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Digiron [165]
3 years ago
12

Which element in each of the following sets would you expect to have the lowest IE₃?

Chemistry
2 answers:
AysviL [449]3 years ago
3 0

Answer:

(a) AL

(b) Sc

(c)Al

Explanation:

Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.

The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

<h3>(a)Na, Mg, Al</h3>

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(b) K, Ca, Sc</h3>

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(c) Li, Al, B</h3>

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

eimsori [14]3 years ago
3 0

Answer:

A. Al

B. Sc

C. Al

Explanation:

The third ionisation energy is the energy required to an extra electron from a +2 ion or the energy required to remove the third electron from an element.

Lithium - 1s2 2s1

Sodium - 1s2 2s2 2p6 3s1

Magnesium - 1s2 2s2 2p6 3s2

Aluminium - 1s2 2s2 2p6 3s2 3p1

Potassium - 1s2 2s2 2p6 3s2 3p6 4s1

Calcium- 1s2 2s2 2p6 3s2 3p6 4s2

Boron - 1s2 2s2 2p1

Scandium - 1s2 2s2 2p6 3s2 3p6 3d1 4s2

Removing 2 electrons,

Li2+- 1s1

Na2+ - 1s2 2s2 2p5

Mg2+ - 1s2 2s2 2p6

Al2+ - 1s2 2s2 2p6 3s1

K2+ - 1s2 2s2 2p6 3s2 3p5

Ca2+ - 1s2 2s2 2p6 3s2 3p4

Boron - 1s2 2s1

Scandium - 1s2 2s2 2p6 3s2 3p6 4s1

So comparing,

A. Na, Mg, Al

The third electron is lost from a p- orbital and the energy level of p- is less than s- orbital but 3s is way less than the 2p so the lowest third ionisation energy is Al

B. K, Ca, Sc

The third electrons are lost from the 3p orbital in K and Ca but in 4s in Sc and if you remember, 4s has a lesser energy level than 3p orbital. So, Sc has the lowest third ionisation energy.

C. Li, Al, B

Al has the lowest third ionisation energy because Li loses its from 1s which is closest to the nucleus and B from 2s which is also close to the nucleus.

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xenn [34]

Answer:

Mass of the salt:  105.6g of KCl.

Mass water: 958.9g of water.

Molality: 1.478m.

Explanation:

<em>Mass of the salt:</em>

In 1L, there are 1.417 moles. In grams:

1.417 moles KCl * (74.54g / mol) = 105.6g of KCl

<em>Mass of the water:</em>

We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:

1L * (1064.5g / L) = 1064.5g - Mass solution.

Mass water = 1064.5g - 105.6g = 958.9g of water

<em>Molality:</em>

Moles KCl = 1.417 moles KCl.

kg Water = 958.9g = 0.9589kg.

Molality = 1.417mol / 0.9589kg = 1.478m

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Answer:

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Explanation:

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If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, what is the heat flow abs
Arisa [49]

If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, then heat flow absorbed (in kj) by the system is -147.026kJ.

<h3>What is entropy? </h3>

The entropy of particle is defined as how random it move. It shows the randomness of the system or may be disorders of the system. It is used to measure the unavailable energy for performing useful work.

Unit of entropy = J/K

<h3>Formula:</h3>

∆s = ∆Q/T

where,

∆s = change in entropy of the surrounding = -326J/K

∆Q = heat absorbed from surrounding

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∆Q = ∆s × T

∆Q = -326 × 451

∆Q = 147,026 J

∆Q = 147.026 kJ

Thus we find that the heat absorbed by the system is 147.026 kJ.

learn more about entropy:

brainly.com/question/14131507

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