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oksian1 [2.3K]
2 years ago
12

A 0.334 g sample of an unknown compound occupies 245 ml at 298 K and 1.22 atm. What is the molar mass of the unknown compound.

Chemistry
1 answer:
marysya [2.9K]2 years ago
6 0

Answer:

27.4 g/mol

Explanation:

Assuming the compound is a gas and that it behaves ideally, we can solve this problem by using the <em>PV=nRT formula</em>, where:

  • P = 1.22 atm
  • V = 245 mL ⇒ 245 mL / 1000 = 0.245 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 298 K

<u>Inputting the data</u>:

  • 1.22 atm * 0.245 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

<u>Solving for n</u>:

  • n = 0.0122 mol

With the <em>calculated number of moles and given mass</em>, we <u>calculate the molar mass</u>:

  • 0.334 g / 0.0122 mol = 27.4 g/mol
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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;

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  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

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<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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