Math 8H

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## Math 8H

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**Math 8H**Solving Chemical Mixture Problems Algebra 1 Glencoe McGraw-Hill JoAnn Evans**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. + = 1st ingredient 2nd ingredient mixture + = cost · amount cost · amount cost · amount 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? x 40 y + = 50% ACID SOLUTION 10% ACID SOLUTION PURE WATER 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**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**y**x 40 + = 50% ACID SOLUTION PURE WATER 10% ACIDSOLUTION + = % · amount % · amount % · amount .50 · 40 + .00 · x = .10 · y 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.**Solve the system using the substitution method.**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? x 4 y + = 10% ACID SOLUTION 80% ACID SOLUTION PURE ACID 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**The first equation only addressed the amount of the liquids.**4 + x = y weak acid + pure acid = stronger acid The second equation in the system needs to address the strength of each solution (percentage of acid).**x**4 y + = 10% ACID SOLUTION 80% ACID SOLUTION PURE ACID + = % · amount % · amount % · amount .10 · 4 + 1.00 · x = .80 · y Why is the percentage on the acid 100%? Because the % tells what percentage of ACID is in each solution. Pure acid is 100% acid.**Solve 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? 50 x y - = 10% SALT SOLUTION 20% SALT SOLUTION PURE WATER Let x = amount of water lost Let y = amount of stronger salt solution First equation: 50 - x = y**y**50 x - = 10% SALT SOLUTON 20% SALT SOLUTION Pure water - = % · amount % · amount % · amount .10 · 50 - .00 · x = .20 · y Why is the percentage on the water 0%? Because the % tells what percentage of SALT is in each solution. Pure water has 0% salt.**Solve the system using the substitution method.**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? x y 36 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 Cream with 27% butterfat Half-and-Half with 11% butterfat Milk with 3% butterfat + = + = % · amount % · amount % · amount .03·x + .27·y = .11·36 Solve the system: 12 L of cream and 24 L of milk are needed.**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? x y 40 + = 25% solution 60% solution 30% solution + = % · amount % · amount % · amount Let x = amount 60% solution Let y = amount of 30% solution 40 + x = y .25(40) + .60(x) = .30y**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.