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:
0.78 atm
Explanation:
Step 1:
Data obtained from the question. This includes:
Mass of CO2 = 5.6g
Volume (V) = 4L
Temperature (T) =300K
Pressure (P) =?
Step 2:
Determination of the number of mole of CO2.
This is illustrated below:
Mass of CO2 = 5.6g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Number of mole CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 5.6/44
Number of mole of CO2 = 0.127 mole
Step 3:
Determination of the pressure in the container.
The pressure in the container can be obtained by applying the ideal gas equation as follow:
PV = nRT
The gas constant (R) = 0.082atm.L/Kmol
The number of mole (n) = 0.127 mole
P x 4 = 0.127 x 0.082 x 300
Divide both side by 4
P = (0.127 x 0.082 x 300) /4
P = 0.78 atm
Therefore, the pressure in the container is
Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).
Answer:
2850 grams I hope this help
