The reactants have one C atom and four O and H atoms. Hope this helps :)
. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:
E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ
Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ
2. The wavelength of a photon with this energy would be:
Energy = hc/wavelength
wavelength = hc/energy
wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m
Converting to nanometers gives: 91.16 nm
3. Repeat the calculation in 1, but using n=5.
4. Repeat the calculation in 2 using the energy calculated in 3.
There is a very simple relationship between the three. First off, power is the amount of energy used over a certain amount of time. Energy is the capacity of carrying out that power. Lastly, time depends on how much energy you have to exert the work.
Hope this helps :)
The given statement is true .
<h3>What is Rutherford’s gold foil experiment?</h3>
- A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
- The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
- For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.
To learn more about Rutherford’s gold foil experiment, refer to:
brainly.com/question/4113533
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