If $18 \mathrm{~g}$ of a radioactive substance are present initially and $2 \mathrm{yr}$ later only $9.0 \mathrm{~g}$ remain, how much of the substance, to the nearest tenth of a gram, will be present after $5 \mathrm{yr}$ ?

After $5 \mathrm{yr}$, there will be $\square \mathrm{g}$ of the radioactive substance
(Do not round untl the final answer. Then round to the nearest tenth as needed.)

Step 1 ：Given that the half-life of the substance is 2 years, which means that in 2 years, half of the substance has decayed.

Step 2 ：The formula for radioactive decay is \(N = N0 * (1/2)^{t/T}\), where \(N\) is the final amount of the substance, \(N0\) is the initial amount of the substance, \(t\) is the time elapsed, and \(T\) is the half-life of the substance.

Step 3 ：We are asked to find the amount of the substance after 5 years, so we substitute the given values into the formula: \(N = 18g * (1/2)^{5/2}\)

Step 4 ：Solving the equation gives us \(N = 18g * (1/2)^{2.5} = 18g * 0.17677669529663688110021109052621\)

Step 5 ：Rounding to the nearest tenth of a gram, we get \(N = 3.2 g\)

Step 6 ：So, after 5 years, there will be approximately \(\boxed{3.2 g}\) of the radioactive substance remaining.