
Concept explainers
(a)
Interpretation:
The Lewis dot structure for TeCl2 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule using valence electrons shown as dots.
Te is a p block element of group 16 with 6 valence electrons.
(a)

Answer to Problem 5E
Lewis dot structure for TeCl2 is given below.
.
Explanation of Solution
In TeCl2, Te is the central atom which is bonded to two Cl atoms. There are three lone pairs of electrons on each Cl atom and two lone pair of electrons on Te atom.
Thus, the Lewis dot structure is as follows:
(b)
Interpretation:
The Lewis dot structure for HI must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule using valence electrons shown as dots.
I is a p block element of group 17 with 7 valence electrons.
(b)

Answer to Problem 5E
Lewis dot structure for HI is given below.
.
Explanation of Solution
In HI, I is bonded to one H atom. There are three lone pairs of electrons on I atom and hydrogen has zero lone pairs.
The structure is represented as follows:
(c)
Interpretation:
Lewis dot structures for AsBr3 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule using valence electrons shown as dots.
As is a p block element of group 15 with 5 valence electrons.
(c)

Answer to Problem 5E
Lewis dot structure for AsBr3 is given below.
.
Explanation of Solution
In AsBr3, As is the central atom which is bonded to three Br atoms. There are one lone pair of electrons on As atom and three lone pairs of electrons on Br atom.
The structure is represented as follows:
(d)
Interpretation:
Lewis dot structures for SiF4 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule using valence electrons shown as dots.
Si is a p block element of group 14 with 4 valence electrons.
(d)

Answer to Problem 5E
Lewis dot structure for SiF4 is given below.
.
Explanation of Solution
In SiF4, Si is the central atom which is bonded to four F atoms. There are three lone pairs of electrons on F atom.
Structure is represented as follows:
(e)
Interpretation:
Lewis dot structures for F2 must be drawn.
Concept Introduction :
Lewis dot structure is the representation of a molecule using valence electrons shown as dots.
F is a p block element of group 17 with 7 valence electrons.
(e)

Answer to Problem 5E
Lewis dot structure for F2 is given below.
.
Explanation of Solution
In F2, both the F atoms are bonded to each other with a single bond. There are three lone pairs of electrons on each F atom.
The structure is represented as follows:
Chapter U2 Solutions
Living by Chemistry
Additional Science Textbook Solutions
Human Anatomy & Physiology (2nd Edition)
Campbell Essential Biology with Physiology (5th Edition)
Human Biology: Concepts and Current Issues (8th Edition)
Campbell Biology (11th Edition)
Microbiology: An Introduction
Biology: Life on Earth with Physiology (11th Edition)
- 12. CH3 OH OH H&C CH3 H₂C N OH H₂C CH3 H&C CH3 H₂C' CH3 H.C CH3OH H.C CH2CH3OH CH3CEN Which one of these 17 compounds is represented by this IR and this 'H NMR spectrum? IR Spectrum 3000 4000 3000 NMR Spectrum 2000 £500 RAVENUMBER 2000 1500 9 8 6 5 10 HP-00-290 ppm m 1000 500 1000 4 °arrow_forwardDraw the structure of (E,6R) 6-methoxy-4-hepten-2-one. Give the IUPAC name of this compound, including stereochemistry. Draw the most stable chair conformation of (cis) 1,3-isobutylcyclohexane. H HC=CCH₂ CH2CH3 EN(CH3)2 -CN(CH3)2arrow_forward10. Write out the mechanism (intermediate/transition state) for this reaction; indicate stereochemistry in product. H3C CH₂OH CH3 SN1 Harrow_forward
- Write "most" under the member of each trio which is most stable. Write "least under the member of each trio which is least stable. b) Draw a Fischer projection of a pair of enantiomers with three chiral carbons. Which of these two would you expect to be more soluble in water? Why? 1-butanol 1-heptanol Which of these two would you expect to have the higher boiling point? Why? hexyl methyl ether 1-heptanolarrow_forwardWrite "most" under the most acidic compound. Write "least" under the least acidic compound. OH NO₂ OCH3 Br 9. Compound X, C50H84F2, reacts with excess H2/Pd to give a C50H88F2 compound. How many rings are in X? How many double bonds are in X? Show your work.arrow_forward4. State whether these two are: a) the same molecule b) c) d) different compounds that are not isomers constitutional isomers diastereomers e) enantiomers CH3 CH₁₂ H OH HO H H OH HO H CH, CH₂ 5. a) How many stereocenters does this compound have? b) How many stereoisomers are possible for this compound? CH₂ OH CHCHarrow_forward
- Calculating the pH at equivalence of a titration A chemist titrates 210.0 mL of a 0.1003 M hydrobromic acid (HBr) solution with 0.7550M KOH solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = ] ☑ o0o 18 Ararrow_forwardDo you do chemistry assignmentsarrow_forwardUsing the conditions of spontaneity to deduce the signs of AH and AS Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A This reaction is always spontaneous, but proceeds slower at temperatures above 120. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous except above 117. °C. AS is (pick one) ΔΗ is (pick one) This reaction is slower below 20. °C than C above. AS is |(pick one) ? 18 Ar 1arrow_forward
- Calculating the pH at equivalence of a titration Try Again Your answer is incorrect. 0/5 a A chemist titrates 70.0 mL of a 0.7089 M hydrocyanic acid (HCN) solution with 0.4574M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = 11.43] G 00. 18 Ar B•arrow_forwardBiological Macromolecules Naming and drawing the products of aldose oxidation and reduction aw a Fischer projection of the molecule that would produce L-ribonic acid if it were subjected to mildly oxidizing reaction conditions. Click and drag to start drawing a structure. X AP ‡ 1/5 Naor Explanation Check McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibilarrow_forward● Biological Macromolecules Identifying the parts of a disaccharide Take a look at this molecule, and then answer the questions in the table below it. CH2OH O H H H OH OH OH H H CH2OH H O OH H OH H H H H OH Is this a reducing sugar? Does this molecule contain a glycosidic bond? If you said this molecule does contain a glycosidic bond, write the symbol describing it. If you said this molecule does contain a glycosidic bond, write the common names (including anomer and enantiomer labels) of the molecules that would be released if that bond were hydrolyzed. If there's more than one molecule, separate each name with a comma. Explanation Check O yes X O no ○ yes O no Uarrow_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





