Given following electron configurations, 1s22s22p63s23p64s23d104p65s24d9 and 1s22s22p63s23p64s23d104p65s14d10.
part 1: name the element.
part 2: both electron configurations are correct, and the second one is what is found most often in nature. suggest a reason for why silver most often exists as an exception to the normal electron configuration rules.

Explanations:- Part 1: We could count the total number of electrons by looking at the electron configurations. Both of these electrons configurations have 47 electrons. If we look at the periodic table then 47 is the atomic number of silver. So, the name of the element is silver and its represented as Ag.

Part 2: As per the rule, Completely filled and half filled orbitals are more stable. First electron configuration has 9 electrons in 4d and we know that d is more stable if it has 5 electrons(half filled) or it has 10 electrons(full filled).

For stability reasons, one of the electron from 5s goes to 4d and for this reason the second electron configuration is found most often in nature for silver.

Few other examples are Cr and Cu.

take a better picture and ill do it

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