Chaptr 4 HW Packet Answer Key

Transcription

1 Chaptr 4 HW Packet Answer Key Multiple Choice 1 point; Calculations three points unless otherwise specified 1 for formula, 1 for substitution, 1 for calculation. 1. An unknown substance dissolves readily in water but not in benzene (a nonpolar solvent). Molecules of what type are present in the substance? a) nonpolar b) none of these c) neither polar nor nonpolar d) polar e) either polar or nonpolar Remember that it s all about interparticle forces. In order for a solute to be able to mix in a liquid, the particles of the liquid have to be able to exert forces of attraction on the particles of the solute, breaking the interparticle forces of the solute, and then, to keep the particles of solute from reconnecting, the particles of solvent have to form bonds with the solvent particles that are more stable than the bonds between the solvent particles themselves. If a solute can dissolve in water, which is polar, the solute must therefore either be polar or ionic, because these are the types of substances with which water can develop forces of attraction. Benzene, which is non- polar, will not be able to develop forces of attraction with polar or ionic substances, and therefore, cannot dissolve these types of substances. In this list of answers, the substance must therefore be polar. Answer D (1 point/1) 2. The interaction between solute particles and water molecules, which tends to cause a salt to fall apart in water, is called a) dispersion b) coagulation c) hydration d) polarization e) conductivity The action of polar water molecules being able to develop forces of attraction with particles of ionic solids, and then being able to surround individual ions with a number of water particles that effectively separates the solute particles and keeps them mixed with the water particles is called hydration. Answer C (1 point/2) 3. Consider two organic molecules, ethanol and benzene. One dissolves in water and the other does not. Why? a) They have different molar masses. b) One is ionic, the other is not.

2 Chapter Error! Unknown document property name.: Error! Unknown document property name. 70 c) One is an electrolyte, the other is not. d) Ethanol contains a polar O H bond, and benzene does not. e) Two of these are correct. Again, in order for a solute to be able to mix in a liquid, the particles of the liquid have to be able to exert forces of attraction on the particles of the solute, breaking the interparticle forces of the solute, and then, to keep the particles of solute from reconnecting, the particles of solvent have to form bonds with the solvent particles that are more stable than the bonds between the solvent particles themselves. If a solute can dissolve in water, which is polar, the solute must therefore either be polar or ionic, because these are the types of substances with which water can develop forces of attraction. Because of its OH group, ethanol is polar and can therefore dissolve in water. Benzene, which is entirely hydrocarbon, is non- polar and will not be able to dissolve in water. (a) Although the substances have different molar masses, and this could potentially play a role in solubility, in this case it does not. (b) Both substances are covalent so neither substance is ionic. (c) Because neither substance can dissociate into ions neither substance is an electrolyte. (d) Ethanol does contain a polar OH bond while benzene does not and this is why, as explained above, ethanol can dissolve while benzene cannot. (e) only one answer suffices. Answer D (1 point/3) 4. T F Polar molecules have an unequal distribution of charge within the molecule. The reason that some molecules are polar is because there is an unequal distribution of charge within the molecule. Remember that this is due to differences in electronegativity between the atoms forming covalent bonds. Answer True (1 point/4) 5. Which of the following is a strong acid? a) HF b) HClO c) HBrO d) KOH e) HClO4 You should have memorized the six strong acids those that can completely dissociate in water memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Therefore, (a) HF (hydrofluoric acid) ), as stated above, is not one of our six strong acids and so is a weak acid. (b) HClO (hypochlorous acid) is not HClO4 as stated above, it is not one of our six strong acids and so is a weak acid.. (c) HBrO (hypobromous acid) is not HBr as stated above, it is not one of our six strong acids and so is a weak acid.. (d) KOH

3 Chapter Error! Unknown document property name.: Error! Unknown document property name. 71 (potassium hydroxide) is a strong base. (e) HClO4 is one of our strong acids. Answer E (1 point/5) 6. All of the following are weak acids except a) HCNO b) HBr c) HF d) HNO2 e) HCN You should have memorized the six strong acids those that can completely dissociate in water memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Therefore, (a) The CNO - ion is called the fulminate ion. It is highly unstable and explosive. Its corresponding acid is called fulminic acid this follows our convention of ate polyatomic ions adding the ic ending to the name of the acid. In any case, it is a weak acid. (b) HBr (hydrobromic acid) is one of our strong acids, so for this problem, it is the exception and the correct answer. (c) HF (hydrofluoric acid), as stated above, is not one of our six strong acids and so is a weak acid. (d) HNO2 (nitrous acid), as stated above, is not one of our strong acids and so is a weak acid. (e) HCN (hydrocyanic acid), as stated above, is not one of our strong acids and so is a weak acid. Answer B (1 point/6) 7. Which of the following is not a strong base? a) Ca(OH)2 b) KOH c) ioh d) Sr(OH)2 e) NH3 Strong bases will either (1) directly give up OH - to a solution when they dissociate, or (2) consist of a soluble metal oxide, which, when it dissociates, will form 2 OH - ions as the dissociated oxygen will pull a hydrogen ion away from a water molecule (thus creating two hydroxide ions). Weak bases are those substances that can remove a hydrogen ion from a water molecule leaving a single OH - behind, but they can only accomplish this weakly. (a) Ca(OH)2 (calcium hydroxide) will give up OH - ions directly to solution, and so is a strong base. (b) KOH (potassium hydroxide) will give up OH - ions directly to solution, and so is a strong base. (c) ioh (lithium hydroxide) will give up OH - ions directly to solution, and so is a strong base. (d) Sr(OH)2 (strontium hydroxide) will give up OH - ions directly to solution, and so is a strong base. (e) NH3 can only weakly remove a hydrogen ion from a water molecule, and so is a weak acid. This is the exception in the list. Answer E (1 point/7) 8. Which of the following is paired incorrectly? a) H2SO4 strong acid b) HNO3 weak acid

4 Chapter Error! Unknown document property name.: Error! Unknown document property name. 72 c) Ba(OH)2 strong base d) HCl strong acid e) NH3 weak base You should have memorized the six strong acids those that can completely dissociate in water memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Strong bases will either (1) directly give up OH - to a solution when they dissociate, or (2) consist of a soluble metal oxide, which, when it dissociates, will form 2 OH - ions as the dissociated oxygen will pull a hydrogen ion away from a water molecule (thus creating two hydroxide ions). Weak bases are those substances that can remove a hydrogen ion from a water molecule leaving a single OH - behind, but they can only accomplish this weakly. Therefore, (a) H2SO4 is a strong acid and so is paired correctly. (b) HNO3 is a strong acid and so is not paired correctly. (c) Ba(OH)2 will give up OH - ions directly to solution, and so is a strong base, and so, is paired correctly. (d) HCl is a strong acid and so is paired correctly. (e) NH3 can only weakly remove a hydrogen ion from a water molecule, and so is a weak acid. Answer B (1 point/8) 9. The man who discovered the essential nature of acids through solution conductivity studies is a) Priestly b) Boyle c) Einstein d) Mendeleev e) Arrhenius Arrhenius discovered that the essential nature of acids was in the dissociation of the acid molecule into a hydrogen ion (which we now know is donated to a water molecule to form the hydronium ion) and the anion corresponding to whatever the specific acid is. This dissociation was discovered in the larger context of discovering that when ionic substances in general dissociate completely, they easily conduct electricity. So, acids become one type of a larger class of compounds called electrolytes. Acids (or ionic substances) that completely dissociate in water are strong electrolytes. Acids (or ionic substances) that only incompletely dissociate in water are called weak electrolytes. Answer E (1 point/9) 10. A solid acid HX is mixed with water. Two possible solutions can be obtained. Which of the following is true? I. II.

5 Chapter Error! Unknown document property name.: Error! Unknown document property name. 73 a) In case I, HX is acting like a weak acid, and in case II, HX is acting like a strong acid. b) In case I, HX is acting like a strong acid, and in case II, HX is acting like a weak acid. c) In both cases, HX is acting like a strong acid. d) In both cases, HX is acting like a weak acid. e) HX is not soluble in water. These diagrams test your true understanding of the concept of complete vs partial dissociation. In the first diagram, all of the HX particles have completely dissociated to form only individual ions, H + (which we know in fact are not individual protons because they will immediately form a covalent bond with a water molecule to form a hydronium ion) and X -. This is what a strong acid does completely dissociates. A weak acid only partially dissociates, and so, while there will be some H+ ions (which again will immediately bond with a water molecule to form a hydronium ion) and an equal amount of X - ions, some of the HX particles will remain intact. So, diagram I depicts a strong acid while diagram II depicts a weak acid. Answer B (1 point/10) 11. T F An acid is a substance that produces OH ions in water. An acid is a substance that produces H3O + ions in water. Answer False (1 point/11)

6 Chapter Error! Unknown document property name.: Error! Unknown document property name A g sample of HF is dissolved in water to give m of solution. The concentration of the solution is: (See Stoichiometry Skills above) You are asked to find the concentration so you immediately write down. M = mol Upon inspection you realize that you do not have moles because you know mol = g you immediately then substitute g for moles. mm mm The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. Plug in the given values and solve. 2 SF. Answer 4.2 M (3 points/14) M = mol = g mm ( )g/mol 17.0g =.20 = 4.2M m of a M solution of oleic acid is diluted with 9.00 m of petroleum ether, forming solution A. Then 2.00 m of solution A is diluted with 8.00 m of petroleum ether, forming solution B. What is the concentration of solution B? (See Dilution Skills above) In this problem, there is a sequence of dilutions and you should take them one at a time, the answer to the first becoming the given information for the next. In the first dilution you are given V1 (1.00 m=.001 ) and M1 (3.95 x 10-4 M) and told that it is diluted with 9.00 m of petroleum ether to form solution A. Be careful 9.00 m IS NOT V2. V2 is the total final volume of the solution, and this is 1.00 m m which is 10.0 m (.0100). Therefore, you can plug these three values into the dilution relationship and find the molarity of solution A. A: M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 = (3.95x10 4 mol )(0.001) = 3.95x10 5 M.0100

7 Chapter Error! Unknown document property name.: Error! Unknown document property name. 75 This molarity then becomes M1 of the next dilution. You are given V1 of this next dilution 2.00 m (.002 ). You are also told that this is mixed with 8.00 m (.008 ) of petroleum ether to give solution B. Be careful again, 8.00 IS NOT V2. V2 is the total final volume of the solution, and this is 2.00 m m which is 10.0 m (.0100). Therefore, again, you can plug these three values into the dilution relationship and find the molarity of solution B. (6 points/20) B : M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 = (3.95x10 5 mol )(0.002) = 7.90x10 6 M m of a M solution of oleic acid is diluted with 9.00 m of petroleum ether, forming solution A. Then 2.00 m of solution A is diluted with 8.00 m of petroleum ether, forming solution B. How many grams of oleic acid are 5.00 m of solution B? (Molar mass for oleic acid = 282 g/mol) (Note: this problem is different from the previous problem in two ways. First, you are not asked to find the concentration of the solution B, you are asked to find the number of grams of oleic acid in a certain amount of solution B. But, to do this, you still have to go through the complete process of finding the concentration of solution B before you can calculate the number of grams. Second, recognize that the molarity of the given solution is slightly different than what it was in question 13. You cannot just use the value of the concentration of solution B in this question again, you have to go through the whole process of calculating it again with this different given concentration.) (See Dilution Skills and Stoichiometry Skills above) In this problem, there is a sequence of dilutions and you should take them one at a time, the answer to the first becoming the given information for the next. In the first dilution you are given V1 (1.00 m=.001 ) and M1 (3.60 x 10-4 M) and told that it is diluted with 9.00 m of petroleum ether to form solution A. Be careful 9.00 m IS NOT V2. V2 is the total final volume of the solution, and this is 1.00 m m which is 10.0 m (.0100). Therefore, you can plug these three values into the dilution relationship and find the molarity of solution A. A: M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 = (3.60x10 4 mol )(0.001) = 3.60x10 5 M.0100 This molarity then becomes M1 of the next dilution. You are given V1 of this next dilution 2.00 m (.002 ). You are also told that this is mixed with 8.00 m (.008 ) of petroleum ether to give solution B. Be careful again, 8.00 IS NOT V2. V2 is the total final volume of the solution, and this is 2.00 m m which is 10.0 m (.0100). Therefore, again, you can plug these three values into the dilution relationship and find the molarity of solution B. B : M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 = (3.60x10 5 mol )(0.002) = 7.20x10 6 M.0100

8 Chapter Error! Unknown document property name.: Error! Unknown document property name. 76 Finally, you are asked for grams of oleic acid, so immediately write down g = (mol)(mm). You are given the molar mass so you do not have to calculate this or include it within the rest of your calculation you simply use it. However, upon inspecting this formula you do not have mol, but because you know mol = g mm you immediately then substitute g mm for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been calculated molarity of solution B and you have been given a volume (5.00 m = ) so plug these values in and solve. (9 points/29) g = (mol)(mm) = ( M )()(mm) = (7.20x10 6 mol )( )(282 g / mol) = 1.02x10 5 g 15. How many grams of NaCl are contained in 350. m of a M solution of sodium chloride? (See Stoichiometry Skills above) You are asked for grams of NaCl, so immediately write down g = (mol)(mm). As you inspect this equation you know you can find the mm of NaCl easily just include its calculation in your final calculation. You are not given the number of moles, but because you know mol = ( M )() you immediately then substitute ( M )() for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given the molarity (.334 M) and volume (350 m=.350). Plug in the values and solve. (3 points/32) g = (mol)(mm) = ( M )()(mm) = (.334 mol )(.350 )(( ) g / mol) = 6.83g 16. Which of the following aqueous solutions contains the greatest number of ions? a) m of 0.10 M NaCl b) m of 0.10 M CaCl2 c) m of 0.10 M FeCl3 d) m of 0.10 M KBr e) m of 0.10 M sucrose Remember that for every one particle of an ionic substance that dissociates completely, you will obtain a number of particles that is equal to the number of individual particles present within the original particle. Therefore, If 1 particle of NaCl dissociates, you will obtain 2 particles, one Na + and one Cl -. If 1 particle of CaCl2 dissociates, you will obtain 3 particles, one particle of Ca 2+ and two particles of Cl -. If 1 particle of FeCl3 dissociates, you will obtain 4 particles, one particle of Fe 3+ and three particles of Cl -. If 1 particle of KBr dissociates, you will obtain 2 particles, one K + and one Br -. Sucrose is not an ionic compound so this answer can be eliminated.

9 Chapter Error! Unknown document property name.: Error! Unknown document property name. 77 If you have a certain number of moles of NaCl particles, when it all dissociates you will have twice the number of moles of ions in solution. If you have a certain number of moles of CaCl2 particles, when it all dissociates you will have three times the number of moles of ions in solution. If you have a certain number of moles of FeCl3 particles, when it all dissociates you will have four the number of moles of ions in solution. If you have a certain number of moles of KBr particles, when it all dissociates you will have twice the number of moles of ions in solution. You know a formula to find the number of moles of particles: mol = ( M )(). You have been given the molarity and volume of each substance simply find the number of moles for each and multiply times the factor you have determined for each substance: ions for NaCl = (2)(mol NaCl) = (2)( M )() = (2)(.10 mol )(.400) =.08 mol ions ions for CaCl 2 = (3)(mol CaCl 2 ) = (3)( M )() = (3)(.10 mol )(.300) =.09 mol ions ions for FeCl 3 = (4)(mol NaCl) = (4)( M )() = (4)(.10 mol )(.200) =.08 mol ions ions for NaCl = (2)(mol NaCl) = (2)( M )() = (2)(.10 mol )(.200) =.04 mol ions Answer B (1 point/33) 17. What mass of calcium chloride, CaCl2, is needed to prepare of a 1.75 M solution? (See Stoichiometry Skills above) You are asked for grams of NaCl, so immediately write down g = (mol)(mm). As you inspect this equation you know you can find the mm of CaCl2 easily just include its calculation in your final calculation.. You are not given the number of moles, but because you know mol = ( M )() you immediately then substitute ( M )() for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given the molarity (1.75 M) and volume (3.650 ). Plug in the values and solve. (3 points/36) g = (mol)(mm) = ( M )()(mm) = (1.75 mol )(3.650 )(( (2)(35.45))g / mol) = 709 g 18. A g sample of SrCl2 is dissolved in m of solution. Calculate the molarity of this solution. (See Stoichiometry Skills above) You are asked to find molarity. Immediately write down M = mol. You are given volume (112.5 m=.1125 ), but not moles. but because you know mol = g you immediately then substitute g for moles. The mm mm calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have grams and molar mass can be calculated in the problem. Plug in the values and solve:

10 Chapter Error! Unknown document property name.: Error! Unknown document property name. 78 M = mol = g mm ((87.62+(2)(35.45))g/mol) 30.1g =.1125 = 1.69 M (3 points/39) 19. What mass of solute is contained in 256 m of a M ammonium chloride solution? (See Stoichiometry Skills above) You are asked to find grams. Immediately write down g = (mol)(mm). You realize that you haven t been given moles, but because you know mol = ( M )() you immediately then substitute ( M )() for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given liters (.256 ) and molarity (.820 M) so, plug in values and solve. (3 points/42) g = (mol)(mm) = ( M )()(mm) = (.820 mol )(.256)(( )g / mol) = 11.2g 20. A g sample of Ba(OH)2 is dissolved in enough water to make liters of solution. How many m of this solution must be diluted with water in order to make of M Ba(OH)2? (See Dilution Skills and Stoichiometry Skills above) As this is a dilution problem, immediately write down the dilution relationship. You are asked to find V1 ( How many m of solution must be diluted ), so isolate this value. Upon inspecting this formula you realize that you have been given M2 (.100 M) and V2 (1.000 ), but have not been given M1 but you have been given the information to calculate it. You know that M 1 = mol so write this down separately. Inspecting this formula you do not have mol but know that mol = g so substitute this in for moles. mm You have been given grams and can easily find mm as part of the calculation. Once you find the value for M1 plug this value into the original dilution relationship and solve for V1. Remember to turn into m. (6 points/48) M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 M 1 = mol = g mm = = g ( ) (.100M )(1.000).1854 M =.1854M =.539 = 539 m 21. What volume of 18 M sulfuric acid must be used to prepare 1.80 of M H2SO4? (See Dilution Skills above) As this is a dilution problem, immediately write down the dilution relationship. You have been asked to find what volume... must be used, which indicates that you need to find V1, so isolate this variable. Inspecting the

11 Chapter Error! Unknown document property name.: Error! Unknown document property name. 79 formula, find that you have been given M1 (18 M), V2 (1.80 ), and M2 (.215 M). Simply plug the values in and solve. (3 points/51) M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 = (.215M )(1.80) 18M =.022 = 22m 22. How many grams of NaOH are contained in m of a 0.77 M sodium hydroxide solution? (See Stoichiometry Skills above) You are asked for number of grams so you should immediately write out g = (mol)(mm) and let this lead you through the rest of the problem. On inspecting this formula you know that you can easily find mm and show this within the final calculation. You realize that you haven t been given moles, but because you know mol = ( M )() you immediately then substitute ( M )() for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given M (.77M) and volume (5.0 x 10 2 m=.50 ). Plug these values in and solve. (3 points/54) g = (mol)(mm) = ( M )()(mm) = (.77 M )(.50)( ) = 15g 23. An analytical procedure requires a solution of chloride ions. How many grams of CaCl2 must be dissolved to make 2.15 of M Cl? (See Stoichiometry Skills above) You are asked for number of grams so you should immediately write out g = (mol)(mm) and let this lead you through the rest of the problem. On inspecting this formula you know that you can easily find mm and show this within the final calculation.. You realize that you haven t been given moles, but because you know mol = ( M )() you immediately then substitute ( M )() for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. Note you have one additional complicating factor at this point. You have been asked to find the number of g of CaCl2 that must be dissolved to give you.0520 M chloride ion (Cl - ), NOT.0520 CaCl2 M solution. Every formula unit of CaCl2, when it dissociates, will give 2 particles of Cl -. Therefore, you would need only ½ the number of moles of CaCl2 to give.0520 M Cl -. Add this into your factor for moles in the equation. You have been given M (.0520 M) and volume (2.15 ). Plug these values in and solve. (3 points/57) g = (mol)(mm) = (.5)( M )()(mm) = (.5)(.0520 M )(2.15 )( (2)(35.45)) = 6.20g 24. T F The concentration of a salt water solution that sits in an open beaker decreases over time. False as water evaporates and the number of water particles relative to the number of salt particles decreases, the concentration will increase. Answer False (1

12 Chapter Error! Unknown document property name.: Error! Unknown document property name. 80 point/52) 25. You have two solutions of chemical A. To determine which has the highest concentration of A in molarity, what is the minimum number of the following you must know? I. the mass in grams of A in each solution II. III. IV. the molar mass of A the volume of water added to each solution the total volume of the solution a) 0 b) 1 c) 2 d) 3 e) You must know all of them. Understand that while we are seeking which solution has the highest concentration, we don t need to know the actual concentration. Remember, concentration is moles per liter: M = mol. From this you can see that to determine which solution had the greatest concentration you would need to know the number of moles and number of. Note, for molarity it is total liters of solution, not liters of water added so, IV is a definite, while III is not necessary. Moles is not given as a choice or we would choose that. Normally, to find moles, you would also need to have the number of grams and the molar mass. However, as the molar mass would be the same for each solute (chemical A for both), just knowing the number of grams for each, given a certain volume, would be enough. So, I is a definite while II is not necessary. Therefore, 2 of the 4 values would need to be known. Answer C (1 point/58) 26. Diabetics often need injections of insulin to help maintain the proper blood glucose levels in their bodies. How many moles of insulin are needed to make up 45 m of M insulin solution? (See Stoichiometry Skills above) You are asked to find moles. You have two formulas that can help you do this: mol = g or mol = ( M )() mm You should be able to quickly decide that in this case you need the latter formula because you are given M and. You immediately write this formula down, plug in the values and solve. (1 point/59) mol = ( M )() = (.0059 mol )(.045) = 2.7x10 4 mol

13 Chapter Error! Unknown document property name.: Error! Unknown document property name You have two solutions of sodium chloride. One is a 2.00 M solution, the other is a 4.00 M solution. You have much more of the 4.00 M solution and you add the solutions together. Which of the following could be the concentration of the final solution? a) 2.50 M b) 3.00 M c) 3.70 M d) 6.00 M e) 7.50 M First, reason that if you have two solutions of different known concentrations and you mix them together, you would be combining the number of moles present in both solutions and combining the volumes of both solutions and the result could never be less than 2.00 or more than 4.00 mole per liter. This eliminates d and e. Then, if you were to mix equal amounts of solution together it would be easy to see that you should get a solution that is exactly halfway between that is, 3.0 M. However, since you have much more of the 4.00 M solution rather than equal amounts, the only valid choice here would be 3.70 M. Answer C (1 point/60) 28. You have equal masses of different solutes dissolved in equal volumes of solution. Which of the solutes would make the solution having the highest molar concentration? a) NaOH b) KCl c) KOH d) ioh e) all the same As mol = g, saying that you have the same mass of different substances would mm mean that the less the molar mass, the greater the number of moles and the greater the concentration. Therefore, the substance that has the lowest molar mass would have the greatest concentration. The chloride ion has a mm of g/mol while the hydroxide ion has a molar mass of = g/mol. Because there is both a KOH and a KCl, the KOH would have the lesser molar mass between these two. The others are also both hydroxides, so the cation with the least molar mass will give the substance with the least molar mass this would be i. So, ioh would give the least molar mass, and so, the greatest concentration. Answer D. (1 point/61) 29. Which of the following do you need to know to be able to calculate the molarity of a salt solution? I. the mass of salt added II. III. the molar mass of the salt the volume of water added

14 Chapter Error! Unknown document property name.: Error! Unknown document property name. 82 IV. the total volume of the solution a) I, III b) I, II, III c) II, III d) I, II, IV e) You need all of the information. Remember, concentration is moles per liter: M = mol. From this you can see that to determine molarity you would need to know the number of moles and number of. Note, for molarity it is total liters of solution, not liters of water added so, IV is a definite, while III is not necessary. Moles is not given as a choice or we would choose that. To find moles, you know mol = g, so you can see that you also need grams (mass of the salt) and the molar mm mass of the salt. Therefore, you need I, II, and IV. Answer D (1 point/62) 30. A m sample of a M solution is left on a hot plate overnight; the following morning the solution is 1.50 M. What volume of solvent has evaporated from the M solution? Although it may not initially look like it, this problem is basically a dilution problem just in reverse. But is still works the same. (See Dilution Skills above) To find the volume lost from the solution you will need to find the final volume and subtract this from the initial volume. (see below) You are given the initial molarity (M1 =.275 M), the initial volume (230.0 m =.2300 ) and the final concentration (M2 = 1.50 M). Write down the dilution relationship, isolate final volume (V2), plug the values in and solve. (4 points/66) V evap = V initial V final = V 1 V 2 = =.1878 = m M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 M 2 = (.275M )(.230) 1.50M = For the reaction 4FeCl2(aq) + 3O2(g) 2Fe2O3(s) + 4Cl2(g), what volume of a M solution of FeCl2 is required to react completely with molecules of O2? You have been asked to find volume (liters) of a solution of FeCl2 so you immediately write down FeCl2 = mol FeCl 2 M FeCl2. (You do not have to use subscripts but it will make like much easier later in the calculation because you will need to say that a certain number of moles of FeCl2 will be equivalent to a certain number of moles of O2, and the two values for moles can easily get confused if they are not labeled.) This formula

15 Chapter Error! Unknown document property name.: Error! Unknown document property name. 83 tells you that in order to find the volume (number of liters) of the.890m solution, you need to know the number of moles of FeCl2 that must be reacted with molecules of O2. According to the mole ratio of FeCl2 to O2 stated to us by the balanced chemical equation, for every 3 mol of O2, 4 mol of FeCl2 are required. 4 Therefore, we can substitute 3 ( )( molo 2 ) in for mol FeCl2 in our calculation (see below). You have not been given the number of moles of oxygen, but you have been given the number of molecules, and you always know that you can turn a number of molecules into a number of moles by dividing by the number of particles in 1 mole, which is Avogadro s number. Substitute molecules O 2 AN in for molo 2 (see below). Then, substitute in the given values and AN and solve: (3 points/69) FeCl2 = mol FeCl 2 M FeCl2 = 4 ( 3) molo 2 ( ) M FeCl2 = ( ) 4 ( 3) molecules O 2 AN M FeCl2 = 4 (8.71x10 ( 3) 21 molecules) (6.022x10 23 molecules/mol).890 mol =.0216 = 21.6m 32. Phosphoric acid, H3PO4, is a triprotic acid. What is the total number of moles of H + available for reaction in 3.50 of M H3PO4? You are asked for mol H + and you have been given and M, and you know that for a triprotic acid like H3PO4, for every mol of acid, 3 mol of H + will dissociate. So, you immediately write down mol H + = (3)( M )(). You already have all of the information so you plug in the values and solve. (3 points/72) mol H + = (3)( M )() = (3)(.400 mol )(3.50) = The following reactions: Pb I PbI2 2Ce I I2 + 2Ce 3+ HOAc + NH3 NH4 + + OAc are examples of a) acid- base reactions b) unbalanced reactions c) precipitation, acid- base, and redox reactions, respectively d) redox, acid- base, and precipitation reactions, respectively e) precipitation, redox, and acid- base reactions, respectively In the first reaction you can see that all is happening is that lead 2+ ions and iodine 1- ions are combining to form the solid compound lead (II) iodide. The atoms are not changing oxidation states in the reaction, they are merely combining to form an ionic solid, which is the essential feature of a precipitation reaction (note oxidation and reduction do not take place during precipitation reactions).

16 Chapter Error! Unknown document property name.: Error! Unknown document property name. 84 In the second reaction we see cerium being reduced from 4+ to 3+ and iodine being oxidized from 1- to 0, so this is an oxidation reduction reaction. These are also called redox reactions. Beside seeing charges change from the reactant side of the equation to the product side, the quickest way to recognize a redox reaction is by seeing an elemental atom be changed into an atom in a compound (or vice versa) because you know that the oxidation state of an elemental substance is 0, while that of an atom in a compound is not 0. In the third reaction we see a molecule with an available hydrogen at the beginning of its formula (meaning that it is an acid not one of the six strong acids, so you also know that it is a weak acid) and the weak base, NH3. The reaction shows the donation of the hydrogen by the weak acid to the weak base to form the conjugate acid, NH4 + and the conjugate base, OAc -. Therefore, the reactions are: precipitation, redox, acid- base, respectively. Answer E (1 point/73) 34. The following reactions 2K(s) + Br2(l) 2KBr(s) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) HCl(aq) + KOH(aq) H2O(l) + KCl(aq) are examples of a) precipitation reactions b) redox, precipitation, and acid- base, respectively c) precipitation (two) and acid- base reactions, respectively d) redox reactions e) none of these In the first reaction we see that two elemental substances (oxidation state 0) are changed to atoms that are part of a compound (oxidation states not zero ). You don t even need to know what the non- zero oxidation states are you know that they have changed from 0 to something else so it must involve oxidation and reduction. As two elements are combining to form a compound this would also be a synthesis reaction. In the second reaction you can see that this is a double replacement reaction, and that the solid compound silver chloride is a precipitate. This is therefore, a precipitation reaction. Note that as consistent with a precipitation reaction, no elements change oxidation states a precipitation reaction does not involve oxidation and reduction. In the third reaction we can see that a strong acid (HCl) is interacting with a strong base (KOH) resulting in neutralization. This is an acid base reaction.

17 Chapter Error! Unknown document property name.: Error! Unknown document property name. 85 Therefore, the reactions are: redox, precipitation, and acid- base respectively. Answer B (1/74) 35. The following reactions ZnBr2(aq) + 2AgNO3(aq) Zn(NO3)2(aq) + 2AgBr(s) KBr(aq) + AgNO3(aq) AgBr(s) + KNO3(aq) are examples of a) oxidation- reduction reactions b) acid- base reactions c) precipitation reactions d) A and C e) none of these In both circumstances you can see that both reactants are ionic compounds and that the cations switch places with each other with respect to their corresponding anions. They are therefore both double replacement reactions. As you can also see, both reactions result in the formation of a solid compound which precipitates out, which means that both are precipitation reactions all double replacement reactions are precipitation reactions. Answer C (1/75) 36. All of the following reactions 2Al(s) + 3Br2(l) 2AlBr3(s) 2Ag2O(s) 4Ag(s) + O2(g) CH4(l) + 2O2(g) CO2(g) + 2H2O(g) can be classified as a) oxidation- reduction reactions b) combustion reactions c) precipitation reactions d) A and B e) A and C In the first reaction we see that two elemental substances (oxidation state 0) are changed to atoms that are part of a compound (oxidation states not zero ). You don t even need to know what the non- zero oxidation states are you know that they have changed from 0 to something else so it must involve oxidation and reduction. As two elements are combining to form a compound this would also be a synthesis reaction. In the second reaction we see the opposite occurring we see atoms that are part of an ionic compound (oxidation states not zero ) being changed into elemental atoms (oxidation state 0). As with the first reaction, this involves oxidation and reduction.

18 Chapter Error! Unknown document property name.: Error! Unknown document property name. 86 However, in contrast to the first reaction, as a compound is being broken into its component elements this is a decomposition reaction. In the third reaction, we have a carbon compound being reacted with oxygen to form carbon dioxide and water. This is the hallmark of a combustion reaction. However, as the other two are not combustion reactions, we cannot say that all the reactions are combustion reactions. But, we also know that all combustion reactions are oxidation reduction reactions especially as they all involve the change of elemental oxygen (oxidation state 0) to oxygen in a compound (oxidation state - 2). Therefore, all of the reactions could be considered to be oxidation reduction reactions. (Note, you also could not say A and B because not all of the reactions are combustion). Answer A (1/76) 37. You have exposed electrodes of a light bulb in a solution of H2SO4 such that the light bulb is on. You add a dilute solution and the bulb grows dim. Which of the following could be in the solution? a) Ba(OH)2 b) NaNO3 c) K2SO4 d) Cu(NO3)2 e) none of these Because sulfuric acid is a strong acid, the first hydrogen will dissociate completely. This means that sulfuric acid is a strong electrolyte and will allow electric current to flow strongly the light bulb is on. If we add a solution and the bulb grows progressively dimmer, this must mean that something is cancelling out the ionic (charged) particles in the solution. The only solution capable of accomplishing this in an acid solution is a basic solution one that provides OH - ions. Therefore, Ba(OH)2 is the logical choice. Answer A (1/77) 38. Aqueous solutions of sodium sulfide and copper(ii) chloride are mixed together. Which statement is correct? a) Both NaCl and CuS precipitate from solution. b) No reaction will occur. c) CuS will precipitate from solution. d) NaCl will precipitate from solution. e) A gas is released. The answer to this question is based on knowledge of solubility rules. We are told that the initial compounds are both soluble, but when mixed together, we might expect something to precipitate out. Except for group one (IA) metals, when combined with other metals sulfides are highly insoluble. Therefore, we would expect copper sulfide to precipitate out. Answer C (1/78) 39. Aqueous solutions of potassium sulfate and ammonium nitrate are mixed together. Which statement is correct?

19 Chapter Error! Unknown document property name.: Error! Unknown document property name. 87 a) NH4SO4 will precipitate from solution. b) KNO3 will precipitate from solution. c) No reaction will occur. d) Both KNO3 and NH4SO4 precipitate from solution. e) A gas is released. The answer to this question is based on knowledge of solubility rules. We are told that the initial compounds are both soluble, but when mixed together, we might expect something to precipitate out. However, ammonium, nitrate, and potassium compounds are almost always soluble, and sulfates are often soluble, so, according to the solubility rules, there is no combination of ions in the mix given that should precipitate out no reaction should occur. Answer C (1 point/79) 40. Which of the following salts is insoluble in water? a) Na2S b) K2CO3 c) Pb(NO3)2 d) CaCl2 e) All of these are soluble in water. (a) Most sulfides are insoluble however, those of 1A metal cations are soluble Na2S is soluble. (b) Most carbonates are insoluble however, those of 1A metal cations are soluble K2CO3 is soluble. (c) Most nitrates, including lead nitrate, are soluble. (d) Most chlorides, including those of 2A metal cations, are soluble. Answer E (1 point/80) 41. How many of the following salts are expected to be insoluble in water? sodium sulfide ammonium sulfate barium nitrate potassium phosphate a) none b) 1 c) 2 d) 3 e) 4 Most sulfides are insoluble however, those of 1A metal cations are soluble Na2S is soluble. Most nitrates, including barium nitrate, are soluble. Most ammonium compound, including ammonium sulfate are soluble. Most phosphates are insoluble. However, those of 1A metal cations are soluble. None of these compounds should be insoluble. Answer A (1 point/81) 42. When NH3(aq) is added to Cu 2+ (aq), a precipitate initially forms. Its formula is: a) Cu(NH)3 b) Cu(NO3)2

20 Chapter Error! Unknown document property name.: Error! Unknown document property name. 88 c) Cu(OH)2 d) Cu(NH3)2 2+ e) CuO This is a tricky question. Recognize that NH3 is a weak base and its mechanism of action is that it pulls a hydrogen ion from H2O to form OH - ions. This means that Cu 2+ ions that are in solution now are exposed to OH - ions. Most hydroxides of transition metal cations are insoluble, including Cu(OH)2. Therefore, this will precipitate out initially. (However, as the reaction proceeds, Cu 2+ ions will secondarily form complex ions with ammonia molecules Cu(NH3)2 2+ ions and these are soluble so after a short time, the precipitate disappears as it is converted to this complex ion.) Answer C (1 point/82) 43. Which of the following ions is most likely to form an insoluble sulfate? a) K + b) Ca 2+ c) S 2 d) i + e) Cl According to solubility rules most sulfates are soluble, except for those of Ba, Pb, Hg and Ca so, Ca 2+ is most likely to form an insoluble sulfate. You should be able to eliminate the anions S 2- and Cl - immediately because these are not going to form a sulfate. Answer B (1 point/83) 44. Which of the following compounds is soluble in water? a) Ni(OH)2 b) K3PO4 c) BaSO4 d) CoCO3 e) PbCl2 According to solubility rules most hydroxides are insoluble except for a few in column 1A and 2A, so (a) will precipitate; most phosphates are insoluble, except for those of the column 1A so, (b) will actually dissolve; most sulfates are soluble, except for those of Ba, Pb, Hg and Ca so, (c) will precipitate; most carbonates are insoluble, except for those of the column 1A, so (d) will precipitate; most chlorides are soluble, except for Ag, Pb, and Hg, so (e) will precipitate. Answer B (1 point/84) 45. Which pair of ions would not be expected to form a precipitate when dilute solutions of each are mixed? a) Al 3+, S 2 b) Pb 2+, Cl c) Ba 2+, PO4 3 d) Pb 2+, OH e) Mg 2+, SO4 2

21 Chapter Error! Unknown document property name.: Error! Unknown document property name. 89 According to solubility rules sulfide precipitates out almost everything so (a) will precipitate; most chlorides are soluble, except for Ag, Pb, and Hg, so (b) will precipitate; most phosphates are insoluble, except for those of the column 1A, so (c) will precipitate; most hydroxides are insoluble except for a few in column 1A and 2A, so (d) will precipitate; most sulfates are soluble, especially those of columns 1A and 2A, so, (e) will not precipitate. Answer E (1 point/85) 46. A solution contains the ions Ag +, Pb 2+, and Ni 2+. Dilute solutions of NaCl, Na2SO4, and Na2S are available to separate the positive ions from each other. In order to effect separation, the solutions should be added in which order? a) Na2SO4, NaCl, Na2S b) Na2SO4, Na2S, NaCl c) Na2S, NaCl, Na2SO4 d) NaCl, Na2S, Na2SO4 e) NaCl, Na2SO4, Na2S Most sulfate salts are soluble, except for those of Ba, Pb, Hg, and Ca. So, add NaSO4 first and only PbSO4 will precipitate out. Filter this out of the solution, leaving Ag and Ni ions in solution. Next, most chloride salts are soluble except for those of Ag, Pb and Hg. There is no more Pb, so add NaCl and AgCl will precipitate out. Filter this out of the solution leaving only Ni ions in solution. Then, add Na2S. This will precipitate almost anything, including the nickel. Answer A (1 point/86) 47. Consider an aqueous solution of calcium nitrate added to an aqueous solution of sodium phosphate. What is the formula of the solid formed in the reaction? a) Ca(PO4)2 b) CaPO4 c) Ca3(PO4)2 d) Ca3(PO3)2 e) none of these Make sure you know how to write the correct formulas for all compounds involved. If necessary to help visualize write out the formulas for the compounds, or even the reactant side of a chemical equation so you can see the correct chemical formulas: Ca(NO 3 ) 2 (aq) and Na 3 PO 4 (aq) In this case, both substances are individually soluble. However, added together your options for possible precipitates are: Ca 3 (PO 4 ) 2 and NaNO 3 From your solubility rules you know that all nitrates are solube especially one combined with column 1A metals. You also know that phosphates (except for those combined with metals in column 1A) are insoluble. So, Ca 3 (PO 4 ) 2 would be the correct answer. Answer C (1 point/87) 48. T F The filtrate is the solid formed when two solutions are mixed.

22 Chapter Error! Unknown document property name.: Error! Unknown document property name. 90 False a solid formed when two solutions are mixed is called the precipitate. The filtrate is liquid that passes through a filter when a liquid mixture containing solid particles is passed over a filter. (1 point/88) Given the following information for questions 49-51: Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. Note these test questions use the term molecular equation to indicate the formula equation we talked about in lecture. 49. The balanced molecular equation contains which one of the following terms? a) AgCl (s) b) 2AgCl (s) c) 2Ba(NO3)2 (aq) d) BaNO3 (aq) e) 3AgCl (aq) Write out the complete balanced formula equation (molecular equation). Remember that this is all of the substances involved, all written in compound form, with states of matter included. You are given the names of all of the compounds so it should be easy to write this equation out and balance it (include states of matter). BaCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Ba(NO 3 ) 2 From the balanced equation you should quickly be able to see that 2AgCl is the correct choice). Answer B (1 point/89) 50. The balanced complete ionic equation contains which of the following terms? a) 2Ba 2+ (aq) b) Cl (aq) c) 2Ag + (aq) d) NO3 (aq) e) AgCl(aq) Remember, this is all substances involved. For dissolved ions, the specific ions are written out, including the aq state of matter. For precipitated substances, the compound formula is written out including the s state of matter. From the formula (molecular) equation you should quickly be able to write out the complete ionic equation, inspect it, and select the correct answer. Answer C (1 point/90) Ba 2+ (aq) + 2Cl (aq) + 2Ag + (aq) + 2 NO 3 (aq) 2AgCl(s) + Ba 2+ (aq) + 2NO 3 (aq) 51. The net ionic equation contains which of the following terms? a) Ag + (aq) b) Ba 2+ (aq)

23 Chapter Error! Unknown document property name.: Error! Unknown document property name. 91 c) NO3 (aq) d) H + (aq) e) AgCl(aq) Remember that the net ionic equation includes only those ions/substances involved in the action. In this case, silver and chloride ions combined to form solid silver chloride. Therefore, these are the only substances listed in the equation. This should appear in reduced form. You should be able to quickly write out these formulas and select the correct answer from the list. Answer A (1 point/91) Ag + (aq) + Cl (aq) AgCl(s) 52. In writing the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of KOH and Mg(NO3)2 are mixed, which of the following would not be written as ionic species? a) KOH b) Mg(NO3)2 c) Mg(OH)2 d) KNO3 e) All of the above would be written as ionic species. You should be able to create a complete ionic equation from the description of a reaction without having to write down the complete formula (molecular) equation. You know that on the reactant side, both substances will be in solution. Therefore, all four ions making up these substances will be written as an ionic species on the reactant side. K + (aq) + OH (aq) + Mg 2+ (aq) + 2 NO 3 (aq) (At this point, you wouldn t know it, but after balancing, the K + and OH - ions would have coefficients of 2) Then, because you know your solubility rules, you would know that Mg(OH)2 will precipitate out, but the other ions would remain in solution. Add these substances on the product side, and balance the equation. 2K + (aq) + 2OH (aq) + Mg 2+ (aq) + 2 NO 3 (aq) Mg(OH ) 2 (s) + 2K + (aq) + 2 NO 3 (aq) Then, select the correct answer Mg(OH)2 would not be written as an ionic species. Answer C (1 point/92) 53. The net ionic equation for the reaction of calcium bromide and sodium phosphate contains which of the following species? a) 2Br (aq) b) PO4 3 (aq) c) 2Ca3(PO4)2(s) d) 6NaBr(aq) e) 3Ca 2+ (aq) You should be able to create a net ionic equation from the description of an equation. You understand that from the description, before a reaction occurs, Ca 2+, Br -, Na +, and