Atom is the smallest indivisible particle of matter.
Answer:
ethyl - 3- heptanol mouthing
The energy of the 434 nm emission line is 4.58×10¯¹⁹ J (Option A)
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Wavelength (λ) = 434 nm = 434×10¯⁹ m
- Planck's constant (h) = 6.626×10¯³⁴ Js
- Speed of light (v) = 3×10⁸ m/s
- Energy (E) =?
<h3>How to determine the energy </h3>
The energy of the 434 nm emission line can be obtained as follow:
E = hv / λ
E = (6.626×10¯³⁴ × 3×10⁸) / 434×10¯⁹ 9.58×10¹⁴
E = 4.58×10¯¹⁹ J
Thus, the energy of the 434 nm emission line is 4.58×10¯¹⁹ J
Learn more about energy:
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Answer:
3.75
Explanation:
it's gonna be a decimal number,cus from the formula 5moles produce 4moles of NO
so if 3 moles we're to be produced,it will be 3x5 divide by 4
