Which structure diagram completes the table of physical structures of the main functional groups? A 6 column table with 1 row. Column 1 is labeled Alchol with entry a molecule R single bonded to O H. Column 2 is labeled Ether with entry a molecule R single bonded to O single bonded to R prime. Column 3 is labeled Ketone with entry blank. Column 4 is labeled Aldehyde with entry Molecule H single bonded to C, which is double bonded to O and single bonded to R. Column 5 is labeled Carboxylic Acid with entry molecule O H single bonded to C, which is double bonded to O and single bonded to R. Column 5 is labeled Ester with entry molecule R prime single bonded to O single bonded to C, which is double bonded to O and single bonded to R. A molecule H single bonded to C, which is double bonded to O and single bonded to R. A molecule R prime single bonded to C, which is double bonded to O and single bonded to R. A molecule O single bonded to C, which is double bonded to R prime and single bonded to R. A molecule R prime single bonded to C, which is single bonded to O and single bonded to R.

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

i think it's c. i'm not sure but i think so.

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year