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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Answer: O2+6H12O6=CO2+ENERGY(ATP)
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Explanation:
Answer:
5.85 x 10⁻² mol Au
Explanation:
Divide the atoms of gold with Avogadro's number to find the solution.