Answer:
2.28 m
Explanation:
Use the relationship
λ=vf
to solve for wavelength λ. Substituting the known quantities yields:
λ=vf
λ=2.99 x 108 m/s1.31 x 108 Hz
λ=2.28 m
Answer:
Activation energy of phenylalanine-proline peptide is 66 kJ/mol.
Explanation:
According to Arrhenius equation- 
, where k is rate constant, A is pre-exponential factor, 
is activation energy, R is gas constant and T is temperature in kelvin scale.
As A is identical for both peptide therefore-
![\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bala-pro%7D%7D%7Bk_%7Bphe-pro%7D%7D%3De%5E%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-E_%7Ba%7D%5E%7Bala-pro%7D%5D%7D%7BRT%7D)
Here 
, T = 298 K , R = 8.314 J/(mol.K) and 
So, ![\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.05%7D%7B0.005%7D%3De%5E%7B%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-%2860000J%2Fmol%29%5D%7D%7B8.314J.mol%5E%7B-1%7D.K%5E%7B-1%7D%5Ctimes%20298K%7D%7D)
(rounded off to two significant digit)
So, activation energy of phenylalanine-proline peptide is 66 kJ/mol
Answer:
all of the above
Explanation:
a hypothesis is indeed the second step. it also can be tested since you made it beforehand. it also is supposed to be educated. so all of the above.
Answer:
The higher the excitation state, the more energy the electron contains. When an electron absorbs energy, it jumps to a higher orbital. ... An electron in an excited state can release energy and 'fall' to a lower state.
