Question #e6091

1 Answer
Douglas K.
Jun 5, 2017

#~ 156.6"g"#

Explanation:

The Half-life of #"_6^14C#

Is: #t_"half-life" = 5730"years"#

Using the formula:

#Q(t) = Q(0)(1/2)^(t/t_"half-life")#

Solve for Q(0):

#Q(0) = Q(t)(1/2)^(-t/t_"half-life")#

Substitute, #Q(t) = 2.5"g"#, #t = 34200"years"# and #t_"half-life" = 5730"years"#:

#Q(0) = 2.5"g"(1/2)^((-34200"years")/(5730"years"))#

#Q(0) ~~ 156.6"g"#

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