When an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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The energy of the electron drops when it transitions levels, as well as the atom releases photons. The emission of the photon occurs as the electron transitions from an energy state to a lower state. The photon energy represents precisely the energy that would be lost when an electron moves to a level with less energy.
When such an excited electron transitions from one energy level to another, this could emit a photon. The energy drop would be equivalent to the power of the photon that is released. In electron volts, the energy of an electron, as well as its associated photon (emitted or absorbed) has been stated.
Therefore, when an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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Answer:
[SO2Cl2] = = 0.015 M
[SO2] = = 0.0027 M
[Cl2] = = 0.0027 M
Q = = = 4.8 × 10−4
No. Q < Kc, so reaction will shift to the right.
Explanation:
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n= 
= 
= density
P = 
density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 = 
M = 34.61
to calculate the Kc
Kc=![\frac{ [NO] [O2]}{NO2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BNO%5D%20%5BO2%5D%7D%7BNO2%7D)
x M NO2 + 
M NO+ 
M O2
Putting the values as molecular weight of NO2, NO,O2
34.61= 
x= 0.33
Kc= 
putting the values in the above equation
Kc = 0.00662
24. F:H :) because Fluorine is the most electronegative
The scientist observes at what rate is the concentration increasing or decreasing.
