EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119227946
Author: Willard
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 14, Problem 41RQ
Interpretation Introduction
Interpretation:
Wilted and limp appearance of lettuce leaf in contact with salad dressing of salt and vinegar and its crisp appearance in contact with plain water have to be explained.
Concept Introduction:
Solution is used for description of any system that consists of mixture of substances that is dissolved in another substance. It has two components, solute and solvent. Component that is present in less quantity and is dissolved in another component is called solute while the one that dissolves solute in it and is present in more amounts is called solvent.
Expert Solution & Answer
Trending nowThis is a popular solution!
Chapter 14 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
Ch. 14.1 - Prob. 14.1PCh. 14.2 - Prob. 14.2PCh. 14.3 - Prob. 14.3PCh. 14.4 - Prob. 14.4PCh. 14.4 - Prob. 14.5PCh. 14.4 - Prob. 14.6PCh. 14.4 - Prob. 14.7PCh. 14.4 - Prob. 14.8PCh. 14.4 - Prob. 14.9PCh. 14.4 - Prob. 14.10P
Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Starch contains CC, CH, CO, and OH bonds. Hydrocarbons have only CC and CH bonds. Both starch and hydrocarbons can form colloidal dispersions in water. Which dispersion is classified as hydrophobic? Which is hydrophilic? Explain briefly.arrow_forwardIn your own words, explain why (a) seawater has a lower freezing point than fresh water. (b) salt is added to the ice in an ice cream maker to freeze the ice cream faster.arrow_forwardRefer to Figure 13.10 ( Sec. 13-4b) to answer these questions. (a) Does a saturated solution occur when 65.0 g LiCl is present in 100 g H2O at 40 C? Explain your answer. (b) Consider a solution that contains 95.0 g LiCl in 100 g H2O at 40 C. Is the solution unsaturated, saturated, or supersaturated? Explain your answer. (c) Consider a solution that contains 50. g Li2SO4 in 200. g H2O at 50 C. Is this solution unsaturated, saturated, or supersaturated? Explain your answer. Figure 13.10 Solubility of ionic compounds versus temperature.arrow_forward
- Fluoridation of city water supplies has been practiced in the United States for several decades. It is done by continuously adding sodium fluoride to water as it comes from a reservoir. Assume you live in a medium-sized city of 150,000 people and that 660 L (170 gal) of water is used per person per day. What mass of sodium fluoride (in kilograms) must be added to the water supply each year (365 days) to have the required fluoride concentration of 1 ppm (part per million)that is, 1 kilogram of fluoride per 1 million kilograms of water? (Sodium fluoride is 45.0% fluoride, and water has a density of 1.00 g/cm3.)arrow_forward6-111 As noted in Section 6-8C, the amount of external pressure that must be applied to a more concentrated solution to stop the passage of solvent molecules across a semipermeable membrane is known as the osmotic pressure The osmotic pressure obeys a law similar in form to the ideal gas law (discussed in Section 5-4), where Substituting for pressure and solving for osmotic pressures gives the following equation: RT MRT, where M is the concentration or molarity of the solution. (a) Determine the osmotic pressure at 25°C of a 0.0020 M sucrose (C12H22O11) solution. (b) Seawater contains 3.4 g of salts for every liter of solution. Assuming the solute consists entirely of NaCl (and complete dissociation of the NaCI salt), calculate the osmotic pressure of seawater at 25°C. (c) The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose (C6H12O6) will be isotonic with blood? (d) Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of this enzyme in 210. mL of solution has an osmotic pressure of 0.953 torr at 25°C. What is the molar mass of lysozyme? (e) The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine the protein's molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.arrow_forwardA pharmacist prepares an isotonic saline solution for intravenous infusion. Instead of preparing a 0.15 M solution, a 1.5 M solution is prepared. What would happen to the red blood cells if this erroneously prepared solution is infused?arrow_forward
- Match each of the following statements about the dissolving of the ionic solid NaCl in water with the term hydrated ion, hydrogen atom, or oxygen atom. a. A Na+ ion surrounded with water molecules b. A Cl ion surrounded with water molecules c. The portion of a water molecule that is attracted to a Na+ ion d. The portion of a water molecule that is attracted to a Cl ionarrow_forwardFor each of the following pairs of solutions, select the solution for which solute solubility is greatest. a. Oxygen gas in water with P = 1 atm and T = 10C Oxygen gas in water with P = 1 atm and T = 20C b. Nitrogen gas in water with P = 2 atm and T = 50C Nitrogen gas in water with P = 1 atm and T = 70C c. Table salt in water with P = 1 atm and T = 40C Table salt in water with P = 1 atm and T = 70C d. Table sugar in water with P = 3 atm and T = 30C Table sugar in water with P = 1 atm and T = 80Carrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningIntroductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY