Answer:
The longest wavelength of radiation that passesses the necessary energy for breaking the Cl- Cl bond (in Cl2) is approximately 494.2 nm, which corresponds to the visible spectrum.
Explanation:
In order to answer this question we need to recall that the energy of a photon is given by:
E = hc/lambda, where
E = energy
h = Planck's constant
c = speed of light in vacuum
lambda = associated photon wavelength
In order to perform the calculations, first we need to change the units of 242kJ/mol to J. For doing this, we to divide by Avogadro's number and multiply by a 1000:
242kJ/mol = (242kJ/mol)*(1mol/6.022x10^23 particles)*(1000J/1kJ)= 4.0186x10^-19 J
Now, we simply solve for lambda and substitute the appropriate values in the energy equation:
lambda = hc/E = (6.626x10^-34 J s)*(3x10^8 m/s)/(4.0186x10^-19 J) = (1.986x10^-25 J m)/(4.0186x10^-19 J) = 4.942x10^-7 m = 494.2x10^-9 m = 494.2 nm
Therefore, the wavelength for a photon to break the Cl-Cl bond in a Cl2 molecule should be 494.2 nm at most, which corresponds to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).
Answer:
I₃/Io % = 0.8.59
Explanation:
A polarizer is a complaint sheet for light in the polarization direction and blocks the perpendicular one. When we use two polarizers the transmission between them is described by Malus's law
I = I₀ cos² θ
Let's apply the previous exposures in our case, the light is indicatively not polarized, so the first polarized lets half of the light pass
I₁ = ½ I₀
The light transmitted by the second polarizer
I₂ = I₁ cos² θ
I₂ = (½ I₀) cos2 28
The transmission by the polarizing third is
I₃ = I₂ cos² θ₃
The angle of the third polarizer with respect to the second is
θ₃ = 90-28
θ₃ = 62º
I₃ = (½ I₀ cos² 28 cos² 62)
Let's calculate
I₃ = Io ½ 0.7796 0.2204
I₃ = Io 0.0859
I₃/Io= 0.0859 100
I₃/Io % = 0.8.59
Answer:
26.822 m/s
Explanation:
60 mi/hr * 5280 ft/mile * 1 hr / 3600 sec * 12 in / foot * 1 meter / 39.37 in = <u>26.822 m/s</u>
