Question

Nobelium-254 has a half life of 3 seconds. How many half-lives would have to occur to
reduce a 15g sample down to 0.09375g?

Answer 1

the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g is 7.3

To calculate the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g, we use the formula below.

Formula:

• 2ⁿ = R/R'........... Equation 1

Where:

• n = Number of half-lives that have occured
• R = Original mass of Nobelium-254
• R' = Mass of Nobelium-254 after decay

From the question,

Given:

• R = 15 g
• R' = 0.09375 g

Substitute these values into equation 1

• 2ⁿ = 15/0.09375
• 2ⁿ = 160
• n = log160/log2
• n = 7.3 half-lives

Hence, the number of half-lives that would have occur to reduce a 15 g sample of Nobelium-254 down to 0.09375 g is 7.3

Learn more about half-lives here: https://brainly.com/question/25750315