1cg is equal to what

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

kupik [55]

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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6 0

3 years ago

The reader can infer from the article that __________.

marin [14]

The answer is definitely D

5 0

2 years ago

Can you solve question 5 and 6 only

Yanka [14]

Answer:

oil floats on water because it is heavier than water

Salt is no longer visible after stirring as salt dissolves in water

by filtration

4 0

3 years ago

12. A voltaic cell consists of a chromium electrode dipped in a 1.20 M chromium (III) nitrate

Lilit [14]

For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as

Ecell = 0.577 V

<h3 /><h3>What is the cell potential at 298 K?</h3>

Generally, the equation for the Oxidation and Reduction is mathematically given as

Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O

Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R

Reaction

2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)

Therefore

Eicell = - 0.14 - ( - 0.74)

Eicell = 0.60

In conclusion

$Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}$

$Ecell = 0.60 - \frac{0.0591 }{6} \frac{log( 1.20)^2}{ ( 0.200)^3}$

Ecell = 0.577 V