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How many liters of acid should be added to 60 l of a 25% acid solution in order to produce a 40% acid solution?
You need to add 15 L of pure acid.
What quantity of a 60% acid solution must be mixed with a 30% acid solution to produce 300ml of 50% acid solution?
At the end, the amount of acid you want to end up with is 50% of 300ml, so that’s 150 ml. Since x + y = 300, that means that x = 200. So you will need 200 ml of 60% solution and 100 ml of 30% solution.
How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?
So, we add 3 L of 90 % acid to 6 L of 15 % acid and get 9 L of 40 % acid .
How many liters of water must be added to 50l of a 30% acid solution in order to produce a 20% acid solution?
Therefore 25 L of water must be added to get a 20% acid solution.
How many Litres of acid are there in 12 Litres of a 20% solution?
∴ Acid is 2.4 liters in 12 liters solution
Learn now!
How many Litres of pure acid are there in 8 Litre of a 20% solution?
Hence 1. 6 l of pure acid are there in 8 litres of a 20% solution.
How many milliliters of a 20% acid solution must be added to 40 milliliters of a 75% acid solution to create a 30% acid solution?
Therefore, 180 milliliters of a 20% acid solution must be added to 40 milliliters of a 75% acid solution to create a 30% acid solution.
How do you calculate mixtures and solutions?
Solving a percent mixture problem can be done using the equation Ar = Q, where A is the amount of a solution, r is the percent concentration of a substance in the solution, and Q is the quantity of the substance in the solution.
How many liters is a 90% acid solution?
Thus 6 litres of a 90% solution of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 l solution of 78% concentrated acid.
How many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
Ernest Z. You would add 24 L of water.
How much of a 20% acid solution would a chemist have to mix with 1 liter of a 40% acid solution to yield a 36% acid solution?
How much of a 20% acid solution would a chemist have to mix with one liter of a 40% acid solution to yield a 36% acid solution? The chemist must add 250 milliliters of the 20% acid solution.
How much water must be added to a 30 liters of a 75% acid solution to reduce it to a 15% solution?
Detailed Solution
Hence, we need to mix 7.5 liters of water with 30 liters of 15% solution.
How many liters of water should be added to 50 liters of a 10% sugar solution so as to bring down the sugar percentage to 5?
The correct answer is 16 (2/3)%.
How many Litres of water should be added to 50 Litres of a 10% sugar solution so as to bring down the sugar percentage to 5?
(10/100)*5 + (50/100)*V = (5 + V)*(25/100) or 50*V + 50 = 25*V + 125. or 25*V = 75 or V = 75/25 = 3 (L) [Ans]
How many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
Ernest Z. You would add 24 L of water.
How much pure alcohol should we add to 18 gallons of 35% acid solution to obtain a solution that is 55 %?
How much pure alcohol should we add to 18 gallons of 35% acid solution to obtain a solution that is 55%? 8% of 84 is 0.08 (84) φ 6.72. Thus our solution has the right concentration.
How many liters is a 90 of concentrated acid?
How many litres of 90% concentrated acid needs to be mixed with 75% solution of concentrated acid to get the result? Solution: Let’s apply the weighted-average formula. So 30 litres needed to be divided in the ratio of 1:4, which gives us 6 litre as the answer.
SOLUTION: How many liters of a 60% acid solution must be mixed with a 75% acid solution to get 20L of a 72% solution?
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Error 403 (Forbidden)
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How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution? | Socratic
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- Most searched keywords: Whether you are looking for How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution? | Socratic Updating You need to add 15 L of pure acid. The idea behind this problem is that you’re adding pure acid to a solution that is 60% v/v acid, so the added acid will change the percent concentration of the starting solution by increasing the volume of solute and the volume of solvent by the same amount. So, start by calculating how much acid you have in your starting solution 25color(red)(cancel(color(black)(“L solution”))) * “60 L acid”/(100color(red)(cancel(color(black)(“L solution”)))) = “15 L acid” Let’s say that x denotes the volume of pure acid that you must add to your starting solution. Since you’re dealing with pure acid, the volume of acid will Increase by x V_”acid” = 15 + x At the same time, the volume of the solution will also incrase by x V_”sol” = 25 + x This means that the target solution’s percent concetration by volume will be equal to V_”acid”/V_”soL” * 100 = 75% ((15 + x)color(red)(cancel(color(black)(“L”))))/((25 + x)color(red)(cancel(color(black)(“L”)))) * 100 = 75% Solve this equation for x to get (15+x) * 100 = (25 + x) * 75 1500 + 100x = 1875 + 75x 25x = 375 implies x = 375/25 = color(green)(“15 L”) So, if you add 15 L of pure acid to 25 L of 60% v/v acid solution, you’ll get 40 L of 75% v/v acid solution.
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What amount of a 60% acid must be mixed with a 30% solution to produce 300ml of a 50% solution? | Wyzant Ask An Expert
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How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution? | Socratic
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- Most searched keywords: Whether you are looking for How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution? | Socratic Updating You must add 3 L of the 90 % acid. For this problem, we can use the relation color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = “amount of solute” color(white)(a/a)|)))” ” Let the “volume of 90 % acid” = xcolor(white)(l) “L” Then, after mixing, we have (6 + x”) L of 40 % acid”. This is made up of “6 L of 15 % acid” and xcolor(white)(l) “L of 90 % acid”. “Moles before = moles after” c_1V_1 = c_2V_2 + c_3V_3 40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)(“L”))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)(“L”))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)(“L”))) 240 + 40x = 90 + 90x 50x = 150 x = 150/50 = 3 So, we add “3 L of 90 % acid” to “6 L of 15 % acid” and get “9 L of 40 % acid”. Check: 3 × 90 + 6 × 15 = 9 × 40 270 + 90 = 360 360 = 360 It checks!
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How many liters of a 60% acid solution must be mixed with a 20% acid solution to get 400 L of a 50% acid solution? | Wyzant Ask An Expert
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how many liters of a 60 acid solution
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How many liters of a 60% acid solution must be mixed with a 30% acid solution to get 270 L of a 50% acid solution? ____ liters of a 60% acid solution must be used. – Math Homework Answers
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- Most searched keywords: Whether you are looking for How many liters of a 60% acid solution must be mixed with a 30% acid solution to get 270 L of a 50% acid solution? ____ liters of a 60% acid solution must be used. – Math Homework Answers How many liters of a 60% ac solution must be mixed with a 30% ac solution to get 270 L of a 50% ac solution? ____ liters of a 60% ac … I’m taking a test that’s due in two hours please help!!!!Algebra 1 Answers,word problems,algebra,story problem,word problem
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SOLUTION: How many liters of a 60% acid solution must be mixed with a 75% acid solution to get 20L of a 72% solution?
How many liters of a 60% acid solution must be mixed with a 75% acid solution to get 20L of a 72% solution?
How many liters of a 60% acid solution must be mixed with a 75% acid solution to get 20L of a 72% solution?
.
Let x = liters of 60% acid solution to be added
then
20-x = liters of 75% acid solution to be added
.
Our equation:
.60x + .75(20-x) = .72(20)
.60x + 15 – .75x = 14.4
15 – .15x = 14.4
– .15x = -.6
x = 4 L (60% solution)
.
75% solution:
20-x = 20-4 = 16 L
You can put this solution on YOUR website!
How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution?
The idea behind this problem is that you’re adding pure acid to a solution that is 60% v/v acid, so the added acid will change the percent concentration of the starting solution by increasing the volume of solute and the volume of solvent by the same amount.
So, start by calculating how much acid you have in your starting solution
#25color(red)(cancel(color(black)(“L solution”))) * “60 L acid”/(100color(red)(cancel(color(black)(“L solution”)))) = “15 L acid”#
Let’s say that #x# denotes the volume of pure acid that you must add to your starting solution. Since you’re dealing with pure acid, the volume of acid will Increase by #x#
#V_”acid” = 15 + x#
At the same time, the volume of the solution will also incrase by #x#
#V_”sol” = 25 + x#
This means that the target solution’s percent concetration by volume will be equal to
#V_”acid”/V_”soL” * 100 = 75%#
#((15 + x)color(red)(cancel(color(black)(“L”))))/((25 + x)color(red)(cancel(color(black)(“L”)))) * 100 = 75%#
Solve this equation for #x# to get
#(15+x) * 100 = (25 + x) * 75#
#1500 + 100x = 1875 + 75x#
#25x = 375 implies x = 375/25 = color(green)(“15 L”)#
So, if you add 15 L of pure acid to 25 L of 60% v/v acid solution, you’ll get 40 L of 75% v/v acid solution.
What amount of a 60% acid must be mixed with a 30% solution to produce 300ml of a 50% solution?
Start by defining variables to represent the amounts of each of the solutions that are going to be mixed.
So let x = the amount of the 60% solution, and y = the amount of the 30% solution.
First of all, you know that x + y = 300, because after they are mixed you want to have 300 ml of the 50% solution.
Second, look at how much acid is in each of the starting solutions and how much is in the ending solution.
The 60% solution is 60% acid, so the amount of acid in it is 60% of x, or .6x. Similarly, the amount of acid in the 30% solution is 0.3y
At the end, the amount of acid you want to end up with is 50% of 300ml, so that’s 150 ml.
When you mix the two solutions, you’re not going to change the total amount of acid you started with. This means that 0.6x + 0.3y = 150
So you have two solutions with two unknowns, and that’s something you can solve:
x + y = 300
0.6x + 0.3y = 150
Let’s use elimination. Multiply the first equation through by 0.6:
0.6x + 0.6y = 0.6*300 = 180
Now you have:
0.6x + 0.6y = 180
0.6x + 0.3y = 150
Subtract:
0.3y = 30
Divide by 0.3
y = 30 / 0.3 = 100
Since x + y = 300, that means that x = 200.
So you will need 200 ml of 60% solution and 100 ml of 30% solution.
So you have finished reading the how many liters of a 60 acid solution topic article, if you find this article useful, please share it. Thank you very much. See more: how many liters of pure acid must be added, how much water is to be mixed with 25l of 75 acid solution to produce 25 acid solution, a solution is made of water and pure acid, how many cc of 80 alcohol must be added to 200 cc of 65 alcohol to get a mixture that is 70% alcohol, pure acid formula, mixture problems with solutions pdf, problem-solving examples with solutions, example of problem solving in math with solution and answer
