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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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1 year ago
1. In the investigation of an unknown alcohol, there was a positive Jones test and a negative Lucas test. What deductions may be
pogonyaev

Answer:

1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.

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3 years ago
When 1.95 g of co(no3)2 is dissolved in 0.350 l of 0.220 m koh, what are [ co2+], [ co(oh)42−], and [ oh−] if kf of co(oh)42− =
Airida [17]
Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
             Co²⁺(aq) + 4 OH⁻    ⇄      Co(OH)₄²⁻
I (M)      0.03045      0.22                   0
C (M)   - 0.03045   - 4 (0.03045)      0.03045
E (M)       - x         0.22 - 4(0.03045)   0.03045
                              = 0.0982
Kf  = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M



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