The wavelength of the characteristic, bright yellow-orange flame test color of sodium is 590 nm. calculate the average energy (δe) associated with this atomic emission line.

The energy associated with the 590nm atomic emission line is 3.369 x 10^49 Joules. This can be determined by multiplying Plank’s constant (6.626 x 10^34 Js) by the speed of light (3x 10^8 m/s), and then dividing the result by the wavelength (590 x 10^-9 m).

During photosynthesis, plants transform electromagnet energy from the sun into chemical energy.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.