
Concept explainers
(a)
Interpretation: Among the given terms aldoses, ketoses, hexoses and aldohexoses, the term that is applied to both
Concept introduction: The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Monosaccharide is categorized into two types. The first type of monosaccharide is aldoses and the second type of monosaccharide is ketoses. Aldoses sugar possesses
Monosaccharides are also classified on the basis of the number of carbon atoms present in it. For example, the sugar that contains five carbon atoms is known as pentoses sugar and the sugar that contains six carbon atoms is known as hexoses sugar.
(b)
Interpretation: Among the given terms aldoses, ketoses, hexoses and aldohexoses, the term that is applied to both
Concept introduction: The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Monosaccharide is categorized into two types. The first type of monosaccharide is aldoses and the second type of monosaccharide is ketoses. Aldoses sugar possesses aldehyde as a functional group and the ketoses sugar possesses ketone as a functional group.
Monosaccharides are also classified on the basis of the number of carbon atoms present in it. For example, the sugar that contains five carbon atoms is known as pentoses sugar and the sugar that contains six carbon atoms is known as hexoses sugar.
(c)
Interpretation: Among the given terms aldoses, ketoses, hexoses and aldohexoses, the term that is applied to both
Concept introduction: The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Monosaccharide is categorized into two types. The first type of monosaccharide is aldoses and the second type of monosaccharide is ketoses. Aldoses sugar possesses aldehyde as a functional group and the ketoses sugar possesses ketone as a functional group.
Monosaccharides are also classified on the basis of the number of carbon atoms present in it. For example, the sugar that contains five carbon atoms is known as pentoses sugar and the sugar that contains six carbon atoms is known as hexoses sugar.
(d)
Interpretation: Among the given terms aldoses, ketoses, hexoses and aldohexoses, the term that is applied to both
Concept introduction: The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Monosaccharide is categorized into two types. The first type of monosaccharide is aldoses and the second type of monosaccharide is ketoses. Aldoses sugar possesses aldehyde as a functional group and the ketoses sugar possesses ketone as a functional group.
Monosaccharides are also classified on the basis of the number of carbon atoms present in it. For example, the sugar that contains five carbon atoms is known as pentoses sugar and the sugar that contains six carbon atoms is known as hexoses sugar.

Trending nowThis is a popular solution!

Chapter 18 Solutions
General, Organic, and Biological Chemistry Seventh Edition
- Look at the following pairs of structures carefully to identify them as representing a) completely different compounds, b) compounds that are structural isomers of each other, c) compounds that are geometric isomers of each other, d) conformers of the same compound (part of structure rotated around a single bond) or e) the same structure.arrow_forwardGiven 10.0 g of NaOH, what volume of a 0.100 M solution of H2SO4 would be required to exactly react all the NaOH?arrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward
- 3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forwardConcentration Trial1 Concentration of iodide solution (mA) 255.8 Concentration of thiosulfate solution (mM) 47.0 Concentration of hydrogen peroxide solution (mM) 110.1 Temperature of iodide solution ('C) 25.0 Volume of iodide solution (1) used (mL) 10.0 Volume of thiosulfate solution (5:03) used (mL) Volume of DI water used (mL) Volume of hydrogen peroxide solution (H₂O₂) used (mL) 1.0 2.5 7.5 Time (s) 16.9 Dark blue Observations Initial concentration of iodide in reaction (mA) Initial concentration of thiosulfate in reaction (mA) Initial concentration of hydrogen peroxide in reaction (mA) Initial Rate (mA's)arrow_forwardDraw the condensed or line-angle structure for an alkene with the formula C5H10. Note: Avoid selecting cis-/trans- isomers in this exercise. Draw two additional condensed or line-angle structures for alkenes with the formula C5H10. Record the name of the isomers in Data Table 1. Repeat steps for 2 cyclic isomers of C5H10arrow_forward
- Explain why the following names of the structures are incorrect. CH2CH3 CH3-C=CH-CH2-CH3 a. 2-ethyl-2-pentene CH3 | CH3-CH-CH2-CH=CH2 b. 2-methyl-4-pentenearrow_forwardDraw the line-angle formula of cis-2,3-dichloro-2-pentene. Then, draw the line-angle formula of trans-2,3-dichloro-2-pentene below. Draw the dash-wedge formula of cis-1,3-dimethylcyclohexane. Then, draw the dash-wedge formula of trans-1,3-dimethylcyclohexane below.arrow_forwardRecord the amounts measured and calculate the percent yield for Part 2 in the table below. Dicyclopentadiene measured in volume Cyclopentadiene measured in grams 0 Measured Calculated Mol Yield Mass (g) or Volume (mL) Mass (g) or Volume (ml) 0.6 2.955 Part 2 Measurements and Results Record the amounts measured and calculate the percent yield for Part 2 in the table below. 0.588 0.0044 2.868 0.0434 N/A Table view List view Measured Calculated Mol $ Yield Melting Point (C) Mass (g) or Volume (ml) Mass (g) or Volume (ml.) Cyclopentadiene 0.1 0.08 0.001189 measured in volume Maleic Anhydride 0.196 N/A cis-norbornene-5,6-endo- dicarboxylic anhydride 0.041 0.0002467 N/A N/A N/A 0.002 N/A N/A 128arrow_forward
- Draw the condensed structural formula and line-angle formula for each: 2,3-dimethylheptane 3-bromo-2-pentanol 3-isopropyl-2-hexene 4-chlorobutanoic acidarrow_forwardRecord the IUPAC names for each of the structures shown below. a) b) c) OH d) OH e)arrow_forwardA solution of 14 g of a nonvolatile, nonelectrolyte compound in 0.10 kg of benzene boils at 81.7°C. If the BP of pure benzene is 80.2°C and the K, of benzene is 2.53°C/m, calculate the molar mass of the unknown compound. AT₁ = Km (14)arrow_forward
- Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningGeneral, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
- Organic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co




