
Concept explainers
(a)
Interpretation:
Calculations have to be carried out for 15.2-2.75+15.67 and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Significant figures: The digits that are measured in a number including one estimated digit are called as significant figures.
The answers for those calculations that involve addition and subtraction will have the same number of decimal places as the original number with the fewest decimal places.
(a)

Answer to Problem 14PE
The answer for 15.2-2.75+15.67 is 28.1.
Explanation of Solution
The calculation for 15.2-2.75+15.67 is,
15.2-2.7515.67_28.12
The number with the least precision is 15.2. Therefore, the answer is rounded off to the nearest tenth: 28.1. The answer for 15.2-2.75+15.67 is 28.1.
(b)
Interpretation:
Calculations have to be carried out for (4.68)(12.5) and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Significant figures: The digits that are measured in a number including one estimated digit are called as significant figures.
The answers for those calculations that involve multiplication and division will have the same number of significant figures as the original number with the fewest number of significant figures.
(b)

Answer to Problem 14PE
The answer for the product of (4.68)(12.5) is 58.5.
Explanation of Solution
The product of (4.68)(12.5) is 58.5. The answer should have three significant figures because both the numbers (4.68)(12.5), with the fewest significant figures, has only three significant figures.
The answer for the product of (4.68)(12.5) is 58.5.
(c)
Interpretation:
Calculations have to be carried out for 182.64.6 and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Refer to part (b).
(c)

Answer to Problem 14PE
The answer for 182.64.6 is 40..
Explanation of Solution
The answer for 182.64.6 is 39.69. The answer should have only two significant figures, since the number 4.6, with the fewest significant figures, has only two significant figures.
In the number 39.69, the last two digits are dropped and one is added to the next digit 9. The answer for 182.64.6 is 40..
(d)
Interpretation:
Calculations have to be carried out for 1986+23.84+0.012 and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Refer to part (a).
(d)

Answer to Problem 14PE
The answer for 1986+23.84+0.012 is 2.010×103.
Explanation of Solution
The sum of 1986+23.84+0.012 is 2009.852. The number with the least precision is 1986. The answer for 1986+23.84+0.012 is 2.010×103.
(e)
Interpretation:
Calculations have to be carried out for 29.3(284)(415) and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Refer to part (b).
(e)

Answer to Problem 14PE
The answer for 29.3(284)(415) is 2.49×10-4.
Explanation of Solution
The answer for 29.3(284)(415) is 2.486×10-4. The answer should have only three significant figures because all the three numbers with fewest significant figures, have only three significant figures.
In the number 2.486×10-4, the last digit is dropped and one is added to the next digit 8. The rounded off three significant number is 2.49×10-4. The answer for 29.3(284)(415) is 2.49×10-4.
(f)
Interpretation:
Calculations have to be carried out for (2.92×10-3)(6.14×105) and the answer has to be provided using proper number of significant figures.
Concept Introduction:
Refer to part (b).
(f)

Answer to Problem 14PE
The answer for (2.92×10-3)(6.14×105) is 1.79×103.
Explanation of Solution
The answer for (2.92×10-3)(6.14×105) is 1792.88. The answer should have only three significant figures, since both the numbers 2.92×10-3,6.14×105 with the fewest significant figures, has only three significant figures.
In the number 1792.88, the last three digits are dropped and the rounded off three significant number is 1.79×103. The answer for (2.92×10-3)(6.14×105) is 1.79×103.
Want to see more full solutions like this?
Chapter 2 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
- please solve this, and help me know which boxes to check. Thank you so much in advance.arrow_forwardElectronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Describe how electronegativity is illustrated on the periodic table including trends between groups and periods and significance of atom size.arrow_forwardDefine the term “transition.” How does this definition apply to the transition metals?arrow_forward
- Describe how the properties of the different types of elements (metals, nonmetals, metalloids) differ.arrow_forwardUse a textbook or other valid source to research the physical and chemical properties of each element listed in Data Table 1 using the following as a guideline: Ductile (able to be deformed without losing toughness) and malleable (able to be hammered or pressed permanently out of shape without breaking or cracking) or not ductile or malleable Good, semi, or poor conductors of electricity and heat High or low melting and boiling points Occur or do not occur uncombined/freely in nature High, intermediate, or low reactivity Loses or gains electrons during reactions or is not reactivearrow_forwardProvide the Physical and Chemical Properties of Elements of the following elements listedarrow_forward
- Questions 4 and 5arrow_forwardFor a titration of 40.00 mL of 0.0500 M oxalic acid H2C2O4 with 0.1000 M KOH, calculate the pH at each of the following volume of KOH used in the titration: 1) before the titration begin;2) 15 mL; 3) 20 mL; 4) 25 mL; 5) 40 mL; 6) 50 mL. Ka1 = 5.90×10^-2, Ka2 = 6.50×10^-5 for oxalic acid.arrow_forwardPredict the major organic product(s), if any, of the following reactions. Assume all reagents are in excess unless otherwise indicated.arrow_forward
- Predict the major organic product(s), if any, of the following reactions. Assume all reagents are in excess unless otherwise indicated.arrow_forwardHow many signals would you expect to find in the 1 H NMR spectrum of each given compound? Part 1 of 2 2 Part 2 of 2 HO 5 ☑ Х IIIIII***** §arrow_forwardA carbonyl compound has a molecular ion with a m/z of 86. The mass spectra of this compound also has a base peak with a m/z of 57. Draw the correct structure of this molecule. Drawingarrow_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





