SlideServe
Télécharger Obtenir Pro
Menu
Accueil Pro
Préférences
ApparenceLight Aide et commentaires
© SlideServe · v2026

Solving Mixture Problems in Chemistry: A Step-by-Step Guide

nayda-cobb avatar
nayda-cobb
20 diapositives · 369 Views · Oct 31, 2014

This article provides a comprehensive guide on how to tackle mixture problems involving different concentrations of substances. We explore examples using solutions with varying acid percentages to demonstrate clear problem-solving steps. From setting up charts to utilizing algebraic equations, we break down each stage to ensure understanding. Readers will learn how to determine the correct amounts of each solution needed to create a desired concentration. Engage with practical examples, from acid mixtures to coffee blends, and enhance your problem-solving skills in chemistry.

Solving Mixture Problems in Chemistry: A Step-by-Step Guide - slide 1
01 / 20Solving Mixture Problems in Chemistry: A Step-by-Step Guide
1 / 17

Télécharger la présentation

Solving Mixture Problems in Chemistry: A Step-by-Step Guide
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation

  1. 10.8 Mixture Problems Goal: To solve problems involving the mixture of substances

  2. Mixture Problems One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid?

  3. Steps to Solve Mixture Problems • Set up a chart (4x4)

  4. Steps to Solve Mixture Problems • Convert the percentages to decimals and fill out the chart • Multiply going across the chart • Add going down the chart • Set up 2 equations with 2 variables (system) • Solve the system by substitution or addition

  5. Let x = y = x 0.80 .80(x) One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? y 0.30 .30(y) .62(200) 200 0.62 124

  6. 8x + 3y =1240 Y= 200-x 8x + 3 (200-x) =1240 Y = 200 -128 8x +600 -3x =1240 x 0.80 .80(x) One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? 5x +600 =1240 Y = 72 L 5x = 640 y 0.30 .30(y) X= 128 L .62(200) 200 0.62 124

  7. x 0.60 .60(x) A chemist has one solution that is 60% acid and another that is 30% acid. How much of each solution is needed to make a 750ml solution that is 50% acid? y 0.30 .30(y) .50(750) 750 0.50 375

  8. x 0.28 .28(x) A chemist has one solution that is 28% oil and another that is 40% oil. How much of each solution is needed to make a 300 L solution that is 36% oil? y 0.40 .4(y) .36(300) 300 0.36 108

  9. Try to make your own chart • How many gallons of a 50% salt solution must be mixed with 60 gallons of a 15% solution to obtain a solution that is 40% salt?

  10. How many gallons of a 50% salt solution must be mixed with 60 gallons of a 15% solution to obtain a solution that is 40% salt?

  11. System x + 60 =y 0 .5x + 9 = 0.4y 5x +90 = 4y 5x + 90 = 4 (x +60) 5x + 90 = 4x + 240 x + 90 =240 x =150 gallons 150 + 60 = y 210 gallons =y

  12. Coffee Beans • How many pounds of coffee beans selling for $2.20 per pound should be mixed with 2 pounds of coffee beans selling for $1.40 per pound to obtain a mixture selling for $2.04 per pound?

  13. How many pounds of coffee beans selling for $2.20 per pound should be mixed with 2 pounds of coffee beans selling for $1.40 per pound to obtain a mixture selling for $2.04 per pound?

  14. System X + 2 =y 2.20x +2.80 = 2.04 y

  15. Your Turn • Come up with your own mixture word problem. Make it interesting! • Remember to include:

  16. Assignment:Page 462 (1 -9) odd

  17. 1 .20 .20 x .00 .00(x) .20 x+1 .06

  18. Vocabulary • Mixture- two substances combined • Concentrate or Solution- how much non-water is mixed (juice) • 10% solution -10% concentration and 90% water