
Concept explainers
(a)
Interpretation:
The equation for the decay of
Concept Introduction:
Radioactive decay of unstable nuclei leads to formation of smaller nuclei that are stable and during this course alpha, beta and gamma radiations are emitted. When alpha particles are released, the new nuclei will display a decrease of 4 units in mass number and decrease of 2 units in
(a)

Answer to Problem 11E
The equation for the decay of
Explanation of Solution
The decay of
While computing the mass numbers and
Mass number: 241 =? + 4
x = 237
Atomic number: 95 =? + 2
y= 93
Hence unknown element is
Hence, the decay of the decay of
(b)
Interpretation:
The balanced equation for the complete decay of
Concept Introduction:
Alpha decay involves the release of alpha particle, namely,
Beta decay involves the release of beta particle, namely,
(b)

Answer to Problem 11E
The final product obtained complete decay of
Explanation of Solution
The complete decay of
In other words, it indicates that
Mass number = 8 x 4 + 0 = 32 units
Atomic number = 16 + (-4) = 12 units
While computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 241 = 0 + 32
x = 209
Atomic number: 95 =? + 12
y= 83
Hence unknown element is
The decay of
(c)
Interpretation:
The intermediate products formed by complete decay of
Concept Introduction:
There is a release of alpha particle in alpha decay wherein the emitted is
In Beta decay, there is of a beta particle which is
(c)

Answer to Problem 11E
The various intermediates formed by complete decay of
Explanation of Solution
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 241 =? + 4
x = 237
Atomic number: 95 =? + 2
y= 93
Hence unknown element is
The alpha decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 237 =? + 4
x = 230
Atomic number: 93 =? + 2
y= 91
Hence unknown element is
The alpha decay of
-The beta decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 233 =? + 0
x = 233
Atomic number: 91 =? + (-1)
y= 92
\Hence unknown element is
The beta decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 233 =? + 4
x = 229
Atomic number: 92 =? + 2
y= 90
Hence unknown element is
The alpha decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 229 =? + 4
x = 225
Atomic number: 90 =? + 2
y= 88
Hence unknown element is
The alpha decay of
-The beta decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 225 =? + 0
x = 225
Atomic number: 88 =? + (-1)
y= 89
Hence unknown element is
The beta decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 225 =? + 4
x = 221
Atomic number: 89 =? + 2
y= 87
Hence unknown element is
The alpha decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 221 =? + 4
x = 217
Atomic number: 87 =? + 2
y= 85
Hence unknown element is
The alpha decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 217 =? + 4
x = 213
Atomic number: 85 =? + 2
y= 83
Hence unknown element is
The alpha decay of
-The beta decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 213 =? + 0
x = 213
Atomic number: 83 =? + (-1)
y= 84
Hence unknown element is
The beta decay of
-The alpha decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 213 =? + 4
x = 209
Atomic number: 84 =? + 2
y= 82
Hence unknown element is
The alpha decay of
-The beta decay of
Computing the mass numbers and atomic mass for unknown element, it is found that
Mass number: 209 =? + 0
x = 209
Atomic number: 82 =? + (-1)
y= 83
Hence unknown element is
The beta decay of
Want to see more full solutions like this?
Chapter 20 Solutions
EBK CHEMICAL PRINCIPLES
- What is the stepwise mechanism for this reaction?arrow_forward32. Consider a two-state system in which the low energy level is 300 J mol 1 and the higher energy level is 800 J mol 1, and the temperature is 300 K. Find the population of each level. Hint: Pay attention to your units. A. What is the partition function for this system? B. What are the populations of each level? Now instead, consider a system with energy levels of 0 J mol C. Now what is the partition function? D. And what are the populations of the two levels? E. Finally, repeat the second calculation at 500 K. and 500 J mol 1 at 300 K. F. What do you notice about the populations as you increase the temperature? At what temperature would you expect the states to have equal populations?arrow_forward30. We will derive the forms of the molecular partition functions for atoms and molecules shortly in class, but the partition function that describes the translational and rotational motion of a homonuclear diatomic molecule is given by Itrans (V,T) = = 2πmkBT h² V grot (T) 4π²IKBT h² Where h is Planck's constant and I is molecular moment of inertia. The overall partition function is qmolec Qtrans qrot. Find the energy, enthalpy, entropy, and Helmholtz free energy for the translational and rotational modes of 1 mole of oxygen molecules and 1 mole of iodine molecules at 50 K and at 300 K and with a volume of 1 m³. Here is some useful data: Moment of inertia: I2 I 7.46 x 10- 45 kg m² 2 O2 I 1.91 x 101 -46 kg m²arrow_forward
- K for each reaction step. Be sure to account for all bond-breaking and bond-making steps. HI HaC Drawing Arrows! H3C OCH3 H 4 59°F Mostly sunny H CH3 HO O CH3 'C' CH3 Select to Add Arrows CH3 1 L H&C. OCH3 H H H H Select to Add Arrows Q Search Problem 30 of 20 H. H3C + :0: H CH3 CH3 20 H2C Undo Reset Done DELLarrow_forwardDraw the principal organic product of the following reaction.arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided structures, draw the curved arrows that epict the mechanistic steps for the proton transfer between a hydronium ion and a pi bond. Draw any missing organic structures in the empty boxes. Be sure to account for all lone-pairs and charges as well as bond-breaking and bond-making steps. 2 56°F Mostly cloudy F1 Drawing Arrows > Q Search F2 F3 F4 ▷11 H. H : CI: H + Undo Reset Done DELLarrow_forward
- Calculate the chemical shifts in 13C and 1H NMR for 4-chloropropiophenone ? Write structure and label hydrogens and carbons. Draw out the benzene ring structure when doing itarrow_forward1) Calculate the longest and shortest wavelengths in the Lyman and Paschen series. 2) Calculate the ionization energy of He* and L2+ ions in their ground states. 3) Calculate the kinetic energy of the electron emitted upon irradiation of a H-atom in ground state by a 50-nm radiation.arrow_forwardCalculate the ionization energy of He+ and Li²+ ions in their ground states. Thannnxxxxx sirrr Ahehehehehejh27278283-4;*; shebehebbw $+$;$-;$-28283773838 hahhehdvaarrow_forward
- Plleeaasseee solllveeee question 3 andd thankss sirr, don't solve it by AI plleeaasseee don't use AIarrow_forwardCalculate the chemical shifts in 13C and 1H NMR for 4-chloropropiophenone ? Write structure and label hydrogens and carbonsarrow_forwardPlease sirrr soollveee these parts pleaseeee and thank youuuuuarrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning




