What is the application of the emission spectra regarding elements?

1 answer:

8 0

The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states. There are many possible electron transitions for each atom, and each transition has a specific energy difference. This collection of different transitions, leading to different radiated wavelengths, make up an emission spectrum. Each element's emission spectrum is unique. Therefore, spectroscopy can be used to identify elements in matter of unknown composition. Similarly, the emission spectra of molecules can be used in chemical analysis of substances.

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A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w

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Answer:

A. 0.2395 w/w %

B. 2394ppm

Explanation:

A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:

Mass glycerol / Total mass * 100

<em>Mass glycerol:</em>

The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):

2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol

<em>Mass of water:</em>

998.9mL and density = 0.9982g/mL:

998.9mL * (0.9982g/mL) = 997.1g of water.

That means percent by mass is:

% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %

B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:

2.394g * (1000mg / 1g) = 2394mg:

Parts per million: 2394mg / L = 2394ppm

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A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen. After cooling to room temperature, thr

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In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.

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Molecules move the fastest in gas

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