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Naddik [55]
3 years ago
11

Calculate the frequency and wavelength for a photon of light with an energy of 3.2x10-19 J.

Chemistry
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

wavelength = 6×10⁻⁷m

f = 0.5 ×10¹⁵ Hz

Explanation:

Given data:

Energy of photon = 3.2×10⁻¹⁹ J

Wavelength of photon = ?

Frequency of photon = ?

Solution:

E = h.f

f = frequency

h = planck's constant

E = energy

f = E/h

f = 3.2×10⁻¹⁹ Kg.m².s⁻²/ 6.63×10⁻³⁴ m².Kg/s

f = 0.5 ×10¹⁵ s⁻¹

s⁻¹ = Hz

f = 0.5 ×10¹⁵ Hz

Wavelength:

speed of light = wavelength × frequency

wavelength = speed of light / frequency

wavelength =  3×10⁸ m.s⁻¹ /0.5 ×10¹⁵s⁻¹  

wavelength = 6×10⁻⁷m

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Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

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Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17
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<span>Answer is: mass of burned butane is 11.6 g.</span>

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