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Marizza181 [45]
2 years ago
13

The cycling of carbon between the living and non-living parts of the Earth system is an important result of which of the followi

ng processes?
a. Herbivores digesting plant material to get nutrients
b. Plants and animals exchanging gases with the atmosphere
c. Condensation of water vapor leading to cloud formation
d. Methane entering the atmosphere from volcanic activity
Chemistry
1 answer:
lesantik [10]2 years ago
5 0

Answer:

b. Plants and animals exchanging gases with the atmosphere

Explanation:

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A 14.60g sample of an unknown compound, composed only of carbon, hydrogen, and oxygen, produced 28.6g of CO2 and 14.6g of H2O in
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Answer: The empirical formula for the given compound is C_{2}H_{5}O

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The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=28.6g

Mass of H_2O=14.6g

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Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 28.6 g of carbon dioxide, \frac{12}{44}\times 28.6=7.8g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.6 g of water, \frac{2}{18}\times 14.6=1.6 of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.32 moles.

For Carbon = \frac{0.65}{0.32}=2.03\approx 2

For Hydrogen  = \frac{1.6}{0.32}=5

For Oxygen  = \frac{0.32}{0.32}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is C_{2}H_{5}O_{1}=C_{2}H_{5}O

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