As a distant star moves away from earth, the light given off by the star has a
1 answer:
Answer:
The wavelength of these photons will become longer. The energy of each of these photons will become lower.
Explanation:
<h3>Wavelength</h3>Light can be considered as electromagnetic waves. The wavelength of a wave is equal to the minimum distance between two troughs (lowest points) in this wave. On the other hand, the frequency of a wave is equal to the number of wavelengths that this wave travels in unit time.
Assume that the speed of light stays the same. The distance that this beam of light travels in unit time will be the same. However, with a lower frequency, there would be fewer wavelengths in that same distance. Therefore, the size of each wavelength will become longer.
If represent the speed of light and
represents the frequency, then the wavelength would be:
.
The energy of each proton of a beam of light is proportional to the frequency
of the light. Let
denote Planck's Constant. The numerical relation between
and
would be:
.
Therefore, if the frequency of this light becomes smaller, the energy
of each of its proton will also become proportionally lower.
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Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M