A common way to describe the bonding energy curve for secondary bonding is the '6-12' potential, which states that

Complete question is;

A common way to describe the bonding energy curve for secondary bonding is the "6-12" potential. which states the bonding energy, E = - A/r^(6) + B/r^(12).

Write the expression that best represents the equilibrium bond length.

Equilibrium bond length is;

r = (2B/A)^(1/6)

Explanation:

We are given bonding energy as;

E = - A/r^(6) + B/r^(12)

Now let's find the derivative of this bonding energy with respect to r and equate to zero to find the bonding length(r).

Thus;

dE/dr = 6A/r^(7) - 12B/r^(13)

Equating to zero gives;

6A/r^(7) - 12B/r^(13) = 0

6A/r^(7) = 12B/r^(13)

Divide both sides by 6 to give;

A/r^(7) = 2B/r^(13)

2B/A = (r^(13))/(r^(7))

2B/A = r^(13 - 7)

2B/A = r^(6)

Thus, r = (2B/A)^(1/6)

chemical energy is stored in the chemical bonds of atoms and molecules. it can only be seen when it is released in a chemical reaction. when chemical energy is released, the substance from which the energy came is often changed into an entirely different substance.

this chemical reaction releases energy in the form of light and heat. chemical energy is the potential energy stored in the arrangement of atoms within molecules. breaking chemical bonds requires energy, while forming new chemical bonds releases energy.

the atp molecule can store energy in the form of a high energy phosphate bond joining the terminal phosphate group to the rest of the molecule. in this form,energy can be stored at one location, then moved from one part of the cell to another, where it can be released to drive other biochemical reactions.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years