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valina [46]
3 years ago
5

It takes 498. kJ/mol to break an oxygen-oxygen double bond. Calculate the maximum wavelength of light for which an oxygen-oxygen

double bond could be
broken by absorbing a single photon.
Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
Neporo4naja [7]3 years ago
5 0

The bond energy required to break the oxygen-oxygen double bond is 240 nm

The energy of the incoming photon = 498 * 10^3 J/mol/6.02 * 10^23 moles = 8.27 * 10^-19 J

Recall that;

E = hc/λ

E = energy of the photon

c = speed of light = 3 * 10^8 m/s

λ = wavelength of photon

h = Plank's constant = 6.63 * 10^-34 Js

λ = hc/E

λ = 6.63 * 10^-34 * 3 * 10^8/8.27 * 10^-19

λ = 2.40 * 10^-7 m

λ = 240 nm

Learn more: brainly.com/question/24302171

Lyrx [107]3 years ago
5 0
  • Energy=498KJ =498×10^3J

We know

\\ \sf\bull\longmapsto E=hv

  • h=planks constant
  • v=frequency

\\ \sf\bull\longmapsto v=\dfrac{E}{h}

\\ \sf\bull\longmapsto v=\dfrac{498\times 10^3J}{6.626\times 10^{-34}Js}

\\ \sf\bull\longmapsto v=75.15\times 10^{37}s^{-1}

\\ \sf\bull\longmapsto v=7.5\times 10^{36}s^{-1}

Now

\\ \sf\bull\longmapsto \lambda=\dfrac{C}{V}

\\ \sf\bull\longmapsto \lambda=\dfrac{3\times 10^{8}ms^{-1}}{7.5\times 10^{36}s^{-1}}

\\ \sf\bull\longmapsto \lambda=2.5\times 10^{-28}m

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Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
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Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
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