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Sati [7]
3 years ago
15

When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does

the change in energy level compare to the energy of the emitted photon?
Chemistry
1 answer:
Andrej [43]3 years ago
8 0
The change in energy level is equivalent to the energy of the released photon. This is related by Planck's equation:
E = hcf
The higher the energy change, the greater the frequency, f, of the photon.
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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
Elanso [62]

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

Now put all the given values in this expression, we get:

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

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4 years ago
So guys what the hell is soluable because im looking at my online test and it says soluable everywhere like its a foreign langua
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Explanation:

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I hope this helps :).

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Answer:

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