Answer:
Explanation:
Given that:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser.
the energy of the first ionization energy of argon is 1520 kJ mol-1.
SInce 1 eV = 96.49 kJ/mol
Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV
= 15.75 eV
To find where the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.
Converting Joules (J) to eV ; we get,
E = 2.53 eV
The energy levels of the first exited state = -13.223 eV
Phosphorous is the answer.
Answer:
6782 has 4, 26.2 has 3, 0.3491 has 4 and 55 has 2 significant figures .
<span>The average molar bond enthalpy of the carbon-hydrogen bond in a CH4 molecule is 416 KJ/mol.
(+716.7 + (4 x 218) - (- 74.6) ) / 4
= + 1663.3 / 4
= 416</span>