please help me with this chemistry problem.

Half-Life

A radium-226 sample initially contains 0.112 mol. How much radium-226 is left in the sample after 6400 years? The half-life of radium-226 is 1600 years.

6400 years is 4 half-lives. The amount remaining is

0.112 mol * (1/2)^4
= 0.112/16 = 0.007 mol

0.014

To solve this problem, we can use the formula for exponential decay:

N = (N0) * (1/2)^(t / t1/2)

Where:
- N is the final amount of radium-226
- N0 is the initial amount of radium-226
- t is the time elapsed
- t1/2 is the half-life of radium-226

Given:
- N0 = 0.112 mol
- t = 6400 years
- t1/2 = 1600 years

Substituting these values into the formula:

N = (0.112 mol) * (1/2)^(6400 years / 1600 years)

Let's calculate this value step-by-step:

Step 1: Simplify the exponent:
1/2^(6400/1600) = 1/2^4 = 1/16

Step 2: Compute the final amount N:
N = (0.112 mol) * (1/16)
N = 0.112 mol / 16
N = 0.007 mol

Thus, after 6400 years, there would be 0.007 mol of radium-226 left in the sample.

To solve this problem, you can use the mathematical equation for radioactive decay, which is given by:

N(t) = N₀ * (1/2)^(t/T)

Where:
N(t) is the amount of radium-226 remaining after time t
N₀ is the initial amount of radium-226
t is the time that has passed
T is the half-life of radium-226

Given that the initial amount of radium-226 (N₀) is 0.112 mol, the time that has passed (t) is 6400 years, and the half-life of radium-226 (T) is 1600 years, we can substitute these values into the equation to find N(t):

N(t) = 0.112 * (1/2)^(6400/1600)

Let's break down the calculation step-by-step:

1. Calculate the exponent value inside the parentheses:
t/T = 6400/1600 = 4

2. Raise 1/2 to the power of the calculated exponent:
(1/2)^4 = 1/16

3. Multiply the result by the initial amount of radium-226:
N(t) = 0.112 * 1/16

4. Simplify the calculation:
N(t) = 0.112/16 = 0.007 mol

So, after 6400 years, approximately 0.007 mol of radium-226 is left in the sample.