Answer:
Rate of the reaction= 9.92× 10^-5 M² min-1
Explanation:
Using the equation of reaction
2N2O5 ⟶ 4NO2+O2
Rate = k[N2O5]²
From the question k= 6.2×10-4
[N2O5]= 0.4
Rate = 6.2×10-4[0.4]²= 9.92×10-5M² min-1
Answer:
Explanation:
energy for photon is calculated in same wasy as for electromagnetic radiation
energy for electromagnetic radiation = hf
where f is the frequency of photon
h is Planck's constant = h = 4.14 × 10−15 eV · s.
thus
energy of photon = 4.14 × 10−15 eV · s * 2.36 x 10-19 Hz
energy of photon = 9.77 * 10−(-15+ -19) eV
energy of photon = 9.77 * 10−34eV answer
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928