Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter U4.4, Problem 7E
Interpretation Introduction
Interpretation: How the matter is conserved in case of burning a piece of paper needs to be explained.
Concept Introduction :
According to law of conservation of matter “matter can neither be created nor be destroyed and it changes from one form to another form.”
Expert Solution & Answer
Answer to Problem 7E
The total mass of reactants should be equal to the total mass of products.
Explanation of Solution
Combustion is a
Paper is made of wood which is a mixture of hydrocarbons.
On burning of paper, it turns into ash and release some smoke.
Mass of the paper = Mass of ash + mass of smoke released
The mass of released gases cannot be determined but it does not mean that conservation of mass does not take place.
Paper changed from one form to another and there is no loss of matter.
This proves the law of conservation of matter.
Chapter U4 Solutions
Living By Chemistry: First Edition Textbook
Ch. U4.1 - Prob. 1TAICh. U4.1 - Prob. 1ECh. U4.1 - Prob. 2ECh. U4.1 - Prob. 3ECh. U4.1 - Prob. 4ECh. U4.1 - Prob. 5ECh. U4.1 - Prob. 6ECh. U4.2 - Prob. 1TAICh. U4.2 - Prob. 1ECh. U4.2 - Prob. 2E
Ch. U4.2 - Prob. 3ECh. U4.2 - Prob. 4ECh. U4.2 - Prob. 5ECh. U4.3 - Prob. 1TAICh. U4.3 - Prob. 1ECh. U4.3 - Prob. 2ECh. U4.3 - Prob. 3ECh. U4.3 - Prob. 4ECh. U4.3 - Prob. 6ECh. U4.4 - Prob. 1TAICh. U4.4 - Prob. 1ECh. U4.4 - Prob. 2ECh. U4.4 - Prob. 3ECh. U4.4 - Prob. 4ECh. U4.4 - Prob. 5ECh. U4.4 - Prob. 6ECh. U4.4 - Prob. 7ECh. U4.5 - Prob. 1TAICh. U4.5 - Prob. 1ECh. U4.5 - Prob. 2ECh. U4.5 - Prob. 3ECh. U4.5 - Prob. 4ECh. U4.6 - Prob. 1TAICh. U4.6 - Prob. 1ECh. U4.6 - Prob. 2ECh. U4.6 - Prob. 3ECh. U4.6 - Prob. 5ECh. U4.6 - Prob. 6ECh. U4.6 - Prob. 7ECh. U4.7 - Prob. 1TAICh. U4.7 - Prob. 1ECh. U4.7 - Prob. 2ECh. U4.7 - Prob. 4ECh. U4.7 - Prob. 5ECh. U4.8 - Prob. 1TAICh. U4.8 - Prob. 1ECh. U4.8 - Prob. 2ECh. U4.8 - Prob. 4ECh. U4.8 - Prob. 5ECh. U4.8 - Prob. 6ECh. U4.8 - Prob. 7ECh. U4.8 - Prob. 8ECh. U4.8 - Prob. 9ECh. U4.8 - Prob. 10ECh. U4.8 - Prob. 11ECh. U4.9 - Prob. 1TAICh. U4.9 - Prob. 1ECh. U4.9 - Prob. 2ECh. U4.9 - Prob. 3ECh. U4.9 - Prob. 4ECh. U4.9 - Prob. 5ECh. U4.9 - Prob. 6ECh. U4.9 - Prob. 7ECh. U4.9 - Prob. 8ECh. U4.10 - Prob. 1TAICh. U4.10 - Prob. 1ECh. U4.10 - Prob. 2ECh. U4.10 - Prob. 3ECh. U4.10 - Prob. 4ECh. U4.10 - Prob. 5ECh. U4.10 - Prob. 6ECh. U4.10 - Prob. 7ECh. U4.10 - Prob. 8ECh. U4.11 - Prob. 1TAICh. U4.11 - Prob. 1ECh. U4.11 - Prob. 2ECh. U4.11 - Prob. 3ECh. U4.11 - Prob. 4ECh. U4.11 - Prob. 5ECh. U4.11 - Prob. 6ECh. U4.11 - Prob. 7ECh. U4.11 - Prob. 8ECh. U4.12 - Prob. 1TAICh. U4.12 - Prob. 1ECh. U4.12 - Prob. 2ECh. U4.12 - Prob. 3ECh. U4.12 - Prob. 4ECh. U4.13 - Prob. 1TAICh. U4.13 - Prob. 1ECh. U4.13 - Prob. 2ECh. U4.13 - Prob. 3ECh. U4.13 - Prob. 4ECh. U4.13 - Prob. 5ECh. U4.13 - Prob. 6ECh. U4.13 - Prob. 7ECh. U4.13 - Prob. 8ECh. U4.13 - Prob. 9ECh. U4.13 - Prob. 10ECh. U4.14 - Prob. 1TAICh. U4.14 - Prob. 1ECh. U4.14 - Prob. 2ECh. U4.14 - Prob. 3ECh. U4.14 - Prob. 4ECh. U4.14 - Prob. 5ECh. U4.14 - Prob. 6ECh. U4.14 - Prob. 7ECh. U4.14 - Prob. 8ECh. U4.14 - Prob. 9ECh. U4.15 - Prob. 1TAICh. U4.15 - Prob. 1ECh. U4.15 - Prob. 2ECh. U4.15 - Prob. 3ECh. U4.15 - Prob. 4ECh. U4.15 - Prob. 5ECh. U4.15 - Prob. 6ECh. U4.15 - Prob. 7ECh. U4.15 - Prob. 8ECh. U4.16 - Prob. 1TAICh. U4.16 - Prob. 1ECh. U4.16 - Prob. 2ECh. U4.16 - Prob. 3ECh. U4.16 - Prob. 4ECh. U4.16 - Prob. 5ECh. U4.16 - Prob. 6ECh. U4.16 - Prob. 7ECh. U4.17 - Prob. 1TAICh. U4.17 - Prob. 1ECh. U4.17 - Prob. 2ECh. U4.17 - Prob. 3ECh. U4.17 - Prob. 5ECh. U4.17 - Prob. 6ECh. U4.17 - Prob. 7ECh. U4.18 - Prob. 1TAICh. U4.18 - Prob. 1ECh. U4.18 - Prob. 2ECh. U4.18 - Prob. 3ECh. U4.18 - Prob. 4ECh. U4.18 - Prob. 5ECh. U4.18 - Prob. 6ECh. U4.18 - Prob. 7ECh. U4.18 - Prob. 8ECh. U4.19 - Prob. 1TAICh. U4.19 - Prob. 1ECh. U4.19 - Prob. 2ECh. U4.19 - Prob. 3ECh. U4.19 - Prob. 4ECh. U4.19 - Prob. 6ECh. U4.19 - Prob. 7ECh. U4.19 - Prob. 8ECh. U4.20 - Prob. 1TAICh. U4.20 - Prob. 1ECh. U4.20 - Prob. 2ECh. U4.20 - Prob. 3ECh. U4.20 - Prob. 4ECh. U4.20 - Prob. 5ECh. U4.20 - Prob. 6ECh. U4.20 - Prob. 7ECh. U4.20 - Prob. 8ECh. U4.21 - Prob. 1TAICh. U4.21 - Prob. 1ECh. U4.21 - Prob. 2ECh. U4.21 - Prob. 4ECh. U4.21 - Prob. 5ECh. U4.21 - Prob. 6ECh. U4.21 - Prob. 7ECh. U4.21 - Prob. 8ECh. U4.22 - Prob. 1TAICh. U4.22 - Prob. 1ECh. U4.22 - Prob. 2ECh. U4.22 - Prob. 3ECh. U4.22 - Prob. 4ECh. U4.22 - Prob. 5ECh. U4.22 - Prob. 6ECh. U4.23 - Prob. 1ECh. U4.23 - Prob. 2ECh. U4.23 - Prob. 3ECh. U4.23 - Prob. 4ECh. U4.23 - Prob. 5ECh. U4.23 - Prob. 6ECh. U4.23 - Prob. 7ECh. U4.24 - Prob. 1ECh. U4.24 - Prob. 2ECh. U4.24 - Prob. 3ECh. U4.24 - Prob. 5ECh. U4.24 - Prob. 6ECh. U4.25 - Prob. 1TAICh. U4.25 - Prob. 1ECh. U4.25 - Prob. 2ECh. U4.25 - Prob. 3ECh. U4.25 - Prob. 4ECh. U4.26 - Prob. 1TAICh. U4.26 - Prob. 1ECh. U4.26 - Prob. 2ECh. U4.26 - Prob. 4ECh. U4.26 - Prob. 5ECh. U4.26 - Prob. 6ECh. U4 - Prob. SI3RECh. U4 - Prob. SI4RECh. U4 - Prob. SII1RECh. U4 - Prob. SII2RECh. U4 - Prob. SII3RECh. U4 - Prob. SII5RECh. U4 - Prob. SII6RECh. U4 - Prob. SIII1RECh. U4 - Prob. SIII2RECh. U4 - Prob. SIII3RECh. U4 - Prob. SIII4RECh. U4 - Prob. SIII5RECh. U4 - Prob. SIII6RECh. U4 - Prob. SIII7RECh. U4 - Prob. SIII8RECh. U4 - Prob. SIV1RECh. U4 - Prob. SIV2RECh. U4 - Prob. SIV3RECh. U4 - Prob. SIV4RECh. U4 - Prob. SV1RECh. U4 - Prob. SV2RECh. U4 - Prob. SV3RECh. 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. 12RE
Additional Science Textbook Solutions
Find more solutions based on key concepts
Why is living epithelial tissue limited to a certain thickness?
Human Anatomy & Physiology (2nd Edition)
Modified True/False 9. A giant bacterium that is large enough to be seen without a microscope is Selenomonas.
Microbiology with Diseases by Body System (5th Edition)
2. The three ropes in FIGURE EX6.2 are tied to a small, very light ring. Two of the ropes are anchored to wa...
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
The enzyme that catalyzes the C C bond cleavage reaction that converts serine to glycine removes the substitue...
Organic Chemistry (8th Edition)
All of the following terms can appropriately describe humans except: a. primary consumer b. autotroph c. hetero...
Human Biology: Concepts and Current Issues (8th Edition)
Which of the following traits would you expect to be inherited as quantitative traits? a. body weight in chicke...
Genetic Analysis: An Integrated Approach (3rd Edition)
Knowledge Booster
Similar questions
- Please draw the structure in the box that is consistent with all the spectral data and alphabetically label all of the equivalent protons in the structure (Ha, Hb, Hc....) in order to assign all the proton NMR peaks. The integrations are computer generated and approximate the number of equivalent protons. Molecular formula: C13H1802 14 13 12 11 10 11 (ppm) Structure with assigned H peaks 2.08 3.13arrow_forwardA 0.10 M solution of acetic acid (CH3COOH, Ka = 1.8 x 10^-5) is titrated with a 0.0250 M solution of magnesium hydroxide (Mg(OH)2). If 10.0 mL of the acid solution is titrated with 10.0 mL of the base solution, what is the pH of the resulting solution?arrow_forwardFirefly luciferin exhibits three rings. Identify which of the rings are aromatic. Identify which lone pairs are involved in establishing aromaticity. The lone pairs are labeled A-D below.arrow_forward
- A 0.10 M solution of acetic acid (CH3COOH, Ka = 1.8 x 10^-5) is titrated with a 0.0250 M solution of magnesium hydroxide (Mg(OH)2). If 10.0 mL of the acid solution is titrated with 10.0 mL of the base solution, what is the pH of the resulting solution?arrow_forwardGiven a complex reaction with rate equation v = k1[A] + k2[A]2, what is the overall reaction order?arrow_forwardPlease draw the structure in the box that is consistent with all the spectral data and alphabetically label all of the equivalent protons in the structure (Ha, Hb, Hc....) in order to assign all the proton NMR peaks. The integrations are computer generated and approximate the number of equivalent protons. Molecular formula: C13H1802 14 13 12 11 10 11 (ppm) Structure with assigned H peaks 2.08 3.13arrow_forward
- CHEMICAL KINETICS. One of the approximation methods for solving the rate equation is the steady-state approximation method. Explain what it consists of.arrow_forwardCHEMICAL KINETICS. One of the approximation methods for solving the rate equation is the limiting or determining step approximation method. Explain what it consists of.arrow_forwardCHEMICAL KINETICS. Indicate the approximation methods for solving the rate equation.arrow_forward
- TRANSMITTANCE เบบ Please identify the one structure below that is consistent with the 'H NMR and IR spectra shown and draw its complete structure in the box below with the protons alphabetically labeled as shown in the NMR spectrum and label the IR bands, including sp³C-H and sp2C-H stretch, indicated by the arrows. D 4000 OH LOH H₂C CH3 OH H₂C OCH3 CH3 OH 3000 2000 1500 HAVENUMBERI-11 1000 LOCH3 Draw your structure below and label its equivalent protons according to the peak labeling that is used in the NMR spectrum in order to assign the peaks. Integrals indicate number of equivalent protons. Splitting patterns are: s=singlet, d=doublet, m-multiplet 8 3Hb s m 1Hd s 3Hf m 2Hcd 2Had 1He 鄙视 m 7 7 6 5 4 3 22 500 T 1 0arrow_forwardRelative Transmittance 0.995 0.99 0.985 0.98 Please draw the structure that is consistent with all the spectral data below in the box and alphabetically label the equivalent protons in the structure (Ha, Hb, Hc ....) in order to assign all the proton NMR peaks. Label the absorption bands in the IR spectrum indicated by the arrows. INFRARED SPECTRUM 1 0.975 3000 2000 Wavenumber (cm-1) 1000 Structure with assigned H peaks 1 3 180 160 140 120 100 f1 (ppm) 80 60 40 20 0 C-13 NMR note that there are 4 peaks between 120-140ppm Integral values equal the number of equivalent protons 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 fl (ppm)arrow_forwardCalculate the pH of 0.0025 M phenol.arrow_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