How many atoms of oxygen (O) are represented in the chemical formula shown below?

The age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

<h3>What is the half-life of a radioactive material?</h3>

The half-life of a radioactive material is the time taken for half the atoms present in the material to decay or disintegrate.

The half-life, $t_\frac{1}{2}$ , the age, t, and amount remaining, $A_{r}$ , of a radioactive material are related by the formula below:

$t = \frac{t_\frac{1}{2}*A_{r}}{-ln2}$

Half-life of carbon-14 = 6000 years

For the Egyptian papyrus with 63% of its original carbon-14 atoms: $t = \frac{6000*0.63}{-0.63} = 4000\:years$

For the Aboriginal charcoal with 22% of its original carbon-14 atoms:

$t = \frac{6000*0.22}{-0.63} = 13106.5\:years$

For the Mayan headdress with 79% of its original carbon-14 atoms:

$t = \frac{6000*0.79}{-0.63} = 2040\:years$

Neanderthal skull with 3% of its original carbon-14 atoms:

$t = \frac{6000*0.03}{-0.63} = 30353\:years$

Therefore, the age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

6 0

2 years ago

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0

user100 [1]

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

5 0

3 years ago