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Zarrin [17]
3 years ago
15

What is the energy in joules of one photon of microwaveradiation with a wavelength 0.122m?

Chemistry
1 answer:
pentagon [3]3 years ago
3 0

<u>Answer:</u> The energy of photon is 162.93\times 10^{-26}J

<u>Explanation:</u>

The relation between energy and wavelength of light is given by Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

E = energy of the light  = ?

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of photon = 0.122 m

Putting values in above equation, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{0.122m}\\\\E=162.93\times 10^{-26}J

Hence, the energy of photon is 162.93\times 10^{-26}J

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The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:
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Answer:

12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow  14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)

Explanation:

The steps of the Ostwald process:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)

3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)

Combinning the equations:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

+

(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2

+

(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3

=

4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) +  4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)

Simplifying:

4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow  14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)

12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow  14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)

The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

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