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zlopas [31]
3 years ago
11

Rank these elements according to FIRST ionization energy, from highest to lowest:

Chemistry
2 answers:
Rzqust [24]3 years ago
3 0
The trend of ionization energy in the periodic table is decreasing from right to left and from top to bottom. In this case, we are given with elements <span>K, Ca, Ge, Se, Br, Kr and see the periodic table to check the trend. The answer from highest to lowest Kr, Br, Se, Ge, Ca, and K</span>
Zina [86]3 years ago
3 0

The order of the first ionization energy of the given elements from highest to lowest is \boxed{{\mathbf{Kr > Br > Se > Ge > Ca > K}}}

Further Explanation:

The energy that is needed to remove the most loosely bound valence electrons from the isolated neutral atom of the gas is known as the ionization energy. It is denoted by IE. The value of IE is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.

Ionization energy is further represented as first ionization, second ionization and so on. When the first electron is removed from a neutral, isolated gaseous atom, the energy needed for the purpose is known as the first ionization energy, written as {\text{I}}{{\text{E}}_1} . Similarly, when the second electron is removed from the positively charged species (cation), the ionization energy is called the second ionization energy \left( {{\text{I}}{{\text{E}}_2}} \right) and so on.

Potassium, calcium, germanium, selenium, bromine, and krypton belong to the same period of the periodic table.

Along a period, the atomic number increases but the number of shells remains the same while moving from left to right.

The number of protons, as well as the electrons, increases with the increase in the atomic number but the extra electrons are added in the same shell. Due to the high positive nuclear charge, the effective nuclear charge also increases. This results in an increase in the first ionization energy. So krypton will have the highest first ionization energy, followed by bromine, selenium, germanium, calcium, and potassium will have the lowest ionization energy.

Learn more:  

1. Why is {\text{6 - decene}}  not possible: brainly.com/question/2095052

2. The gold foil experiment: brainly.com/question/1859083

Answer details:  

Grade: Senior School  

Subject: Chemistry  

Chapter: Periodic classification of elements

Keywords: first ionization energy , K, Ca, Ge, Se, Br, Kr, protons, electrons, atomic number, isolated, neutral, atom, nuclear charge.

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3 years ago
An unknown metal displaces Ali from solution but does not displace sodium using the activity series determine the unknown metal
MAXImum [283]

Answer:

The reactivity of metal is determined by the reactivity series. ... The metal which easily displaced aluminium will lie above in the series but that same element cannot displace sodium, so it will lie below in the series. Hence, from the series, we conclude that the unknown metal could be calcium or magnesium.

Explanation:

Hope this helps! :)

8 0
2 years ago
What is the conjugate acid in the following equation:
yanalaym [24]

Answer:

HNO₂

Explanation:

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The conjugate acid is what's formed when the base has accepted a proton.

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NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻

base     acid     conj.      conj.

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5 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
Abbreviation for mole
Sedbober [7]

Answer:

Maybe mol

Explanation:

6 0
3 years ago
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