General, Organic, and Biological Chemistry - 4th edition
General, Organic, and Biological Chemistry - 4th edition
4th Edition
ISBN: 9781259883989
Author: by Janice Smith
Publisher: McGraw-Hill Education
Question
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Chapter 10, Problem 21P
Interpretation Introduction

(a)

Interpretation:

The given table of isotope should be completed.

General, Organic, and Biological Chemistry - 4th edition, Chapter 10, Problem 21P , additional homework tip  1

Concept Introduction:

Isotopes are the compounds having the same atomic number but different atomic mass.

To write an isotope symbol atomic number(Z) is written on the lower left side and atomic mass(A) is written on the upper left side of an element.

Expert Solution
Check Mark

Answer to Problem 21P

The complete table for isotopes symbol is represented as follows:

General, Organic, and Biological Chemistry - 4th edition, Chapter 10, Problem 21P , additional homework tip  2

Explanation of Solution

Atomic number is equal to the number of protons in an element.

Mass number = Number of protons + Number of neutrons

The atomic number for chromium is 24 and mass number is 51.

Since atomic number = number of protons

Thus, the number of protons in chromium is 24 and the number of neutrons in chromium can be calculated by simply subtracting number of protons from mass number as follows:

  Mass number = Number of protons + Number of neutrons51=24+ Number of neutronsNumber of neutrons=5124=27

Therefore, the isotope symbol for Chromium is 2451Cr

Interpretation Introduction

(b)

Interpretation:

The given table of isotopes should be completed.

General, Organic, and Biological Chemistry - 4th edition, Chapter 10, Problem 21P , additional homework tip  3

Concept Introduction:

Isotopes are the compounds having the same atomic number but different atomic mass.

To write an isotope symbol atomic number(Z) is written on the lower left side and atomic mass(A) is written on the upper left side of an element.

Expert Solution
Check Mark

Answer to Problem 21P

The complete table for isotopes symbol is represented as follows:

Atomic number(Z)Mass number(A)Number of protonsNumber of neutronsIsotope symbol
Palladium-10346103465746103Pd

Explanation of Solution

Atomic number is equal to the number of protons in an element.

Mass number = Number of protons + Number of neutrons.

The atomic number for the element listed in b option is given which is equal to 46 and the mass number is 103. So, the element is Palladium (Pd).

Since atomic number = number of protons.

Thus, the number of protons in palladium is 46 and the number of neutrons in palladium can be calculated by simply subtracting number of protons from the mass number as follows:

  Mass number = Number of protons + Number of neutrons103=46+ Number of neutronsNumber of neutrons=10346=57

Therefore, the isotope symbol for Palladium is 46103Pd.

Interpretation Introduction

(c)

Interpretation:

The given table of isotopes should be completed.

General, Organic, and Biological Chemistry - 4th edition, Chapter 10, Problem 21P , additional homework tip  4

Concept Introduction:

Isotopes are the compounds having the same atomic number but different atomic mass.

To write an isotope symbol atomic number(Z) is written on the lower left side and atomic mass(A) is written on the upper left side of an element.

Expert Solution
Check Mark

Answer to Problem 21P

The complete table for isotopes symbol is represented as follows:

Atomic number(Z)Mass number(A)Number of protonsNumber of neutronsIsotope symbol
Potassium-19194219231942K

Explanation of Solution

Atomic number is equal to the number of protons in an element.

Mass number = Number of protons + Number of neutrons

The number of protons in element listed in option c is 19 and the number of neutrons is 23.

Since atomic number = number of protons

Thus, the atomic number of this element is 19 and the element will be Potassium (K) and the mass number will be calculated as follows:

  Mass number = Number of protons + Number of neutrons=19+ 23=42

Therefore, the isotope symbol for Palladium is 1942K.

Interpretation Introduction

(d)

Interpretation:

The given table of isotopes should be completed.

General, Organic, and Biological Chemistry - 4th edition, Chapter 10, Problem 21P , additional homework tip  5

Concept Introduction:

Isotopes are the compounds having the same atomic number but different atomic mass.

To write an isotope symbol atomic number(Z) is written on the lower left side and atomic mass(A) is written on the upper left side of an element.

Expert Solution
Check Mark

Answer to Problem 21P

The complete table for isotopes symbol is represented as follows:

Atomic number(Z)Mass number(A)Number of protonsNumber of neutronsIsotope symbol
Xenon-13354133547954133Xe

Explanation of Solution

Atomic number is equal to the number of protons in an element.

Mass number = Number of protons + Number of neutrons

The number of protons for element listed in d option are 54 and mass number is 133.

Since atomic number = number of protons

Thus, atomic number for this element will be 54 and the element is Xenon (Xe) and the number of neutrons in xenon can be calculated by simply subtracting number of protons from mass number as follows:

  Mass number = Number of protons + Number of neutrons133=54+ Number of neutronsNumber of neutrons=13354=79

Therefore, the isotope symbol for Xenon is 54133Xe.

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Chapter 10 Solutions

General, Organic, and Biological Chemistry - 4th edition

Ch. 10.2 - Prob. 10.3PPCh. 10.2 - Prob. 10.9PCh. 10.2 - Prob. 10.10PCh. 10.2 - Prob. 10.11PCh. 10.3 - Prob. 10.4PPCh. 10.3 - Prob. 10.12PCh. 10.3 - Prob. 10.13PCh. 10.3 - Prob. 10.14PCh. 10.4 - To treat a thyroid tumor, a patient must be given...Ch. 10.4 - A sample of iodine-131 (t1/2=8.0 days) has an...Ch. 10.4 - Prob. 10.15PCh. 10.5 - Prob. 10.16PCh. 10.5 - Prob. 10.17PCh. 10.5 - Prob. 10.18PCh. 10.6 - Prob. 10.7PPCh. 10.6 - Prob. 10.8PPCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - Prob. 50PCh. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 56PCh. 10 - Prob. 57PCh. 10 - Prob. 58PCh. 10 - Prob. 59PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - Prob. 62PCh. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - Prob. 72PCh. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 77PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84PCh. 10 - Prob. 85PCh. 10 - Prob. 86PCh. 10 - Prob. 87PCh. 10 - Prob. 88PCh. 10 - Prob. 89CPCh. 10 - Prob. 90CP
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