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AleksandrR [38]
3 years ago
14

What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?​

Chemistry
2 answers:
Dmitrij [34]3 years ago
8 0

Answer:

B. 4.25 x 10-19J is correct via a p e x

Explanation:

svetlana [45]3 years ago
3 0

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

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Answer:

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Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}

N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2

Consequently, the equilibrium constant is computed as:

Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6

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Explanation:

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Answer:

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