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Sophie [7]
2 years ago
11

13. What is the photon energy associated with visible light with a wavelength of 4.25x10^-7 m?

Chemistry
1 answer:
4vir4ik [10]2 years ago
5 0

\\ \sf\bull\longmapsto E=hv

\\ \sf\bull\longmapsto E=\dfrac{hc}{\lambda}

\\ \sf\bull\longmapsto E=\dfrac{6.626\times 10^{-34}Js\times 3\times 10^8ms^{-1}}{4.25\times 10^{-7}m}

\\ \sf\bull\longmapsto E=\dfrac{19.878\times 10^{-26}Jm}{4.25\times 10^{-7}m}

\\ \sf\bull\longmapsto E=4.677\times 10^{-19}J

\\ \sf\bull\longmapsto E=4677\times 10^{-16}J

\\ \sf\bull\longmapsto E=4677eV

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Explanation:

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To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

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Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0 kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>

ΔH = <span>795.2 kJ</span>

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