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jekas [21]
3 years ago
15

Certain atoms emit photons of light with an energy of 3.820 ✕ 10−19 J. Calculate the frequency (in Hz) and wavelength (in nm) of

one of these photons.
frequency Hz

and

wavelength nm

What is the total energy (in kJ) in 1 mole of these photons?

kJ
Chemistry
1 answer:
charle [14.2K]3 years ago
4 0

Answer:

ν = 5,765x10¹⁴ Hz

λ = 520 nm

230,0 kJ/mole

Explanation:

To convert energy yo frequency you need to use:

E = hν

Where E is energy (3,820x10⁻¹⁹ J)

h is Planck's constant (6,626x10⁻³⁴ Js)

And ν is frequency, replacing ν = 5,765x10¹⁴ s⁻¹ ≡ 5,765x10¹⁴ Hz

To convert frequency to wavelength:

c = λν

Where s is speed of light (2,998x10⁸ ms⁻¹)

ν is frequency (5,765x10¹⁴ s⁻¹)

And λ is wavelength, replacing: λ = 5,200x10⁻⁷ ≡ 520 nm

If 1 photon produce 3,820x10⁻¹⁹ J, in mole of photons produce:

3,820x10⁻¹⁹ J ×\frac{6,022x10^{23}}{1 mole} = 230040 J/mole ≡ 230,0 kJ/mole

I hope it helps!

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Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

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Hey there!

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------------------------------------

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Hope this helps!

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