Which reaction occurs at the anode during the electrolysis of aqueous NaCl solution using inert electrodes?
1 answer:
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Answer:
PART A: 412.98 nm
PART B: 524.92 nm
Explanation:
The equation below can be used for a diffraction grating of nth order image:
n*λ = d*sinθ
Therefore, for first order images, n = 1 and:
λ = d*sinθ.
The angle θ can be calculated as follow:
tan θ = 9.95 cm/15.0 cm = 0.663 and
θ =
(0.663) = 33.56°
Thus: d =λ/sin θ = 461/sin 33.56° = 833.97 nm
PART A:
For a position of 8.55 cm:
tan θ = 8.55 cm/15.0 cm = 0.57 and
θ =
(0.57) = 29.68°
Therefore:
λ =d*sin θ = 833.97*sin 29.68° = 412.98 nm
PART B:
For a position of 12.15 cm:
tan θ = 12.15 cm/15.0 cm = 0.81 and
θ =
(0.81) = 39.01°
Therefore:
λ =d*sin θ = 833.97*sin 39.01° = 524.92 nm