Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
8th Edition
ISBN: 9780134015187
Author: John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 4, Problem 4.47AP
(a)
Interpretation Introduction
Interpretation:
The lone pair of the given molecule is to be drawn.
Concept Introduction:
Lone pairs are the pair of valence electrons that are not participating in bonding.
Lone pairs are represented in Lewis structures which represents lines for the bonding between atoms of a molecule and the dots for the lone pairs of electrons that may exist in the molecule.
Valence electrons of carbon, hydrogen and oxygen are 4, 1 and 6 respectively.
(b)
Interpretation Introduction
Interpretation:
The lone pair of the given molecule is to be drawn.
Concept Introduction:
Lone pairs are the pair of valence electrons that are not participating in bonding.
Lone pairs are represented in Lewis structures which represents lines for the bonding between atoms of a molecule and the dots for the lone pairs of electrons that may exist in the molecule.
Valence electrons of carbon, hydrogen and sulfur are 4, 1 and 6 respectively.
(c)
Interpretation Introduction
Interpretation:
The lone pair of the given molecule is to be drawn.
Concept Introduction:
Lone pairs are the pair of valence electrons that are not participating in bonding.
Lone pairs are represented in Lewis structures which represents lines for the bonding between atoms of a molecule and the dots for the lone pairs of electrons that may exist in the molecule.
Valence electrons of oxygen are 6.
(d)
Interpretation Introduction
Interpretation:
The lone pair of the given molecule is to be drawn.
Concept introduction:
Lone pairs are the pair of valence electrons that are not participating in bonding.
Lone pairs are represented in Lewis structures which represents lines for the bonding between atoms of a molecule and the dots for the lone pairs of electrons that may exist in the molecule.
Valence electrons of nitrogen and carbon are 5 and 4 respectively.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Consider a system consisting of an egg in an incubator. The white and yolk of the egg contain proteins, carbohydrates, and
lipids. If fertilized, the egg transforms from a single cell to a complex organism.
How does the entropy change in both the system (developing chick) and suroundings (the egg environment) drive the
irreversible process of chick development?
☐
The release of glucose from sucrose, which produces energy needed for chick development, decreases entropy in
the surroundings.
Chick development increases entropy in the system, which causes a concominant decrease in entropy in
the surroundings.
Carbohydrates, proteins, and lipids within the egg break down into CO2 and H2O, which increases entropy in
the surroundings.
Chick development decreases entropy in the system, but this is smaller than the concominant increase in entropy in
the surroundings.
The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine,
which has a pKa of 9.6, can exist either in the protonated form -NH or as the free base -NH2, because of the reversible
equilibrium
R-NH =R-NH₂ + H+
In what pH range can glycine be used as an effective
buffer due to its amino group?
pH 8.6 to pH 10.6
In a 0.1 M solution of glycine at pH 9.0, what
fraction of glycine has its amino group in the
-NH form?
Correct Answer
Correct Answer
45
How much 5 M KOH must be added to 1.0 L of
0.1 M glycine at pH 9.0 to bring its pH to 10.0?
10 mL
When 99% of the glycine is in its -NH form, what
is the numerical relation between the pH of the
solution and the pKa of the amino group?
pH = pKa - 2
Correct Answer
Correct Answer
The glycolytic enzyme Phosphofructokinase (PFK) catalyzes the following reaction: Fructose-6-phosphate
(F6P) + ATP → Fructose-1,6-bisphosphate (F1,6BP) + ADP AG"=-14.2 kJ/mol This is considered the
enzymatic step that commits a sugar substrate to glycolysis.
a) Calculate the standard free energy of hydrolysis of fructose-1,6-bisphosphate.
b) What is the equilibrium constant for this coupled reaction?
c) ATP is a known inhibitor of PFK. If the cellular concentrations of ATP and ADP are 5 mM and 1.0mM
respectively, and the concentrations of F6P and F1,6BP are 2mM, what is the free energy change of
the system?
Chapter 4 Solutions
Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
Ch. 4.1 - Prob. 4.1PCh. 4.2 - Prob. 4.2PCh. 4.2 - Prob. 4.3PCh. 4.3 - Prob. 4.4PCh. 4.3 - Prob. 4.5PCh. 4.4 - The BF3 molecule can also react with NH3 by...Ch. 4.5 - Prob. 4.7PCh. 4.7 - Prob. 4.8PCh. 4.7 - Add lone pairs where appropriate to the following...Ch. 4.7 - Prob. 4.10P
Ch. 4.7 - Prob. 4.11PCh. 4.7 - The molecular model shown here is a representation...Ch. 4.7 - Prob. 4.1CIAPCh. 4.7 - Prob. 4.2CIAPCh. 4.7 - Prob. 4.13PCh. 4.8 - Prob. 4.3CIAPCh. 4.8 - Prob. 4.4CIAPCh. 4.8 - Prob. 4.14PCh. 4.8 - Prob. 4.15PCh. 4.8 - Prob. 4.16PCh. 4.8 - Prob. 4.17KCPCh. 4.9 - The elements H, N, O, P, and S are commonly bonded...Ch. 4.9 - Prob. 4.19PCh. 4.10 - Look at the molecular shape of formaldehyde (CH2O)...Ch. 4.10 - Prob. 4.21PCh. 4.10 - Prob. 4.22KCPCh. 4.11 - Prob. 4.5CIAPCh. 4.11 - Prob. 4.6CIAPCh. 4.11 - Prob. 4.23PCh. 4.11 - Prob. 4.24PCh. 4 - What is the geometry around the central atom in...Ch. 4 - Prob. 4.26UKCCh. 4 - The ball-and-stick molecular model shown here is a...Ch. 4 - Prob. 4.28UKCCh. 4 - Prob. 4.29UKCCh. 4 - Prob. 4.30UKCCh. 4 - What is a covalent bond, and how does it differ...Ch. 4 - Prob. 4.32APCh. 4 - When are multiple bonds formed between atoms and...Ch. 4 - Identify the bonds formed between the following...Ch. 4 - Prob. 4.35APCh. 4 - Prob. 4.36APCh. 4 - Prob. 4.37APCh. 4 - Prob. 4.38APCh. 4 - Prob. 4.39APCh. 4 - Prob. 4.40APCh. 4 - Prob. 4.41APCh. 4 - Prob. 4.42APCh. 4 - Prob. 4.43APCh. 4 - Prob. 4.44APCh. 4 - Prob. 4.45APCh. 4 - Prob. 4.46APCh. 4 - Prob. 4.47APCh. 4 - If a research paper appeared reporting the...Ch. 4 - Consider the following possible structural...Ch. 4 - Prob. 4.50APCh. 4 - Prob. 4.51APCh. 4 - Prob. 4.52APCh. 4 - Prob. 4.53APCh. 4 - Prob. 4.54APCh. 4 - Draw a Lewis structure for the following...Ch. 4 - Prob. 4.56APCh. 4 - Ethanol, or grain alcohol, has the formula C2H6O...Ch. 4 - Prob. 4.58APCh. 4 - Tetrachloroethylene, C2Cl4, is used commercially...Ch. 4 - Prob. 4.60APCh. 4 - The carbonate ion, CO32, contains a double bond....Ch. 4 - Prob. 4.62APCh. 4 - Prob. 4.63APCh. 4 - Prob. 4.64APCh. 4 - Prob. 4.66APCh. 4 - Predict the geometry around each carbon atom in...Ch. 4 - Prob. 4.68APCh. 4 - Prob. 4.69APCh. 4 - Prob. 4.70APCh. 4 - Prob. 4.71APCh. 4 - Prob. 4.72APCh. 4 - Which of the following bonds are polar? If a bond...Ch. 4 - Prob. 4.74APCh. 4 - Based on electronegativity differences, would you...Ch. 4 - Arrange the following molecules in order of the...Ch. 4 - Prob. 4.77APCh. 4 - Prob. 4.78APCh. 4 - Prob. 4.79APCh. 4 - Prob. 4.80APCh. 4 - Prob. 4.81APCh. 4 - Prob. 4.82APCh. 4 - Prob. 4.83APCh. 4 - Prob. 4.84APCh. 4 - Prob. 4.85CPCh. 4 - Prob. 4.86CPCh. 4 - Prob. 4.87CPCh. 4 - Prob. 4.88CPCh. 4 - Prob. 4.89CPCh. 4 - The phosphonium ion, PH4+, is formed by reaction...Ch. 4 - Prob. 4.91CPCh. 4 - Prob. 4.92CPCh. 4 - Prob. 4.93CPCh. 4 - Prob. 4.94CPCh. 4 - Prob. 4.95CPCh. 4 - Prob. 4.96CPCh. 4 - Prob. 4.97CPCh. 4 - Write Lewis structures for molecules with the...Ch. 4 - Prob. 4.99CPCh. 4 - Prob. 4.100GPCh. 4 - Hydrazine is a substance used to make rocket fuel....Ch. 4 - Prob. 4.102GPCh. 4 - Titanium forms both molecular and ionic compounds...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- 2) Consider the following reaction: A + 2B 3C + D At equilibrium the concentration of the reactants and products are: [A] = 20.0 mM [C] = 3.0 mM [B] = 4.0 mM [D] = 50.0 mM Calculate (a) the equilibrium constant and (b) AG". Comment on which side of this reaction is more likely to occur.arrow_forwardGlycine is a diprotic acid, which can potentially undergo two dissociation reactions, one for the a-amino group (NH), and the other for the carboxyl (-COOH) group. Therefore, it has two pK₁ values. The carboxyl group has a pK₁ of 2.34 and the α-amino group has a pK2 of 9.60. Glycine can exist in fully deprotonated (NH2-CH2-COO¯), fully protonated (NH3-CH2-COOH), or zwitterionic form (NH3-CH2-COO¯). Match the pH values with the corresponding form of glycine that would be present in the highest concentration in a solution of that pH. fully deprotonated form NH2-CH2-COO- fully protonated form NH–CH,–COOH zwitterionic form NH–CH,−COO Answer Bank pH 7.0 pH 11.9 pH 6.0 pH 8.0 pH 1.0arrow_forwardThe AG of hydrolysis of a sugar phosphate (S-O-P) to the free sugar (S-OH) is -26.6 kJ/mol in a hypothetical cell in which the steady-state concentrations of sugar phosphate, free sugar, and inorganic phosphate are 1.0 mM, 0.20 mM, and 50.0 mM, respectively. S-O-P + H2O S-OH + Pi (a) What is the AG°' for this reaction? (b) In the cell, S-O-P is formed by the transfer of a phosphate group from ATP. What would the AG be for the transfer of the g-phosphate from ATP to this sugar (S-OH)? [AG for ATP hydrolysis is -31 kJ/mol.]arrow_forward
- 1) Consider the reaction: A B + C (a) What is the Keg for this reaction? AG= -8.80 kJ/mol (b) The reverse reaction is initiated by creating a solution containing 20mM B, 1mM A and 150mM C. At the instant these are mixed, what is the free energy change associated with the reaction?arrow_forwardWhat is the chemical importance of the negative charge on the phosphate group? Be asspecific as possible. In what ways might this negative charge have beenthermodynamically useful during the evolution of ATP-binding proteins?arrow_forwardOne prominent theory on life origins was that RNA enzymes came into existence early inthe prebiotic history of Earth and were able to do basic chemical catalyses. Eventually,this “RNA-world” was overtaken by the stability of DNA as an information carrier and thediversity of catalytic functions capable of being performed by polypeptides. Is the RNA world hypothesis is a well-founded model?arrow_forward
- The AG" of hydrolysis (ATP + H2O --> ADP + Pi) is -31.0 kJ/mol. Answer the following questions assuming that the steady-state concentrations in the cell are as indicated below. (Note: Steady-state refers to a non-equilibrium situation that exists due to a balance between reactions that supply and remove these substances.) [ADP] = 0.40 mM, [ATP] = 4.0 mM, and [Pi] = 40.0 mM a) Calculate the equilibrium constant for this reaction. b) What would the AG' for ATP hydrolysis be in the cell? c) Is this reaction at equilibrium in the cell? Briefly explain your answer.arrow_forward5) Theoretically, ATP did not have to become our bodies' main energy currency. Two alternative carriers, acetyl phosphate and S-adenosylmethionine could have been utilized, rather than ATP. AG" for acetyl phosphate hydrolysis is -43.3 kJ/mol and AG" for S- adenosylmethionine hydrolysis is -25.6 kJ/mol. (a) Calculate the weight of each alternative energy carrier that would need to be consumed by humans on a 2000 calorie per day diet if our bodies could not recycle it. Assume a 50% absorption of energy from our diet. (b) If our bodies contain 25g of each alternative energy carrier and they CAN be recycled, how many times would each molecule of each energy carrier need to be recycled? (c) Comment on the special properties of ATP and why it is unlikely that these alternative carriers would be utilized biologically.arrow_forwardGive three reasons why evolution may have selected for phosphates compared to othersimilar leaving groups such as conjugated carboxylic acids or amides. Explain whatbenefit each of your reasons has granted to the living organism.arrow_forward
- The preferred substrate is T because the enzyme half-saturates at 7.00 mM for T, but requires 28.0 mM for U, and 112 mM for S. b Question Content Area The rate constant k 2 with substrate S is 9.60×107 sec-1, with substrate T, k 2 = 6.00×104 sec-1, and with substrate U, k 2 = 2.40×106 sec-1. Calculate the catalytic efficiency with S, T, and U. Catalytic efficiency with S = Catalytic efficiency with T = Catalytic efficiency with U = Does enzyme A use substrate S, substrate T, or substrate U with greater catalytic efficiency?arrow_forwardFumerase catalyzes the conversion of fumerate to malate. fumerate + H2O ⇋ malate The turnover number, kcat, for fumerase is 8.00×102 sec-1. The Km of this enzyme for fumerate is 5.00×10-3 μmol mL-1. a In an experiment using 2.00×10-3 μmol·mL-1, what is Vmax?arrow_forwardSuppose you wanted to make a buffer of exactly pH 7.00 using KH2PO4 and Na2HPO4. If the final solution was 0.18 M in KH2PO4, you would need 0.25 M Na2HPO4. Now assume you wish to make a buffer at the same pH, using the same substances, but want the total phosphate molarity ([HPO42−]+[H2PO−4]) to equal 0.20 M. What concentration of the Na2HPO4 would be required?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Principles Of Radiographic Imaging: An Art And A ...Health & NutritionISBN:9781337711067Author:Richard R. Carlton, Arlene M. Adler, Vesna BalacPublisher:Cengage Learning
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning
Principles Of Radiographic Imaging: An Art And A ...
Health & Nutrition
ISBN:9781337711067
Author:Richard R. Carlton, Arlene M. Adler, Vesna Balac
Publisher:Cengage Learning
Macromolecules | Classes and Functions; Author: 2 Minute Classroom;https://www.youtube.com/watch?v=V5hhrDFo8Vk;License: Standard youtube license