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Ede4ka [16]
3 years ago
15

A farmer had an accident and spilled a chemical into his pond. Several days later, he notices many dead frogs around the pond. T

hat summer, the number of gnats around the pond was higher than ever. This is because
a. the chemical is good for the gnats.
b. the frogs are a limiting factor for the gnats.
c. the gnats are a limiting factor for the frogs.
d. the chemical is a limiting factor for the gnats.
Chemistry
2 answers:
quester [9]3 years ago
6 0

Answer:

The correct answer is option b.

Explanation:

A limiting factor or limiting reagent is a compound of reactant which gets used up first and effects the formation of product in a chemical reaction.

Around the pond frog would be feeding on gnat, and due to their presence population of gnats were in control .But due to accidental spilling of chemical in pond resulted in death of frogs. Which were actually responsible for less population of gnats around the pond but due to dead action of chemical on them (frogs) their population in pond declined due to which increase in population of gnats was observed.

On generalizing this fact:

  • Higher the population of frogs lower will be the population of gnats.
  • lower the population of frogs higher will be the population of gnats.

So, from this we can conclude that frogs are the limiting factors for the gnats as more the number of frogs lessor will be number of gnats.

andrezito [222]3 years ago
3 0
B. the frogs are a limiting factor for the gnats
the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase
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Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

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The balanced chemical reaction is,

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Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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