Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

How many days does it take 16.Og of Gold-198 to decay to 1.0g? (each half-
life = 2.69 days) *
O 10 days
0 2.69 days
O 10.76 days
16 days
PLZ HELP NOW

Sagot :

drpelezo

Answer:

10.8 days (3 sig.figs.)

Explanation:

All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt

which is solved for time of decay (t) => t = ln(A/A₀) / -k

A = final weight = 1.0 gram

A₀ = initial weight = 16.0 grams

k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹

t = ln(1/16) / -0.258da⁻¹ = (-2.77/-0.258) days = 10.74646792 days (calculator)

≅ 10 days (1 sig. fig. based on given 1 gram mass)

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.