10 gallons..
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)
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
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.
mordant black 2 is used as an indicator in comleometric titrations, usuall against Di Na EDTA Solution. the indicator is mixed with NaCl/ KI in appropriate quantity...
2.5 is the answer....5(.04)+.10(x)= .06(5+x).2+.1x=.3+.06x.04x=.1or 4x=10 so x =2.5
25 gallons
50 gallons @ 3% must be added.
614
630
70gallons
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)
x=45
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
pH less than 7
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.
x = 10% solution y = 35% solution .1x+.35y=.20*12 x+y=12 y=12-x .1x+.35(12-x)=.20*12 .1x+4.2-.35x=2.4 -.25x=-1.8 .25x=1.8 x=7.2 gallons y=4.8 gallons
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.