I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to
heat the reaction in order to double the equilibrium constant?
1 answer:
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Answer:
spacing between the slits is 405.32043 × m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4 × cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 × 0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
Answer:
1.6 x 10⁻⁶ J/m³
Explanation:
I = Intensity of electromagnetic radiation = 475 Wm⁻²
U = average total energy density of electromagnetic radiation
c = speed of electromagnetic radiation = 3 x 10⁸ m/s
Intensity of electromagnetic radiation is given as
Inserting the values
U = 1.6 x 10⁻⁶ J/m³