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.
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Answer:
0.188mol
Explanation:
Using the formula;
mole = mass/molar mass
Molar mass of hypomanganous acid. (H3MnO4) = 1(3) + 55 + 16(4)
= 3 + 55 + 64
= 122g/mol
According to this question, there are 22.912g of H3MnO4
mole = 22.912g ÷ 122g/mol
mole = 0.188mol
all the elements in group 18 are Nobel gases or inert gases . all the elements such as neon , helium, argon etc. ,their outermost shell is completely filled . The noble gases have the largest ionization energies, reflecting their chemical inertness
Answer:
6 x 10⁶ g Fe
Explanation:
Step 1: Set up dimensional analysis
7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)
Step 2: Multiply, divide, and cancel out units
atoms Fe and atoms Fe cancel out.
mol Fe and mol Fe cancel out.
We should be left with g Fe.
7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe
116279(55.85) = 6.49 x 10⁶ g Fe
Step 3: Sig figs
There is only 1 sig fig in this problem.
6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe