The wavelength of the photon required to excite this molecule from its ground state, to its first excited state is 1240 nm.
This is given by the equation:
wavelength = hc/(E_homo - E_lumo)
where h is Planck's constant =6.626070 * 10^-34 J.m , c is the speed of light = 3.0 x 10^8 m/s^2, and E_homo and E_lumo are the energies of the highest occupied molecular orbital and the lowest unoccupied molecular orbital, respectively.
In this particular case, the wavelength of the required photon would be:
wavelength = hc/(-2.42 hartree - 0.65 hartree)
= 6.626070 * 10^-34 X 3.0 x 10^8 / (-3.07)
= 1240 nm
Hence , The wavelength of the photon required to excite this molecule from its ground state, to its first excited state is 1240 nm.
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Answer: Option C) 4 moles
Explanation:
The reaction below shows the decomposition of 2 moles of water to yeild 2 moles of hydrogen and 1 mole of oxygen respectively.
2H2O(l) --> 2H2(g) + O2 (g)
So, if 2 moles of H2O produce 1 mole of O2
8 moles of H2O will produce Z mole of O2
To get the value of Z, cross multiply
2 x Z = 1 x 8
2Z = 8
Z = 8/2
Z = 4
Thus, 4 moles of O2 is produced from the complete decomposition of 8 moles H2O .
Answer:
B. burning a piece of wood
Explanation:
The Chemical Would Be The Air Coming From The Wood While Burning It
Tell Me If Im Correct
Answer:
e.iv, iii, i, ii
Explanation:
Be has electron configuration of 1s^2 2s^2. During bond formation, one 2s electron is promoted to a 2p orbital. Hence two hybrid sp beryllium orbitals are formed. These two hybrid orbitals are used in sigma bond formation to fluorine atoms.
Recall that fluorine has filled 2p orbitals containing lone pairs of electrons. These filled orbitals can interact with the unhybridized 2p-orbitals on beryllium to form pi bonds. This is the reason for the sequence of events chosen in the answer.
