Answer:
13.5 years
Explanation:
Initial Concentration [Ao] = 10g
Final Concentration [A] = 0.768g
Time t= 50 years
Half life t1/2 = ?
These quantities are related by the following equations;
ln[A] = ln[Ao] - kt ......(i)
t1/2 = ln(2) / k ...........(ii)
where k = rate constant
Inserting the values in eqn (i) and solving for k, we have;
ln(0.768) = ln(10) - k (50)
-0.2640 = 2.3026 - 50k
50k = 2.3026 + 0.2640
k = 2.5666 / 50 = 0.051332
Insert the value of k in eqn (ii);
t1/2 = ln(2) / k
t1/2 = 0.693 / 0.051332 = 13.5 years
Answer:
in particular 69-189 084 = 9283983
Answer:
As
Explanation:
For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.
The only pentavalent element among the options in arsenic, hence the answer.