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VARVARA [1.3K]
2 years ago
6

Hydrogen peroxide is sold commercially as an aqueous solution in brown bottles to protect it from light. Calculate the longest w

avelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide.

Physics
2 answers:
JulsSmile [24]2 years ago
7 0

Answer:

the longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide is 842nm

Explanation:

Δ H = 142 kJ/mol  

This is the energy for 1 mol of molecules.

For 1 molecule,

E = \frac{142000J}{1mol} \frac{1mol}{6.022 \times 10^2^3molecules} \\ = 2.358 \times 19^-^1^9J

E = \frac{h_c}{\lambda}

\lambda = \frac{h_c}{E} \\ = \frac{6.626 \times 10^-^3^4Js \times 2.998 \times 10^8m/s}{2.358 \times 19^-1^9J} \\= 8.42 \times 10^-^7\\\\=842nm

the longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide is 842nm

snow_lady [41]2 years ago
5 0

Explanation:

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Answer: 0.258

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s=\pi{(\frac{d}{2})}^{2}  (2) Where d  is the diameter of the circumference.

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So, in this problem we have two transversal areas:

<u>For aluminium:</u>

s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}

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<u>For copper:</u>

s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}

s_{Cu}=0.00000201m^{2}    (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}     (5)

R_{Al}=0.0647\Omega     (6)  Resistance of aluminium wire

<u>Copper wire:</u>

R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}     (6)

R_{Cu}=0.250\Omega     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

Ratio=\frac{R_{Al}}{R_{Cu}}   (8)

\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}   (9)

Finally:

\frac{R_{Al}}{R_{Cu}}=0.258  This is the ratio

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