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Standardized Test PrepSolutions Preview Understanding Concepts Reading Skills Interpreting Graphics.

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Presentation on theme: "Standardized Test PrepSolutions Preview Understanding Concepts Reading Skills Interpreting Graphics."— Presentation transcript:

1 Standardized Test PrepSolutions Preview Understanding Concepts Reading Skills Interpreting Graphics

2 Standardized Test PrepSolutions Understanding Concepts 1. An industrial chemist stirs some crystals into water. The resulting liquid appears completely uniform under a microscope. What is the liquid? A. a colloid B. a solution C. a suspension D. a heterogeneous mixture

3 Standardized Test PrepSolutions Understanding Concepts, continued 1. An industrial chemist stirs some crystals into water. The resulting liquid appears completely uniform under a microscope. What is the liquid? A. a colloid B. a solution C. a suspension D. a heterogeneous mixture

4 Standardized Test PrepSolutions Understanding Concepts, continued 2. After the lungs take in oxygen during respiration, oxygen gas is dissolved in the bloodstream. Under what conditions can the most oxygen be dissolved in the blood? F. high blood pressure and high body temperature G. low blood pressure and low body temperature H. low blood pressure and high body temperature I. high blood pressure and low body temperature

5 Standardized Test PrepSolutions Understanding Concepts, continued 2. After the lungs take in oxygen during respiration, oxygen gas is dissolved in the bloodstream. Under what conditions can the most oxygen be dissolved in the blood? F. high blood pressure and high body temperature G. low blood pressure and low body temperature H. low blood pressure and high body temperature I. high blood pressure and low body temperature

6 Standardized Test PrepSolutions Understanding Concepts, continued 3. Potassium sulfate, K 2 SO 4, has a molar mass of 174 g. If potassium sulfate is the solute in 2 L of a solution that has a concentration of 0.25 M, how many grams of potassium sulfate are in the solution? A.1,392 g B.348 g C.87 g D.43.5 g

7 Standardized Test PrepSolutions Understanding Concepts, continued 3. Potassium sulfate, K 2 SO 4, has a molar mass of 174 g. If potassium sulfate is the solute in 2 L of a solution that has a concentration of 0.25 M, how many grams of potassium sulfate are in the solution? A.1,392 g B.348 g C.87 g D.43.5 g

8 Standardized Test PrepSolutions Understanding Concepts, continued 4. The molecules of compound A have unevenly distributed electric charge. The molecules of compound B have evenly distributed electric charge. Which compound is more likely to dissolve easily in water? Explain your reasoning.

9 Standardized Test PrepSolutions Understanding Concepts, continued 4. The molecules of compound A have unevenly distributed electric charge. The molecules of compound B have evenly distributed electric charge. Which compound is more likely to dissolve easily in water? Explain your reasoning. Answer: Compound A is polar, so it will be most likely to dissolve in water because water is polar.

10 Standardized Test PrepSolutions Reading Skills SUPERCRITICAL DECAFFEINATION One of the most common ways to remove caffeine from coffee and to preserve the flavor of the beverage is to use a solvent that dissolves the caffeine but leaves the rest of the plant material undissolved. One of the main difficulties with removing caffeine is that caffeine is a nonpolar compound. Therefore, a nonpolar solvent is required to dissolve caffeine. But most effective nonpolar solvents are poisonous to humans. Although carbon dioxide, CO 2, is a safe nonpolar compound, it is a gas under normal conditions and cannot act as a solvent for caffeine.

11 Standardized Test PrepSolutions Reading Skills, continued When both the pressure and temperature of a fluid are increased beyond a specific (or critical) point, a fluid has some properties of liquids and some properties of gases. Fluids under these conditions are called supercritical fluids. In the 1960s, coffee companies began using supercritical CO 2 to extract caffeine. The gaslike behavior of CO 2 allows its molecules to penetrate into the plant material. The liquid aspects of CO 2 allow it to dissolve caffeine molecules.

12 Standardized Test PrepSolutions Reading Skills, continued 5. Once the caffeine is removed from the plant materials, how might the caffeine be recovered from the supercritical CO 2 ? F. by forcing the CO 2 back through the plant materials G. by lowering the temperature and pressure of the CO 2 H. by changing the polarity of the caffeine molecules I. by using a harsher solvent

13 Standardized Test PrepSolutions Reading Skills, continued 5. Once the caffeine is removed from the plant materials, how might the caffeine be recovered from the supercritical CO 2 ? F. by forcing the CO 2 back through the plant materials G. by lowering the temperature and pressure of the CO 2 H. by changing the polarity of the caffeine molecules I. by using a harsher solvent

14 Standardized Test PrepSolutions Reading Skills, continued 6. Why can’t water be used to dissolve the caffeine found in coffee and tea?

15 Standardized Test PrepSolutions Reading Skills, continued 6. Why can’t water be used to dissolve the caffeine found in coffee and tea? Answer: Caffeine is nonpolar and water is polar.

16 Standardized Test PrepSolutions Interpreting Graphics The graphic below represents three beakers that contain 500 g of water with different amounts of sugar. One is unsaturated, one is saturated, and one is supersaturated. Use this graphic to answer questions 7–8.

17 Standardized Test PrepSolutions Interpreting Graphics, continued 7. How many grams of sugar are there altogether in the three cylinders? A. 2,670 g C. 534 g B. 1,602 g D. 267 g

18 Standardized Test PrepSolutions Interpreting Graphics, continued 7. How many grams of sugar are there altogether in the three cylinders? A. 2,670 g C. 534 g B. 1,602 g D. 267 g

19 Standardized Test PrepSolutions Interpreting Graphics, continued 8. Which solution would show the most dramatic results if an additional crystal of sugar were dropped into it? What would those results be?

20 Standardized Test PrepSolutions Interpreting Graphics, continued 8. Which solution would show the most dramatic results if an additional crystal of sugar were dropped into it? What would those results be? Answer: The solution that contains 220 g of sugar per 100 g of water is the most concentrated solution, so it is the supersaturated solution. If a sugar crystal were added, all of the excess sugar would crystallize out of the solution until the solution was saturated again.

21 Standardized Test PrepSolutions Interpreting Graphics, continued The graph below shows how the solubility of a mystery solid in water depends on temperature. Use this graph to answer questions 9–10.

22 Standardized Test PrepSolutions Interpreting Graphics, continued 9. The solubility of the solid is 88 g/100 g H 2 O at 20 ˚C. What is the solubility of the substances at 60 ˚C? F. 88 g/100 g H 2 O H. 122 g/100 g H 2 O G. 100 g/100 g H 2 O I. 264 g/100 g H 2 O

23 Standardized Test PrepSolutions Interpreting Graphics, continued 9. The solubility of the solid is 88 g/100 g H 2 O at 20 ˚C. What is the solubility of the substances at 60 ˚C? F. 88 g/100 g H 2 O H. 122 g/100 g H 2 O G. 100 g/100 g H 2 O I. 264 g/100 g H 2 O

24 Standardized Test PrepSolutions Interpreting Graphics, continued 10. Water normally boils at 100 ˚C. The scientists performing this experiment were able to measure the solubility at 100 ˚C because the solution was not boiling. Why was the solution not boiling at 100 ˚C ?

25 Standardized Test PrepSolutions Interpreting Graphics, continued 10. Water normally boils at 100 ˚C. The scientists performing this experiment were able to measure the solubility at 100 ˚C because the solution was not boiling. Why was the solution not boiling at 100 ˚C ? Answer: The dissolved solute raised the boiling point of the solution.


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