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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The balanced equation for the reaction is as follows;
Li₂O + H₂O ---> 2LiOH
Stoichiometry of Li₂O to H₂O is 1:1
Mass of Li₂O reacted - 18.9 g
Number of Li₂O moles reacted - 18.9 g / 30 g/mol = 0.63 mol
An equivalent amount of moles of water have reacted - 0.63 mol
mass of water required - 0.63 mol x 18 g/mol = 11.34 g
A mass of 11.34 g of water is required
Answer:
15.52 kj
Explanation:
H2 O mole wt ~ 18 gm
6.35 gm is 6.35 gm /18 gm/mole = .353 moles of H2O
.353 moles * 44 kj/mole =
In warmer weather gases tend to expand and take up more room, thus increasing pressure. but in colder weather they will condense or contract and take up less space, therefore lowering the pressure of the tire in this situation.
