CHEMISTRY:MOLECULAR NATURE...-ALEKS 360
CHEMISTRY:MOLECULAR NATURE...-ALEKS 360
8th Edition
ISBN: 9781259916083
Author: SILBERBERG
Publisher: MCG
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Chapter 13, Problem 13.157P

(a)

Interpretation Introduction

Interpretation:

The kH for O2 in acrylic acid is to be calculated.

Concept introduction:

Henry’s law gives the quantitative relationship between the pressure of the gas and its solubility. It states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

Sgas=kH×Pgas (1)

Here, Sgas is the solubility of the gas.

kH is Henry’s constant.

Pgas is the partial pressure of the gas.

(a)

Expert Solution
Check Mark

Answer to Problem 13.157P

The kH for O2 in acrylic acid is 7.83×103mol/Latm.

Explanation of Solution

Rearrange equation (1) to calculate kH as follows:

kH=SgasPgas (2)

Substitute 1.64×103mol/L for Sgas and 0.2095atm for Pgas in equation (2).

kH=1.64×103mol/L0.2095atm=7.8282×103mol/Latm=7.83×103mol/Latm

Conclusion

The kH for O2 in acrylic acid is 7.83×103mol/Latm.

(b)

Interpretation Introduction

Interpretation:

The molarity of O2 is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

Sgas=kH×Pgas (1)

(b)

Expert Solution
Check Mark

Answer to Problem 13.157P

The molarity of O2 is 4×105M.

Explanation of Solution

Substitute 7.8282×103mol/Latm for kH and 0.005atm for Pgas in equation (1).

Sgas=(7.83×103mol/Latm)(0.005atm)=3.9141×105mol/L=4×105mol/L

The solubility is equal to the molarity so the molarity of O2 is 4×105M.

Conclusion

The molarity of O2 is 4×105M.

(c)

Interpretation Introduction

Interpretation:

The mole fraction of O2 is to be calculated.

Concept introduction:

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by X.

The formula to calculate the mole fraction is as follows:

Molefraction(X)=amount(mol)ofsoluteamount(mol)ofsolute+amount(mol)ofsolvent (3)

(c)

Expert Solution
Check Mark

Answer to Problem 13.157P

The mole fraction of O2 is 3×106.

Explanation of Solution

4×105M of O2 indicates that 4×105mol of O2 is dissolved in 1L of the solution. 14.6M of acrylic acid means that 14.6mol of acrylic acid are dissolved in 1L of the solution.

Substitute 4×105mol for the moles of solute and 14.6mol for the moles of solvent in equation (3).

Molefraction(X)=4×105mol4×105mol+14.6mol=2.73972×106=3×106

Conclusion

The mole fraction of O2 is 3×106.

(d)

Interpretation Introduction

Interpretation:

The concentration of O2 in ppm is to be calculated.

Concept introduction:

The ppm or parts per million is a concentration term that is defined as the mass of any substance divided by the mass of the solution, multiplied by 106.

The formula to calculate the concentration of an ion in ppm is as follows:

ppm=(massof solutemassofsolution)106 (4)

(d)

Expert Solution
Check Mark

Answer to Problem 13.157P

The concentration of O2 is 1 ppm.

Explanation of Solution

The formula to calculate the mass of the compound is as follows:

Mass of compound=(Moles of compound)(Molar mass) (5)

Substitute 4×105mol for the moles of O2 and 32g/mol for the molar mass in equation (5) to calculate the mass of O2.

Mass of O2=(4×105mol)(32g1mol)=0.00128g

Substitute 14.6mol for the moles of acrylic acid and 72.06g/mol for the molar mass in equation (5) to calculate the mass of acrylic acid.

Mass of acrylic acid=(14.6mol)(72.06g1mol)=1052.076g

The formula to calculate the mass of the solution is as follows:

Massofsolution=Massofsolute+Massofsolvent (6)

Substitute 0.00128g for the mass of solute and 1052.076g for the mass of solvent in equation (6).

Massofsolution=0.00128g+1052.076g=1052.07728g

Substitute 0.00128g for the mass of solute and 1052.07728g for the mass of solution in equation (4).

ppm=(0.00128g1052.07728g)106=1.2165ppm=1ppm

Conclusion

The concentration of O2 is 1 ppm.

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Chapter 13 Solutions

CHEMISTRY:MOLECULAR NATURE...-ALEKS 360

Ch. 13.5 - Prob. 13.6AFPCh. 13.5 - Prob. 13.6BFPCh. 13.6 - Calculate the vapor pressure lowering of a...Ch. 13.6 - Prob. 13.7BFPCh. 13.6 - Prob. 13.8AFPCh. 13.6 - Prob. 13.8BFPCh. 13.6 - Prob. 13.9AFPCh. 13.6 - Prob. 13.9BFPCh. 13.6 - A solution is made by dissolving 31.2 g of...Ch. 13.6 - Prob. 13.10BFPCh. 13.7 - Prob. B13.1PCh. 13.7 - Prob. B13.2PCh. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Which would you expect to be more effective as a...Ch. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10PCh. 13 - Prob. 13.11PCh. 13 - What is the strongest type of intermolecular force...Ch. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - What is the relationship between solvation and...Ch. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Use the following data to calculate the combined...Ch. 13 - Use the following data to calculate the combined...Ch. 13 - State whether the entropy of the system increases...Ch. 13 - Prob. 13.39PCh. 13 - Prob. 13.40PCh. 13 - Prob. 13.41PCh. 13 - Prob. 13.42PCh. 13 - Prob. 13.43PCh. 13 - Prob. 13.44PCh. 13 - For a saturated aqueous solution of each of the...Ch. 13 - Prob. 13.46PCh. 13 - Prob. 13.47PCh. 13 - Prob. 13.48PCh. 13 - Prob. 13.49PCh. 13 - Prob. 13.50PCh. 13 - Prob. 13.51PCh. 13 - Prob. 13.52PCh. 13 - Prob. 13.53PCh. 13 - Prob. 13.54PCh. 13 - Prob. 13.55PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - Calculate the molarity of each aqueous...Ch. 13 - Prob. 13.58PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - How would you prepare the following aqueous...Ch. 13 - Prob. 13.61PCh. 13 - Prob. 13.62PCh. 13 - Prob. 13.63PCh. 13 - Prob. 13.64PCh. 13 - Prob. 13.65PCh. 13 - Prob. 13.66PCh. 13 - Prob. 13.67PCh. 13 - Prob. 13.68PCh. 13 - Prob. 13.69PCh. 13 - Prob. 13.70PCh. 13 - Prob. 13.71PCh. 13 - Prob. 13.72PCh. 13 - Prob. 13.73PCh. 13 - Prob. 13.74PCh. 13 - Prob. 13.75PCh. 13 - Prob. 13.76PCh. 13 - Prob. 13.77PCh. 13 - Prob. 13.78PCh. 13 - Prob. 13.79PCh. 13 - Prob. 13.80PCh. 13 - Prob. 13.81PCh. 13 - What are the most important differences between...Ch. 13 - Prob. 13.83PCh. 13 - Prob. 13.84PCh. 13 - Prob. 13.85PCh. 13 - Prob. 13.86PCh. 13 - Prob. 13.87PCh. 13 - Prob. 13.88PCh. 13 - Classify each substance as a strong electrolyte,...Ch. 13 - Prob. 13.90PCh. 13 - Prob. 13.91PCh. 13 - Which solution has the lower freezing point? 11.0...Ch. 13 - Prob. 13.93PCh. 13 - Prob. 13.94PCh. 13 - Prob. 13.95PCh. 13 - Prob. 13.96PCh. 13 - Prob. 13.97PCh. 13 - Prob. 13.98PCh. 13 - Prob. 13.99PCh. 13 - The boiling point of ethanol (C2H5OH) is 78.5°C....Ch. 13 - Prob. 13.101PCh. 13 - Prob. 13.102PCh. 13 - Prob. 13.103PCh. 13 - Prob. 13.104PCh. 13 - Prob. 13.105PCh. 13 - Prob. 13.106PCh. 13 - Prob. 13.107PCh. 13 - Prob. 13.108PCh. 13 - Prob. 13.109PCh. 13 - Prob. 13.110PCh. 13 - Prob. 13.111PCh. 13 - In a study designed to prepare new...Ch. 13 - The U.S. Food and Drug Administration lists...Ch. 13 - Prob. 13.114PCh. 13 - Prob. 13.115PCh. 13 - Prob. 13.116PCh. 13 - In a movie theater, you can see the beam of...Ch. 13 - Prob. 13.118PCh. 13 - Prob. 13.119PCh. 13 - Prob. 13.120PCh. 13 - Prob. 13.121PCh. 13 - Gold occurs in seawater at an average...Ch. 13 - Prob. 13.123PCh. 13 - Prob. 13.124PCh. 13 - Prob. 13.125PCh. 13 - Prob. 13.126PCh. 13 - Pyridine (right) is an essential portion of many...Ch. 13 - Prob. 13.128PCh. 13 - Prob. 13.129PCh. 13 - Prob. 13.130PCh. 13 - Prob. 13.131PCh. 13 - Prob. 13.132PCh. 13 - Prob. 13.133PCh. 13 - Prob. 13.134PCh. 13 - Prob. 13.135PCh. 13 - Prob. 13.136PCh. 13 - Prob. 13.137PCh. 13 - Prob. 13.138PCh. 13 - Prob. 13.139PCh. 13 - Prob. 13.140PCh. 13 - Prob. 13.141PCh. 13 - Prob. 13.142PCh. 13 - Prob. 13.143PCh. 13 - The release of volatile organic compounds into the...Ch. 13 - Although other solvents are available,...Ch. 13 - Prob. 13.146PCh. 13 - Prob. 13.147PCh. 13 - Prob. 13.148PCh. 13 - Prob. 13.149PCh. 13 - Prob. 13.150PCh. 13 - Prob. 13.151PCh. 13 - Suppose coal-fired power plants used water in...Ch. 13 - Urea is a white crystalline solid used as a...Ch. 13 - Prob. 13.154PCh. 13 - Prob. 13.155PCh. 13 - Prob. 13.156PCh. 13 - Prob. 13.157PCh. 13 - Prob. 13.158PCh. 13 - Prob. 13.159PCh. 13 - Prob. 13.160PCh. 13 - Prob. 13.161PCh. 13 - Prob. 13.162PCh. 13 - Figure 12.11 shows the phase changes of pure...Ch. 13 - KNO3, KClO3, KCl, and NaCl are recrystallized as...Ch. 13 - Prob. 13.165PCh. 13 - Prob. 13.166PCh. 13 - Prob. 13.167P
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY