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Papessa [141]
2 years ago
11

Select the statement(s) that explain(s) the relationship between the arrangement of elements by size and first ionization energy

.a. The two arrangements are the same because IE of elements increases from left to right and the radii of elements decreases from left to right.b. The first set of paired electrons in a p subshell experience a large amount of electron-electron repulsion, making the fourth electron in a p subshell easy to remove.c. A full s subshell is able to shield a newly filled p subshell from the nucleus, making the first electron in a p subshell easy to remove.d. The third unpaired electron in a p subshell causes a lot of electron-electron repulsion among the orbitals, making the third electron in a p subshell easy to remove.e. A p subshell with only one unpaired electron is extremely stable, making the fifth electron in a p subshell easy to remove.
Chemistry
2 answers:
kow [346]2 years ago
8 0

c. A full s subshell is able to shield a newly filled p subshell from the nucleus, making the first electron in a p subshell easy to remove.

Explanation:

From the given options, a full s-sublevel is able to shield a newly filled p-subshell from the nucleus thereby making the first electron in a p-subshell easy to remove is correct.

What is ionization energy?

Ionization energy is a measure of the readiness of an atom to lose an electron.

First ionization energy is the energy required to remove the most loosely held electron in the gas phase.

The size of an atom/element depends on the number of electrons it contains. The more the electrons, the larger its size.

  • The larger an atom becomes the lesser the ionization energy needed to remove the first electron from its outermost shell.

Electron - electron repulsion occurs when two electrons in the same sub-level repels one another.

Shielding effect is the ability of the inner electrons to protect the outer electrons from the pull of the nuclear charge.

In option C, a s-subshell has a greater shielding effect than the p,d and f sub-shell in that order.

A newly introduced electron in the p-sublevel will be loosely held and easier to remove.

Learn more:

First ionization energy brainly.com/question/2153804

#learnwithBrainly

quester [9]2 years ago
5 0

c. A full s subshell is ready to shield a newly filled p subshell from the nucleus, making the primary electron in an exceedingly p subshell easy to get rid of.

<h2>Further explanation </h2>

From the given options, a full s-sublevel {is able|is in an exceedingly position|is ready} to shield a newly filled p-subshell from the nucleus thereby making the primary electron in a p-subshell easy to get rid of is correct.

<h3>What is ionization energy? </h3>

Ionization energy may be a measure of the readiness of an atom to lose an electron.  First ionization energy is that the energy required to get rid of the foremost loosely held electron within the gas phase.  The size of an atom/element depends on the number of electrons it contains. The more the electrons, the larger its size.  The larger an atom becomes the lesser the ionization energy needed to get rid of the primary electron from its outermost shell.  Electron-electron repulsion occurs when two electrons within the same sub-level repel each other.  The shielding effect is the ability of the inner electrons to shield the outer electrons from the pull of the nuclear charge.

In option C, an s-subshell contains a greater shielding effect than the p,d, and f sub-shell therein order. A newly introduced electron within the p-sublevel is going to be loosely held and easier to get rid of.

Learn More

ionization energy  brainly.com/question/13811443

Electron-electron brainly.com/question/13811443

Details

Class: College

Subject: Chemistry

Keywords: ionization, energy, electron

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Answer:

8.9 KJ

Explanation:

Given data:

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Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

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Answer:

Given:

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New pressure: approximately 3.4\times 10\; \rm atm (34\; \rm atm.) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

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By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, V_0 = 45\; \rm m^{3} while V_1 = 5.0\; \rm m^{3}.

Let P_0 denote the pressure of this gas before the volume change (P_0 = 3.65\; \rm atm.) Let P_1 denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between P_1\! and P_0\!:

\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0.

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\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm.

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K.

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Let P_1 denote the pressure of this gas before this temperature change (P_1 = 32.85\; \rm atm.) Let P_2 denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at V_2 = V_1 = 5.0\; \rm m^{3}.

Apply Amonton's Law to find the ratio between P_2 and P_1:

\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}.

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Answers:

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