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motikmotik
3 years ago
9

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions _____, and the anion becomes larg

er.
2. The reverse is true for the cation, which becomes ____ than the neutral atom.
Chemistry
2 answers:
Ray Of Light [21]3 years ago
3 0

Answer:

1) increases

2) smaller

Explanation:

Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.

For a cation, the converse is true and the cation is found to be smaller than the neutral atom.

PilotLPTM [1.2K]3 years ago
3 0

This question is incomplete, I got the complete one from google as below:

I−>I>I+

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.

2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.

Answer:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.

2.The reverse is true for the cation, which becomes smaller than the neutral atom.

Explanation:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.

2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.

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27/13Al+4/2He >>>> 30/15P+
Veseljchak [2.6K]

Answer

Explanation:

27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

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27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

5 0
3 years ago
Fill in the blanks to complete each statement about the heating of Earth's surface.
Aleonysh [2.5K]

Earth takes in thermal energy from the Sun in a process called thermal radiation.

Sunlight strikes Earth's surface at different angles. This angle is called the angle of insolation.

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

Thermal energy refers to the energy contained within a system that is responsible for its temperature.

Earth takes in thermal energy from the Sun in a process called thermal radiation.

Sunlight strikes Earth's surface at different angles. This angle is called the angle of insolation.

Learn more about thermal energy here:

brainly.com/question/11278589

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7 0
2 years ago
For most atoms, a stable configuration of electrons is attained when the atom __________. hints hint 1. (click to open) for most
sveticcg [70]

Correct answer: has a completely filled outermost shell

Atoms of the element with complete outermost shells are stable. So, in order to attain stability the atom either loses electrons or gains electrons to completely fill the outermost shell. The stable electronic configuration for the s and p-block elements is exhibited by the noble gases or the group 8 elements. All the unstable atoms try to attain the electronic configuration of the nearest noble gas with completely filled outermost shell.

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3 years ago
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ladessa [460]
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7 0
3 years ago
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
3 0
3 years ago
Read 2 more answers
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