right now you have 7 gallons of solution, 50% acid,
Let S = amount of solution, and A = amount of acid.
A1 = (50%)*S1 = 0.5 * 7 gal = 3.5 gal
the desired % is 80% = 0.8
we add x amount of pure (100% solution), so for each x gallons we add:
A2 = A1 + x and S2 = S1 + x ; also, A2 = 0.8 * S2
so we have 0.8 * S2 = A1 + x --> 0.8 * ( S1 + x) = A1 + x
Substitute S1 = 7 and A1 = 3.5, and solve for x.
0.8 * ( 7 + x) = 3.5 + x ---> 5.6 + 0.8*x = 3.5 + x
2.1 - 0.2*x = 0 : x = 10.5 gallons
x=45
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.
2 gallons.
0.25 gallons of water (or 1 quart)
.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
x=45
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.
pH less than 7
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
2 gallons.
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.
0.25 gallons of water (or 1 quart)
2%
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.
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.