Arrange the following substances in order of decreasing magnitude of lattice energy.
2 answers:
The following substances in order of decreasing magnitude of lattice energy are as follows: SrO > BaO > NaF > KCl
<h3>Explanation: </h3>Ionic compounds are more stable because their have an electrostatic force between the two opposite ions. After the formation of ions, they combine together to form the ionic compound. The lattice energy or lattice enthalpy is the energy released in this process. It means the energy released when a cation and a anion combine together to form one mole of an ionic compound that known as the lattice energy or lattice enthalpy.
To calculation of lattice energy we can done by using Hess’s law (or called Born Haber cycle). Hess' law of constant or the heat summation, also known as Hess' law is a relationship in physical chemistry that named after Germain Hess. Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
The formula is
The compounds BaO and SrO, have ions' charges ( + 2 ) and ( - 2 ) respectively will have greater lattice energies that the compounds NaF and KCl that have charges ( + 1 ) and ( - 1 ) respectively
( BaO and SrO ) > ( NaF and KCl )
In NaF, the ions are smaller than the KCl ions so it has the greater lattice energy. Sr is smaller than Ba therefore SrO has a greater lattice energy than BaO.
Therefore SrO > BaO > NaF > KCl
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Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, =
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density,
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s