The half life for the radioactive decay of potassium- 40 to argon- 40 is 1.26 x 109 years. suppose nuclear chemical analysis shows that there is 0.771 mmol of argon-40 for every 1.000 mmol of potassium-40 in a certain sample of rock. calculate the age of the rock. round your answer to 2 significant digits.

Given:

Half life t1/2 for K-40 = 1.26*10⁹ years

Initial amount of K-40, A₀ = 1.000 mmol

Amount of Ar-40 formed = 0.771 mmol

To determine:

The age of the rock

Explanation:

The radioactive decay can be expressed mathematically as:

A = A₀exp(-kt)(1)

where A₀ = initial amount

A = amount after time t

k = decay constant = 0.693/t1/2 (2)

Here: k = 0.693/1.26*10⁹ yr = 0.55 *10⁻⁹ yr⁻¹

Now:

A₀ of K-40= 1.000 mmol

A of K-40 = 1.000 - 0.771 = 0.229 mmol

Based on eq(1)

0.229 = 1.000 exp(-0.55*10⁻⁹t)

t = 2.68*10⁹ years

Ans: Age of rock = 2.7*10⁹ yrs

Oxide of M is and sulfate of

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

Moles of hydrogen gas produced = 0.01225 mol

Moles of metal =

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

x = 2.9 ≈ 3

Formulas for the oxide and sulfate of M will be:

Oxide of M is and sulfate of .