7x/10 (That's 7/10 times x). So if x = 10 gallons, then there are 7 gallons of bleach in the solution. (7gal/10gal = 0.70 = 70%). This answer if correct if the percentage is by volume. Otherwise, the question can not be answered without knowledge of the basis of the percentage.
4314.9 grams
135 gallons. Let the multiple of 90 gallons to add be x, then using the percentages of antifreeze in the solutions: (25 + 50x) ÷ (1 + x) = 40 ⇒ 25 + 50x = 40 + 40x ⇒ 10x = 15 ⇒ x = 1.5 Therefore add 90 x 1.5 = 135 gallons.
You will need 3.2 gallons.
Suppose G gallons are required. At 70% concentration, this will contain 0.7*G gallons of antifreeze. The 70 gallons of 25% contains 17.5 gallons of the active ingredient. In total, then, there are (G + 70) gallons containing 0.7G + 17.5 gallons of antifreeze and this represents 60%. So (0.7G + 17.5) / (G + 70) = 60/100 10*(0.7G + 17.5) = 6*(G + 70) 7G + 175 = 6G + 420 G = 420 - 175 = 245. Answer: 245 gallons.
add 25ml more of solution x * 20 = 100 * 25 x = 25
10 gallons..
25 gallons
50 gallons @ 3% must be added.
132 gallons 16 : 84 of 60 gallons = 9.6 : 50.4 gallons 80 : 20 of 132 gallons = 105.6 : 26.4 gallons added = 115.2 : 76.8 gallons = 1.5:1 ( 60:40)
614
630
70gallons
420 gallons. Here's how to find it; If X = Gallons of 80% mix, 0.8*X + 0.1*70 = 0.7*(X+70) Which we can then solve, 0.8X + 7 = 0.7X + 49 0.1X = 42 X = 420
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
70 gallons of 20% solution contains 70*0.2 = 14 gallons of antifreeze. Suppose you need G gallons of the 80% antifreeze solution. This will contain 0.8*G gallons of antifreeze. Total volume of solution = G + 70 gallons Volume of antifreeze required in this solutions to make it a 70% solution is 0.7*(G + 70) = 0.7G + 49 gallons. Volume of antifreeze = 14 + 0.8G gallons So 0.7G + 49 = 14 + 0.8G 0.7G + 35 = 0.8G 35 = 0.1G 350 = G Answer: 350 gallons.
The mass of sodium hypochlorite in 2,5 kg solution is 131,25 g.Sodium chloride is only the product of a decomposition.
4314.9 grams