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Vladimir [108]
3 years ago
8

What explains the change in ionization energy that occurs between removing the first and second electrons from an atom

Chemistry
1 answer:
Jobisdone [24]3 years ago
5 0
Options are as follow,

A. The ionization energy decreases because the ratio of the protons to electrons increases.
B. The ionization energy increases because the ratio of the protons to electrons increases.
C. The ionization energy decreases because the ratio of the protons to electrons decreases.
<span>D. The ionization energy increases because the ratio of the protons to electrons decreases.
</span>
Answer:
            Option-B (The ionization energy increases because the ratio of the protons to electrons increases).

Explanation:
                  Ionization energy is the amount of energy required to remove one electron from the valence shell of an atom.
                  Also, first IE is energy required for removing electron from neutral atom, 2nd IE is energy required for is for removing electron from a mono positive atom and so on.
                  Example:
                                  1st IE of Na is 495.8 kJ/mol

                                   2nd IE of Na is 4562.4 kJ/mol

Second ionization energy is greater because after the removal of first electron the ratio of protons to electrons increases, resulting in more nuclear effect. Also, the second IE is greater because the Na⁺ having noble gas configuration is a stable state, hence more energy is required to knock out electron from it.
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A sample of gasoline has a mass of 595.0 lb and a volume of 14.0 ft^3. What is its specific gravity?
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Explanation:

The given data is as follows.

            mass of gasoline = 595.0 lb

            volume of gasoline = 14.0 ft^{3}

As it is known that density is the amount of mass of a substance divided by its volume.

So, density of gasoline will be as follows.

                Density = \frac{mass}{volume}

                              = \frac{595.0 lb}{14.0 ft^{3}}

                              = 42.5 lb/ft^{3}

As, 1 lb = 0.4536 kg and 1 ft = 0.3048 m. Now, putting these values into the above equation (1) as follows.

                       Density = 42.5 lb/ft^{3}

                                     = 42.5 \times \frac{0.4536 kg}{(0.3048 m)^{3}}

                                     = 680.792 kg/m^{3}    

                                     = 680.8 kg/m^{3} (approx)

Density of water is 1000 kg/m^{3}. To measure specific gravity of gasoline, the formula will be as follows.

               specific gravity of gasoline = \frac{\text{density of gasoline}}{\text{density of water}}    

                                         = \frac{680.8 kg/m^{3}}{1000 kg/m^{3}}

                                         = 0.681

Thus, we can conclude that the specific gravity of gasoline is 0.681.

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1) accepts a proton

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