Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter U4.77, Problem 6E
(a)
Interpretation Introduction
Interpretation:
The mass of carbon atom in grams in one mole of methane should be calculated.
Concept introduction:
Mole is an SI unit that is used to measure the quantity of the substance. It is the quantity of a substance that contains the same number of atoms as accurately present in 12.00 g of carbon-12 is known as mole.
The number of moles is defined as the ratio of mass to the molar mass.
Number of moles = Mass (g)Molar mass
The ratio of the mass of the given compound to the number of moles of the compound is known as molar mass.
The one mole of any substance is equal to the
(a)
Expert Solution
Answer to Problem 6E
Mass of carbon atom in one mole of methane = 12 g.
Explanation of Solution
The given molecule is methane ( CH4 ).
Number of moles of methane = 1 mole
In one mole of methane, one mole of carbon is present.
Molar mass of carbon = 12 g/mole.
Thus, the mass of carbon atom is one mole of methane = 1 mole×12 g/mole
Mass of carbon atom in one mole of methane = 12 g.
(b)
Interpretation Introduction
Interpretation:
The mass of carbon atoms in grams in one mole of methanol should be calculated.
Concept introduction:
Mole is an SI unit that is used to measure the quantity of the substance. It is the quantity of a substance that contains the same number of atoms as accurately present in 12.00 g of carbon-12 is known as mole.
The number of moles is defined as the ratio of mass to the molar mass.
Number of moles = Mass (g)Molar mass
The ratio of the mass of the given compound to the number of moles of a compound is known as molar mass.
The one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.
(b)
Expert Solution
Answer to Problem 6E
12 g.
Explanation of Solution
The given molecule is methanol ( CH4O ).
Number of moles of methanol = 1 mole
In one mole of methanol, one mole of carbon is present.
The molar mass of carbon atom is 12 g/mol that is 1 mol of C atom has 12 g.
Therefore, the mass of carbon atom in 1 mol of methanol will be 12 g.
(c)
Interpretation Introduction
Interpretation:
The mass of carbon atoms in grams in one mole of ethanol should be calculated.
Concept introduction:
Mole is an SI unit that is used to measure the quantity of the substance. It is the quantity of a substance that contains the same number of atoms as accurately present in 12.00 g of carbon-12 is known as mole.
Number of moles is defined as the ratio of mass to the molar mass.
Number of moles = Mass (g)Molar mass
The ratio of the mass of the given compound to the number of moles of the compound is known as molar mass.
The one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.
(c)
Expert Solution
Answer to Problem 6E
Mass of carbon atom in one mole of ethanol = 24 g.
Explanation of Solution
The given molecule is ethanol ( C2H6O ).
Number of moles of ethanol = 1 mole
In one mole of ethanol, 2 moles of carbon are present.
The mass of 1 mol of carbon atom is 12 g thus, mass of 2 mol will be:
m=2×12=24 g
Chapter U4 Solutions
Living by Chemistry
Ch. U4.68 - Prob. 1TAICh. U4.68 - Prob. 1ECh. U4.68 - Prob. 2ECh. U4.68 - Prob. 3ECh. U4.68 - Prob. 4ECh. U4.68 - Prob. 5ECh. U4.68 - Prob. 6ECh. U4.69 - Prob. 1TAICh. U4.69 - Prob. 1ECh. U4.69 - Prob. 2E
Ch. U4.69 - Prob. 3ECh. U4.69 - Prob. 4ECh. U4.69 - Prob. 5ECh. U4.70 - Prob. 1TAICh. U4.70 - Prob. 1ECh. U4.70 - Prob. 2ECh. U4.70 - Prob. 3ECh. U4.70 - Prob. 4ECh. U4.70 - Prob. 6ECh. U4.71 - Prob. 1TAICh. U4.71 - Prob. 1ECh. U4.71 - Prob. 2ECh. U4.71 - Prob. 3ECh. U4.71 - Prob. 4ECh. U4.71 - Prob. 5ECh. U4.71 - Prob. 6ECh. U4.71 - Prob. 7ECh. U4.72 - Prob. 1TAICh. U4.72 - Prob. 1ECh. U4.72 - Prob. 2ECh. U4.72 - Prob. 3ECh. U4.72 - Prob. 4ECh. U4.73 - Prob. 1TAICh. U4.73 - Prob. 1ECh. U4.73 - Prob. 2ECh. U4.73 - Prob. 3ECh. U4.73 - Prob. 5ECh. U4.73 - Prob. 6ECh. U4.73 - Prob. 7ECh. U4.74 - Prob. 1TAICh. U4.74 - Prob. 1ECh. U4.74 - Prob. 2ECh. U4.74 - Prob. 4ECh. U4.74 - Prob. 5ECh. U4.75 - Prob. 1TAICh. U4.75 - Prob. 1ECh. U4.75 - Prob. 2ECh. U4.75 - Prob. 4ECh. U4.75 - Prob. 5ECh. U4.75 - Prob. 6ECh. U4.75 - Prob. 7ECh. U4.75 - Prob. 8ECh. U4.75 - Prob. 9ECh. U4.75 - Prob. 10ECh. U4.75 - Prob. 11ECh. U4.76 - Prob. 1TAICh. U4.76 - Prob. 1ECh. U4.76 - Prob. 2ECh. U4.76 - Prob. 3ECh. U4.76 - Prob. 4ECh. U4.76 - Prob. 5ECh. U4.76 - Prob. 6ECh. U4.76 - Prob. 7ECh. U4.76 - Prob. 8ECh. U4.77 - Prob. 1TAICh. U4.77 - Prob. 1ECh. U4.77 - Prob. 2ECh. U4.77 - Prob. 3ECh. U4.77 - Prob. 4ECh. U4.77 - Prob. 5ECh. U4.77 - Prob. 6ECh. U4.77 - Prob. 7ECh. U4.77 - Prob. 8ECh. U4.78 - Prob. 1TAICh. U4.78 - Prob. 1ECh. U4.78 - Prob. 2ECh. U4.78 - Prob. 3ECh. U4.78 - Prob. 4ECh. U4.78 - Prob. 5ECh. U4.78 - Prob. 6ECh. U4.78 - Prob. 7ECh. U4.78 - Prob. 8ECh. U4.79 - Prob. 1TAICh. U4.79 - Prob. 1ECh. U4.79 - Prob. 2ECh. U4.79 - Prob. 3ECh. U4.79 - Prob. 4ECh. U4.80 - Prob. 1TAICh. U4.80 - Prob. 1ECh. U4.80 - Prob. 2ECh. U4.80 - Prob. 3ECh. U4.80 - Prob. 4ECh. U4.80 - Prob. 5ECh. U4.80 - Prob. 6ECh. U4.80 - Prob. 7ECh. U4.80 - Prob. 8ECh. U4.80 - Prob. 9ECh. U4.80 - Prob. 10ECh. U4.81 - Prob. 1TAICh. U4.81 - Prob. 1ECh. U4.81 - Prob. 2ECh. U4.81 - Prob. 3ECh. U4.81 - Prob. 4ECh. U4.81 - Prob. 5ECh. U4.81 - Prob. 6ECh. U4.81 - Prob. 7ECh. U4.81 - Prob. 8ECh. U4.81 - Prob. 9ECh. U4.82 - Prob. 1TAICh. U4.82 - Prob. 1ECh. U4.82 - Prob. 2ECh. U4.82 - Prob. 3ECh. U4.82 - Prob. 4ECh. U4.82 - Prob. 5ECh. U4.82 - Prob. 6ECh. U4.82 - Prob. 7ECh. U4.82 - Prob. 8ECh. U4.83 - Prob. 1TAICh. U4.83 - Prob. 1ECh. U4.83 - Prob. 2ECh. U4.83 - Prob. 3ECh. U4.83 - Prob. 4ECh. U4.83 - Prob. 5ECh. U4.83 - Prob. 6ECh. U4.83 - Prob. 7ECh. U4.84 - Prob. 1TAICh. U4.84 - Prob. 1ECh. U4.84 - Prob. 2ECh. U4.84 - Prob. 3ECh. U4.84 - Prob. 5ECh. U4.84 - Prob. 6ECh. U4.84 - Prob. 7ECh. U4.85 - Prob. 1TAICh. U4.85 - Prob. 1ECh. U4.85 - Prob. 2ECh. U4.85 - Prob. 3ECh. U4.85 - Prob. 4ECh. U4.85 - Prob. 5ECh. U4.85 - Prob. 6ECh. U4.85 - Prob. 7ECh. U4.85 - Prob. 8ECh. U4.86 - Prob. 1TAICh. U4.86 - Prob. 1ECh. U4.86 - Prob. 2ECh. U4.86 - Prob. 3ECh. U4.86 - Prob. 4ECh. U4.86 - Prob. 6ECh. U4.86 - Prob. 7ECh. U4.86 - Prob. 8ECh. U4.87 - Prob. 1TAICh. U4.87 - Prob. 1ECh. U4.87 - Prob. 2ECh. U4.87 - Prob. 3ECh. U4.87 - Prob. 4ECh. U4.87 - Prob. 5ECh. U4.87 - Prob. 6ECh. U4.87 - Prob. 7ECh. U4.87 - Prob. 8ECh. U4.88 - Prob. 1TAICh. U4.88 - Prob. 1ECh. U4.88 - Prob. 2ECh. U4.88 - Prob. 4ECh. U4.88 - Prob. 5ECh. U4.88 - Prob. 6ECh. U4.88 - Prob. 7ECh. U4.88 - Prob. 8ECh. U4.89 - Prob. 1TAICh. U4.89 - Prob. 1ECh. U4.89 - Prob. 2ECh. U4.89 - Prob. 3ECh. U4.89 - Prob. 4ECh. U4.89 - Prob. 5ECh. U4.89 - Prob. 6ECh. U4.90 - Prob. 1ECh. U4.90 - Prob. 2ECh. U4.90 - Prob. 3ECh. U4.90 - Prob. 4ECh. U4.90 - Prob. 5ECh. U4.90 - Prob. 6ECh. U4.90 - Prob. 7ECh. U4.91 - Prob. 1ECh. U4.91 - Prob. 2ECh. U4.91 - Prob. 3ECh. U4.91 - Prob. 5ECh. U4.91 - Prob. 6ECh. U4.92 - Prob. 1TAICh. U4.92 - Prob. 1ECh. U4.92 - Prob. 2ECh. U4.92 - Prob. 3ECh. U4.92 - Prob. 4ECh. U4.93 - Prob. 1TAICh. U4.93 - Prob. 1ECh. U4.93 - Prob. 2ECh. U4.93 - Prob. 4ECh. U4.93 - Prob. 5ECh. U4.93 - Prob. 6ECh. U4 - Prob. C13.3RECh. U4 - Prob. C13.4RECh. U4 - Prob. C14.1RECh. U4 - Prob. C14.2RECh. U4 - Prob. C14.3RECh. U4 - Prob. C14.5RECh. U4 - Prob. C14.6RECh. U4 - Prob. C15.1RECh. U4 - Prob. C15.2RECh. U4 - Prob. C15.3RECh. U4 - Prob. C15.4RECh. U4 - Prob. C15.5RECh. U4 - Prob. C15.6RECh. U4 - Prob. C15.7RECh. U4 - Prob. C15.8RECh. U4 - Prob. C16.1RECh. U4 - Prob. C16.2RECh. U4 - Prob. C16.3RECh. U4 - Prob. C16.4RECh. U4 - Prob. C17.1RECh. U4 - Prob. C17.2RECh. U4 - Prob. C17.3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RECh. U4 - Prob. 1STPCh. U4 - Prob. 2STPCh. U4 - Prob. 3STPCh. U4 - Prob. 4STPCh. U4 - Prob. 5STPCh. U4 - Prob. 6STPCh. U4 - Prob. 7STPCh. U4 - Prob. 8STPCh. U4 - Prob. 9STPCh. U4 - Prob. 10STPCh. U4 - Prob. 11STPCh. U4 - Prob. 12STPCh. U4 - Prob. 13STPCh. U4 - Prob. 14STPCh. U4 - Prob. 15STPCh. U4 - Prob. 16STPCh. U4 - Prob. 17STPCh. U4 - Prob. 18STPCh. U4 - Prob. 19STPCh. U4 - Prob. 20STP
Additional Science Textbook Solutions
Find more solutions based on key concepts
Which one of the following is not a fuel produced by microorganisms? a. algal oil b. ethanol c. hydrogen d. met...
Microbiology: An Introduction
Which elements would you expect to lose electrons in chemical changes? a. potassium b. sulfur c. fluorine d. ba...
Introductory Chemistry (6th Edition)
Pigeons may exhibit a checkered or plain color pattern. In a series of controlled matings, the following data w...
Concepts of Genetics (12th Edition)
What is the difference between cellular respiration and external respiration?
Human Physiology: An Integrated Approach (8th Edition)
An atom with a formal charge does not necessarily have more or less electron density than the atoms in the mole...
Organic Chemistry (8th Edition)
Explain how the behavior of homologous chromosomes in meiosis parallels Mendels law of segregation for autosoma...
Genetic Analysis: An Integrated Approach (3rd Edition)
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
- Is the structural form shown possible given the pKa constraints of the side chains?arrow_forwardon x Fina X Sign X Sign x lab X Intro X Cop X chat X My x Grac x Laur x Laur x ashes x S Shox S SHE x a eve.macmillanlearning.com/ihub/assessment/f188d950-dd73-11e0-9572-0800200c9a66/d591b3f2-d5f7-4983-843c-0d00c1c0340b/f2b47861-07c4-4d1b-a1ee-e7db27d6b4ee?actualCourseld=d591b3f2-c stions estion. ct each urces. +95 Macmillan Learning Draw the product formed by the reaction of potassium t-butoxide with (15,25)-1-bromo-2-methyl-1-phenylbutane (shown). Clearly show the stereochemistry of the product. H BH (CH3)3CO-K+ +100 H3CW (CH3)3COH +85 H3CH2C +95 ossible ↓ Q Search Select Draw Templates More C H 0 bp A Erase 2Q 112 Resouarrow_forwardIdentify the structure of the PTH derivative generated after two rounds of Edman degradation.arrow_forward
- Use the data below from an electron impact mass spectrum of a pure compound to deduce its structure. Draw your structure in the drawing window. Data selected from the NIST WebBook, https://webbook.nist.gov/chemistry/ m/z Relative intensity 31 0.5 30 26 29 22 28 100 27 33 26 23 15 4 • You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. 妊 n ? Previous Nextarrow_forwardfor this question. Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M+ = 98.1106. Exact Masses of the Most Abundant Isotope of Selected Elements Isotope Natural abundance (%) Exact mass 1H 99.985 1.008 12C 98.90 12.000 14N 99.63 14.003 160 99.76 15.995 Molecular formula (In the order CHNO, with no subscripts)arrow_forwardPLEASE READ!!! I DONT WANT EXAMPLES, I DONT WANT WORDS OR PARAGRAPHS!!! PLEASE I UNDERSTAND THE BASICS BUT THIS IS AN EXCEPTION THAT EVEN THE INTERNET CANT HELP!!!! THIS IS THE THIRD TIME I'VE SENT THOSE QUESTIONS SO PLEASE DONT RESEND THE SAME STUFF, ITS NOT HELPING ME!!! I ALSO ALREADY TRIED TO DRAW THE MECHANISM MYSELF, SO IF ITS RIGHT PLEASE TELL ME OR TELL ME WHAT I HAVE TO CHANGE!!! First image: I have to SHOW (DRAWING) the mechanism (with arows and structures of molecules) NOT WORDS PLEASE! of the reaction at the bottom. Also I have to show by mecanism why the reaction wouldn't work if the alcohol was primary Second image: I have to show the mechanism (IMAGE) (with arrows and structures of the molecules) NOT WORDS PLEASE !! for the reaction on the left, where the alcohol A is added fast in one portion HOMEWORK, NOT EXAM!! ALL DETAILS ARE IN THE IMAGES PLEASE LOOK AT THE IMAGES, DONT LOOK AT THE AI GENERATED TEXT!!!arrow_forward
- Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M+ = 85.0899. Exact Masses of the Most Abundant Isotope of Selected Elements Isotope Natural abundance (%) Exact mass 1H 99.985 1.008 12C 98.90 12.000 14N 99.63 14.003 160 99.76 15.995 Molecular formula (In the order CHNO, with no subscripts)arrow_forwardUse the data below from an electron impact mass spectrum of a pure compound to deduce its structure. Draw your structure in the drawing window. Data selected from the NIST WebBook, https://webbook.nist.gov/chemistry/ m/z Relative intensity 59 3.0 58 64 43 100 15 23 • You do not have to consider stereochemistry. •You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one. + n[] 85 // ? CH4 Previous Nextarrow_forwardWrite the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M* = 128.0632. Exact Masses of the Most Abundant Isotope of Selected Elements Isotope Natural abundance (%) Exact mass 1H 99.985 12C 98.90 14N 99.63 160 99.76 Molecular formula 1.008 12.000 14.003 15.995 (In the order CHNO, with no subscripts)arrow_forward
- Can I please get help with this? And can I please the lowest possible significant number?arrow_forwardWhat is the molar mass of a gas that takes three times longer to effuse than helium?arrow_forwardFirst image: I have to show the mecanism (with arows and structures) of the reaction at the bottom. Also I have to show by mecanism why the reaction wouldn't work if the alcohol was primary Second image: I have to show the mecanism (with arrows and structures) for the reaction on the left, where the alcohol A is added fast in one portion its not an examarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles 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 Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY