At the end of 2019, Blossom Company has accounts receivable of $756,900 and an allowance for doubtful accounts of $59,600. On January 24, 2020, the company learns that its receivable from Megan Gray is not collectible, and management authorizes a write-off of $5,800. On March 4, 2020, Blossom Company receives payment of $5,800 in full from Megan Gray. Prepare the journal entries to record this transaction. (Credit account titles are automatically indented when amount is entered. Do not indent manually.) Account Titles and Explanation Debit Credit (To reverse write-off) (To record collection from Gray)

Explanation:

Adjusting entries:      

S.no.Accounts title and explanation  Debit $       Credit $

a.         Accounts receivable           5800  

      Allowance for doubtful accounts                                   5800

       (for reversing the write off)    

b.  Cash account                          5800  

         Accounts receivable                                             5800

         (for cash collection from write off account)  

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