<u>Answer:</u> The energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)
<u>Explanation:</u>
To calculate the energy of one photon, we use Planck's equation, which is:
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
= wavelength of light =
(Conversion factor:
)
h = Planck's constant = ![6.625\times 10^{-34}J.s](https://tex.z-dn.net/?f=6.625%5Ctimes%2010%5E%7B-34%7DJ.s)
c = speed of light = ![3\times 10^8m/s](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E8m%2Fs)
Putting values in above equation, we get:
![E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{5.25\times 10^{-7}mm}\\\\E=3.79\times 10^{-19}J](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B6.625%5Ctimes%2010%5E%7B-34%7DJ.s%5Ctimes%203%5Ctimes%2010%5E8m%2Fs%7D%7B5.25%5Ctimes%2010%5E%7B-7%7Dmm%7D%5C%5C%5C%5CE%3D3.79%5Ctimes%2010%5E%7B-19%7DJ)
Hence, the energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)
Answer:
The molecular formula of the compound is
. The molecular formula is obtained by the following expression shown below
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Explanation:
Given molecular mass of the compound is 176 g/mol
Given empirical formula is
Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.
Empirical formula mass of the compound = ![\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}](https://tex.z-dn.net/?f=%5Cleft%20%28%202%5Ctimes12%2B4%2B16%20%5Cright%20%29%20%5Ctextrm%7B%20u%7D%20%3D%2044%20%5Ctextrm%7B%20g%2Fmol%7D)
![n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4](https://tex.z-dn.net/?f=n%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5Ctextrm%7BMolecular%20formula%20mass%7D%7D%7B%5Ctextrm%7BEmpirical%20formula%20mass%7D%7D%20%5C%5Cn%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B176%7D%7B44%7D%20%3D%204)
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Molecular formula = 4
Molecular formula is ![C_{8}H_{16}O_{4}](https://tex.z-dn.net/?f=C_%7B8%7DH_%7B16%7DO_%7B4%7D)
Metallic bonding
Metals consist of giant structures of atoms arranged in a regular pattern. The electrons from the outer shells of the metal atoms are delocalised , and are free to move through the whole structure. This sharing of delocalised electrons results in strong metallic bonding .
Answer:
1s², 2s², 2p³
Explanation:
The atomic number of Nitrogen is seven. So it contains seven protons and seven electrons in neutral form. Also, the electronic configuration cited above contains seven electrons among which two electrons are present in first shell and five electrons are present in valence shell respectively.