Interpretation:
The statement “not all mixtures are solutions” has to be explained.
Concept Introduction:
Mixture: Mixture is a combination of two or more pure components in which each pure component retains its individual chemical properties. Substances tend to mix naturally, it is difficult to keep any substance in pure form.
Types of Mixtures: Depending upon the nature of the components that form a mixture, we can have different types of mixtures and can be classified as either heterogeneous or homogeneous.
Homogeneous mixture: A homogeneous mixture is a mixture that has a uniform composition throughout; it always has a single phase. A solution is an example of homogeneous mixtures where two components form a single phase. Components cannot be distinguishable in the mixture.
Solution: The solution is a homogeneous mixture of two components. One of them is the solute and the other is the solvent. The solvent is present usually in a higher proportion than the solute in the solution. Components of the solution cannot be distinguishable in the mixture.
Explanation of Solution
The solution is a homogeneous mixture of two components and it always has a single phase. Mixtures can also be either heterogeneous or homogeneous, where heterogeneous mixture forms at least two phases that remain separate from each other and homogeneous mixture forms a single phase. So, the solution is an example of mixtures. In other words, we can say all solutions are mixtures but all mixtures are not solutions.
Want to see more full solutions like this?
Chapter 14 Solutions
Glencoe Chemistry: Matter and Change, Student Edition
Additional Science Textbook Solutions
Campbell Biology in Focus (2nd Edition)
Organic Chemistry (8th Edition)
Genetic Analysis: An Integrated Approach (3rd Edition)
Campbell Biology (11th Edition)
Applications and Investigations in Earth Science (9th Edition)
Anatomy & Physiology (6th Edition)
- Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/zarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardPredicting the pro Predict the major products of this organic reaction. Explanation Check m ☐ + 5 1.03 Click and drag t drawing a stru 2. (CH₂)₂S 3 2 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Centerarrow_forward
- starting material target If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area. Be sure you follow the standard ALEKS rules for submitting syntheses. + More... X Explanation Check C टे Br T Add/Remove step ☐ Br Br © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacarrow_forwardDon't used hand raitingarrow_forwardRelative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/zarrow_forward
- Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100arrow_forwardPart IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.arrow_forwardFor each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY