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MariettaO [177]
3 years ago
5

A river has different populations of lampreys and paddlefish. The paddlefish feed on zooplankton, while the lampreys feed on the

paddlefish. Humans living around the river hunt the paddlefish. What will happen to the ecosystem if the paddlefish are hunted in excess?
Chemistry
2 answers:
Serjik [45]3 years ago
8 0

The answer is: The population of paddlefish will decrease, while the population of zooplankton will increase.

The paddlerfish is predator to zooplankton. An decrease in predator (in this example the paddlerfish) will increase the number and genetic variation in a population of zooplankton.

An decrease of food (in this example the paddlerfish) will decrease population of the lampreys.

Hman activity (hunting) decrease population of the paddlerfish.

ivolga24 [154]3 years ago
4 0
A couple of things will happen. One, the paddlefish may be hunted into nonexistent numbers, meaning that that amount of zooplankton would increase. Two, due to the decreased amount of the paddlefish, the lampreys would moves elsewhere or begin to feed on something else. 
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Consider the reaction between a solution of molecule A and a solid block of molecule B. In general, for a reaction to occur, the
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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
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Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

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The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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