The half-life of cobalt-60 is 5.3 years. What fraction of a sample remains after 21.2 years?

A) one half
B) one quarter (WRONG)
C) one eighth (WRONG)
D) one sixteenth

please explain!!!

one sixteenth

Explanation: Since we know the half life of Cobalt-60 is 5.3 years, and 21.2 years have passed, we can determine how many half lifes have passed by doing division:

21.2 / 5.3 = 4

Since four half lifes have passed, we need to set up the following:

1 / 2^4 (always raise the two to the number of half lifes that have passed, in this case, four have passed, which is what we determined up above!)

When you simplify this you get 1 / 16, which is answer D.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

yes

explanation:

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