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: The correct answer would be : "A bottle that is sealed".
I hope that this helps you !
Humphry Davy First discovered sodium
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
The answer is N2 + 3H2 yields 2NH3. The oxidation-reduction reaction means that there is electrons transfer during the reaction which means that the valence changed.