Wiki User
∙ 12y ago6 ounces
Wiki User
∙ 12y agoTo find the amount of pure acid to add, set up an equation based on the amount of acid in the original solution and the final solution. Let x be the amount of pure acid to add. The amount of acid in the original solution is 0.4 × 12 = 4.8 ounces. The amount of acid in the final solution is 0.6 × (12 + x) = 4.8 + 0.6x ounces. Therefore, to get a 60% acid solution, you would need to add 10.67 ounces of pure acid.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
how many grams of glucose must be added to 525g of 2.5 percent leg mass glucose solution?and give the furmela?
The final percent concentration of the solution would be approximately 12.0% methanol. This is calculated by dividing the volume of methanol by the total volume of the solution (600 ml / 5000 ml) and then multiplying by 100 to get the percentage.
Acids produce hydrogen ions as the only positive ions in aqueous solution. When an acid dissolves in water, it donates a hydrogen ion (H+) to the solution, resulting in the formation of hydrogen ions.
Acetic acid is added to the solution, but the pH of the solution does not change. Sodium hydroxide, a base, is added to the solution, but the pH of the solution does not change.
add 25ml more of solution x * 20 = 100 * 25 x = 25
NOT: 15 ounces to make 10%Correct answer:Twenty ouncesCalculus:Let it be M ounces. Then:20%*(M ounce) + 5%*(40 ounce) = 10%*(M+40 ounce)20*M + 200 = 10*M + 400(20-10)*M = 400-20010M = 200M = 20
2%
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
12*(90%)=10.8=X*.25 10.8/.25=X=43.2 43.2-12=31.2 31.2 grams of water must be added
10 liters.
add 4 parts water per part solution
20%
The original mixture contains 41.4 ounces of glycol. for this to be 30 percent of the mixture, the total mixture must be 138 ounces, so 46 ounces of water must be added.
4 litres
50 gallons @ 3% must be added.
98 mL