Question

reacting with water in an acidic solution at a particular temperature, compound a decomposes into compounds b and c according to the law of uninhibited decay. an initial amount of 0.60 m of compound a decomposes to 0.57 m in 30 minutes. how much of compound a will remain after 3 hours? how long will it take until 0.10 m of compound a remains?
after 3 hours, the amount of compound a remaining will be .44 m.
(do not round until the final answer. then round to the nearest hundredth as needed.)
there will be 0.10 m of compound a remaining in \\(\square\\) minutes.
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Find the decay constant \(k\)

The law of uninhibited decay is \(A = A_0e^{-kt}\). Given \(A_0=0.60\), \(A = 0.57\) when \(t = 30\) minutes.
Substitute into the formula: \(0.57=0.60e^{-k\times30}\).
Divide both sides by \(0.60\): \(\frac{0.57}{0.60}=e^{- 30k}\).
Simplify \(\frac{0.57}{0.60}=0.95\), so \(0.95 = e^{-30k}\).
Take the natural logarithm of both sides: \(\ln(0.95)=\ln(e^{-30k})\).
Since \(\ln(e^{x})=x\), we have \(\ln(0.95)=-30k\).
Solve for \(k\): \(k=-\frac{\ln(0.95)}{30}\approx-\frac{- 0.051293}{30}\approx0.00171\).

Step2: Find the amount after \(t = 180\) minutes

Now, find \(A\) when \(t = 180\) minutes (\(3\) hours), \(A_0 = 0.60\), and \(k\approx0.00171\).
Use the formula \(A = A_0e^{-kt}\), so \(A=0.60e^{-0.00171\times180}\).
Calculate \(-0.00171\times180=-0.3078\).
Then \(A = 0.60e^{-0.3078}\).
Since \(e^{-0.3078}\approx0.735\), \(A=0.60\times0.735 = 0.441\approx0.44\) M.

Step3: Find the time \(t\) when \(A = 0.10\)

Use the formula \(A = A_0e^{-kt}\), with \(A = 0.10\), \(A_0 = 0.60\), and \(k\approx0.00171\).
\(0.10=0.60e^{-0.00171t}\).
Divide both sides by \(0.60\): \(\frac{0.10}{0.60}=e^{-0.00171t}\).
Simplify \(\frac{0.10}{0.60}=\frac{1}{6}\approx0.1667\), so \(0.1667 = e^{-0.00171t}\).
Take the natural logarithm of both sides: \(\ln(0.1667)=\ln(e^{-0.00171t})\).
Since \(\ln(e^{x})=x\), we have \(\ln(0.1667)=-0.00171t\).
Solve for \(t\): \(t=-\frac{\ln(0.1667)}{0.00171}\).
Since \(\ln(0.1667)\approx - 1.7918\), \(t=\frac{1.7918}{0.00171}\approx1047.8\) minutes.

Answer:

After \(3\) hours, the amount of compound \(A\) remaining will be \(0.44\) M. It will take approximately \(1047.8\) minutes until \(0.10\) M of compound \(A\) remains.