To find the amount of the rare isotope remaining after 9 seconds, we can use the formula provided:
\[ y = 5 \times (2.7)^{-0.12t} \]
Here, \( y \) is the remaining amount of the isotope, \( 5 \) is the initial amount in grams, and \( t \) is the time in seconds. We need to substitute \( t = 9 \) into the formula.
Step-by-Step Calculation:
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Substituting \( t = 9 \): \[ y = 5 \times (2.7)^{-0.12 \times 9} \]
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Calculating the exponent: \[ -0.12 \times 9 = -1.08 \]
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Calculating \( (2.7)^{-1.08} \): \[ (2.7)^{-1.08} \approx \frac{1}{(2.7)^{1.08}} \approx \frac{1}{3.007} \approx 0.332 \]
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Calculating \( y \): \[ y = 5 \times 0.332 \approx 1.66 \]
So, approximately 1.66 grams of the isotope remains 9 seconds after 5 grams of the isotope is created.
Final Answer
There would be 1.66 grams of the isotope remaining after 9 seconds.