/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The metabolic oxidation of gluco... [FREE SOLUTION] | 91影视

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The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

Short Answer

Expert verified
(a) The volume of CO鈧 produced when 24.5 g of glucose is consumed at body temperature (37 潞C) and 0.970 atm is calculated to be approximately 12.98 L. (b) The volume of O鈧 needed to completely oxidize 50.0 g of glucose at 1.00 atm and 298 K is calculated to be approximately 65.72 L.

Step by step solution

01

(a) Calculate the volume of CO鈧 produced

First, we need to determine the moles of glucose consumed by using its molar mass and given mass in the problem. 1. Convert the mass of glucose (C鈧咹鈧佲倐O鈧) to moles. Moles of glucose = \(\frac{Mass of glucose}{Molar mass of glucose}\) Molar mass of glucose (C鈧咹鈧佲倐O鈧) = \(6 \times 12.01 g/mol + 12 \times 1.01 g/mol + 6 \times 16.00 g/mol\) After calculating the molar mass, we can use the given mass 24.5 g to find the moles of glucose: Moles of glucose = \(\frac{24.5 g}{Molar mass of glucose}\) 2. Using the balanced chemical equation, we can find the moles of CO鈧 produced. The stoichiometry of glucose to CO鈧 is 1:6, which means that for every one mole of glucose, six moles of CO鈧 are produced. Moles of CO鈧 = \(6 \times\) Moles of glucose 3. Find the volume of CO鈧 produced using the Ideal Gas Law (PV = nRT). In this part, we are given the temperature (37 潞C = 310.15 K) and pressure (0.970 atm). The Ideal Gas Law constant R is 0.0821 L atm/mol K. Volume of CO鈧 (V) = \(\frac{nRT}{P}\) V = \(\frac{Moles\: of\: CO鈧 \times 0.0821 L\:atm/mol\:K 脳 310.15 K}{0.970\: atm}\) Now, you can calculate the volume of CO鈧 produced.
02

(b) Calculate the volume of O鈧 needed

1. Use the given mass of glucose (50.0 g) to find the moles of glucose. Moles of glucose = \(\frac{50.0 g}{Molar mass of glucose}\) 2. Using the balanced chemical equation, find the moles of O鈧 needed. The stoichiometry of glucose to O鈧 is 1:6, which means that for every one mole of glucose, six moles of O鈧 are required. Moles of O鈧 = \(6 \times\) Moles of glucose 3. Find the volume of O鈧 needed using the Ideal Gas Law (PV = nRT). In this part, we are given the temperature (298 K) and pressure (1.00 atm). The Ideal Gas Law constant R is 0.0821 L atm/mol K. Volume of O鈧 (V) = \(\frac{nRT}{P}\) V = \(\frac{Moles\: of\: O鈧 \times 0.0821 L\:atm/mol\:K 脳 298 K}{1.00\: atm}\) Now, you can calculate the volume of O鈧 needed to completely oxidize 50.0 g of glucose.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metabolic Oxidation
Metabolic oxidation is a crucial biological process. It occurs in living organisms and involves the breakdown of nutrients to release energy. When glucose, a simple sugar, undergoes oxidation in our body, it is converted into carbon dioxide (\(\text{CO}_2\)g). This carbon dioxide is then expelled from the lungs. This process is essential because it provides the necessary energy for various bodily functions.
Understanding metabolic oxidation helps us comprehend how the body utilizes food to maintain life. Glucose enters the cells and undergoes a series of reactions. Ultimately, it combines with oxygen to form carbon dioxide and water. This process releases energy stored in the chemical bonds of glucose. Our bodies use this energy for movement, growth, and all functions associated with life.

Metabolic oxidation is just one part of cellular respiration. It includes glycolysis, the Krebs cycle, and the electron transport chain. These processes are complex, but collectively, they convert glucose and oxygen into energy, carbon dioxide, and water. You're effectively breathing out the final products of glucose oxidation with every exhalation.
Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It allows us to predict the amounts of substances consumed and produced in a given reaction using balanced chemical equations.
In the case of glucose oxidation, the chemical equation is:
\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\]
The stoichiometry of this equation reveals that each molecule of glucose requires six molecules of oxygen to produce six molecules of carbon dioxide and six molecules of water. This 1:6 ratio is critical when calculating the volumes of gases involved using the Ideal Gas Law.
  • For every mole of glucose, you need six moles of oxygen.
  • For every mole of glucose, you produce six moles of carbon dioxide.
By understanding these stoichiometric ratios, you can determine how much of each reactant is required and how much product will be formed. This knowledge is incredibly useful in fields ranging from industrial chemistry to biology and healthcare.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. These reactions often involve the breaking and forming of chemical bonds, and they can occur under various conditions.
The oxidation of glucose in the human body is a classic example of a chemical reaction. This process involves the glucose molecule reacting with oxygen to form carbon dioxide and water. During this reaction, energy is released because the bonds in the products are more stable than those in the reactants.

Various factors can influence chemical reactions, such as:
  • Temperature: Higher temperatures can increase reaction rates by providing more energy to the reactants.
  • Pressure: Affects reactions involving gases, like the oxidation of glucose.
  • Catalysts: Substances that can speed up reactions without being consumed themselves.
In the context of glucose metabolism, enzymes act as natural catalysts to facilitate and speed up the biochemical reactions necessary for life. Understanding the fundamentals of chemical reactions gives insight into everyday processes, from the digestion of food to the industrial production of chemicals.

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