MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
Question
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Chapter 4, Problem 4.34P

(a)

Interpretation Introduction

Interpretation:

The molarity of each ion is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  Molarity=moles of solute(mol)volume of solution(L)        (1)

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is kg/m3. The formula to calculate density is,

  Density=MassVolume        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 4.34P

The molarity of Na+, Cl, Mg2+, SO42, Ca2+, K+, HCO3 and Br ions are 0.470M, 0.553M, 0.0544M, 0.0285M, 0.0111M, 0.0144M, 0.00383M and 0.00098M respectively.

Explanation of Solution

Rearrange the equation (2) to calculate the volume.

  Volume=MassDensity        (3)

Substitute 1.025g/cm3 for density and 1 kg for mass in the equation (3) to calculate the volume of seawater.

  Volumeof sea water=(1 kg1.025g/cm3)(1000g1kg)(1mL1cm3)(1L1000mL)=0.97560976L

The formula to calculate the moles of NaCl is as follows:

  Moles of NaCl=[given massof NaClmolecular mass of NaCl]        (4)

Substitute 26.5g for the mass of NaCl and 58.44 g/mol for the molar mass of NaCl in the equation (4).

  Moles of NaCl=[26.5g58.44 g/mol]=0.4534565mol

One mole of NaCl dissociates to form one mole of Na+ and one mole of Cl. Therefore, 0.4534565mol of NaCl dissociates to form 0.4534565mol of Na+ and 0.4534565mol of Cl.

The formula to calculate the moles of MgCl2 is as follows:

  Moles of MgCl2=[given massof MgCl2molecular mass of MgCl2]        (5)

Substitute 2.40g for the mass of MgCl2 and 95.21 g/mol for the molar mass of MgCl2 in the equation (5).

  Moles of MgCl2=[2.40g95.21 g/mol]=0.025207mol

One mole of MgCl2 dissociates to form one mole of Mg2+ and two moles of Cl. Therefore, 0.025207mol of MgCl2 dissociates to form 0.025207mol of Mg2+ and 0.050415mol of Cl.

The formula to calculate the moles of MgSO4 is as follows:

  Moles of MgSO4=[given massof MgSO4molecular mass of MgSO4]        (6)

Substitute 3.35g for the mass of MgSO4 and 120.37 g/mol for the molar mass of MgSO4 in the equation (6).

  Moles of MgSO4=[3.35g120.37 g/mol]=0.0278308mol

One mole of MgSO4 dissociates to form one mole of Mg2+ and one mole of SO42. Therefore, 0.0278308mol of MgSO4 dissociates to form 0.0278308mol of Mg2+ and 0.0278308mol of SO42.

The formula to calculate the moles of CaCl2 is as follows:

  Moles of CaCl2=[given massof CaCl2molecular mass of CaCl2]        (7)

Substitute 1.20g for the mass of CaCl2 and 110.98 g/mol for the molar mass of CaCl2 in the equation (7).

  Moles of CaCl2=[1.20g110.98 g/mol]=0.0108128mol

One mole of CaCl2 dissociates to form one mole of Ca2+ and two moles of Cl. Therefore, 0.0108128mol of CaCl2 dissociates to form 0.0108128mol of Ca2+ and 0.0216255mol of Cl.

The formula to calculate the moles of KCl is as follows:

  Moles of KCl=[given massof KClmolecular mass of KCl]        (8)

Substitute 1.05g for the mass of KCl and 74.55 g/mol for the molar mass of KCl in the equation (4).

  Moles of KCl=[1.05g74.55 g/mol]=0.0140845mol

One mole of KCl dissociates to form one mole of K+ and one mole of Cl. Therefore, 0.0140845mol of KCl dissociates to form 0.0140845mol of K+ and 0.0140845mol of Cl.

The formula to calculate the moles of NaHCO3 is as follows:

  Moles of NaHCO3=[given massof NaHCO3molecular mass of NaHCO3]        (9)

Substitute 0.315g for the mass of NaHCO3 and 84.01 g/mol for the molar mass of NaHCO3 in the equation (9).

  Moles of NaHCO3=[0.315g84.01 g/mol]=0.00374955mol

One mole of NaHCO3 dissociates to form one mole of Na+ and one mole of HCO3. Therefore, 0.00374955mol of NaHCO3 dissociates to form 0.00374955mol of Na+ and 0.00374955mol of HCO3.

The formula to calculate the moles of NaBr is as follows:

  Moles of NaBr=[given massof NaBrmolecular mass of NaBr]        (10)

Substitute 0.098g for the mass of NaBr and 102.98 g/mol for the molar mass of NaBr in the equation (10).

  Moles of NaBr=[0.098g102.98 g/mol]=0.0009524735mol

One mole of NaBr dissociates to form one mole of Na+ and one mole of Br. Therefore, 0.0009524735mol of NaBr dissociates to form 0.0009524735mol of Na+ and 0.0009524735mol of Br.

The Cl ions come from NaCl, MgCl2, CaCl2 and KCl. Therefore the total moles of Cl ions are calculated as follows:

  Total moles ofClions=(Cl)NaCl+(Cl)MgCl2+(Cl)CaCl2+(Cl)KCl=(0.4534565mol+0.050415mol+0.0216255mol+0.0140845mol)=0.5395815mol

The Na+ ions come from NaCl, NaHCO3 and NaBr. Therefore the total moles of Na+ ions are calculated as follows:

  Total moles ofNa+ions=(Na+)NaCl+(Na+)NaHCO3+(Na+)NaBr=(0.4534565mol+0.00374955mol+0.0009524735mol)=0.458158523mol

The Mg2+ ions come from MgCl2 and MgSO4. Therefore the total moles of Mg2+ ions are calculated as follows:

  Total moles ofMg2+ions=(Mg2+)MgCl2+(Mg2+)MgSO4=(0.025207mol+0.0278308mol)=0.0530355mol

The formula to calculate the molarity of an ion is as follows:

  Molarity=moles of ionsvolume of sea water        (11)

Substitute 0.5395815mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of Cl ions.

  (Molarity)Cl=0.5395815mol0.97560976L=0.55307M0.553M

Substitute 0.458158523mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of Na+ ions.

  (Molarity)Na+=0.458158523mol0.97560976L=0.469612M0.470M

Substitute 0.0530355mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of Mg2+ ions.

  (Molarity)Mg2+=0.0530355mol0.97560976L=0.054361M0.0544M

Substitute 0.0278308mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of SO42 ions.

  (Molarity)SO42=0.0278308mol0.97560976L=0.028526M0.0285M

Substitute 0.0108128mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of Ca2+ ions.

  (Molarity)Ca2+=0.0108128mol0.97560976L=0.011083M0.0111M

Substitute 0.0140845mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of K+ ions.

  (Molarity)K+=0.0140845mol0.97560976L=0.014437M0.0144M

Substitute 0.00374955mol for the moles of ion and 0.97560976L for the volume of sea water in the equation (11) to calculate the molarity of HCO3 ions.

  (Molarity)HCO3=0.00374955mol0.97560976L=0.003843M0.00384M

Substitute 0.0009524735mol for the moles of ion and 0.97560976L for the volume of seawater in the equation (11) to calculate the molarity of Br ions.

  (Molarity)Br=0.0009524735mol0.97560976L=0.0009763M0.00098M.

Conclusion

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

(b)

Interpretation Introduction

Interpretation:

The total molarity of alkali metal ions is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  Molarity=moles of solute(mol)volume of solution(L)        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

(b)

Expert Solution
Check Mark

Answer to Problem 4.34P

The total molarity of alkali metal ions in sea water is 0.484M.

Explanation of Solution

The alkali metal ions include from Na+ and K+. Therefore the molarity of alkali metal ions is calculated as follows:

  Molarity ofalkali metal ions=(molarity of Na+)+(molarity ofK+)=(0.469612M+0.014437M)=0.484049M0.484M.

Conclusion

Na+ and K+ lie in the first group and hence they are alkali metal ions.

(c)

Interpretation Introduction

Interpretation:

The total molarity of alkaline earth metal ions is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  Molarity=moles of solute(mol)volume of solution(L)        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

(c)

Expert Solution
Check Mark

Answer to Problem 4.34P

The total molarity of alkaline earth metal ions in sea water is 0.065M.

Explanation of Solution

The alkaline earth metal ions include from Mg2+ and Ca2+. Therefore the molarity of alkaline earth metal ions is calculated as follows:

  Molarity ofalkaline earth metal ions=(molarity of Mg2+)+(molarity ofCa2+)=(0.054361M+0.011083M)=0.065444M0.065M.

Conclusion

Mg2+ and Ca2+ lie in the second group and hence they are alkaline earth metal ions.

(d)

Interpretation Introduction

The total molarity of anions is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  Molarity=moles of solute(mol)volume of solution(L)        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts large amount of electricity.

(d)

Expert Solution
Check Mark

Answer to Problem 4.34P

The total molarity of anions in seawater is 0.586M.

Explanation of Solution

The anions include come from Cl, SO42, HCO3 and Br. Therefore the total molarity of anions is calculated as follows:

  Molarity ofanions=[(molarity of Cl)+(molarity of SO42)+(molarity of HCO3)+(molarity of Br)]=0.55307M+0.028526M+0.003843M+0.0009763M=0.5864153M0.586M.

Conclusion

Cl, SO42, HCO3 and Br have negative charge on them and hence, they are anions.

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Chapter 4 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 4.1 - A chemist dilutes 60.0 mL of 4.50 M potassium...Ch. 4.1 - Prob. 4.6BFPCh. 4.1 - Prob. 4.7AFPCh. 4.1 - Prob. 4.7BFPCh. 4.3 - Prob. 4.8AFPCh. 4.3 - Prob. 4.8BFPCh. 4.3 - Prob. 4.9AFPCh. 4.3 - Molecular views of the reactant solutions for a...Ch. 4.3 - It is desirable to remove calcium ion from hard...Ch. 4.3 - To lift fingerprints from a crime scene, a...Ch. 4.3 - Despite the toxicity of lead, many of its...Ch. 4.3 - Mercury and its compounds have uses from fillings...Ch. 4.4 - How many OH−(aq) ions are present in 451 mL of...Ch. 4.4 - Prob. 4.12BFPCh. 4.4 - Prob. 4.13AFPCh. 4.4 - Prob. 4.13BFPCh. 4.4 - Prob. 4.14AFPCh. 4.4 - Prob. 4.14BFPCh. 4.4 - Another active ingredient in some antacids is...Ch. 4.4 - Prob. 4.15BFPCh. 4.4 - What volume of 0.1292 M Ba(OH)2 would neutralize...Ch. 4.4 - Calculate the molarity of a solution of KOH if...Ch. 4.5 - Prob. 4.17AFPCh. 4.5 - Prob. 4.17BFPCh. 4.5 - Prob. 4.18AFPCh. 4.5 - Prob. 4.18BFPCh. 4.5 - Prob. 4.19AFPCh. 4.5 - Prob. 4.19BFPCh. 4.6 - Prob. 4.20AFPCh. 4.6 - Prob. 4.20BFPCh. 4 - Prob. 4.1PCh. 4 - What types of substances are most likely to be...Ch. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Which of the following scenes best represents how...Ch. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - A mathematical equation useful for dilution...Ch. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Does an aqueous solution of each of the following...Ch. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Calculate each of the following quantities: Mass...Ch. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Calculate each of the following quantities: Volume...Ch. 4 - Prob. 4.30PCh. 4 - Concentrated sulfuric acid (18.3 M) has a density...Ch. 4 - Prob. 4.32PCh. 4 - Muriatic acid, an industrial grade of concentrated...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Write two sets of equations (both molecular and...Ch. 4 - Why do some pairs of ions precipitate and others...Ch. 4 - Use Table 4.1 to determine which of the following...Ch. 4 - The beakers represent the aqueous reaction of...Ch. 4 - Complete the following precipitation reactions...Ch. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - If 25.0 mL of silver nitrate solution reacts with...Ch. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - With ions shown as spheres and solvent molecules...Ch. 4 - The precipitation reaction between 25.0 mL of a...Ch. 4 - A 1.50-g sample of an unknown alkali-metal...Ch. 4 - Prob. 4.54PCh. 4 - The mass percent of Cl− in a seawater sample is...Ch. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Write a general equation for a neutralization...Ch. 4 - Prob. 4.59PCh. 4 - (a) Name three common weak acids. (b) Name one...Ch. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - How many moles of H+ ions are present in each of...Ch. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Complete the following acid-base reactions with...Ch. 4 - Limestone (calcium carbonate) is insoluble in...Ch. 4 - Prob. 4.68PCh. 4 - Prob. 4.69PCh. 4 - How many grams of NaH2PO4 are needed to react with...Ch. 4 - Prob. 4.71PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - A mixture of bases can sometimes be the active...Ch. 4 - Describe how to determine the oxidation number of...Ch. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - Why must every redox reaction involve an oxidizing...Ch. 4 - Prob. 4.82PCh. 4 - Identify the oxidizing agent and the reducing...Ch. 4 - Give the oxidation number of carbon in each of the...Ch. 4 - Prob. 4.85PCh. 4 - Give the oxidation number of nitrogen in each of...Ch. 4 - Give the oxidation number of sulfur in each of the...Ch. 4 - Prob. 4.88PCh. 4 - Give the oxidation number of phosphorus in each of...Ch. 4 - Give the oxidation number of manganese in each of...Ch. 4 - Give the oxidation number of chromium in each of...Ch. 4 - Identify the oxidizing and reducing agents in the...Ch. 4 - Identify the oxidizing and reducing agents in the...Ch. 4 - Identify the oxidizing and reducing agents in the...Ch. 4 - Identify the oxidizing and reducing agents in the...Ch. 4 - The active agent in many hair bleaches is hydrogen...Ch. 4 - A person’s blood alcohol (C2H5OH) level can be...Ch. 4 - Which type of redox reaction leads to each of the...Ch. 4 - Why do decomposition redox reactions typically...Ch. 4 - Which of the types of reactions discussed in...Ch. 4 - Are all combustion reactions redox reactions?...Ch. 4 - Give one example of a combination reaction that is...Ch. 4 - Prob. 4.103PCh. 4 - Prob. 4.104PCh. 4 - Prob. 4.105PCh. 4 - Prob. 4.106PCh. 4 - Prob. 4.107PCh. 4 - Predict the product(s) and write a balanced...Ch. 4 - Prob. 4.109PCh. 4 - Predict the product(s) and write a balanced...Ch. 4 - Prob. 4.111PCh. 4 - Predict the product(s) and write a balanced...Ch. 4 - How many grams of O2 can be prepared from the...Ch. 4 - How many grams of chlorine gas can be produced...Ch. 4 - Prob. 4.115PCh. 4 - Prob. 4.116PCh. 4 - A mixture of KClO3 and KCl with a mass of 0.950 g...Ch. 4 - Prob. 4.118PCh. 4 - Before arc welding was developed, a displacement...Ch. 4 - Iron reacts rapidly with chlorine gas to form a...Ch. 4 - A sample of impure magnesium was analyzed by...Ch. 4 - Why is the equilibrium state said to be...Ch. 4 - Prob. 4.123PCh. 4 - Describe what happens on the molecular level when...Ch. 4 - When either a mixture of NO and Br2 or pure...Ch. 4 - Prob. 4.126PCh. 4 - Nutritional biochemists have known for decades...Ch. 4 - Limestone (CaCO3) is used to remove acidic...Ch. 4 - The brewing industry uses yeast to convert glucose...Ch. 4 - A chemical engineer determines the mass percent of...Ch. 4 - Prob. 4.131PCh. 4 - You are given solutions of HCl and NaOH and must...Ch. 4 - The flask represents the products of the titration...Ch. 4 - To find the mass percent of dolomite [CaMg(CO3)2]...Ch. 4 - On a lab exam, you have to find the concentrations...Ch. 4 - Nitric acid, a major industrial and laboratory...Ch. 4 - Prob. 4.137PCh. 4 - In 1995, Mario Molina, Paul Crutzen, and F....Ch. 4 - Sodium peroxide (Na2O2) is often used in...Ch. 4 - A student forgets to weigh a mixture of sodium...Ch. 4 - Prob. 4.141PCh. 4 - Prob. 4.142PCh. 4 - Physicians who specialize in sports medicine...Ch. 4 - Thyroxine (C15H11I4NO4) is a hormone synthesized...Ch. 4 - Over time, as their free fatty acid (FFA) content...Ch. 4 - Prob. 4.146PCh. 4 - Calcium dihydrogen phosphate, Ca(H2PO4)2, and...Ch. 4 - Prob. 4.148PCh. 4 - Prob. 4.149PCh. 4 - Prob. 4.150PCh. 4 - In 1997 and 2009, at United Nations conferences on...Ch. 4 - In a car engine, gasoline (represented by C8H18)...Ch. 4 - Prob. 4.153PCh. 4 - Prob. 4.154PCh. 4 - Prob. 4.155PCh. 4 - Prob. 4.156PCh. 4 - Prob. 4.157P
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