The half life of sulfur-38 is 2.87 hours. after 8.61 hours, what would the percent of the original radiation is left?

12.5% remaining after 8.61 hrs or 3 half-lives.

Explanation:

Method I: 1st Order Decay Equation

Given t(1/2) = 2.87 hrs => k = 0.693/t(1/2) = (0.693/2.87) hr⁻¹ = 0.242 hr⁻¹

C(final) = C(initial)e^-k·t => ln(C/C₀) = -k·t => lnC - lnC₀ = - k·t => lnC = lnC₀ - k·t

lnC = ln(100%) - [(0.242 hr⁻¹)(8.63 hr)] = (4.605 - 2.09)% = 2.517%

lnC = 2.517% => C = (e²°⁵¹⁷)% = 12.5% remaining

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Method 2 - Decay Trend Graph => see attachment

Note: All radio decay trends follow 1st order kinetics. A tracing of the decay trend for a 1st order process with half-life of 2.87 hrs shows S-38 decays over 3 half-lives (= 8.61 hrs). Therefore ...

after 1st half-life => 50% remains

after 2nd half-life => 25% remains

after 3rd half-life => 12.5% remains

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.