The more polar bond has to be indicated using partial charge of δ + o r δ − . C − O a n d C − N Concept Introduction: The partial negative ( δ – ) end of a bond is the atom with the larger electronegativity. The partial positive ( δ + ) end of a bond is the atom with the smaller electronegativity.
The more polar bond has to be indicated using partial charge of δ + o r δ − . C − O a n d C − N Concept Introduction: The partial negative ( δ – ) end of a bond is the atom with the larger electronegativity. The partial positive ( δ + ) end of a bond is the atom with the smaller electronegativity.
Solution Summary: The author explains that the more polar bond can be indicated using partial charge of delta –) and the positive end of a bond is the smaller electronegativity.
The more polar bond has to be indicated using partial charge of δ+orδ−.
C−OandC−N
Concept Introduction:
The partial negative (δ–) end of a bond is the atom with the larger electronegativity. The partial positive (δ+) end of a bond is the atom with the smaller electronegativity.
(b)
Interpretation Introduction
Interpretation:
The more polar bond has to be indicated using partial charge of δ+orδ−.
B−OandP−S
Concept Introduction:
Refer part (a).
(c)
Interpretation Introduction
Interpretation:
The more polar bond has to be indicated using partial charge of δ+orδ−.
P−HandP−N
Concept Introduction:
Refer part (a).
(d)
Interpretation Introduction
Interpretation:
The more polar bond has to be indicated using partial charge of δ+orδ−.
Laser. Indicate the relationship between metastable state and stimulated emission.
The table includes macrostates characterized by 4 energy levels (&) that are
equally spaced but with different degrees of occupation.
a) Calculate the energy of all the macrostates (in joules). See if they all have
the same energy and number of particles.
b) Calculate the macrostate that is most likely to exist. For this macrostate,
show that the population of the levels is consistent with the Boltzmann
distribution.
macrostate 1 macrostate 2 macrostate 3
ε/k (K) Populations
Populations
Populations
300
5
3
4
200
7
9
8
100
15
17
16
0
33
31
32
DATO: k = 1,38×10-23 J K-1
Don't used Ai solution
Chapter 6 Solutions
OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
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