Assuming that "half" refers to "half-life", 360/60 = 6 so fraction left = 1/26 = 1/64
3/8
6 hours = 2 half lives, thus 25 % would remain. 0.25 x 2 mg = 0.5 mg.Done another way...fraction remaining = 0.5^n where n = number of half lives = 6hr/3hr = 2fraction remaining = 0.5^2 = 0.250.25 x 2 mg = 0.5 mg
1/2 (50%) of them.
0.62 and a half as a fraction is 5/8.
If a radioisotope undergoes six half-lives, only (1/64) or (0.015625) of the original radioisotope remains, because half of the remaining material decays at each half-life.
To solve this problem, you can use the formula for radioactive decay: N(t) = N0 * (1/2)^(t/T), where N(t) is the amount of the isotope remaining at time t, N0 is the initial amount, t is the elapsed time, and T is the half-life. Plugging in the values: 1.50 g = 12.0 g * (1/2)^(t/8.07), solve for t to find the time it takes for the iodine-131 to decay to 1.50 g.
After 5 half-lives, 3.125% (or 1/2^5) of a radioactive sample remains. Each half-life reduces the sample by half, so after 5 half-lives, there is only a small fraction of the original sample remaining.
After three half-lives, 12.5% of the radioactive isotope is remaining. This is because each half-life reduces the amount of radioactive material by half.
After three half-lives, only 1/8 (or 12.5%) of the original radioactive sample remains. This is because each half-life reduces the amount of radioactive material by half, so after three half-lives, you would have (1/2) * (1/2) * (1/2) = 1/8 of the original sample remaining.
After 60 years, there will be 2.5 grams of hydrogen-3 remaining from the original 80.0 grams sample. This is calculated by dividing the time elapsed by the half-life multiple times (60 years ÷ 12 years = 5 half-lives, 1/2^5 = 1/32). Multiplying 80.0 grams by 1/32 gives 2.5 grams remaining.
Assuming that "half" refers to "half-life", 360/60 = 6 so fraction left = 1/26 = 1/64
After two half lives, 25% of the original carbon-14 would remain. This is because half of the remaining carbon-14 decays during each half life, leaving you with 50% after one half life and 25% after two half lives.
After 32 days, approximately 5 milligrams of the 80-milligram sample of Iodine-131 would be left. Iodine-131 has a half-life of about 8 days, so after each 8-day period, half of the remaining sample will decay.
91.16% of the daughter product has formed after 3.5 half lives.
To calculate the amount of thorium remaining after 2 half-lives, you use the formula: amount = initial amount * (1/2)^n, where n is the number of half-lives. If we assume the initial amount is 1 gram, after 2 half-lives, there would be 0.25 grams of thorium remaining.
An eighth remains.