Explanation: Given The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability density curve for the function Ψ 2 ( x ) = sin 2 x contains all the positive values of the given function over the whole range. Therefore, the probability density curve for the given function is, Figure 1 (b) Explanation: Given The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability of finding electron for the given function is maximum on the values of x where the probability density curve has the maximum value. For the given function the value of sin x is maximum at the values x = π 2 and x = 3 π 2 . Therefore, the probability density curve for the given function has a peak at these values of x where probability of finding an electron is maximum. (c) Explanation: The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability of finding electron for the given function is minimum on the values of x where the probability density curve has the minimum value. For the given function the value of sin x is zero at the value of x = π . Therefore, the probability density curve for the given function has a node at this value of x where probability of finding electron is nil. Conclusion: (a) The probability density curve for the given function is as follows: (b) The values of x is maximum at x = π 2 and x = 3 π 2 . (c) The probability of finding an electron at x = π is zero and this point is called node.
Explanation: Given The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability density curve for the function Ψ 2 ( x ) = sin 2 x contains all the positive values of the given function over the whole range. Therefore, the probability density curve for the given function is, Figure 1 (b) Explanation: Given The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability of finding electron for the given function is maximum on the values of x where the probability density curve has the maximum value. For the given function the value of sin x is maximum at the values x = π 2 and x = 3 π 2 . Therefore, the probability density curve for the given function has a peak at these values of x where probability of finding an electron is maximum. (c) Explanation: The wave function for the electron in one dimensional system is, Ψ ( x ) = sin x The probability of finding electron for the given function is minimum on the values of x where the probability density curve has the minimum value. For the given function the value of sin x is zero at the value of x = π . Therefore, the probability density curve for the given function has a node at this value of x where probability of finding electron is nil. Conclusion: (a) The probability density curve for the given function is as follows: (b) The values of x is maximum at x = π 2 and x = 3 π 2 . (c) The probability of finding an electron at x = π is zero and this point is called node.
Given The wave function for the electron in one dimensional system is,
Ψ(x)=sinx
The probability density curve for the function Ψ2(x)=sin2x contains all the positive values of the given function over the whole range. Therefore, the probability density curve for the given function is,
Figure 1
(b)
Explanation:
Given The wave function for the electron in one dimensional system is,
Ψ(x)=sinx
The probability of finding electron for the given function is maximum on the values of x where the probability density curve has the maximum value. For the given function the value of sinx is maximum at the values x=π2 and x=3π2 . Therefore, the probability density curve for the given function has a peak at these values of x where probability of finding an electron is maximum.
(c)
Explanation: The wave function for the electron in one dimensional system is,
Ψ(x)=sinx
The probability of finding electron for the given function is minimum on the values of x where the probability density curve has the minimum value. For the given function the value of sinx is zero at the value of x=π . Therefore, the probability density curve for the given function has a node at this value of x where probability of finding electron is nil.
Conclusion:
(a) The probability density curve for the given function is as follows:
(b) The values of x is maximum at x=π2 and x=3π2 . (c) The probability of finding an electron at x=π is zero and this point is called node.
What are the IUPAC Names of all the compounds in the picture?
1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following
compounds. Please show your work. (8) SF2, CH,OH, C₂H₂
b) Based on your answers given above, list the compounds in order of their Boiling Point
from low to high. (8)
19.78 Write the products of the following sequences of reactions. Refer to your reaction road-
maps to see how the combined reactions allow you to "navigate" between the different
functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18
roadmaps along with your new Chapter 19 roadmap for these.
(a)
1. BHS
2. H₂O₂
3. H₂CrO4
4. SOCI₂
(b)
1. Cl₂/hv
2. KOLBU
3. H₂O, catalytic H₂SO4
4. H₂CrO4
Reaction
Roadmap
An alkene 5. EtOH
6.0.5 Equiv. NaOEt/EtOH
7. Mild H₂O
An alkane
1.0
2. (CH3)₂S
3. H₂CrO
(d)
(c)
4. Excess EtOH, catalytic H₂SO
OH
4. Mild H₂O*
5.0.5 Equiv. NaOEt/EtOH
An alkene 6. Mild H₂O*
A carboxylic
acid
7. Mild H₂O*
1. SOC₁₂
2. EtOH
3.0.5 Equiv. NaOEt/E:OH
5.1.0 Equiv. NaOEt
6.
NH₂
(e)
1. 0.5 Equiv. NaOEt/EtOH
2. Mild H₂O*
Br
(f)
i
H
An aldehyde
1. Catalytic NaOE/EtOH
2. H₂O*, heat
3. (CH,CH₂)₂Culi
4. Mild H₂O*
5.1.0 Equiv. LDA
Br
An ester
4. NaOH, H₂O
5. Mild H₂O*
6. Heat
7.
MgBr
8. Mild H₂O*
7. Mild H₂O+
Chapter 14 Solutions
Chemistry: The Central Science, Books a la Carte Plus MasteringChemistry with eText -- Access Card Package (13th Edition)
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