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2 years ago

The electron configuration that belongs to the atom with the lowest first ionization energy is?

Chemistry

1 answer:

sasho [114]2 years ago

4 0

The electron configuration that belongs to the atom with the lowest first ionization energy is francium.

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

Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.

<h3>Ionization energy across period</h3>

Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.

<h3>Ionization energy along group</h3>

Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.

Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.

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How do you measure for acidity or alkalinity?

natita [175]

By measuring its PH.

7 0

3 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

HURRY I NEED TO GET OFF SOON PELASEE THIS IS ALSO FOR A 60 POONT GRADE ITS SO IMPORTANT WILL GIVE BRAIN LIEST

loris [4]

Answer:

C is the right answer

Explanation:

7 0

3 years ago

Beverly and Carl are in a race. Their graphs show the data.

Nikitich [7]

Given the data, the correct statement is

Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

<h3>How to determine the initial acceleration of Beverly</h3>

Initial velocity (u) = 0 m/s
Final velocity (v) = 15 m/s
Time (t) = 10 s
Initial acceleration (a₁) =?

a₁ = (v – u) / t

a₁ = (15 – 0) / 10

a₁ = 1.5 m/s²

<h3>How to determine the final acceleration of Beverly</h3>

Initial velocity (u) = 15 m/s
Final velocity (v) = 35 m/s
Time (t) = 50 - 10 = 40 s
Final acceleration (a₂) =?

a₂ = (v – u) / t

a₂ = (35 – 15) / 40

a₂ = 0.5 m/s²

<h3>How to determine the initial acceleration of Carl</h3>

Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 10 s
Initial acceleration (a₁) =?

a₁ = (v – u) / t

a₁ = (10 – 0) / 10

a₁ = 1 m/s²

<h3>How to determine the final acceleration of Carl</h3>

Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Time (t) = 50 - 10 = 40 s
Final acceleration (a₂) =?

a₂ = (v – u) / t

a₂ = (30 – 10) / 40

a₂ = 0.5 m/s²

SUMMARY

Initial acceleration of Beverly = 1.5 m/s²
Final acceleration of Beverly = 0.5 m/s²
Initial acceleration of Carl = 1 m/s²
Final acceleration of Carl = 0.5 m/s²

From the above calculations, we can see that Beverly's initial acceleration is higher than that of Carl's and their final acceleration is the same.

Therefore, the correct answer to the question is:

Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s

Complete question

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