Math 8H Algebra 1 Glencoe McGraw-Hill JoAnn Evans Solving Chemical Mixture Problems.
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Transcript of Math 8H Algebra 1 Glencoe McGraw-Hill JoAnn Evans Solving Chemical Mixture Problems.
Math 8H
Algebra 1 Glencoe McGraw-Hill JoAnn Evans
Solving
Chemical Mixture Problems
Mixture problems were introduced earlier this year. In those problems we saw
different solid ingredients like prunes and apricots, each at their own price,
combined together to form a mixture at a new price.
+ =cost · amount
cost · amount cost · amount
1st ingredie
nt
2nd ingredient
mixture
+ =Chemical mixture problems are another type of mixture problem. Instead of a cost for each ingredient, we’ll consider
the strength of the solution, measured in percents.
% · amount % · amount % · amount+ =
Mr. Williams has 40 ml of a solution that is 50% acid. How much water should he add to make a solution that is 10% acid?
+ =50% ACID SOLUTION
PURE WATE
R
10% ACID SOLUTION
The problem asks for the amount of water Mr. Williams should add.
Let x = amount of water added
We also need to know how much of the 10% solution he’ll end up with.
Let y = amount of new solution
First equation: 40 + x = y
40 x y
The first equation only addressed the amount of the liquids.
40 + x = y
The equation says that Mr. Williams started with 40 ml of a strong acid solution, then added x ml
of water and ended up with y ml of a weaker acid solution.
The second equation in the system needs to address the strength of each solution (percentage
of acid).
% · amount % · amount % · amount+ =
+ =50% ACID SOLUTION
PURE WATER
10% ACID SOLUTION
% · amount
% · amount =+ % · amount
.50 · 40 + .00 · x = .10 · y
y10.x0)40(50.
yx40
Why is the percentage on the water 0%?
Because the % tells what percentage of
ACID is in each solution. There is no
acid in the water added.
40 x y
y10.x0)40(50.
yx40Solve the system using the substitution method.
)x40(10.x0)40(50.
x10.4020 44
x10.16
x160
Mr. Williams needs to add 160 ml of water.
How many liters of acid should Mrs. Bartley add to 4 L of a 10% acid solution to make a solution that is 80%
acid?
+ =10% ACID SOLUTION
PURE ACID
80% ACID SOLUTION
The problem asks for the amount of acid Mrs. Bartley should add.
Let x = amount of acid added
We also need to know how much of the new solution she’ll end up with.
Let y = amount of stronger acid solution
First equation: 4 + x = y
4 x y
4 + x = y
The first equation only addressed the amount of the liquids.
weak acid + pure acid = stronger acid
The second equation in the system needs to address the strength of
each solution (percentage of acid).
+ =10% ACID SOLUTION
PURE ACID
80% ACID SOLUTION
% · amount % · amount=+ % · amount
.10 · 4 + 1.00 · x = .80 · y
y80.x00.1)4(10.
yx4Why is the
percentage on the acid 100%?
Because the % tells what
percentage of ACID is in each
solution. Pure acid is 100% acid.
4 x y
)x4(80.x00.1)4(10.
x80.20.3x40. 40.40.
x80.80.2x x80.x80.
80.2x20. 14x
y80.x00.1)4(10.
yx4Solve the system using the substitution method.
Mrs. Bartley needs to add 14L of acid.
How many liters of water must Miss Elias EVAPORATE from 50 L of a 10% salt solution
to produce a 20% salt solution?
- =10% SALT SOLUTION
PURE WATER
20% SALT SOLUTION
Let x = amount of water lost
Let y = amount of stronger salt solution
First equation: 50 - x = y
50 x y
- =10% SALT SOLUTON
Pure water
20% SALT SOLUTION
% · amount
% · amount
=- % · amount
.10 · 50 - .00 · x = .20 · y
y20.x00.)50(10.
yx50Why is the percentage
on the water 0%? Because the % tells what percentage of
SALT is in each solution. Pure water has 0% salt.
50 x y
Solve the system using the substitution method.
y20.x00.)50(10.
yx50
)x50(20.x00.)50(10.
x20.1005
x20.105
1010
x20.5
x25
25 L of water must be evaporated.
Milk with 3% butterfat was mixed with cream with 27% butterfat to produce 36 L of Half-and-Half with
11% butterfat content. How much of each was used?
+ =Milk with 3%
butterfat
Cream with 27% butterfat
Half-and-Half with 11% butterfat
Let x = amount of milk added
Let y = amount of cream added
First equation:
x + y = 36
This equation says we started with x liters of milk and are adding y liters of cream to produce 36 liters of Half-and-Half.
x y 36
+ =Milk with
3% butterfat
Cream with 27% butterfat
Half-and-Half with 11% butterfat
% · amount
% · amount
=+ % · amount
.03·x + .27·y = .11·36
)36(11.y27.x03.
36yx
12 L of cream and 24 L of milk are needed.
12y
88.2y24.
96.3y24.08.1
96.3y27.y03.08.1
)36(11.y27.)y36(03.
)36(11.y27.x03.
Solve the system:
x y 36
A chemistry experiment calls for a 30% solution of copper sulfate. Mr. McGhee has 40 milliliters of 25% solution. How many milliliters of 60% solution should
he add to make a 30% solution?
+ =25% solution
60% solution
30% solution
% · amount% · amount % · amount+ =Let x = amount 60% solution
Let y = amount of 30% solution40 + x = y
.25(40) + .60(x) = .30y
40 x y
40 + x = y
.25(40) + .60(x) = .30y
10 + .60x = .30(40 + x)
10 + .60x = 12 + .30x
.60x = 2 + .30x
.30x = 2
x ≈ 6.67
Mr. McGhee needs to add 6.67 milliliters of the 60% solution.