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 hours, how many ng of 55-co were originally present?

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 .

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answer:

technetium is the most reactive. this is followed by silver. the least reactive out of these three elements is strontium.