Four atoms are arbitrarily labeled d, e, f, and g. their electronegativities are as follows: d = 3.8, e = 3.3, f = 2.8, and g = 1.3. if the atoms of these elements form the molecules de, dg, eg, and df, how would you arrange these molecules in order of increasing covalent bond character? (use the appropriate < , =, or > symbol to separate substances in the list.)

Thus the order of covalent character will be

DG < EG < DF < DE

Explanation:

The electronegativity decides the nature of bond formed between two atoms.

More the difference in electronegativity more the ionic character in the bond formed.

Let us check the electronegativity difference between the elements forming molecules

a) DE :

3.8-3.3 = 0.5

b) DG :

3.8-1.3 = 2.5

c) EG

3.3-1.3 = 2

d) DF

3.8-2.8 = 1.3

So the maximum ionic character will be in DG and minimum in DE

Thus the order of covalent character will be

DG < EG < DF < DE

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

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.