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melisa1 [442]
3 years ago
7

The minimum energy required to break the oxygen-oxygen bond in oxygen is 495 kJ/mol. What is the longest wavelength of radiation

that possesses the necessary energy to break this bond? What type of electromagnetic radiation is this?
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

241 nm, it is UV light range

Explanation:

The minimum energy needed required to break the oxygen-oxygen bond = 495 kJ/mol

Energy needed for 1 molecule of oxygen = 495 × 10³ J / avogadro's constant = 495 × 10³ J  / ( 6.02 × 10²³) = 8.223 × 10⁻¹⁹ J

Energy = hv

where h = Planck constant = 6.626 × 10 ⁻³⁴ m²kg/s and v = frequency

c speed of light = vλ

c / λ = v

E = hc / λ

λ  = hc / E = (6.626 × 10 ⁻³⁴ m²kg/s × 3.0 × 10 ⁸ m/s) / (8.223 × 10⁻¹⁹ J)  = 2.41 × 10⁻⁷m = 241 nm

UV light wavelength is between 400 nm - 10 nm

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Which event would be impossible to explain by using John Dalton’s model of the atom?
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Does pb(no3)2 + Na3(PO4) = Pb3(PO4)2 + Na(NO3) have a precipitate?
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Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

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In general, the more _ a metal has, the stronger its metallic bonds will be because...
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The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:
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The allowable combination for the atomic orbital is n=3, l=1 or 2, m_{l}=+1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; m_{l} values are integers from -l to =l(l limits ml)

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Therefore, the allowable condition is n=7, l=1 or 2, m_{l}=+3.

To know more about quantum numbers, visit: brainly.com/question/16979660

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