Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol
To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.
The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.
Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .
Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."
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Lithium has 1 valence electron available for bonding. So its A.
Answer:
300 kg
Explanation:
The give parameters obtained from a similar question are;
The mass of oseltamivir per capsule = 75 mg
The mass of oseltamivir required, <em>M</em>, is given as follows;
M = The mass of oseltamivir per capsule × The mass taken per person per day × 5 days × The number of people in the city
M = 0.75 mg/capsule × 2 capsule/(day·person) × 5 days × 400,000 people
M = 300,000,000 mg = 300 kg
