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  1. ...not have those amounts of energy leaving blank spaces in the spectrum .

    2 Answers · Science & Mathematics · 18/10/2020

  2. ...out all the C5H12O compounds and compare their HNMR spectra .  The spectra on that site tell you which peak belongs to...

    1 Answers · Science & Mathematics · 01/09/2020

  3. From the MS the parent ion has a value of 114.  Since you have indicated it is a ketone, then it could be Heptan-2/3/4-one.  The large abundance at '43' would indicate a 'CH3CH2CH2^+ ion. or CH3C(=O)^+ on ; the ketone...

    1 Answers · Science & Mathematics · 02/04/2020

  4. ...diagram" is called a 13C-NMR spectrum . It is always important to name "diagrams...integration in the aromatic region (~65-85 ppm) of the NMR- spectrum , plus 2 remaining aromatic carbon signals...

    2 Answers · Science & Mathematics · 05/11/2019

  5. No spectrum .  Since b gives m/z the type of spectrum must be mass spec.

    1 Answers · Science & Mathematics · 01/09/2020

  6. ... of light, many of them being in the visible part of the spectrum . And the list goes on

    5 Answers · Science & Mathematics · 28/05/2020

  7. ν=3.2881⋅1015s−1⋅(1/2²−1/n²) where n=9 v=3.2881⋅10^15s−1⋅(1/4 − 1/81) = 7.814311728395061 × 10^14 s-1 λ = c/v so 299792458 ms-1 /(7.814311728395061 × 10^14s-1) in metres λ = 383.645378403112 rounds to 383.6 nm in nanometres

    2 Answers · Science & Mathematics · 14/09/2020

  8. 1. Use the Rydberg equation: 1 / ((1.09737×10^7 m^−1) × (1/2^2 - 1/5^2)) = 4.33938 × 10^-7 m = 434 nm 2. (299792458 m / s) / (4.33938 × 10^-7 m) = 6.91 × 10^14 s^-1 = 6.91 × 10^14 Hz It is violet visible light.

    2 Answers · Science & Mathematics · 07/10/2020

  9. The smaller the transition, the less energy the photon will have. Since longer wavelength means lower energy, and since shorter wavelength means greater energy, 3-->2: 656.3 nm 4-->2: 468.3 nm 5-->2: 432.4 nm 6-->2: 410.3 nm

    1 Answers · Science & Mathematics · 18/10/2020

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