A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by: En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

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oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

explanation:

0.07 g/cm³

1.37 g/cm³