A 7.500-mg sample of chromium-55 is analyzed after 14.0 min. and found to contain 0.469 mg of chromium-55. what is the half-life of chromium-55?

ln(No/N) = kt

No = 7.500 mg
N = 0.469 mg
t = 14 min
solve for k.
Substitute k into the following:
k = 0.693/t1/2

tanx 4 the help:)

To find the half-life of chromium-55, we can make use of the formula for radioactive decay:

N(t) = N₀ * (1/2)^(t / t½)

Where:
N(t) is the amount of chromium-55 remaining at time t,
N₀ is the initial amount of chromium-55,
t is the time that has passed, and
t½ is the half-life of chromium-55.

In this case, we have the initial amount (N₀) as 7.500 mg, the time passed (t) as 14.0 min, and the final amount (N(t)) as 0.469 mg.

We can rearrange the equation to solve for half-life (t½):

t½ = t * (log(N₀) - log(N(t))) / log(2)

Substituting the given values:

t½ = 14.0 * (log(7.500) - log(0.469)) / log(2)

Using a scientific calculator, we can evaluate the expression to find the half-life of chromium-55.