Question 385992: I hope I got the right section. I'm having trouble with "Chemical Word Problems" (that's what my textbook says). The original problem is "One solution was 5% bromide,, and the other was 40% bromide. How much of each should be used to get 60 milliliters of a solution that is 12% bromide?"
The 5% bromide "container" I labelled N. The 40% I labelled D (per the instructions of my text).
What I did: I realized that I have two equations: N+d=100 (according to the textbook)which means that n= 100-d ; So I wrote this: ( ) + ( ) = 100 I plugged "n" into the first paranthes, and d into the second. I turned the percents into decimals (5%= 0.5 and 40%= 0.4). I plugged it in, to get:
0.5(N) + 0.4(D) = (100)0.12 (0.12=12%) Since N=100-D, I replaced N with 100-D to get:
0.5(100-D) + 0.4D = 12 I solved to get 50 - 0.5D + 0.4D = 12 --> 50 - .1D =12 ------> Subtracted 50 from both sides to get ----->-.1DN= -38 ----> divided both sides by -.1 ----> and I got 380. Logically, that can't be, because N+D=100. If D = 380 which is greater than 100... What am I doing wrong?!? I've done other problems of this nature, using the SAME process, and they have all been wrong! Please Help!
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Try this!
Let x = the required number of ml. of the 5% bromide solution. Then (60-x) will be the required number of ml. of the 40% bromide solution and the sum of these two amounts (x+60-x) will equal the 60 ml. of the final 12% bromide solution.
Here's the equation after changing the percentages to their decimal equivalents:
0.05x+0.4(60-x) = 0.12(60)
Simplify and solve for x.
0.05x+24-0.4x = 7.2 Combine like-terms.
0.05x-0.4x+24 = 7.2
-0.35x+24 = 7.2 Subtract 24 from both sides.
-0.35x = -16.8 Divide both sides by -0.35
x = 48 and 60-x = 12
So you'll need to mix 48 ml. of 5% bromide solution with 12 ml. of 40% bromide solution to obtain 60 ml. of 12% bromide solution.
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