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OleMash [197]
2 years ago
8

15)A photon has a frequency of 2.68 x 106 Hz. Calculate its energy.

Chemistry
2 answers:
Verizon [17]2 years ago
4 0

Answer:

Explanation:

A photon has a frequency (ν) of 2.68 x 10. 6. Hz. ... Calculate the energy (E) and wavelength (λ) of a photon of light with a frequency (ν) of 6.165 x 10. 14 ... Calculate the wavelength and energy of light that has a frequency of 1.5 x 10. 15. Hz.

Lelechka [254]2 years ago
3 0
Hello!

To solve that, you can use Planck’s equation, ok?

E = hf
E = [6.63*10^(-34)]*[2.68*10^(6)]
E = 17,77*10^(-28) = 1,77*10^(-27) J

So its energy is 1,77*10^(-27) J.
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Gases emitted during volcanic activity often contain high concentrations of hydrogen sulfide and sulfur dioxide. these gases may
kompoz [17]
There can be three possible answers to this question: the amount of moles of SO₂ gas needed to react with 6.41 mol H₂S, and the amount of S and H₂O gas produced.

Amount of SO₂:
6.41 mol H₂S (1 mol SO₂/2 mol 2 mol H₂S) = <em>3.205 moles SO₂ gas</em>

Amount of S:
6.41 mol H₂S (3 mol S/2 mol 2 mol H₂S) =<em> 9.615 moles S solid</em>

Amount of H₂O:
6.41 mol H₂S (2 mol H₂O/2 mol 2 mol H₂S) = <em>6.41 moles H₂O gas</em>
7 0
2 years ago
Titration reveals that 11.6 mL of 3.0M sulfuric acid are required to neutralize the sodium hydroxide in 25.00mL of NaOH solution
Deffense [45]

Answer:

Explanation:

Molarity of acid(volume of acid)(# of H ions)= molarity of base(volume of base)(# of OH ions)

M(v)(#)=M(v)(#)

sulfuric acid    sodium hydroxide

H2SO4           NaOH

(3)(11.6)(2)=M(25)(1)

M=2.784

6 0
2 years ago
Which of the following was most clearly, even uniquely perhaps, revealed by the discovery of the photoelectric effect?
nika2105 [10]

Answer : Option 3) Wave/Particle duality.

Explanation : The experiment on discovery of photoelectric effect revealed about the photoelectrons of light that can behave as particle or waves.

The photoelectric effect is observed when the emission of electrons or other free carriers occurs on shining a light radiation on a material. The electrons emitted from this can be called photo electrons. These photoelectrons may behave as wave or particle in duality which holds that light and matter exhibit properties of both waves and of particles.

7 0
2 years ago
Read 2 more answers
A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0
2 years ago
Read 2 more answers
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
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