Light with a wavelength of 614.5 nm looks orange. what is the energy, in joules, per photon of this orange light? what is the energy in ev (1 ev = 1.602 × 10−19 j)?

The energy is 2.00eV

Explanation:

From Planck's theory

E=hf = hc/k

Where h is the planck constant =

6.6×10^-34m2kg/s

C= speed of light =3.0 ×10^8m/s

k= wavelength = 651.4nm=651.4 ×10^-9

E=(6.6×10^-34 × 3.0 ×10^8)/(651.4×10^-9)

E=3.22 × 10^-19J

1eV = 1.602×10^-19J

E=2.00eV

The energy is 2.00eV

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.