grade 10 mixture problems
Download
Skip this Video
Download Presentation
Grade 10 Mixture Problems

Loading in 2 Seconds...

play fullscreen
1 / 9

Grade 10 Mixture Problems - PowerPoint PPT Presentation


  • 85 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Grade 10 Mixture Problems' - olathe


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
grade 10 mixture problems

Grade 10 Mixture Problems

A simple presentation by Mr. Agostini

the problem
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.
the idea
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.
defining your variables first
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
creating your two equations equation 1
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)
creating your two equations equation 2
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)
so what do we have
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
when solving you can use any method you wish yay
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

when solving you can use any method you wish yay1
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

ad