“The mass of H 2 O ” should be calculated Concept Introduction: Here we have used the concept of ideal gases and its equation. ► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law. ► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume Pα 1 V ► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature VαT ► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules Volume of the gas α Number of molecules α Moles of the gas Vα n ► Combining these three laws, we get the ideal gas equation V α n T P V = nRT P PV = nRT where, R is universal gas constant V is the volume n is the number of moles P is the pressure T is the temperature ► The relation between number of moles and mass of the gas is n = m M Where m is the mass of the gas M is molar mass of the gaseous compound ► Molar mass is the number of times a molecule of the substance is heavier than 1 1 2 th the mass of an atom of carbon − 12 . ► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction. Thus, in stoichiometric calculation we come across chemical formula and chemical equation. ► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction. ► Limiting reagent is the substance which gets consumed completely when the reaction is carried out. Given: Pressure of O 2 gas, P = 1 atm Temperature of O 2 gas, T=37 0 C Volume of O 2 = 7 .5 L Mass of glucose = 18 g

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Answer 1

Question

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

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Answer 2

Question

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Answer 6

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Answer 7

Chapter 11, Problem 11.113CQ

Interpretation Introduction

Interpretation:

“The mass of $H_{2} O$ ” should be calculated

Concept Introduction:

Here we have used the concept of ideal gases and its equation.

► Ideal gases are the gases which obeys the ideal gas equation under all conditions of temperature and pressure. The ideal gas equation can be obtained by combining the Boyle’s Law, Charles’s Law and Avogadro Law.

► Boyle’s Law: It states that at constant temperature, the pressure has an inverse relation with volume

$Pα \frac{1}{V}$

► Charles’s Law: It states that at constant pressure, the volume of the gas has direct relation with temperature

$VαT$

► Avogadro Law: It states that equal volumes of all gases under same conditions of temperature and pressure contains the same number of molecules

Volume of the gas $α$ Number of molecules

$α$ Moles of the gas

$Vα n$

► Combining these three laws, we get the ideal gas equation

$V α \frac{n T}{P}$

$V = \frac{nRT}{P}$

$PV = nRT$

where,

$R$ is universal gas constant

$V$ is the volume

$n$ is the number of moles

$P$ is the pressure

$T$ is the temperature

► The relation between number of moles and mass of the gas is

$n = \frac{m}{M}$

Where $m$ is the mass of the gas

$M$ is molar mass of the gaseous compound

► Molar mass is the number of times a molecule of the substance is heavier than $\frac{1}{1 2^{th}}$ the mass of an atom of carbon $- 12$ .

► We have used the concept of stoichiometry where when reactants undergo a reaction, they form product in certain ratio based on the number of moles associated with the elements which participated in the reaction.

Thus, in stoichiometric calculation we come across chemical formula and chemical equation.

► A chemical equation directly states that the substance and the number of moles of each substance involved in the chemical reaction.

► Limiting reagent is the substance which gets consumed completely when the reaction is carried out.

Given:

Pressure of $O_{2}$ gas, $P = 1 atm$

Temperature of $O_{2}$ gas, ${T=37}^{0} C$

Volume of $O_{2} = 7 .5 L$

Mass of glucose = $18 g$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 11.1 - Prob. 11.1QAP Ch. 11.1 - Prob. 11.2QAP Ch. 11.1 - Prob. 11.3QAP Ch. 11.1 - Prob. 11.4QAP Ch. 11.1 - Prob. 11.5QAP Ch. 11.1 - Prob. 11.6QAP Ch. 11.1 - Prob. 11.7QAP Ch. 11.1 - Prob. 11.8QAP Ch. 11.2 - Why do scuba divers need to exhale air when they...Ch. 11.2 - Why does a sealed bag of chips expand when you...

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