Answered

In fireworks, the heat of the reaction of an oxidizing agent, such as KClO4, with an organic compound excites certain salts, which emit specific colors. Strontium salts have an intense emission at 641 nm. What is the energy (in kJ) of this emission for 5.33 g of the chloride salt of strontium? Assume that all the heat produced is converted to emitted light. Enter to 2 decimal places.

Answer :

SebasAime

Answer:

6.27 kJ

Explanation:

You need to do a conversion between nm to kJ/mol:

[tex]E=h*\frac{1}{wavelength}[/tex]

If you express the wavelength in meter resulting energy will be in joules so you continue with the conversion:

[tex]E=h*\frac{1}{wavelength}*\frac{1 kJ}{1000J} *\frac{6.023*10^{23}}{mol}[/tex]

With 641 nm:

[tex]E=h*\frac{1}{641*10^{-9}m}*\frac{1 kJ}{1000J} *\frac{6.023*10^{23}}{mol}=186.62 kJ/mol[/tex]

Now to calcule the energy released by 5.33 g of SrCl2 (M=158.53 g/mol) you take the previous calculated energy:

[tex]E=5.33g*\frac{1molSrCl2}{158.53g}*\frac{186.62kJ}{mol}  = 6.27 kJ[/tex]

mickymike92

Based on the wavelength of light emitted, the energy emitted as heat is 6.28 kJ/mol.

What is the energy of emitted light?

The energy of emitted light or a photon is calculated using the formula below:

E = hc/λ

Where;

  • h is Planck's constant = 6.63 * 10⁻³⁴ m²kg/s
  • c is speed of light = 3 * 10⁸ m/s
  • λ is wavelength of light

Calculating the energy of a photon of emitted light of wavelength 641 nm

λ = 641 nm = 6.41 * 10⁻⁷ m

E = 6.63 * 10⁻³⁴ * 3 * 10⁸ / 6.41 * 10⁻⁷ m

E = 3.103 * 10⁻¹⁹ J

Molar mass of Strontium chloride = 158.53 g/mol

Moles of strontium chloride in 5.33 g = 5.33 g/158.53 g/mol = 0.0336 moles

1 mole of strontium chloride produces 6.02 * 10²³ photons

Number of photons in 0.03362 moles =  6.02 * 10²³ * 0.3362

Number of photons = 2.0239 * 10²² photons

Energy emitted as heat = 3.103 * 10⁻¹⁹ J * 2.0239 * 10²² = 6280 J

Energy emitted as heat  = 6.28 kJ/mol

Learn more about energy of photons at: https://brainly.com/question/4177755

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