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Luden [163]
2 years ago
5

The first four ionization energies for an imaginary element, Xz, are E1 = 102 kcal, E2 = 186 kcal, E3 = 4021 kcal, and E4 = 4862

kcal. The number of valence electrons most probable as indicated by this ionization data is:
Chemistry
2 answers:
baherus [9]2 years ago
7 0

Answer: Two valence electrons.

Explanation: Ionization energy is the energy required to remove electrons from valence shell of an atom in its gaseous state.

More closer are the electrons to the nucleus, more energy is required to remove them. Ionization energy increases as the effective nuclear charge increases.

First two ionization energy values are low and there isn't a much difference between them but the third ionization energy is very much high as compared to first and second ionization energy values. The fourth ionization energy value is also closer to third ionization energy value.

It indicates that first and second electrons are removed from same shell. Similarly, third and fourth electrons are also removed from same shell which is different than the shell from which first and second electrons are removed.

So, outer most shell that is valence shell has only two electrons. Third ionization energy has increased a lot as the third electron is removed from the inner shell and first two electrons are removed from valence shell.

Anni [7]2 years ago
4 0
Hi :) If your answers choices are what I'm thinking, the answer is two :)
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When a gas is cooled at constant pressure, what happens to its molecules and volume?
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J. Explain how an atom's valence electron configuration determines its place on the periodic table.​
Darya [45]

Answer:

Elements having same valence electrons are placed in <u>same group.</u>

Explanation:

First, let's start with some basic concepts of modern periodic table:

1. Modern Periodic table : It is the arrangement of element in the increasing order of their atomic numbers

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5 0
3 years ago
1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

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