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erastovalidia [21]
3 years ago
7

1) The spectrum of lithium has a red line of 670.8 nanometers. (Remember 1 m = 1 X 109 nm) a. Convert the nanometer to meter usi

ng dimensional analysis. b. Calculate the frequency of the wave. c. Calculate the energy of a photon with this wavelength.
Chemistry
1 answer:
puteri [66]3 years ago
5 0
<span>E = ħc / λ
 ħ = plancks constant = 6.626x10^-34 Js
c = speed of light = 2.999x10^8 m/s
λ = wavelength of light = 670.8x10^-9 m
 
E = (6.626x10^-34 Js) x (2.999x10^8 m/s) x (1 / 670.8x10^-9 m)
E = 2.962x10^-19 J

</span><span>3x10^8 / (670.8 * 10^-9) =4.47x10^14 Hz
 
4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19
</span><span>
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
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Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2
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Answer:

<u>Option 1</u>:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

Explanation:

<u>Ionic radius is the radius of an atom´s ion in ionic crystal structure</u>.<u> </u><u><em>In an ion that lose an electron, to form a cation, the radius of the ion gets smaller</em></u><em>, </em>because the repulsion between electrons decrease because fewer electrons are present. Conversely, <u><em>adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.</em></u>                  

<u><em>Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers</em></u>.                        

<u><em>In a isolelectronic series (same number of electrons),</em></u> <u><em>the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius </em></u>beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

<u>We will use this principle to solve our problem</u>.  

In our case, the given ions are:  

  • N⁻³ :    Z = 7,  e⁻ = 10
  • O⁻²:     Z= 8,   e⁻ =10
  • F⁻:       Z = 9,  e⁻ = 10
  • Na⁺:    Z= 11,   e⁻ = 10
  • Mg⁺²:  Z=12,   e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

<em><u>Hence the decreasing order of ionic radius is:</u></em>

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

Have a nice day!

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3 years ago
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3 years ago
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Answer:

\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅,  then convert to molecules of P₂O₅.

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\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}

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\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}

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