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How many gallons of a 10 percent ammonia solution should be mixed with 50 gallons of a 30 percent ammonia solution to make a 15 percent ammonia solution?
Wiki User
∙ 12y ago
Make an equation:
x(.10) = 50gal(.15)
Solve algebraically:
x = 75 gal
Wiki User
∙ 12y ago
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How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent solution?
600ml.
What percent of starch solution is used for H2S gas analysis with Iodine?
0.5 to 1 percent starch solution should be used
How much pure acid should be mixed with 4 gallons of a 50 percent acid sollution in order to get an 80 percent acid solution?
Assuming that you mean that your pure acid is twice as concentrated as your 50% acid. Pretending your pure acid is at 1 mol/gallon and the 50% acid is 0.5 mol/gallon1 molgallon-1 * x gallons = x mol0.5 molgallon-1 * 4 gallons = 2 mol(x + 2) mol for (x+4) gallons = 0.8(x+2) mol / (x+4) gallons= 0.8 molgallon-1x + 2 = 0.8x + 3.20.2x = 1.2x = 6add 6 gallons of pure acid to the 4 gallons to make 10 gallons of 80% acid solution
What is sudsy ammonia?
Sudsy ammonia is just a mixture of regular household ammonia solution, water, and liquid detergent. One recipe is 2/3 water, 1/3 "full-strength" household ammonia, and a squirt of liquid detergent. A link for full instructions should be below this.
If you have a 300ml of a 15 percent hydrochloric acid solution in dilute the solution to 1000ml with sterile water what percent should appear on the label?
The concentration of the diluted solution will be 15(300/1000) = 4.5 %, if the percent is expressed on a weight/volume basis.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze 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.
How much pure acid should be mixed with 2 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?
pH less than 7
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 much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be mixed with 2 gallons of a 50 percent acid solution in order to get an 2 percent acid solution?
tv=tv, so 0,25x50=0,1v; v=12,5/0,1; v=125ml of water. So the solution must have 125ml of water for the title to be 10%, then we must add 125-50ml(75ml) of water to it.
What happens when ammonium chloride solution is heated with caustic soda solution?
ammonia should release on heating.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How many gallons of 80 percent solution must be mixed with 100 gallons of a 15 percent antifreeze solution to get a mixture that is 70 percent antifreeze?
For these types of problems, I recommend finding the initial amount of antifreeze. To find this value, multiply the percentage of antifreeze (15%) by the amount of solution (100 gallons). This gets:0.15*100=15 gallons of antifreeze*Now we need make an equation that represents this problem. We let x be the number of gallons of 80% solution.(.80x+15)/(x+100)=0.70The top of the equation represents the amount of antifreeze, while the bottom is the total volume of the solution. If we divide these values, we should get the final concentration %, .70.Now we just need to manipulate the equation algebraically:.80x+15=.70x+70.10x=55x=550 gallons of 80% solution*If the concentration of a solution is given in percentage, it is probably referring to mass, not volume. Because of this, the values for the volume of antifreeze are actually untrue. But because the density of both substances remains the same, the final answer should be correct.
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
How much water should be added to 30 gallons of a solution that is 70 percent antifreeze in order to get a mixture that is 60 percent antifreeze?
First off do not use tap water. Use only distilled water. You will need to add 3 gallons of distilled water to the solution to get 60% antifreeze and 40% distilled water or a 60/40 mix.
