How does the emission spectrum support the idea of quantized energy levels
1 answer:
Explanation:
Different atoms absorb and emit specific wavelengths of electromagnetic radiation and nothing in between. These absorption and emission spectra are actually used to identify atoms of elements in a substance. This phenomenon is explained by Bohr's theory of quantized energy levels in an atom – called orbital levels. When an electron 'jumps' from a lower to higher orbital level, it absorbs a specific wavelength of electromagnetic radiation specific to the ‘jump’. Vice versa, when an electron 'jumps' to a lower orbital level is emits an equivalent and specific wavelength of electromagnetic radiation.
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Answer:
a.
1. NO2(g): Reactant.
2. CO(g): Reactant.
3. NO3(g): Intermediate.
4. CO2(g): Product.
5. NO(g): Product.
b. See attached picture.
Explanation:
Hello,
a. In this case, given the reactions, we can identify each species as:
1. NO2(g): Reactant because it remains at the left side in the overall reaction.
2. CO(g): Reactant because it remains at the left side in the overall reaction.
3. NO3(g): Intermediate because it is a product in the step 1 and a reactant in step 2, for that reason it is not present in the overall reaction.
4. CO2(g): Product because it remains at the right side in the overall reaction.
5. NO(g): Product because it remains at the right side in the overall reaction.
b. In this case, given that the first step is slow and endothermic, it has a high activation energy and the products will have more energy than the reactants, for that reason the final energy is above the initial point. Moreover, since the second step is fast and exothermic, it has a low activation energy and the products will have less energy than the reactants, for that reason, the reaction coordinate diagram is shown on the attached file.
Regards.
The initial temperature is 137.34 °C.
<u>Explanation:</u>
As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.
Q = mcΔT
So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.
ΔT =
ΔT =
Since the difference in temperature is 2.66, then the initial temperature will be
Final temperature - Initial temperature = Change in temperature
140-Initial temperature = 2.66
Initial temperature = 140-2.66 = 137.34 °C
Thus, the initial temperature is 137.34 °C.