In chemistry class, 8 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
2007-11-27 06:43:11 · 4 answers · asked by burban 2 in Science & Mathematics ➔ Mathematics
Let n be the number of liters of 10% solution to be added:
Write an equation for:
"8 liters of 4% + n liters of 10% = (n + 8) liters of 6%"
8 x 0.04 + n x 0.10 = (n + 8) x 0.06
Expand out:
0.32 + 0.10n = 0.06n + 0.48
Subtract 0.32 from both sides:
0.10n = 0.06n + 0.16
Subtract 0.06n from both sides:
0.04n = 0.16
Divide both sides by 0.04
n = 0.16 / 0.04
n = 4 liters
Double-check:
8000 ml x 4/100 = 320 ml
4000 ml x 10/100 = 400 ml
--------------------------------------
12000 ml x 6/100 = 720 ml
2007-11-27 06:49:07 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
x=#liters of 10% solution
.04*8+.1*x=.06(x+8)
.32+.1x=.06x+.48
32+10x=6x+48
4x=16
x=4
2007-11-27 06:49:11 · answer #2 · answered by chasrmck 6 · 0⤊ 0⤋
volume of 10% solution needed
= {(6 -- 4)/(10 -- 6)}*8L
= 4L
2007-11-27 06:56:23 · answer #3 · answered by sv 7 · 0⤊ 0⤋
8L = 8000ml
0.04*8000ml = 320
0.1 * 8000ml = 800
800-320 = 480
0.1*X = 480
x = 4800ml
4.8L
2007-11-27 06:48:49 · answer #4 · answered by science_guy 5 · 0⤊ 1⤋