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Andrew [12]
3 years ago
14

Calculate the wavelength of a photon having energy of 1.257 X 10-24 joules. (Planck’s constant is 6.626 x 10-34 joule seconds; t

he speed of light is 2.998 x 108 m/s)
Chemistry
1 answer:
maw [93]3 years ago
5 0
E=h\nu\\\\
\nu=\frac{c}{\lambda} \ \ \ \Rightarrow E=\dfrac{hc}\lambda}\\\\
\lambda=\dfrac{hc}{E}=\dfrac{6,626*10^{-34}Js*2,998*10^{8}\dfrac{m}{s}}{1,257*10^{-24}J}=15,803*10^{\frac{-34+8}{-24}}m=\\\\\\=15,803*10^{-2}m=1,5803*10^{-1}m
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Explanation:

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2 years ago
A container of oxygen with a volume of 60 L is heated from 300 K to 400 k, What is the new volume?
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80L

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Answer:

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Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

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