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Math 8H Algebra 1 Glencoe McGraw-Hill JoAnn Evans Solving Chemical Mixture Problems

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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 1 st ingredient 2 nd 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 + =

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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 WATER 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

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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 + =

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+= 50% ACID SOLUTION PURE WATER 10% ACID SOLUTION % · 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. 40 x y

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Solve the system using the substitution method. Mr. Williams needs to add 160 ml of water.

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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

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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).

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+= 10% ACID SOLUTION PURE ACID 80% ACID SOLUTION % · 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. 4 x y

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Solve the system using the substitution method. Mrs. Bartley needs to add 14L of acid.

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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 50x y

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-= 10% SALT SOLUTON Pure water 20% SALT SOLUTION % · 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. 50x y

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Solve the system using the substitution method. 25 L of water must be evaporated.

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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

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+= Milk with 3% butterfat Cream with 27% butterfat Half- and -Half with 11% butterfat % · amount = +.03 · x +.27 · y =.11 · 36 12 L of cream and 24 L of milk are needed. Solve the system: x y36

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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 + = Let x = amount 60% solution Let y = amount of 30% solution 40 + x = y.25(40) +.60(x) =.30y 40 x y

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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.

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10.8 Mixture Problems Goal: To solve problems involving the mixture of substances.

10.8 Mixture Problems Goal: To solve problems involving the mixture of substances.

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