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alekssr [168]
2 years ago
6

A photon has an energy of 2.93 × 10 to the power of -25 J. What is its frequency? What type of electromagnetic radiation is this

photon?
Chemistry
2 answers:
Svet_ta [14]2 years ago
6 0
You need to know the energy frequency relationship for photons, which is thanks to Max Planck:

Photon Energy = Planck constant x Frequency

Rarranged:
Photon Energy / Planck Constant = Frequency

Planck Constant = 6.63x10^-34

2.93x10^-25 / 6.63x10^-34 = Frequency

valkas [14]2 years ago
3 0

<u>Answer:</u> The radiation has a frequency of 4.43\times 10^{9}Hz and is a type of radio wave.

<u>Explanation:</u>

The equation given by Planck's follows:

E=h\nu

where,

E = energy of the light  = 2.93\times 10^{-25}J

h = Planck's constant  = 6.62\times 10^{-34}Js

\nu = frequency of light = ?

Putting values in above equation, we get:

2.93\times 10^{-25}J=6.62\times 10^{-34}Js\times \nu\\\\\nu=\frac{2.93\times 10^{-25}J}{6.62\times 10^{-34Js}}=4.43\times 10^{9}Hz

The relation between frequency and wavelength is given as:

\nu=\frac{c}{\lambda}

where,

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

\nu = frequency of radiation = 4.43\times 10^{8}s^{-1}

\lambda = wavelength of the radiation = ?

Putting values in above equation, we get:

4.43\times 10^{8}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=\frac{3\times 10^8m/s}{4.43\times 10^8}s^{-1}}=0.677m

The radiation having wavelength 0.677 m belongs to radio waves.

Hence, the radiation has a frequency of 4.43\times 10^{9}Hz and is a type of radio wave.

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225,563,910 m/s

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At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl &lt;----&gt; Br2 + Cl2 is 0.141. If the initial c
ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

                   2BrCl\rightleftharpoons Br_2+Cl_2

<u>Initial:</u>          0.02               0.03

<u>At eqllm:</u>    0.02-2x     x     0.03+x

The expression of K_c for above equation follows:

K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}

We are given:

K_c=0.141

Putting values in above equation, we get:

0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0
3 years ago
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