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Grade 10 Mixture Problems

Grade 10 Mixture Problems. A simple presentation by Mr. Agostini. The Problem.

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Grade 10 Mixture Problems

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  1. Grade 10 Mixture Problems A simple presentation by Mr. Agostini

  2. The Problem • A chemistry teacher needs to make 20 L of 40% sulfuric acid solution. The two containers of acid solutions available contain 30% sulfuric acid and 50% sulfuric acid. How many litres of each solution must be mixed to make a container with 40% sulfuric acid solution.

  3. The Idea • Did you know: 30% sulfuric acid contain is made up of 30% acid and the rest is water to dilute the solution. So in a 100 L container, 30L is acid and 70 L is water. • You are going to mix the containers together to obtain 20 litres of a solution that 40% of it will be sulfuric acid. How many litres from each container will we need to use to make this mixture. • If you had to guess, what would you say: Recall 30% acid in one container and 50% in the other. You need 40% in the mixture. How many litres from each mixture do you think we would need.

  4. Defining your Variables First • Define your variables first: (Remember the last sentence in the question usually tells us what we are looking for) • Let x be the amount in L that is poured from container 1 at 30% acid solution • Let y be the amount in L that is poured from container 2 at 50% acid solution

  5. Creating your two equations:Equation 1 • Equation 1 is usually build on the amount of liquid (in this case) that is required. We need 20 L. • Therefore: x amount is being poured in from container 1 and y amount is being poured in from container 2, to make 20 L. Their sum is 20 then. • So the equation is: • x + y = 20 (Equ: 1)

  6. Creating your two equations:Equation 2 • We create equation 2 from the amount of pure acid that is needed from each container and how much pure acid is in the 20L container. (Recall: 40% of 20 L of pure acid is required) • The amount(L) of pure acid from container 1 will be 30% of x amount(L) poured into the beaker or 0.30x. • The amount(L) of pure acid from container 2 will be 50% of y amount(L) poured into the beaker or 0.50y • Therefore, we combine 30% of x with 50% of y to obtain 40% of 20 L • The second equation is: • 0.30x + 0.50y = 0.40(20) (Equ: 2)

  7. So what do we have: • Let x = amount in L of 30% sulphuric acid soluton. • Let y = amount in L of 50% sulphuric acid soluton. • (1) x + y = 20 • (2) 0.30x + 0.50y = 0.40(20) • Or • (2) 0.3x + 0.5y = 8

  8. When Solving you can use any method you wish. Yay!!!!! • Elimination Method: ( x -3) ( x 10) -3x – 3y = -60 3x + 5y = 80 Back sub into x + y = 20 x + 10 = 20 x = 20 – 10 x = 10 Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution. x + y = 20 0.3x + 0.5y = 8 + 2y = 20 y = 10

  9. When Solving you can use any method you wish. Yay!!!!! • Substitution Method: (1) x + y = 20 Back sub into x + y = 20 (2) 0.3x + 0.5y = 8 10 + y = 20 y = 20 – x sub into (2) y = 20 - 10 0.3x + 0.5(20 – x) = 8 y = 10 0.3x + 10 – 0.5x = 8 Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution. 0.3x – 0.5x = 8 - 10 – 0.2x = - 2 x = 10

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