Answer:
After 47.9 days, will remain 14.5mg of the isotope
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
<em>Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration. </em>
We can find rate constant from half-life as follows:
K = ln 2 / 27.7 days
K = 0.025 days⁻¹
Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)
ln [A] = 2.6726
[A] = e^ (2.6726)
[A] = 14.5mg
<h3>After 47.9 days, will remain 14.5mg of the isotope</h3><em />
Answer:
= -457.9 kJ and reaction is product favored.
Explanation:
The given reaction is associated with 2 moles of
Standard free energy change of the reaction () is given as:
, where T represents temperature in kelvin scale
So,
So, for the reaction of 1.57 moles of ,
As, is negative therefore reaction is product favored under standard condition.
Answer:
a)
b)
Explanation:
a) The reaction:
The free-energy expression:
The element wich is reduced is the Fe and the one that oxidates is the Mg:
The electrons transfered (n) in this reaction are 2, so:
b) If you have values of enthalpy and enthropy you can calculate the free-energy by:
with T in Kelvin