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natta225 [31]
4 years ago
8

How much energy will a photon with a frequency of 6.8 x 104 Hz emit?

Chemistry
1 answer:
Advocard [28]4 years ago
8 0

Answer:

Energy emitted by photon = 45.06 × 10−32 J

Explanation:

Energy of photon is given by  E = hf

where

h is the Planck's constant whose value is 6.626×10−34J⋅s

f is the frequency of photon

________________________________________

given that

f = 6.8 x 104 Hz

1 HZ = 1 s^-1

E =  6.626×10−34J⋅s * 6.8 x 10^4 Hz

E = 45.06 × 10−32 J

Thus,

Energy emitted by photon is 45.06 × 10−32 J

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The distance from one point to another is the distance traveled by a light beam between them.

If you have the coordinates of the 2 points, then the distance between them is

√ of [ (the difference in their x-values)² + (the difference in their y-values)² ]
6 0
3 years ago
Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Brilliant_brown [7]

Answer:

0.364

Explanation:

Let's do an equilibrium chart for the reaction of combustion of ammonia:

2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)

4.8atm 1.9atm 0 0 Initial

-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)

4.8-2x 1.9-(3/2)x x 3x Equilibrium

At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.

If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium

pNH₃ = 4.8 - 2*0.63 = 3.54 atm

pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm

pH₂O = 3*0.63 = 1.89 atm

The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:

Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*(pO_2)^{3/2}]

Kp = [0.63*(1.89)³]/[(3.54)²*(0.955)^{3/2}]

Kp = 4.2533/11.6953

Kp = 0.364

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3 years ago
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