Question

PLEASE HELP

Answer 1

Answer:

$A\approx2.255 \text{ grams}$

Step-by-step explanation:

We will use the standard formula for half-life:  

$\displaystyle A=A_0(a)^\frac{t}{d}$

Where A₀ is the initial amount, a is the half-life, d is the time for one half-life, and t is the number of days.

So, we will substitute 1/2 for a and 24.3 for d. This yields:

$\displaystyle A=A_0(\frac{1}{2})^\frac{t}{24.3}$

We have a 3 gram sample. Therefore, our initial amount A₀ is 3:

$\displaystyle A=(3)(\frac{1}{2})^\frac{t}{24.3}$

And we want to find the remaining sample after 10 days. So, t=10. Substitute:

$\displaystyle A=(3)(\frac{1}{2})^\frac{10}{24.3}$

Use a calculator. So:

$A\approx2.255 \text{ grams}$

Therefore, for a 3 gram sample of Phosphorous-32 with a half-life of 24.3 days, after 10 days, there will be approximately 2.255 grams remaining.