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Leto [7]
4 years ago
12

Electrospray ionization and atmospheric pressure chemical ionization are two methods that are used to introduce the eluate from

the liquid chromatography column into a mass spectrometer. Which ionization method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer?
Chemistry
1 answer:
Monica [59]4 years ago
4 0

Answer:

Electrospray ionization

Explanation:

Electrospray ionization Is a soft ionization technique used in producing ions from macromolecules, it is employed especially in spectrometry where high voltage is used to create an aerosol from a liquid.It can be employed in Knowing molecular weights of molecules and biological macromolecules such as Peptides and proteins. Therefore, electrospray ionization is the method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer

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8 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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