# Question #a4462

##### 1 Answer

#### Explanation:

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(A_t = A_0 * (1/2)^n)))#

Here

#A_t# is the amount of the radioactive substance that remainsundecayedafter a time period#t# #A_0# is theinitial amountof said substance#n# is thenumber of half-livesthat pass in the time period#t#

Your first goal here is to figure out the value of

#A_t/A_0 = (1/2)^n#

This will be equivalent to

#ln(A_t/A_0) = ln[(1/2)^n]#

#ln(A_t/A_0) = n * ln(1/2)#

Since

#ln(1/2) = ln(1) - ln(2) = - ln(2)#

you can say that

#n = -ln(A_t/A_0)/ln(2)#

Plug in your values to find

#n = - ln((0.100 color(red)(cancel(color(black)("g"))))/(0.750color(red)(cancel(color(black)("g")))))/ln(2) = 2.907#

Now, you know that **half-lives** must pass in order for the sample to be reduced from **days**.

The number of half-lives that pass in a given period of time can be written as

#color(blue)(ul(color(black)(n = "total time that passes"/"half-life" = t/t_"1/2")))#

The total time

#n = t/t_"1/2" implies t = n * t_"1/2"#

In your case, you will have

#color(darkgreen)(ul(color(black)(t = 2.907 * "3.823 days" = "11.1 days")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the initial and final mass of the sample.