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Make a graph of [A] versus time for zero-, first-, and

Chemistry | 8th Edition | ISBN: 9780547125329 | Authors: Steven S. Zumdahl ISBN: 9780547125329 153

Solution for problem 3 Chapter 12

Chemistry | 8th Edition

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Chemistry | 8th Edition | ISBN: 9780547125329 | Authors: Steven S. Zumdahl

Chemistry | 8th Edition

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Problem 3

Make a graph of [A] versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.

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Chapter 16: Carbohydrates 16.3 Some Important Oligosaccharides  Oligomers of sugars­ disaccharides­ formed by linking two monosaccharides units by glyosidic bonds  What makes Sucrose an Important Compound o Sucrose­ able sugar­ from sugarcane and sugar beets o Monosaccharide units­ alpha D glucose and beta­D­fructose o Alpha C1 carbon of glucose is linked to beta C2 carbon of fructose in a glyosidic linkage o Sucrose is not reducing sugar because both anomeric groups are involved in glycosidic linkages o Free glucose is reducing sugar­ free fructose can give positive test (ketone rather than aldehyde in opening chain form) o When sucrose is consumed­ hydrolyzed to glucose and fructose­ degraded by metabolic processes to provide energy o Some artificial sweeteners derived from sucrose, less dangerous o Splenda­ differs from sucrose because of three hydroxyl groups replaced with chlorines and configuration at carbon 4 of six­membered pyranose is inverted  galactose derivative o Hydroxyl groups replaced by chlorine are bonded to carbon 1 and 6 of fructose and carbon4 of galactose o Sucralose is not metabolize by bond­ does not provide calories  safe sugar substitute  Are any other Disaccharides important to us o Lactose­ disaccharide with beta­d­glucose o Galactose is C­4 epimer of glucose  only difference from glucose is C4 configuration is inverted o Anomeric carbon of glucose not involved in glycosidic linkage o 2 anomeric forms of lactose  designated by glucose residue o Lactose is reducing sugar because the group at the anomeric carbon of glucose portion is not involved in a glyosidic linkage  free to react with oxidizing agents o Maltose­ disaccharide from hydrolysis of starch  Had two residues of D­glucose in a alpha (1­4) linkage  Differences from cellobiose by glycosidic linkage  Mammals can digest maltose but not cellobiose  Maltose is used in milk and beer 16.4 Structures and Functions of Polysaccharides  Polysaccharide­ many monosaccharides are linked together  Occur in organisms that are compose of very few types of monosaccharide compontes  Homopolysaccharide­ only one type of monomer  Heteropolysaccharide­ more than one type of monomer  Glucose is most common monomer  Polysaccharides include specification of which monomers are present and sequence of monomers  Requires type of glycosidic linkage  Cellulose and chitin­ polysaccharides with beta glycosidic linkages­ structural materials  Starch and glycogen­ polysaccharides with alpha glycosidic linkages  serve as carbs storage polymers in plants and animals  How Do Cellulose and Starch differ From Each other o Cellulose is major structural component in plants o Linear homopholysaccharide of beta­D­glucose, all residues are linked in a beta 1­4 glycosidic bond o Individual chains are hydrogen bonded together o Animals cant hydrolyze cellulose o Cellulases­ hydrolyze cellulose by attacking the alpha and beta linkages between glucose  Is there more than one form of starch o Importance of carbs as energy sources is that there are uses for some polysaccharides in metabolism o Starches­ polymers of alpha­d­glucose that occurs in plant cells o Types of starches can be distinguished from one another by chain branching o Starches are storage molecules­ mechanism for releasing glucose from starch when organism needs energy o Alpha and beta amylase­ attack alpha 1­4 linkages o Amylose can be completely degraded to glucose and maltose  How is Glycogen related to starch o Glycogen­ branched chain polymer of alpha­d­glucose o Has chain of alpha 1­4 linkages with alpha 1­6 linkages at branch points o Main difference is that glycogen is more highly branched o Every glycogen molecule there is a protein called glycogenin o Number of branch points is important­ more branched polysaccharide­ more water soluble o Organism needs energy quickly­ glycogen phosphorylase has more potential targets if there are more branches  quicker mobilization of glucose  What is Chitin o Polysaccharide that is similar to cellulose in both structure and function, also has linear homopolysaccharide with all the residues linked in beta 1­4 linkages o Differs from cellulose because of monosaccharide unit o Has N­acetyl­beta­d­glucosamine o Plays structural role and has mechanical strength because of how the individual strands are held together by hydrogen bonds o Component of exoskeletons of invertebrates, occurs in cell walls  What roles do Polysaccharides play in structure of cell walls o Heteropolysaccharides are major components of bacterial cell walls o Distinguishing feature­ polysaccharides are cross linked by peptides o Cross links of bacterial cell walls have small peptides o Tetrapeptides are cross linked by another small peptide and have five amino acids o Occurrence of d amino acids and nactylmuramic acid shows a biochemical and structural difference between prokaryotes and eukaryotes o Extensive cross linking produces three dimensional network of mechanical strength o Peptidoglycan­ formed by cross linking of polysaccharides by peptides o Pectin­ polymer made up of d­galacturonic acid­ derivative of galactose o Extracted from plants because of commercial importance o Lignin­ polymer of coniferyl alcohol­ tough and durable material  Do Polysaccharides play any specific roles in connective tissues o Glycosaminoglycans­ polysaccharide based on repeating disaccharide where one of the sugars is an amino sugar and one of them has negative charge o Polysaccharides involved in wide variety of cell functions 16.5 Glycoproteins  Have carb residue and polypeptide chain  Antibodies­ bind to and immobilize antigens  Carbs are important in antigenic determinants­ portions of an antigenic molecule that antibodies recognize and to which they bind  How are Carbs important in the immune response o Super important in determinates of blood groups o A,B, AB, O blood groups o Distinctions between groups depend on oligosaccharide portions of glycoprotiens on surface of blood cells o All blood types­ oligosaccharides contains sugar L­fuctose o Type A­ has N­Acetylglactosamine on the non­reducing end of oligosaccharide o Type B­ alpha­d­Glactose o Type O­ has neither terminal residues present o AB­ has both terminal residues o Glydoprotiens are also in eukaryotic cell membranes o Proteoglycans­ glycoproteins with high carbohydrate content o Constantly being synthesized and broken down

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Chapter 12, Problem 3 is Solved
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Textbook: Chemistry
Edition: 8
Author: Steven S. Zumdahl
ISBN: 9780547125329

The answer to “Make a graph of [A] versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.” is broken down into a number of easy to follow steps, and 19 words. This full solution covers the following key subjects: Compare, graph, Graphs, half, Lives. This expansive textbook survival guide covers 22 chapters, and 2897 solutions. This textbook survival guide was created for the textbook: Chemistry, edition: 8. The full step-by-step solution to problem: 3 from chapter: 12 was answered by , our top Chemistry solution expert on 11/15/17, 04:25PM. Chemistry was written by and is associated to the ISBN: 9780547125329. Since the solution to 3 from 12 chapter was answered, more than 252 students have viewed the full step-by-step answer.

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