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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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3 years ago
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(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b)
Gala2k [10]

Answer:

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

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Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻  with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

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4 0
3 years ago
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Brrunno [24]

Answer:

The molecular formula of the compound is C_{8}H_{16}O_{4}. The molecular formula is obtained by the following expression shown below

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is  C_{2}H_{4}O

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Empirical formula mass of the compound = \left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}

n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Molecular formula = 4 \times C_{2}H_{4}O

Molecular formula is C_{8}H_{16}O_{4}

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