Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
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Explanation:
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I found a presentation of Food web of a pond that will greatly connect with the above problem. http://www.eduweb.com/portfolio/earthsystems/food/foodweb4.html
Normal setting:
Species Pop. size
Blue heron Medium
Perch Medium
Bass Medium
Minnows Medium
Inverts Medium
Algae Medium
If PERCH population size decreases,
Species Pop. sizeBlue heron LowPerch LowBass MediumMinnows HighInverts HighAlgae Low
As you can see, 4 other species are affected when the Perch population size decreased. Bass is not affected.
Digestive I think, that transfers food.
