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OpenStudy (anonymous):
A chemist needs to mix a 12% acid solution with a 32% acid solution to obtain an 8-liters mixture consisting of 20% acid. How many liters of each of the acid solutions must be used?
OpenStudy (anonymous):
I need to know, pliz? Anyone that's good enough at math?
OpenStudy (anonymous):
we have 2 different Acid . yea ?
OpenStudy (anonymous):
K
OpenStudy (anonymous):
Again ;)
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OpenStudy (anonymous):
Alright....
OpenStudy (anonymous):
did you understand question ?
OpenStudy (anonymous):
I did.
OpenStudy (anonymous):
What things question wants ?
OpenStudy (anonymous):
i AM OUT OF HERE! What do you think? I'll make a new question and tag better helpers...
tnx for the help.
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OpenStudy (anonymous):
!!! sorry me !
OpenStudy (anonymous):
I want to help u ;)
OpenStudy (anonymous):
You are wasting my time like crazy....
OpenStudy (anonymous):
No no ...!
never !
Sorry me friend ...
OpenStudy (zpupster):
.12ofxLiters
.32of y liters
total volume is 8 L
so x+y=8
and
.12x+.32y =
.2 of 8L
we have 2 equations
x+y=8
.12x+.32y =.1.6
make it easier mult 2nd equation by 100
x+y=8
12x+32y =160
lets use the elimination methos mult first equaiotn by 12 and subtract one from the other
12x+12y=96
12x+32y =160
20y=64
y=3.2
y= 3.2L
do the check
and subbing
8-3.2=x
x=4.8L
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