10 g of a radioactive substance was placed in an air tight container 50 years ago. When the container was opened, only 2.5 g remains. What was the half life of this material?

25 years.

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 10 g

Time (t) = 50 years

Amount remaining (N) = 2.5 g

Half-life (t½) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 10 g

Amount remaining (N) = 2.5 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

2.5 = 1/2ⁿ × 10

Cross multiply

2.5 × 2ⁿ = 10

Divide both side by 2.5

2ⁿ = 10 / 2.5

2ⁿ = 4

Express 4 in index form with 2 as the base.

2ⁿ = 2²

n =2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the half-life of the radioactive substance. This can be obtained as follow:

Time (t) = 50 years

Number of half-lives (n) = 2

Half-life (t½) =?

n = t / t½

2 = 50 / t½

Cross multiply

2 × t½ = 50

Divide both side by 2

t½ = 50 / 2

t½ = 25 years.

Thus, the half-life of the radioactive substance is 25 years.

the dependant variable is spilled water

the independent is the heat

the answers would be

the first one and the second one