1. Iodine-131 is used to find leaks in water pipes it has a half life of 8.14 days. If 36 days have passed and there is now 2.34g left.

What is the initial amount?

2. Sr-85 is used in bone scans and has a half life of 64.9 days. If there is 23 percent of the sample remaining how much time has passed?

3. A bacteria increases fivefold every 6 hours how much bacteria will there be if the initial amount is 4000 and 48 hours have passed?

1. 2.34 = x (1/2)^(36 / 8.14) ... log(2.34) - [(36 / 8.14) log(1/2)] = log(x)

2. .23 = (1/2)^(t / 64.9) ... log(.23) = (t / 64.9) log(1/2)

3. x = 4000 * 5^(48 / 6)

To solve these questions, we can use the concept of radioactive decay for the first two questions and exponential growth for the third question. I'll explain the process step by step for each question:

1. To find the initial amount of Iodine-131, we can use the formula for radioactive decay.

Half-life (T1/2) = 8.14 days
Remaining amount (N) = 2.34g
Time passed (t) = 36 days

The formula for radioactive decay is:
N = N0 * (1/2)^(t / T1/2)

Where N0 is the initial amount.

Rearranging the formula gives us:
N0 = N / (1/2)^(t / T1/2)

Substituting the given values, we get:
N0 = 2.34g / (1/2)^(36 / 8.14)

Calculating the value results in an initial amount (N0) of approximately 64.08 grams.

So, the initial amount of Iodine-131 is approximately 64.08 grams.

2. To find the time passed for Sr-85 bone scan, we can use the formula for radioactive decay again.

Half-life (T1/2) = 64.9 days
Remaining amount (N) = 23% (0.23) of the sample
Initial amount (N0) = 100% (1) of the sample

The formula for radioactive decay remains the same:
N = N0 * (1/2)^(t / T1/2)

We need to find the time passed (t), so we rearrange the formula:
t = T1/2 * log2(N / N0)

Substituting the given values, we get:
t = 64.9 * log2(0.23 / 1)

Calculating the expression, the time passed (t) is approximately 33.48 days.

Therefore, approximately 33.48 days have passed for Sr-85 bone scan.

3. To find the amount of bacteria after a given time with exponential growth, we can use the formula:

N = N0 * (growth rate)^(t / time interval)

Initial amount (N0) = 4000 bacteria
Growth rate = fivefold increase = 5
Time interval = 6 hours
Time passed (t) = 48 hours

Using the formula, we obtain:
N = 4000 * 5^(48 / 6)

Simplifying the expression, we get:
N = 4000 * 5^8

Evaluating the expression yields the result:
N ≈ 51,200,000 bacteria

Therefore, after a time of 48 hours, there will be approximately 51,200,000 bacteria.