Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 23E
Interpretation Introduction

Interpretation:

The table with atomic notations is to be completed by providing the missing information.

Concept introduction:

The atomic number of an atom is equal to the number of protons in an atom. Mass number or atomic mass of an atom can be calculated by taking the sum of atomic number and number of neutrons.

Expert Solution & Answer
Check Mark

Answer to Problem 23E

The table with atomic notations with complete information is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80

Explanation of Solution

In the atomic notation, ZASy the ‘Sy’ represents the symbol of the element, the subscript represents the atomic number and the superscript represents the mass number. The atomic number is equal to the number of protons present in the atom. For a neutral atom, the number of electrons is same as the atomic number.

The number of neutrons is calculated by the formula shown below.

Numberofneutrons=MassnumberAtomicnumber …(1)

The given atomic notation is 511B. The atomic number for the given notation is 5. Thus, the number of electrons and protons are 5. The mass number is 11.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=115=6

Thus, the number of neutrons in atom of 511B is 6.

The given atomic notation is 715N.The atomic number for the given notation is 7. Thus, the number of electrons and protons are 7. The mass number is 15.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=157=8

Thus, the number of neutrons in atom of 715N is 8.

The given atomic notation is 2040Ca.The atomic number for the given notation is 20. Thus, the number of electrons and protons are 20. The mass number is 40.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=4020=20

Thus, the number of neutrons in atom of 2040Ca is 20.

The given atomic notation is 80200Hg.The atomic number for the given notation is 80. Thus, the number of electrons and protons are 80. The mass number is 200.

Substitute the values of mass number and atomic number in the equation (1).

Numberofneutrons=20080=120

Thus, the number of neutrons in atom of 80200Hg is 120.

The table with the information of atomic number, mass number, number of neutrons, number of protons and number of electrons for given atomic notations is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80
Conclusion

The table with atomic notations with complete information is shown below.

Atomic
Notation
Atomic
Number
Mass
Number
Number of protons Number of neutrons Number of electrons
511B 5 11 5 6 5
715N 7 15 7 8 7
2040Ca 20 40 20 20 20
80200Hg 80 200 80 120 80

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m
Determine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +
Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWL

Chapter 4 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 4 - Prob. 11CECh. 4 - Prob. 12CECh. 4 - Prob. 1KTCh. 4 - Prob. 2KTCh. 4 - Prob. 3KTCh. 4 - Prob. 4KTCh. 4 - Prob. 5KTCh. 4 - Prob. 6KTCh. 4 - Prob. 7KTCh. 4 - Prob. 8KTCh. 4 - Prob. 9KTCh. 4 - Prob. 10KTCh. 4 - Prob. 11KTCh. 4 - Prob. 12KTCh. 4 - Prob. 13KTCh. 4 - Prob. 14KTCh. 4 - Prob. 15KTCh. 4 - Prob. 16KTCh. 4 - Prob. 17KTCh. 4 - Prob. 18KTCh. 4 - Prob. 19KTCh. 4 - Prob. 20KTCh. 4 - Prob. 21KTCh. 4 - Prob. 22KTCh. 4 - Prob. 23KTCh. 4 - Prob. 24KTCh. 4 - Prob. 25KTCh. 4 - Prob. 1ECh. 4 - Prob. 2ECh. 4 - Prob. 3ECh. 4 - Prob. 4ECh. 4 - Prob. 5ECh. 4 - Prob. 6ECh. 4 - Prob. 7ECh. 4 - Prob. 8ECh. 4 - Prob. 9ECh. 4 - Prob. 10ECh. 4 - Prob. 11ECh. 4 - Prob. 12ECh. 4 - Prob. 13ECh. 4 - Prob. 14ECh. 4 - Prob. 15ECh. 4 - Prob. 16ECh. 4 - Prob. 17ECh. 4 - Prob. 18ECh. 4 - Prob. 19ECh. 4 - Prob. 20ECh. 4 - Prob. 21ECh. 4 - Prob. 22ECh. 4 - Prob. 23ECh. 4 - Prob. 24ECh. 4 - Prob. 25ECh. 4 - Prob. 26ECh. 4 - Prob. 27ECh. 4 - Prob. 28ECh. 4 - Prob. 29ECh. 4 - Prob. 30ECh. 4 - Prob. 31ECh. 4 - Prob. 32ECh. 4 - Prob. 33ECh. 4 - Prob. 34ECh. 4 - Prob. 35ECh. 4 - Prob. 36ECh. 4 - Prob. 37ECh. 4 - Prob. 38ECh. 4 - Prob. 39ECh. 4 - Prob. 40ECh. 4 - Prob. 41ECh. 4 - Prob. 42ECh. 4 - Prob. 43ECh. 4 - Prob. 44ECh. 4 - Prob. 45ECh. 4 - Prob. 46ECh. 4 - Prob. 47ECh. 4 - Prob. 48ECh. 4 - Prob. 49ECh. 4 - Prob. 50ECh. 4 - Prob. 51ECh. 4 - Prob. 52ECh. 4 - Prob. 53ECh. 4 - Prob. 54ECh. 4 - Prob. 55ECh. 4 - Prob. 56ECh. 4 - Prob. 57ECh. 4 - Prob. 58ECh. 4 - Prob. 59ECh. 4 - Prob. 60ECh. 4 - Prob. 61ECh. 4 - Prob. 62ECh. 4 - Prob. 63ECh. 4 - Prob. 64ECh. 4 - Prob. 65ECh. 4 - Prob. 66ECh. 4 - Prob. 67ECh. 4 - Prob. 68ECh. 4 - Prob. 69ECh. 4 - Prob. 70ECh. 4 - Prob. 71ECh. 4 - Prob. 72ECh. 4 - Prob. 73ECh. 4 - Prob. 74ECh. 4 - Prob. 75ECh. 4 - Prob. 76ECh. 4 - Prob. 77ECh. 4 - Prob. 78ECh. 4 - Prob. 79ECh. 4 - Prob. 80ECh. 4 - Prob. 81ECh. 4 - Prob. 82ECh. 4 - Prob. 83ECh. 4 - Prob. 84ECh. 4 - Prob. 85ECh. 4 - Prob. 86ECh. 4 - Prob. 87ECh. 4 - Prob. 88ECh. 4 - Prob. 89ECh. 4 - Prob. 90ECh. 4 - Prob. 91ECh. 4 - Prob. 92ECh. 4 - Prob. 93ECh. 4 - Prob. 94ECh. 4 - Prob. 95ECh. 4 - Prob. 96ECh. 4 - Prob. 97ECh. 4 - Prob. 98ECh. 4 - Prob. 1STCh. 4 - Prob. 2STCh. 4 - Prob. 3STCh. 4 - Prob. 4STCh. 4 - Prob. 5STCh. 4 - Prob. 6STCh. 4 - Prob. 7STCh. 4 - Prob. 8STCh. 4 - Prob. 9STCh. 4 - Prob. 10STCh. 4 - Prob. 11STCh. 4 - Prob. 12STCh. 4 - Prob. 13STCh. 4 - Prob. 14STCh. 4 - Prob. 15STCh. 4 - Prob. 16STCh. 4 - Prob. 17STCh. 4 - Prob. 18ST
Knowledge Booster
Background pattern image
Chemistry
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
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
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Chemistry In Focus
Chemistry
ISBN:9781337399692
Author:Tro, Nivaldo J.
Publisher: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