Final answer:
The percentages given in the question are indicative of the amount of nobelium remaining after certain numbers of half-lives (55 seconds each for 254^102No): 25.0% corresponds to two half-lives, 50.0% to one, 75.0% to less than one, and 12.5% to three half-lives.
Step-by-step explanation:
The question asks about the percentage of 254^102No remaining after certain intervals, given its half-life. To calculate this, we can use the concept of half-lives which tells us that after one half-life, 50% of an isotope remains; after two half-lives, 25% remains; after three half-lives, 12.5% remains, and so on.
For the percentages provided:
- 25.0% would be after two half-lives (55 seconds x 2), so it takes 110 seconds to have 25% remaining.
- 50.0% is just one half-life, so it takes 55 seconds to have 50% remaining.
- 75.0% cannot be directly related to a whole number of half-lives, but it indicates that it's between the original amount and one half-life.
- 12.5% is after three half-lives (55 seconds x 3), taking 165 seconds to leave 12.5% remaining.
Note: For exact percentage calculations like 75.0%, more complex decay equations involving logarithms would be needed, but this is beyond the scope of the question as it asks only for the listed percentages which correspond to whole half-lives.