Determine how the following bonds are made using lewis dots structures

Answer :

1.)

Calcium has 2 valence electrons and chlorine has 7 valence electrons. For neutral molecule, 2 chlorine atoms share their valence electrons with 1 calcium atom. The bonds are formed between the calcium and chlorine atom.

2.)

Aluminium has 3 valence electrons and fluorine has 7 valence electrons. For neutral molecule, 3 fluorine atoms share their valence electrons with 1 aluminium atom. The bonds are formed between the aluminium and fluorine atom.

3.)

Hydrogen has 1 valence electron and chlorine has 7 valence electrons. For neutral molecule, 1 chlorine atom share their valence electrons with 1 hydrogen atom. The bonds are formed between the hydrogen and chlorine atom.

4.)

Hydrogen has 1 valence electron and sulfur has 6 valence electrons. For neutral molecule, 2 hydrogen atoms share their valence electrons with 1 sulfur atom. The bonds are formed between the hydrogen and sulfur atom.

5.)

Chlorine has 7 valence electrons. For neutral molecule, 1 chlorine atom share their valence electrons with 1 chlorine atom. The bonds are formed between chlorine and chlorine atom.

The lewis-dot structure image is shown below.

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