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WARRIOR [948]
3 years ago
5

What is the wavelength of a photon with an energy of 3.38 x 10-19 J?

Chemistry
2 answers:
Mars2501 [29]3 years ago
5 0

C. 588

I just did the test

irina1246 [14]3 years ago
4 0

Answer:

C. 588 nm

Explanation:

Given parameters:

Energy of the photon  = 3.38 x 10⁻¹⁹J

Unknown:

Wavelength of the photon   = ?

Solution:

The energy of a photon can be expressed as;

     E = \frac{hc}{wavelength}

    hc  = E x wavelength

     Wavelength  = \frac{hc }{E}  

  h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s

  c is the speed of light  = 3 x 10⁸m/s

  E is the energy  

   Wavelength = \frac{6.63 x 10^{-34} x 3 x 10^{8} }{3.38 x 10^{-19} }    = 5.89 x 10⁻⁷m

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3 years ago
Read 2 more answers
Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.
mario62 [17]

Answer:

The order will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

Explanation:

According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.

For example: between C and H, we rank Carbon first.

If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.

For example:

Among CH₃ and C₂H₅, the priority will be given to C₂H₅.

If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.

Hence CH=CH₂ means, that there are two carbons attached to CH carbon.

So the order based on above selection rules will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

7 0
3 years ago
Which of these household items is powered by alternating current?
AlladinOne [14]

Answer:

C. Hair dryer

Explanation:

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3 0
3 years ago
PLEASE HELP! I really don’t understand! Thanks!
PIT_PIT [208]

http://century.rochester.k12.mn.us/cms/One.aspx?portalId=3086882&pageId=6133921

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6 0
3 years ago
Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h
scZoUnD [109]
44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
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4 0
3 years ago
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