A. The radioactive decay equation is N = N0
where T is the
half-life (5730 years), N0 is the number of atoms at time t = 0 and
N is the number at time t.
Rewriting this as:
(N/N0) = 
Since N = (1/8) N0 and
substituting known values:
1/8 = 
Taking ln of both
sides:
ln(1/8)= -ln(2)*t/5730
t = - 5730 * ln(1/8) /
ln (2)
t = 17,190 years
The tree was cut down 17,190
years ago.
B. N0 = 1,500,000 carbon-14 atoms
Since N = (1/8) N0
N = 187,500 carbon
atoms left
The answer would be 371 because it has multiple complete digits
Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of
at 298 K will be as follows.
=
+
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that
at 298 K for the given process is -394 kJ.