The half-life of a pesticide determines its persistence in the environment. a common pesticide degrades in a first-order process with a half-life of 2 days. what fraction (in decimal notation) of the pesticide remains in the environment after 18 days? enter to 4 decimal places.

0.0020 fraction of the pesticide remains in the environment after 18 days

Explanation:

For a first order reaction,

where N is remaining mass after "t" time , is initial mass,   represents fraction of mass remains after "t" time and is half-life

Here t is 18 days and is 2 days

So,

Hence 0.0020 fraction of the pesticide remains in the environment after 18 days

The fraction (in decimal notation) of the pesticide that remains in the environment after 18 days is 0.0020

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2 days

Time (t) = 18 days

n = t / t½

n = 18 / 2

Thus, 9 half-lives has elapsed.

Original amount (N₀) = 1

Number of half-lives (n) = 9

N = 1/2ⁿ × N₀

Divide both side by N₀

N / N₀ = 1/2ⁿ

N / N₀ = 1/2⁹

Thus, the fraction of the pesticide that remains in the environment is 0.0020

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each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

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