The concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock

5.0 x 10⁹ years.

Explanation:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

Explanation:

The half-life of potassium-40 (1.248 ×10⁹ y) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

We see that after four half-lives, ¹/₁₆ or 6.25 % of the original mass remains.

Conversely, if 6.25 % of the sample remains, the age of the sample must be four half-lives.

the answer is c. nacl