Why is it easier to remove an electron from a sodium atom than a magnesium atom

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

2 years ago

A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How

frozen [14]

Explanation:

The given data is as follows.

Mass of a lead atom = $3.14 \times 10^{-22}$

Volume = 2.00 $cm^{3}$

Density = 11.3 $g/cm^{3}$

As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.

So, in 2 cubic centimeter there will be $2 \times 11.3 g = 22.6 g$ of lead atoms.

One lead atom has a mass of $3.14 \times 10^{-22}$ . Therefore, number of atoms present in 22.6 g of lead will be as follows.

$\frac{22.6 g}{3.14 \times 10^{-22}}$

= $7.197 \times 10^{22} atoms$

Thus, we can conclude that there are $7.197 \times 10^{22} atoms$ of lead are present.

4 0

2 years ago

A vacuum gage connected to a chamber reads 35 kpa at a location where the atmospheric pressure is 92 kpa. determine the absolute

stealth61 [152]

Vacuum gage connected to a chamber reads 35 kpa at a location where the atmospheric pressure is 92 kpa then the absolute pressure in the chamber and express is 57kpa

Gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures and absolute pressure is the sum of gauge pressure and atmospheric pressure

Here to find absolute pressure atmospheric pressure=92kpa and vacuum gage=35 kpa

Pabsolute pressure=Patmospehric pressure - Pvaccume

92-35=57kpa

Know more about absolute pressure

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6 0

1 year ago

Which of the following would lose heat the fastest?

Molodets [167]

Answer:

15 grams of water

Explanation:

15 grams of water of water would lose heat the faster compared to higher masses of water.

Water generally is a poor conductor heat.

To heat up a unit of water, significant amount of energy must be added to the body of water.
With time, the body continues to increase in temperature.
A 500g mass of water will take more time to lose heat.

5 0

2 years ago

Calculate the volume of a box with the following dimensions: length=8 cm

wel

Answer: 240 cm³

Explanation:

Volume of the box = l × w × h

= (8 cm × 6 cm × 5 cm)

= 240 cm³

8 0

1 year ago