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Ede4ka [16]
2 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]2 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]2 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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2 years ago
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Both hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you open
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Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass.

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2 years ago
What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?
Thepotemich [5.8K]

Answer:

CHCl₃

Explanation:

Given parameters:

Carbon = 5.03g

Hydrogen = 0.42g

Chlorine = 44.5g

The empirical formula shows the simplest formula of a compound.

To deduce the empirical, we need two pieces of information:

> Mass of the elements or the percentage composition of the compound

>The relative atomic masses of the elements

In order to derive the empirical formula from these parameters,

>>> find the number of moles of elements by dividing the mass given by the relative atomic mass of the respective atom

>>> Divide through by the smallest mole

>>> Approximate or multiply by a factor that would make it possible for whole numbers to be obtained

From the question, we have been given the mass of each element.

Now using the period table, we can obtain the relative atomic masses of each atom:

Carbon = 12gmol⁻¹

Hydrogen = 1gmol⁻¹

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C H Cl

Mass(in g) 5.03 0.42 44.5

Moles 5.03/12 0.42/1 44.5/37.5

0.42 0.42 1.19

Dividing

by

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Mole ratio 1 1 2.83

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The empirical formula is CHCl₃

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2 years ago
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6 0
2 years ago
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The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

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Explanation:

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density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

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  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
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