A parchment fragment was discovered that had about 68% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment. (Round your answer to the nearest hundred years.)

The parchments is 3200 years old.

Explanation: The carbon-14 is radioactive with a half-life (t1/2) of 5730 years. This means that it will take 5730 years for an initial sample of the isotope is reduced to its half. Because of it, Carbon-14 can be used to determine the age of an object.

From the question, "68% as much 14C radioactivity as plant material today" means that there is still 68% of the initial sample of the isotope. To determine how many half-lives passed in that period of time, you use:

A = A₀· , in which

A is the amount left after a period of time;

A₀ is the initial amount;

n is the number of half-lives passed in the same period

It's known the remaining amount: A = ·A₀.

Substituing and calculating:

·A₀ = A₀·

0.68 =

㏑0.68 = ㏑

- 0.387 = n·( - 0.693)

n =

n = 0.558

n is the amount of half-lives passed and it can be written as:

n = , where:

t is the period of time;

t1/2 is the half-life;

n =

t = n · t1/2

t = 0.558 . 5730

t = 3200 years

The parchment is approximately 3200 years of age.

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

i believe it is c.) 1575 j.