Wiki User
∙ 13y agoright 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
Wiki User
∙ 13y agox=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?
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
Let x represent the gallons of 10% ammonia solution. The total volume of the mixture is x + 50 gallons. The equation for the mixture is: 0.10x + 0.30(50) = 0.15(x + 50). Solving this equation gives x = 50 gallons of the 10% ammonia solution needed.
2 gallons.
To get a 2% acid solution, you need to dilute the 50% acid solution with water. Since the final volume is 2 gallons, you will need to mix 2 gallons of water with the 2 gallons of 50% acid solution to get a 2% acid solution.
0.25 gallons of water (or 1 quart)
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.
.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300
2%