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How Many Liters Of A 60 Acid Solution? Update New

How Many Liters Of A 60 Acid Solution

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How Many Liters Of A 60 Acid Solution
How Many Liters Of A 60 Acid Solution

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.

How much of a 60% acid solution must be mixed with 50 gallons?

1 Expert Answer

We will call these solutions 1, 2, and 3. We have a 60% acid solution (solution 1), then we have a 50 gallon, 24% solution (solution 2).


How many liters of each solution should be mixed to obtain 10 liters of a 40% acid solution?4.2.65

How many liters of each solution should be mixed to obtain 10 liters of a 40% acid solution?4.2.65
How many liters of each solution should be mixed to obtain 10 liters of a 40% acid solution?4.2.65

Images related to the topicHow many liters of each solution should be mixed to obtain 10 liters of a 40% acid solution?4.2.65

How Many Liters Of A 60 Acid Solution
How Many Liters Of Each Solution Should Be Mixed To Obtain 10 Liters Of A 40% Acid Solution?4.2.65

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 .

What amount of a 60 acid solution must be mixed with 30 solution to produce 300ml of a 50 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 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.


A vessel has 60 litres of solution of acid and water having 80% acid. How much water be added

A vessel has 60 litres of solution of acid and water having 80% acid. How much water be added
A vessel has 60 litres of solution of acid and water having 80% acid. How much water be added

Images related to the topicA vessel has 60 litres of solution of acid and water having 80% acid. How much water be added

A Vessel Has 60 Litres Of Solution Of Acid And Water Having 80% Acid. How Much Water Be Added
A Vessel Has 60 Litres Of Solution Of Acid And Water Having 80% Acid. How Much Water Be Added

How many mL of 60 percent acid should be added to pure acid to make 50 mL of 85 percent acid?

You need to add 18.75 mL of 60% solution to pure acid to make 50 mL of 85% acid.

How many liters of a 90 of concentrated acid needs to be mixed with a 75?

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.

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.


A vessel has 60 litres of solution of acid and water having 80%acid. How much water be added

A vessel has 60 litres of solution of acid and water having 80%acid. How much water be added
A vessel has 60 litres of solution of acid and water having 80%acid. How much water be added

Images related to the topicA vessel has 60 litres of solution of acid and water having 80%acid. How much water be added

A Vessel Has 60 Litres Of Solution Of Acid And Water Having 80%Acid. How Much Water Be Added
A Vessel Has 60 Litres Of Solution Of Acid And Water Having 80%Acid. How Much Water Be Added

How many Litres of water must be added to 20 Litre of 10% solution of salt and water to produce a 6% solution?

Total Mixture = (20+x) Litre . Salt = 6% . x = (40/3) = 13(1/3) Litre. Hence, 13(1/3) Litre water must be added to make the salt 6% of the Solution..

How many Litres of 90% concentrated acid do I need?

The answer is (C) 1 6 L. Explore more such questions and answers at BYJU’S. Was this answer helpful?

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