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 .
The first one is abundant
If ALL individuals of a species were not able to have offspring, then that species would eventually become extinct.
<span>79.70 grams
First, calculate the molar mass of Na2CO3 by looking up the molar mass of the elements used in it.
Sodium = 22.989769
Carbon = 12.0107
Oxygen = 15.999
Multiply each molar mass by the number of atoms used, and sum the results to get the molar mass
2 * 22.989769 + 12.0107 + 3 * 15.999 =105.9872
Finally, multiply the molar mass by the number of moles you need.
0.752 * 105.9872 = 79.7024 grams</span>
