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pashok25 [27]
3 years ago
9

Calculate thr number of photons having a wavelength of 10.0 μm required to produce 1.0 kJ of energy.

Chemistry
2 answers:
slega [8]3 years ago
7 0

Answer:

The number of photons are 5.028\times 10^{27}.

Explanation:

E=\frac{h\times c}{\lambda}

where,

E = energy of photon =  

h = Planck's constant = 6.63\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 10.0 μm  =10^{-5} m

1 μm = 10^{-6} m

E=\frac{6.63\times 10^{-34}Js\times 3\times 10^8m/s}{10^{-5} m}

E=1.989\times 10^{-20} Joules

Let the n number of photons with energy equal to E' = 1.0 kJ = 1000 J

n\times E=E'

n\times 1.989\times 10^{-20} J=1000 J

n=\frac{1000 J}{1.989\times 10^{-20} J}=5.028\times 10^{27}

The number of photons are 5.028\times 10^{27}.

ddd [48]3 years ago
4 0
Ok so first you need to figure out the energy of ONE photon with that wavelength. Using E=hc/lambda, you get E= 1.99 * 10^-20 J/photon. Now, how many photons do you need to add up to get to one kilojoule=1000 joules? 1000J / (1.99 * 10^-20 J/photon) = approximately 5 * 10^22 photons hope this helps
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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
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ivolga24 [154]

Answer & Explanation:

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Which comppund contains both ionic and covalent bonds
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4 0
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Calculate the grams of nitrogen in 125 g of NH4NO3
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Answer:

43.75 g of Nitrogen

Explanation:

We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:

Mole of NH₄NO₃ = 1 mole

Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)

= 14 + 4 + 14 + 48

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Mass of NH₄NO₃ =?

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Mass of NH₄NO₃ = 1 × 80 = 80 g

Next, we shall determine the mass of N in 1 mole of NH₄NO₃.

Mass of N in NH₄NO₃ = 2N

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Thus,

80 g of NH₄NO₃ contains 28 g of N.

Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:

80 g of NH₄NO₃ contains 28 g of N.

Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.

Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen

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