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3 years ago

12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?

Chemistry

1 answer:

bazaltina [42]3 years ago

5 0

Answer:

$u=4.57x10^5GHz$

Explanation:

Hello.

In this case, given the formula:

$E=h*u$

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

$u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\ \\u=4.57x10^{14}s^{-1}$

Or also:

$u=4.57x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ \\u=4.57x10^5GHz$

Best regards.

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Explanation:

The given balanced reaction is as follows.

$2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O$

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of $NH_{4}NO_{3}$ present will be as follows.

Moles of $NH_{4}NO_{3}$ = $\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}$

= $\frac{86000 g}{80.043 g/mol}$

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Therefore, moles of $N_{2}$ , $O_{2}$ and $H_{2}O$ produced by 1074.42 mole of $NH_{4}NO_{3}$ will be as follows.

Moles of $O_{2}$ = $\frac{1}{2} \times 1074.42 mol$

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Moles of $N_{2}$ = $\frac{2}{2} \times 1074.42 mol$

= 1074.42 mol

Moles of $H_{2}O$ = $\frac{4}{2} \times 1074.42 mol$

= 2148.84 mol

Therefore, total number of moles will be as follows.

537.21 mol + 1074.42 mol + 2148.84 mol

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PV = nRT

1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as $307^{o}C$ = 307 + 273 = 580 K)

V = 179066.06 L

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