Chemistry, 08.07.2020 14:01 katier9407

It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single bond could be broken by absorbing a single photon. Round your answer to significant digits.

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Answer from: 1jennator

It takes approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

The maximum wavelength of a photon that can break one such bond is approximately $3.66\times 10^{-7}\; \rm m$ (in vacuum.) That's the same as $3.66 \times 10^{2}\; \rm nm$ (rounded to three significant figures.)

Explanation:

Energy per bond

The standard bond enthalpy of $\rm C-Cl$ single bonds is approximately $\rm 327\; \rm kJ \cdot mol^{-1}$ (note that the exact value can varies across sources.) In other words, it would take approximately $327\; \rm kJ$ of energy to break one mole of these bonds.

The Avogadro Constant $N_A \approx 6.023\times 10^{23}\; \rm mol^{-1}$ gives the number of $\rm C-Cl$ bonds in one mole of these bonds. Based on these information, calculate the energy of one such bond:

$\begin{aligned}& E(\text{one $\mathrm{C-Cl}$ bond}) \\ &= \frac{E(\text{one mole of $\mathrm{C-Cl}$ bonds})}{N_A} = \frac{327\; \rm kJ\cdot mol^{-1}}{6.023\times 10^{23}\; \rm mol^{-1}} \\ &\approx 5.429\times 10^{-22}\; \rm kJ = 5.429\times 10^{-19}\; \rm J \end{aligned}$ .

Therefore, it would take approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

Minimum frequency and maximum wavelength

The Einstein-Planck Relation relates the frequency of a photon to its energy :

$E = h \cdot f$ .

The here represents the Planck Constant:

$h \approx 6.63 \times 10^{-34}\; \rm J \cdot s$ .

A photon that can break one $\rm C-Cl$ single bond should have more than $5.43\times 10^{-19}\; \rm J$ of energy. Apply the Einstein-Planck Relation to find the frequency of a photon with exactly that much energy:

$\begin{aligned}f &= \frac{E}{h}\\ &\approx \frac{5.43\times 10^{-19}\; \rm J}{6.63 \times 10^{-34}\; \rm J\cdot s} \\ &\approx 8.19 \times 10^{14}\; \rm s^{-1} = 8.19 \times 10^{14}\; \rm Hz\end{aligned}$ .

What would be the wavelength $\lambda$ of a photon with a frequency of approximately $8.19 \times 10^{14}\; \rm Hz$ ? The exact answer to that depends on the medium that this photon is travelling through. To be precise, the exact answer depends on the speed of light in that medium:

$\displaystyle \lambda = \frac{(\text{speed of light})}{f}$ .

In vacuum, the speed of light is $c \approx 2.998\times 10^{8}\; \rm m \cdot s^{-1}$ . Therefore, the wavelength of that $8.19 \times 10^{14}\; \rm Hz$ photon in vacuum would be:

$\begin{aligned} \lambda &= \frac{c}{f} \\ & \approx \frac{2.998\times 10^{8}\; \rm m \cdot s^{-1}}{8.19\times 10^{14}\; \rm s^{-1}} \\ &\approx 3.66 \times 10^{-7}\; \rm m = 3.66 \times 10^{2}\; \rm nm\end{aligned}$ .

(Side note: that wavelength corresponds to a photon in the ultraviolet region of the electromagnetic spectrum.)

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$