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Step2247 [10]
3 years ago
14

A certain shade of blue has a frequency of 7.07 × 1014 Hz. What is the energy of exactly one photon of this light?

Chemistry
2 answers:
taurus [48]3 years ago
8 0

Answer:

E=4.68\times 10^{-19}\ J

Explanation:

It is given that,

Frequency of shade of blue, f=7.07\times 10^{14}\ Hz

Let E is the energy of exactly one photon of this light. The relationship between the energy and the frequency is given by :

E=h\times f

h is the Planck's constant

E=6.63\times 10^{-34}\times 7.07\times 10^{14}

E=4.68\times 10^{-19}\ J

So, the energy of exactly one photon of this light is 4.68\times 10^{-19}\ J. Hence, this is the required solution.

IRISSAK [1]3 years ago
4 0
E = HF, where H is Planck's constant, 6.63 x 10 -  34 j.s
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The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.

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1 year ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

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number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

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The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

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GIven that :

Ka = 1.9 x 10^{-5}

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3 years ago
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