Chemistry College

1. Choose the atom with the larger first ionization energy.

Select one:

a. Titanium
b. Manganese


2. Choose the atom with the larger first ionization energy.

Select one:

a. Silicon
b. Tin

Answers

Answer 1

The atom with the larger first ionization energy is Titanium. Option a.

The atom with the larger first ionization energy is Tin. Option b.

Ionization and ionization energy

Ionization is the process of removing one or more electrons from an atom or molecule, resulting in the formation of an ion. This can be achieved through a variety of methods, such as exposure to high-energy radiation or contact with other charged particles.

Ionization energy is the amount of energy required to remove an electron from a neutral atom or molecule, resulting in the formation of a positively charged ion. This energy is typically measured in electron volts (eV) or kilojoules per mole (kJ/mol), and varies depending on the identity of the atom or molecule and the electronic configuration of its valence shell. Ionization energy is an important property of atoms and molecules, as it can provide insight into their reactivity and chemical behavior.

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Related Questions

what does 2NaOH equal

Answers

2NaOH represents the chemical formula for sodium hydroxide.

match each substance correctly to the principal type(s) of intermolecular force(s) present, other than covalent bonding.

Answers

Substance       intermolecular force

CH2OH   --->   Hydrogen bonding

CH3F       -->    Dipole-dipole forces

C3H8       -->    Dispersion forces

CaCL2      -->   Ionic bonding

The intermolecular force present in CH2OH is hydrogen bonding. The intermolecular force present in CH3F is  Dipole-dipole forces. Ionic bonding is defined as a type of chemical bonding that involves the electrostatic attraction between oppositely charged ions or between two atoms with sharply different electronegativities. It is the primary interaction occurring in ionic compounds. Hydrogen bonding results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.

Dipole-dipole forces are defined as a attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.  Dispersion force is defined as a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.

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The complete question is,

Match each substance correctly to the principal type(s) of intermolecular force(s) present, other than covalent bonding.

CH2OH          Ionic bonding

CH3F              Hydrogen bonding

C3H8              Dispersion forces

CaCL2            Dipole-dipole forces

Using C2H4 + 3 O2 -> 2 CO2 + 2 H2O. If 20 moles of fuel are combusted in the above equation, how many moles of CO2 are produced?

Answers

According to the balanced equation:

1 mole of C2H4 produces 2 moles of CO2

Therefore, to determine the number of moles of CO2 produced when 20 moles of C2H4 are combusted:

20 moles C2H4 x (2 moles CO2/1 mole C2H4) = 40 moles CO2

Therefore, 40 moles of CO2 are produced.

Identify each of the following orbitals, and determine the n and quantum numbers. Explain your answers.

Answers

(a) one radial node the Number of radial nodes = n - l - 1

And number of angular nodes = l

n = 3 and l = 1

Orbital is 3p.

(b) It has zero angular node hence s-orbital and there is 1 radial node . 1 = n - 0 - 1 ; n = 2 and l = 0

The orbital is 2s.

(c) the shape of the orbital is that of dz². There is two angular nodes and there is no radial node.

n = 3 and l = 2

Hence the orbital is 3dz².

What is radial node?

In atomic physics, a radial node is a point in space where the probability density of finding an electron in an atom is zero. It is a type of nodal plane that occurs in atomic orbitals, which are regions of space where electrons are most likely to be found.

Radial nodes occur in the radial distribution function of an atomic orbital, which describes the probability density of finding an electron at a given distance from the nucleus. The number of radial nodes in an atomic orbital is equal to n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Radial nodes represent regions of space where the radial wave function of the electron changes sign.

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When a mineral breaks along a weekly bonded plane it is called

Answers

Answer: Cleavage


Explanation:

When a mineral breaks along a weekly bonded plane it is called cleavage

draw a mechanism showing the penta-coordinate intermediate and the formation of the phosphorylated intermediate (which is an anhydride)

Answers

The formation of the phosphorylated intermediate (an anhydride) involves the formation of a penta-coordinate intermediate. This intermediate is formed by a nucleophilic attack of the sulfur on the phosphorus atom of the phosphate group.

In this mechanism, the sulfur atom of the sulfate group nucleophilically attacks the phosphorus atom of the phosphate group to form a penta-coordinate intermediate. This intermediate then rearranges to form a phosphorylated intermediate, which is an anhydride.

Mechanism showing the penta-coordinate intermediate and the formation of the phosphorylated intermediate are given as follows:

Step 1: Alkyl Phosphate Formation : The first step of the mechanism includes the formation of an alkyl phosphate. A proton is abstracted by OH− from the phosphate group to create the alkyl phosphate. The base catalyzes this step.

Step 2: Binding to Mg2+After the alkyl phosphate is created, the magnesium ion binds to it.

Step 3: Nucleophilic attack: Following that, the nucleophilic attack happens, with the nucleophile being the water molecule. It is coordinated with the magnesium ion. It occurs at phosphorus, causing it to be phosphorylated. It results in the creation of a pentacoordinate intermediate.

Step 4: Release of Orthophosphate: Orthophosphate is released as a result of the reaction between pentacoordinate intermediate and water. It results in the creation of a diester intermediate.

Step 5: Subsequent Hydrolysis: In the final step, the intermediate diester is hydrolyzed to form orthophosphate and the final product. This is accomplished via nucleophilic substitution.

The end result is a free phosphate group that is bound to the alcohol's oxygen. A phosphate anhydride is formed in the process.

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