An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hour
1 answer:
Answer:
Explanation:
Let A₀ = the original amount of ⁵⁵Co .
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
The general formula for the amount remaining is:
A =A₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
Data:
A = 1.90 ng
t = 45 h
t_½ = 18.0 h
Calculation:
(a) Calculate n
n = 45/18.0 = 2.5
(b) Calculate A
1.90 = A₀ × (½)^2.5
1.90 = A₀ × 0.178
A₀ = 1.90/0.178 = 10.7 ng
The original mass of ⁵⁵Co was .
You might be interested in
Unfortunately you did not specify the electronic configuration in the question, however since one of the answers must be a halogen, i took the liberty to attach an image with the configuration (both the simple numeric and spdf form) for all the halogen and all you have to do is match the electronic configuration you have in your question to the one in the table attached and you can then deduce the answer.
Hope this helps.
Hope this helps.