There are 3.29g of iodine -126 remaining in a smple originally containing 26.3 g of Iodine - 126 the half life of Iodine - 126 is 13 days. How old is the sample?

At study hall test is after lunch and cant figure out htese 2 on my review packet

WEll, follow the formula...

amountremaining=Original*e^(-.692t/13)

3.29/26/3= e^(-.692t/13)
take the ln of each side
-.207 = -.692t/13

solve for t in days.

To determine the age of the sample, we can use the concept of half-life. The half-life is the time it takes for half of the radioactive material to decay.

Given:
Initial amount of iodine-126 (I-126) in the sample = 26.3 g
Remaining amount of I-126 in the sample = 3.29 g
Half-life of I-126 = 13 days

To solve for the age of the sample, we can use the following formula:

Remaining amount = Initial amount * (1/2)^(time/half-life)

In this case, we know the remaining amount and the initial amount, and we need to solve for time.

Let's plug in the values and solve for time:

3.29 g = 26.3 g * (1/2)^(time/13)

To isolate the exponent, we can take the logarithm (base 1/2) of both sides:

log base (1/2) of (3.29) = log base (1/2) of (26.3 * (1/2)^(time/13))

Using the property of logarithm which states that log base b of (a * b^n) = log base b of a + n:

log base (1/2) of (3.29) = log base (1/2) of (26.3) + (time/13)

To simplify further, we can use the property that log base b of b = 1:

log base (1/2) of (3.29) = log base (1/2) of (26.3) + (time/13)
log base (1/2) of (3.29) - log base (1/2) of (26.3) = time/13

Using a calculator, we can evaluate the logarithms:

-2.999 = -4.7048 + (time/13)
-2.999 + 4.7048 = time/13
1.7058 = time/13

To solve for time, we can multiply both sides by 13:

1.7058 * 13 = time
22.1794 = time

Therefore, the age of the sample is approximately 22.18 days.