A carbon–oxygen double bond within a sugar molecule absorbs electromagnetic radiation at a frequency of 6.0 x 10 12 s -1 . What portion of the electromagnetic spectrum does this represent?

Here we're told that a carbon oxygen double bond within a sugar molecule absorbs electromagnetic radiation at a frequency of 6.0 times 10 to the 12 seconds inverse. What portion of the electromagnetic spectrum does this represent? Alright. So work has been doing math problems and trying to solve here, but remember we talked about the electromagnetic spectrum in its range. If you go back and take a look at that, you'll see that the only type of electromagnetic radiation that exists within this range of oops. Within this range times 10 to the 12 would have to be infrared light. So here the answer is c. Going back and taking a look at the electromagnetic spectrum, we see that infrared can exist comfort comfortably between a range of 10 to 12 to around a little over 10 to the 14 seconds inverse. So in this particular question, there's no math needed. Here, it's option c.

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9. Quantum Mechanics

Electromagnetic Spectrum

General Chemistry

9. Quantum Mechanics

Electromagnetic Spectrum

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Multiple Choice

A carbon–oxygen double bond within a sugar molecule absorbs electromagnetic radiation at a frequency of 6.0 x 10¹² s^-1. What portion of the electromagnetic spectrum does this represent?

Radio Waves

Micowave

Infrared

Green Light

Gamma Ray

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Verified step by step guidance

Understand the electromagnetic spectrum: The electromagnetic spectrum is a range of all types of electromagnetic radiation, which includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

Identify the given frequency: The problem states that the frequency of the radiation absorbed by the carbon–oxygen double bond is \(6.0 \times 10^{12} \text{s}^{-1}\).

Recall the frequency ranges for different portions of the electromagnetic spectrum: Infrared radiation typically falls within the frequency range of \(10^{11} \text{s}^{-1}\) to \(10^{14} \text{s}^{-1}\).

Compare the given frequency to the known ranges: The frequency \(6.0 \times 10^{12} \text{s}^{-1}\) falls within the infrared range, which is between \(10^{11} \text{s}^{-1}\) and \(10^{14} \text{s}^{-1}\).

Conclude the portion of the spectrum: Based on the comparison, the frequency of \(6.0 \times 10^{12} \text{s}^{-1}\) corresponds to the infrared portion of the electromagnetic spectrum.