Two different equations showing the change in Gibbs free energy are given. The derivation of E ° as a function of temperature for the given equations, the graphical determination of Δ H ° and Δ S ° from measurements of E ° at different temperature and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature is to be stated. Concept introduction: Gibbs free energy is basically the maximum amount of non-expansion work done. Therefore, it is represented as, W max = Δ G ° The relationship between Gibbs free energy change and cell potential is given by the formula, Δ G ° = − n F E ° cell The relation between Δ G ° , Δ H ° and Δ S ° is given as, Δ G ° = Δ H ° − T Δ S ° To determine: The derivation of E ° as a function of temperature for the given equations, the graphical determination of Δ H ° and Δ S ° from measurements of E ° at different temperatures and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature. The relation obtained from the given equations is, E ° cell = T ( Δ S ° n F ) + ( − Δ H ° n F )
Two different equations showing the change in Gibbs free energy are given. The derivation of E ° as a function of temperature for the given equations, the graphical determination of Δ H ° and Δ S ° from measurements of E ° at different temperature and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature is to be stated. Concept introduction: Gibbs free energy is basically the maximum amount of non-expansion work done. Therefore, it is represented as, W max = Δ G ° The relationship between Gibbs free energy change and cell potential is given by the formula, Δ G ° = − n F E ° cell The relation between Δ G ° , Δ H ° and Δ S ° is given as, Δ G ° = Δ H ° − T Δ S ° To determine: The derivation of E ° as a function of temperature for the given equations, the graphical determination of Δ H ° and Δ S ° from measurements of E ° at different temperatures and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature. The relation obtained from the given equations is, E ° cell = T ( Δ S ° n F ) + ( − Δ H ° n F )
Two different equations showing the change in Gibbs free energy are given. The derivation of
E° as a function of temperature for the given equations, the graphical determination of
ΔH° and
ΔS° from measurements of
E° at different temperature and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature is to be stated.
Concept introduction:
Gibbs free energy is basically the maximum amount of non-expansion work done. Therefore, it is represented as,
Wmax=ΔG°
The relationship between Gibbs free energy change and cell potential is given by the formula,
ΔG°=−nFE°cell
The relation between
ΔG°,
ΔH° and
ΔS° is given as,
ΔG°=ΔH°−TΔS°
To determine: The derivation of
E° as a function of temperature for the given equations, the graphical determination of
ΔH° and
ΔS° from measurements of
E° at different temperatures and the property used for designing a reference half-cell that would produce a potential relatively stable with respect to temperature.
The relation obtained from the given equations is,
Dr. Mendel asked his BIOL 260 class what their height was and what their
parent's heights were. He plotted that data in the graph below to determine if
height was a heritable trait.
A. Is height a heritable trait? If yes, what is the heritability value? (2 pts)
B. If the phenotypic variation is 30, what is the variation due to additive alleles?
(2 pts)
Offspring Height (Inches)
75
67.5
60
52.5
y = 0.9264x + 4.8519
55
60
65
MidParent Height (Inches)
70
75
12pt v
V
Paragraph B IUA
>
AT2 v
V
Experiment:
Each team will be provided with 5g of a mixture of acetanilide and salicylic acid. You will divide it into three 1.5 g portions in separate 125 mL Erlenmeyer flasks savıng some for melting point analysis. Dissolve the mixture in each flask in ~60mL of DI water by heating to boiling on a hotplate. Take the flasks off the hotplate once you have a clear solution and let them stand on the bench top for 5 mins and then allow them to cool as described below.
Sample A-Let the first sample cool slowly to room temperature by letting it stand on your lab bench, with occasional stirring to promote crystallization.
Sample B-Cool the second sample 1n a tap-water bath to 10-15 °C
Sample C-Cool the third sample in an ice-bath to 0-2 °C
Results:
weight after recrystalization and melting point temp.
A=0.624g,102-115°
B=0.765g, 80-105°
C=1.135g, 77-108
What is the percent yield of A,B, and C.
Rel. Intensity
Q
1. Which one of the following is true of the compound
whose mass spectrum is shown
here? Explain how you decided.
100
a) It contains chlorine.
b) It contains bromine.
c) It contains neither chlorine nor bromine.
80-
60-
40-
20-
0.0
0.0
TT
40
80
120
160
m/z
2. Using the Table of IR Absorptions how could you
distinguish between these two
compounds in the IR?
What absorbance would one compound have that the
other compound does not?
HO
CI
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