The half-life of carbon-14 is 5730 years. therefore, 5730 years after being cut from a tree, a piece of wood will have 50.0% of the carbon-14 that it had originally. what percentage of the original carbon-14 will remain in the wood after another 5730 years?

For this problem, we use the formula for radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We can use the equation above to find the percent of the original substance at a given time. However, the time given here is equal to the half life of C-14. So the percent left would be 50% or half the original.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.