y = 420
thus 525mL of 75% solution and 420mL of 30% solution must be mixed to produce 945mL of a 50% acid solution
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
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.
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
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
How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?
x=45
How much pure acid should be mixed with 6 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?
50% acid in a 6 gallon solution means that 3 gallons are acid. 9 gallons more acid will give you a total of 12 gallons of acid in a 15 gallon solution. 12 is 80% of 15.
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
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.
When an acid solution is mixed with a basic solution?
When an acid solution is mixed with a basic solution, there will be a neutralization reaction in which hydrogen ions from the acid will combine with hydroxide ions from the base, to form 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
Does sulphuric acid solution foam when mixed?
It depends what it is mixed with.
A chemical solution contains 2 of acid. If there is 5 ml of acid the quantity of the solution is?
Name the unit of 2.
What quantity can tell you whether a solution is acid or basic?
ph
How many liters of a 50 percent alcohol solution must be mixed with 40 liters of a 90 percent solution to get a 60 percent solution?
mary mixed 2l of an 80% acid solution with 6l of a 20% acid solution. what was the percent of acid in the resulting mixture
What does an ion have to do with acids and bases?
The quantity of hydrogen ions in a solution indicates whether the solution is an acid or a base.
When the right amounts of acid and alkaline are mixed the solution is?
With the right proportions, the solution will be neutral.
Acetic acid is what type of acid?
Acetic acid is a weak acid because it only ionises to a minimal extent in solution to produce hydrogen ions.CH3COOH CH3COO- + H+By definition, a weak acid is a substance which only ionises to a minimal extent in solution to produce hydrogen ions. A strong acid, is a substance which ionises completely in solution to produce hydrogen ions.
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
