Answer:
1.64x10⁻¹⁸ J
Explanation:
By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.
When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:
E = hc/λ
Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.
The wavelength can be calculated by:
1/λ = R*(1/nf² - 1/ni²)
Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:
1/λ = 1.097x10⁷ *(1/1² - 1/2²)
1/λ = 8.227x10⁶
λ = 1.215x10⁻⁷ m
So, the energy is:
E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)
E = 1.64x10⁻¹⁸ J
X=107 because the sides of a hexagon must add up to 720
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Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

Thus, we solve for the molar mass of the metal to obtain:

For the subsequent problems, we proceed as follows:
a.

b.

c.

Regards!
%yield = 54.6%
<h3>Further explanation</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
(theoretical)
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
<h3 />
Reaction
2ZnS+3O₂ ⇒ 2ZnO+2SO₂
MW ZnS = 97.474 g/mol

MW ZnO = 81.38 g/mol
- mol ZnO (from mol ZnS as limiting reactant, O₂ excess)


Theoretical production = 125.388
