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3 Answers

Although there are several valid calculation procedures for this question (see other answers),
chemists prefer simplify the calculations needed by using a general formula that applies to all mixing operations.
Two quantities (h and l ) can be combined or mixed to obtain a desired intermediate quantity i in accordance with:

L= h-i /(h-i) +(i-l) = h-i / h-l and H= i-l / h-l

Where l and h are the lower and higher quantity and H and L are the parts of the higher and lower quantities respectively to be mixed to obtain one part of the desired quantity.
In our case:
h=24; l= 15; and i =18.3 , hence L=5.7/9 and H-3.3/9
And to prepare 90 parts (oz.) of the product:

90L = 5.7 x 90/9 = 57oz. and 90H = 3.3x 90/9 =33 oz.

Regards

Edmondo C.
x = oz of 15% Cu
y = oz of 24% Cu
 
x + y = 90  =>  y = 90 - x
 
(0.15 x + 0.24 y)/90 = 18.3%
Multiply both sides by 100:
(15 x + 24 y)/90 = 18.3
Multiply both sides by 90:
15 x + 24 y = 90(18.3)
Substitute 90 - x for y:
15 x + 24 (90 - x) = 90 (18.3)
15 x - 24 x = 90 (18.3 - 24)
- 9 x = 90 (-5.7)
x = 10 (5,7) = 57 oz
Let F = amount of 15% alloy
Let T = amount of 24% alloy
F     +     T  =90
.15F +  .24T=90×.183
Rewrite as
F      +     T    =  90
150F+240T    =90×183=16470
multiply first equation by 240 and have
240F  +  240T= 21600
150F  +  240T= 16470
subtract
90F               = 5130
F=57 and so T =33
 
 

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