INTOR TO CHEMISTRY LLF
INTOR TO CHEMISTRY LLF
5th Edition
ISBN: 9781264501731
Author: BAUER
Publisher: MCG
bartleby

Concept explainers

Question
Book Icon
Chapter 15, Problem 30QP

(a)

Interpretation Introduction

Interpretation:

The balanced equation of beta emission by A47114g is to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

A nuclear reaction involves changes at the subatomic level such as changes in the number of protons, number of neutrons or energy state. A nuclear reaction is balanced by two conservation laws. That are,

1. conservation of mass number that conserves the mass number of the elements.

2. conservation of charges that conserves the atomic number of the elements.

The reaction can be written as follows:

A47114gXZA+e10

Where, A is mass number, Z is the atomic number of the unknown product, X .

Apply the conservation conditions for mass numbers and atomic numbers to identify the product, X .

Balance the mass numbers according to the law of conservation of mass numbers. The mass number of the reactant is equal to the sum of mass numbers of the products.

114=A + 0A = 114

Similarly, balance the atomic numbers according to the law of conservation of charges. The atomic number of the reactant is equal to the sum of atomic numbers of the products.

47=Z + 1Z = 48

The element with atomic number 48 is cadmium. Therefore, the product, X is C48114d .

Hence, the balanced equation for the reaction is,

A47114gC48114+e10

(b)

Interpretation Introduction

Interpretation:

The balanced equation of alpha emission by P84210o is to be determined.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction can be written as follows:

P84210oXZA +α24

Where, A is mass number, Z is the atomic number of the unknown product, X .

Apply the conservation conditions for mass numbers and atomic numbers to identify the product, X .

Balance the mass numbers according to the law of conservation of mass numbers. The mass number of the reactant is equal to the sum of mass numbers of the products.

210=A + 4A = 206

Similarly, balance the atomic numbers according to the law of conservation of charges. The atomic number of the reactant is equal to the sum of atomic numbers of the products.

84=Z + 2Z = 82

The element with atomic number 82 is lead. Therefore, the product, X is P82206b .

Hence, the balanced equation for the reaction is,

P84210oP82206b +α24

(c)

Interpretation Introduction

Interpretation:

The balance equation of electron capture by B56133a is to be determined.

(c)

Expert Solution
Check Mark

Explanation of Solution

The reaction can be written as follows:

B56133+e10XZA 

Where, A is mass number, Z is the atomic number of the unknown product, X .

Apply the conservation conditions for mass numbers and atomic numbers to identify the product, X .

Balance the mass numbers according to the law of conservation of mass numbers. The mass number of the reactant is equal to the sum of mass numbers of the products.

133+0=AA = 133

Similarly, balance the atomic numbers according to the law of conservation of charges. The atomic number of the reactant is equal to the sum of atomic numbers of the products.

561=ZZ = 55

The element with atomic number 55 is caesium. Therefore, the product, X is C55133s .

Hence, the balanced equation for the reaction is,

B56133a+e10C55133s

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
In the case of isopilianions, briefly state:- why polymeric species with a defined MW are formed.- why the extent of polymerization is different depending on the metal.- why these polyhedra with such special structures are formed.
A carboxylic acid reacts with water to form a carboxylate ion and H,O+. Complete the reaction. reaction: (CH),CHCH2COOH + H2O (CH), CHCH, COO¯ + H₂O+ Write the IUPAC name of the carboxylate ion formed in the reaction. IUPAC name: BIU X2 SPECIAL GREEK ALPHABET ~ I
Show work. Don't give Ai generated solution

Chapter 15 Solutions

INTOR TO CHEMISTRY LLF

Ch. 15 - Prob. 5PPCh. 15 - Prob. 6PPCh. 15 - Prob. 7PPCh. 15 - Prob. 8PPCh. 15 - Prob. 9PPCh. 15 - Prob. 10PPCh. 15 - Prob. 11PPCh. 15 - Prob. 1QPCh. 15 - Prob. 2QPCh. 15 - Prob. 3QPCh. 15 - Prob. 4QPCh. 15 - Prob. 5QPCh. 15 - Prob. 6QPCh. 15 - Prob. 7QPCh. 15 - Prob. 8QPCh. 15 - Prob. 9QPCh. 15 - Prob. 10QPCh. 15 - Prob. 11QPCh. 15 - Prob. 12QPCh. 15 - Prob. 13QPCh. 15 - Prob. 14QPCh. 15 - Prob. 15QPCh. 15 - Prob. 16QPCh. 15 - Prob. 17QPCh. 15 - Prob. 18QPCh. 15 - Prob. 19QPCh. 15 - Prob. 20QPCh. 15 - Prob. 21QPCh. 15 - Prob. 22QPCh. 15 - Prob. 23QPCh. 15 - Prob. 24QPCh. 15 - Prob. 25QPCh. 15 - Prob. 26QPCh. 15 - Prob. 27QPCh. 15 - Prob. 28QPCh. 15 - Prob. 29QPCh. 15 - Prob. 30QPCh. 15 - Prob. 31QPCh. 15 - Prob. 32QPCh. 15 - Prob. 33QPCh. 15 - Prob. 34QPCh. 15 - Prob. 35QPCh. 15 - Prob. 36QPCh. 15 - Prob. 37QPCh. 15 - Prob. 38QPCh. 15 - Prob. 39QPCh. 15 - Prob. 40QPCh. 15 - Prob. 41QPCh. 15 - Prob. 42QPCh. 15 - Prob. 43QPCh. 15 - Prob. 44QPCh. 15 - Prob. 45QPCh. 15 - Prob. 46QPCh. 15 - Prob. 47QPCh. 15 - Prob. 48QPCh. 15 - Prob. 49QPCh. 15 - Prob. 50QPCh. 15 - Prob. 51QPCh. 15 - Prob. 52QPCh. 15 - Prob. 53QPCh. 15 - Prob. 54QPCh. 15 - Prob. 55QPCh. 15 - Prob. 56QPCh. 15 - Prob. 57QPCh. 15 - Prob. 58QPCh. 15 - Prob. 59QPCh. 15 - Prob. 60QPCh. 15 - Prob. 61QPCh. 15 - Prob. 62QPCh. 15 - Prob. 63QPCh. 15 - Prob. 64QPCh. 15 - Prob. 65QPCh. 15 - Prob. 66QPCh. 15 - Prob. 67QPCh. 15 - Prob. 68QPCh. 15 - Prob. 69QPCh. 15 - Prob. 70QPCh. 15 - Prob. 73QPCh. 15 - Prob. 74QPCh. 15 - Prob. 75QPCh. 15 - Prob. 76QPCh. 15 - Prob. 77QPCh. 15 - Prob. 78QPCh. 15 - Prob. 79QPCh. 15 - Prob. 80QPCh. 15 - Prob. 81QPCh. 15 - Prob. 82QPCh. 15 - Prob. 83QPCh. 15 - Prob. 84QPCh. 15 - Prob. 85QPCh. 15 - Prob. 86QPCh. 15 - Prob. 87QPCh. 15 - Prob. 88QPCh. 15 - Prob. 89QPCh. 15 - Prob. 90QPCh. 15 - Prob. 91QPCh. 15 - Prob. 92QPCh. 15 - Prob. 93QPCh. 15 - Prob. 94QPCh. 15 - Prob. 95QPCh. 15 - Prob. 96QPCh. 15 - Prob. 97QPCh. 15 - Prob. 98QPCh. 15 - Prob. 99QPCh. 15 - Prob. 100QPCh. 15 - Prob. 101QPCh. 15 - Prob. 102QPCh. 15 - Prob. 103QPCh. 15 - Prob. 104QPCh. 15 - Prob. 105QPCh. 15 - Prob. 106QPCh. 15 - Prob. 107QPCh. 15 - Prob. 108QPCh. 15 - Prob. 109QPCh. 15 - Prob. 110QPCh. 15 - Prob. 111QPCh. 15 - Prob. 112QPCh. 15 - Prob. 113QPCh. 15 - Prob. 114QPCh. 15 - Prob. 115QPCh. 15 - Prob. 116QPCh. 15 - Prob. 117QPCh. 15 - Prob. 118QPCh. 15 - Prob. 119QPCh. 15 - Prob. 120QPCh. 15 - Prob. 121QPCh. 15 - Prob. 122QPCh. 15 - Prob. 123QP
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:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:9781559539418
Author:Angelica Stacy
Publisher:MAC HIGHER
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: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning