FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 11, Problem 94CE
Interpretation Introduction
Interpretation:
The volume of sodium and potassium ion has to be calculated. The number of sodium atoms that fits in the volume occupied by a potassium atom has to be identified.
Concept Introduction:
Atomic radii:
The atomic radius of an atom can be defined as the distance between the center of the nucleus and the outermost electron. In the periodic table, across a period atomic radius decreases and atomic radius increases down the group. For isoelectronic species, the atomic radii increase with decrease of charge on the element.
The volume of a sphere with radius r is calculated by the equation,
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Metals can be rolled into sheets and stamped into various forms. In contrast, diamond is very hard and
brittle. Which explanation for these different properties is correct?
Metals have semi-ionic bonds, whereas diamond has covalent bonds.
The electrons of a metal are held more tightly to the parent atom than the electrons of carbon. Hence,
the bonds in a metal are stronger than the bonds in diamond.
Metals are made of metal atoms, whereas diamond is made of non-metal carbon atoms.
The electrons that surround a metal atom are free to move through the metal. The bonding electron
pairs in a diamond are held tightly between two carbon atoms in an overall tetrahedral pattern.
Diamond has strong double bonds between carbon atoms. Metal bonds are normally single covalent
bonds, which bend easily.
Carbon naturally occurs in two forms: diamond and graphite. Why do these two forms have very different properties?
The key difference is that diamonds have other elements bonded within their structure.
The differences are explained by the number of covalent and ionic bonds within each substance.
The differences are explained by the density: graphite is very high and diamond is much lower.
The differences are explained by how the carbon atoms within each substance are covalently bonded together.
NaCl is a composite material because it has both Na and Cl.
True or False?
Chapter 11 Solutions
FOUND.OF COLLEGE CHEMISTRY
Ch. 11.1 - Prob. 11.1PCh. 11.2 - Prob. 11.2PCh. 11.3 - Prob. 11.3PCh. 11.4 - Prob. 11.4PCh. 11.4 - Prob. 11.5PCh. 11.5 - Prob. 11.6PCh. 11.6 - Prob. 11.7PCh. 11.6 - Prob. 11.8PCh. 11.7 - Prob. 11.9PCh. 11.8 - Prob. 11.10P
Ch. 11.9 - Prob. 11.11PCh. 11.10 - Prob. 11.12PCh. 11 - Prob. 1RQCh. 11 - Prob. 2RQCh. 11 - Prob. 3RQCh. 11 - Prob. 4RQCh. 11 - Prob. 5RQCh. 11 - Prob. 6RQCh. 11 - Prob. 7RQCh. 11 - Prob. 8RQCh. 11 - Prob. 9RQCh. 11 - Prob. 10RQCh. 11 - Prob. 11RQCh. 11 - Prob. 12RQCh. 11 - Prob. 13RQCh. 11 - Prob. 14RQCh. 11 - Prob. 15RQCh. 11 - Prob. 16RQCh. 11 - Prob. 17RQCh. 11 - Prob. 18RQCh. 11 - Prob. 19RQCh. 11 - Prob. 20RQCh. 11 - Prob. 21RQCh. 11 - Prob. 22RQCh. 11 - Prob. 23RQCh. 11 - Prob. 24RQCh. 11 - Prob. 25RQCh. 11 - Prob. 26RQCh. 11 - Prob. 28RQCh. 11 - Prob. 30RQCh. 11 - Prob. 31RQCh. 11 - Prob. 33RQCh. 11 - Prob. 36RQCh. 11 - Prob. 1PECh. 11 - Prob. 2PECh. 11 - Prob. 3PECh. 11 - Prob. 4PECh. 11 - Prob. 5PECh. 11 - Prob. 6PECh. 11 - Prob. 7PECh. 11 - Prob. 8PECh. 11 - Prob. 9PECh. 11 - Prob. 10PECh. 11 - Prob. 11PECh. 11 - Prob. 12PECh. 11 - Prob. 13PECh. 11 - Prob. 14PECh. 11 - Prob. 15PECh. 11 - Prob. 16PECh. 11 - Prob. 17PECh. 11 - Prob. 18PECh. 11 - Prob. 19PECh. 11 - Prob. 20PECh. 11 - Prob. 21PECh. 11 - Prob. 22PECh. 11 - Prob. 23PECh. 11 - Prob. 24PECh. 11 - Prob. 25PECh. 11 - Prob. 26PECh. 11 - Prob. 27PECh. 11 - Prob. 28PECh. 11 - Prob. 29PECh. 11 - Prob. 30PECh. 11 - Prob. 31PECh. 11 - Prob. 32PECh. 11 - Prob. 33PECh. 11 - Prob. 34PECh. 11 - Prob. 35PECh. 11 - Prob. 36PECh. 11 - Prob. 37PECh. 11 - Prob. 38PECh. 11 - Prob. 39PECh. 11 - Prob. 40PECh. 11 - Prob. 47PECh. 11 - Prob. 48PECh. 11 - Prob. 49PECh. 11 - Prob. 50PECh. 11 - Prob. 51PECh. 11 - Prob. 52PECh. 11 - Prob. 55AECh. 11 - Prob. 56AECh. 11 - Prob. 57AECh. 11 - Prob. 58AECh. 11 - Prob. 59AECh. 11 - Prob. 63AECh. 11 - Prob. 64AECh. 11 - Prob. 65AECh. 11 - Prob. 66AECh. 11 - Prob. 67AECh. 11 - Prob. 68AECh. 11 - Prob. 76AECh. 11 - Prob. 77AECh. 11 - Prob. 78AECh. 11 - Prob. 81AECh. 11 - Prob. 82AECh. 11 - Prob. 83AECh. 11 - Prob. 84AECh. 11 - Prob. 85AECh. 11 - Prob. 86AECh. 11 - Prob. 87AECh. 11 - Prob. 88CECh. 11 - Prob. 89CECh. 11 - Prob. 90CECh. 11 - Prob. 92CECh. 11 - Prob. 93CECh. 11 - Prob. 94CECh. 11 - Prob. 95CE
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
- You are spending the summer working for a chemical company. Your boss has asked you to determine where a chlorine ion of effective charge -e would situate itself near a carbon dioxide ion. The carbon dioxide ion is composed of 2 oxygen ions each with an effective charge -2e and a carbon ion with an effective charge +3e. These ions are arranged in a line with the carbon ion sandwiched midway between the two oxygen ions. The distance between each oxygen ion and the carbon ion is 3.0 x 10-11 m. Assuming that the chlorine ion is on a line that is perpendicular to the axis of the carbon dioxide ion and that the line goes through the carbon ion, what is the equilibrium distance for the chlorine ion relative to the carbon ion on this line? For simplicity, you assume that the carbon dioxide ion does not deform in the presence of the chlorine ion. Looking in your trusty physics textbook, you find the charge of the electron is 1.60 x 10-19 C.arrow_forwardConsider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is Δ?∘f=−925 kJ/mol. The enthalpy of sublimation of M is Δ?sub=175 kJ/mol. The first, second, and third ionization energies of M are IE1=579 kJ/mol, IE2=1677 kJ/mol, and IE3=2479 kJ/mol. The electron affinity of X is Δ?EA=−369 kJ/mol. (Refer to the hint). The bond energy of X2 is BE=179 kJ/mol. Determine the lattice energy of MX3.arrow_forward1. Below is a list of enthalpy changes for the Born-Haber cycle for the formation of solid LiF from Li(s) and F(g). Use these data to determine the lattice energy for the formation LiF(s). Li(s) → Li(g) ΔH1 = +162 kJ/mol Li(g) → Li+(g) + e- ΔH2 = +520.2 kJ/molF2(g) → 2F(g) ΔH3 = 154 kJ/mol F(g) + e- → F-(g) ΔH4 = -328 kJ/molLi(s) + 1/2F2(g) → LiF(s) ΔHf = -612 kJ/mol Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. 1371 kJ/mol b. -1371 kJ/mol c. 1043 kJ/mol d. -1043 kJ/molarrow_forward
- For salt (NaCl) and sugar (sucrose, C12H11O11) added to water, draw pictures of what you expect the atoms, ions and/or molecules to look like after each substance has been added to water.arrow_forwardA neutral object is sketched below, in black and white. There are positive and negative electric charges inside this object, and at least move around. Now suppose a negative charge comes close to this object, as shown in the sketch. Shade the object red anywhere you expect it to become more negatively charged, and shade the object blue anywhere you expect it to become more positively charged. Start over ol 9arrow_forward84. The ionic radii of the ions S2–, Cl–, and K+ are 184, 181, 138 pm respectively. Explain why these ions have different sizes even though they contain the same number of electrons.arrow_forward
- Answer the questions in the table below about the shape of the carbon tetrachloride (CC14) molecule. How many electron groups are around the central carbon atom? Note: one "electron group" means one lone pair, one single bond, one double bond, or one triple bond. What phrase best describes the arrangement of these electron groups around the central carbon atom? (You may need to use the scrollbar to see all the choices.) 0 (choose one) × Śarrow_forward1. The formula for silver oxide is Ag2O. True False 2. The Cl–, S2–, and Na+ ions have the same electron structure. True False 3. The formula of the compound that contains two atoms of nitrogen and five atoms of oxygen per formula is 2N5O NO N2O5 N5O2 4. The number of valence electrons in a bromate ion, BrO3- is 26 23 25 24arrow_forwardGiven the following information, use the Kapustinskii equation (shown below and included in the additional information sheet I encouraged you to download and have available) to calculate the lattice energy for Cr2S3 in kJ/mol. Ionic radii: Cr3+ (62 pm), S2− (184 pm) d* = 34.5 pm Κ = 1.21 × 105 kJ pm/molarrow_forward
- Some polyatomic anions contain a metal as part of the anion. For example, the anion dichromate has the formula Cr 2O 72 − and the anion permanganate has the formula MnO 4−. Write the formula of the ionic compound formed from each of these anions and a potassium cation. Name each compound.arrow_forwardSome of the batteries currently used in your smartphone contain lithium ions (Li +) as their electricity conductors. Currently, researchers are also developing batteries that use sodium ions (Na +) as a substitute for lithium ions due to their abundance in nature. Of course, lithium (Li) and sodium (Na) atoms have different properties due to their different electronic structures. Compare the radius sizes of the following atoms and ions: -Li and Na -ion Li + and Li -ion Li + and Na + ions. 2. How does the first ionization energy change from lithium to sodium? Is it getting bigger or smaller? Explain why.arrow_forwardTwo neutral objects are sketched below, in black and white. There are positive and negative electric charges inside these objects, and at least some of those charges can move around. Now suppose a negative charge comes close to these objects, as shown in the sketch. Shade the objects red anywhere you expect them to become more negatively charged, and shade the objects blue anywhere you expect them to become more positively charged. + X 5 ola 18 Ar 18arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Introductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningGeneral 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 LearningWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: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
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Atomic Number, Atomic Mass, and the Atomic Structure | How to Pass ChemistryThe Nucleus: Crash Course Chemistry #1; Author: Crash Course;https://www.youtube.com/watch?v=FSyAehMdpyI;License: Standard YouTube License, CC-BY