SAT Chemistry

TEST 1

SAT ChemistryTEST 1

DETAILED EXPLANATIONSOF ANSWERS

I ANSWER KEY IPART A PART A1. IC) 7. IE)2. (D) 8. (C)

13. IE) 19. (E)3. (B) 9. (G)

14. IB) 20. IE) I. (e)4. (B) 10. IB)

15. IC) 21. (D) The metalloids have characteristics of both the metals and the non-

5. (D) 11. (D)16. IA) 22. (B) metals. Metals are conductors while nonmetals are insulators.

6. IA) 17. (D) 23. IE)12. (D) 18. (A) 2. (D)

PARTBThe halogens (Group VII A) el" Br2' and 12 are examples of di-

101.

atomic molecules. None of the other choices occur as diatomic molecules.

T, F 105. F, T102. F, F 106. F, F

109. F, T 113. F, F 3. (B)103. T, T, GE 107. T, T

110. F, T 114. T, T, CE An oxide of formula X10 indicates that X has an oxidation state of104. F, T 108. F, F

111. T, F 115. T, T, GE +1 since oxygen has an oxidation state of -2. The alkali metals (Group IA)112. T, T, GE 116. F, F have an oxidation state of +1. The alkali earth metals (Group IIA) usually

PARTehave an oxidation state of +2, thus giving an oxide with the formula XO.

24. (G) 36.

The metalloids have varying oxidation states as do the rare earths. The

25. (E)halogens usually have an oxidation state of -1.

(D) 48. (A)26.

37. (B) 60. (D)IA) 49. (B)

27.38. (D) 61. (E) 4. (B)

IE) 39. (E)50. (B) 62. (E) The sulfate group has an oxidation number of -2. Thus, an oxidation

28. (D) 51. IB)29.

40. IG) 63. IC)number of +2 is required to produce an acid salt with the formula XS04

(D) 52. IE)30.

41. (D) 64. (D) The alkaline earth metals usually have an oxidation number of +2.(E) 42. (B)

53. IG) 65. (C)31. IE) 54. (G)43. (G) 66. (G) 5. (D)32. (A) 55. IB)33.

44. (E) 67. (E) The halogens have the largest electronegativity values since they re:(A) 56. IB)

34.45. (D) 68. (E) quire only one electron to completely fill their valence shell. The alkali

(A) 57. (B)35.

46. (G) 69. (A) metals have the smallest electronegativity values since they can lose one(A) 47. (B) 58. (E) electron and have a complete valence shell. The alkali earth metals ~so59. (A) have small electronegativities since they can lose two ~lectrons to attal~a

~oblegas configuration. The metalloids have intermediate electronegauv-tty values.

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enzo-!::tenza:

SAT Chemistry

SAT ChemistryTEST 1

I ANSWER KEY IPARTA1. (C) 7. (E) 13. (E) 19. (E)2. (D) B. (C) 14. (8) 20. (E)3. (B) 9. (C) 15. (C) 21. (D)4. (B) 10. (B) 16. (A) 22. (B)5. (D) 11. (D) 17. (D) 23. (E)6. (A) 12. (D) 18. (A)

PART a101. T, F 105. F, T 109. F, T 113. F, F102. F,F 106. F, F 110.F,T 114. T, T, CE103. T, T, CE 107. T, T 111. T, F 115. T, T, CE104. F, T 108. F, F 112. T, T, CE 116. F, F

PARTe24. (C) 36. (E) 48. (A) 60. (D)25. (D) 37. (B) 49. (B) 61. (E)26. (A) 3B. (0) 50. (B) 62. (E)27. (E) 39. (E) 51. (8) 63. (C)28. (D) 40. (C) 52. (E) 64. (D)29. (D) 41. (0) 53. (C) 65. (C)30. (E) 42. (B) 54. (C) 66. (C)31. (E) 43. (C) 55. (8) 67. (E)32. (A) 44. (E) 56. (B) 68 (E)33. (A) 45. (0) 57. (8) 69. (A)34. (A) 46. (C) 58. (E)35. (A) 47. (B) 59. (A)

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DETAILED EXPLANATIONSOF ANSWERS

TEST 1

PART A

L (e)The metalloids have characteristics of both the metals and the non-

metals. Metals are conductors while nonmetals are insulators.

2. (0)The halogens (Group VII A) Cl" Br,. and I, are examples of di-

atomic molecules. None of the other choices occur as diatomic molecules.

3. (B).An oxide of formula X20 indicates that X has an oxidation state of

+1 SInce oxygen has an oxidation state of -2. The alkali metals (Group IA)have an oxidation state of +1. The alkali earth metals (Group I1A) usuallyhave an oxidation state of +2, thus giving an oxide with the formula XO.The metalloids have varying oxidation states as do the rare earths. Thehalogens usually have an oxidation state of-1.

4. (B)The sulfate group has an oxidation number of -2. Thus, an oxidation

number of +2 is required to produce an acid salt with the formula XS04The alkaline earth metals usually have an oxidation number of +2.

5. (0). The halogens have the largest electronegativity values since they re:

quire only one electron to completely fill their valence shell. The alkalimetals have the smallest electronegativity values since they can lose oneelectron and have a complete valence shell. The alkali earth metals alsohave small electronegativities since they can lose two electrons to attai~ a~oble gas configuration. The metalloids have intennediate alectroneganv-tty values.

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6. (AlThe smallest ionization energies are realized when the removal ofone

electron yields a noble gas configuration. Thus, the alkali metals havethelowest ionization energies, then the alkali earth metals, metalloids, andhalogens in order of increasing ionization energy.

7. (E)A buffer is a solution made by the combination of a weak acid anda

salt of its conjugate base. For example, acetic acid combined with sodiumacetate would comprise a buffer. A buffer is intended to be insensitive[0modest changes in pH.

8. (C)Both strong acids and bases dissociate nearly completely in aqueous

solution. For example, Hel is termed a strong acid because in aqueoussolution it dissociates into H+ and CI- and the concentration of Hel itselfis negligible.

9. (C). As a strong acid dissociates essentially completely, its acid dissocia-tion constant, Ka will be extremely high. For HCl:

[H+][CI-]Ka =:[HCI]

Since [H+] and [Cl"] are very large compared with (HCI), K" will be verylarge as well.

10. (B)

The Br0nsted-Lowry notion of an acid-base reaction requires a pro-tontobetranf d I . and ab sterre . n this concept an acid must donate a protonase must accept one.

11. (D)

NH3 is a weak base because in the reaction

NH,(aq) +H20

Detailed Explanations of Answers I Test 1

12. (0)Cl is also considered to be a weak base. This is so because it is the

conjugatebase of a strong acid.

13. (E)NaHC03 is a salt. It is a salt of the weak acid H,C03

14. (B). NaOH is a strong base because it dissociates completely to give OH-IOns.

15. (C)The oxidation number of Na in NaCI is +1. In all monatomic ions, the

oxidationnumber is that of the charge on that ion.

16. (A)The oxidation number of CI in e12 is zero (0). In any elementary

substance. the charge of an element is zero (0). The charge of 0 in 2H inHz. are also 0. for example.

17. (0). The oxidation number of S in Na-S is -2. Because the charge on Na15 + I and there are two of them in order to create a neutral compound. theoxidationnumber of S must be ~2.

18. (A)Solid sulfur is yellow. In addition, it smells like rotten eggs.

19. (E)Aqueous hydrochloric acid is colorless.

20. (E)Aqueous sodium hydroxide is colorless as well.

21. (0)In the reaction given, since there is an excess of oxygen. it is the

amount only of ammonia put in that will determine how m~ch producttherewill be. For every 4 moles of ammonia put into the reacuon- 6 molesof water will be obtained. The first step is to determine how many molesare in 16.00 grams of ammonia. Dividing J 6.00 by the molecular weightof ammonia:

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SAT Chemistry

4 moles ammonia 0.9395 moles ammonia~6 moles water x moles water

16.00 g ~ 0.9395 moles NH317.03 g/mole

Using cross multiplication, if for every 4 moles of NH3 you get 6 molesofwater:

x :;::1.409 moles of water. To convert this to grams, we multiply by themolecular weight of water:

1.409 moles H20X 18.02 glmol ~ 25.39 g H20.

This corresponds to choice (D).

22. (B)Here you must determine which is the limiting reagent. First yo,u

must convert the masses of both the ammonia and the oxygen to therrrespective number of moles by dividing by the molecular weights.

66.00 g NH,17.04 g/mole ~ 3.873 moles NH3

54.00 g 0,32.00 g/rnole ~ 1.688 moles 0,

Oxygen is therefore the limiting reagent.. For every 5 moles of oxygen put in, you get 6 moles of water OUl.

Usmg cross multiplication:

:i.moles oxygen _ 1.688 moles oxygen6 -moles water x moles waterx ~ 2 025 m I th olecul~

.:" 0 es water. Convert to grams by multiplying by e mweight of water:

2.025 moles X 18.02 ~ 36.49 grams O2

This Corresponds to choice (B).

23 (E)For every 4 I . 4 molesof

NO rna es of NH] that are put into the reacuon. 513mol are .Plroduced. Therefore, if 2513 moles of NH

3are reacted, 2,

esw1lbeprod d' .102 uce . ThIS is choice (E).

Detailed Explanations of Answers / Test 1

PARTS

101 T, FAcid rain is generally considered to be destructive to limestone

(CaCO]) because acid rain contains sulfuric acid (H2S04) or nitric acid(HNO]).The reaction that occurs is

CaCO, (5) + H2S04 (aq) -> CaSO, (5) + CO, (g) +H20

102. F, FThe correct net ionic equation is

Ca'+ (aq) + 504'"- (aq) -> CaSO, (5).NaClis soluble in water and does not appear in the final net ionic equation.

103.T, T, CEKspequals the product of the equilibrium concentrations of the ions in

a compound where each concentration is raised to the power of the ioncoefficient.

104. F, TIn a neutral solution, [H+] = 1x 10-7. This is equivalent to saying that

~solution has a pH of 7. This is not true in an acidic or basic solution. Kw

IS, however, always equal to I x 10--14.

105.F, TStrong acids do ionize completely in an aqueous solution, but HF

doesnot ionize completely and is therefore considered a weak acid.

106. F, FNH

4+ is a weak acid because it doesn't dissociate completely in wa-

ter. NH4+ is the conjugate acid of NH3, a weak base.

107. T, TBoth statements are true but the latter is not a correct explanation of

the former. By the Lewis definition, F- is a base because it can donate anelectronpair.

108. F, FBecause normality depends on the number of transferable protons in

a Speciesand molarity doesn't, a IN solution of H2S04 is half as concen-tratedin terms of H2S04 as a 1M solution of H2S04. 103

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SA T Chemistry

109. F, T .

These solutions together would not make a good buffer. A buffer ISmade by a weak acid and the salt of its conjugate base or vice versa. It IStrue that OH- is a common ion, but this is irrelevant to use as a buffer here.

no. F, TLike any other elementary substance, the CI in el

2has an oxidationnumber ofO.

]J I. T, F

Water has a high boiling point for its molecular weight due to hydro.gen bonding. Polar covalent bonding also occurs in methane (CH

4) which

has a boiling point of -1620C.

112. T, T, CE

Benzene is a poor electrolyte because it does not ionize in watersolution.

113. F, F

The reaction of zinc with hydrochloric acid goes to completion be-cause the hYdrogen gas which is evolved is allowed to escape. This is ineffect removing one of the reaction products causing the equilibrium toshift towards completion of the reaction.

114. T, T, CE

Atoms of the same element form Covalent bonds because their elec-tronegativity values are the same. Bonds formed between elements whoseelectronegativities differ from 0.5 to 1.7 form polar covalent bonds. Dif-ferences greater than 1.7 in element electronegativities result in ionicbonds.

115. T, T, CE

Most metals characteristics SUchas malleability, flexibility, strength,and electrical conductivity are characteristic of the positive atomic nucleisurrounded by mobile electrons.

116. F, F

Acetic acid is a weak acid because it is only Partially dissociated in awater SOlution.

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Detailed Explanations of Answers / Test 1

PARTe

24. (C)Referring to the periodic table we see that element 32 is germanium.

~ennanium is a metalloid as are boron, silicon, arsenic, antimony, tellu-num,polonium, and astatine. Chemically, metalloids exhibit both positiveand negative oxidation states and combine with metals and nonmetals.They are characterized by electronegativity values between those of themetalsand the nonmetals.

25. (0)The prime consideration in representing the bonding of a polyatomic

element or compound is that each atom bonded should have a completevalence shell (eight electrons except hydrogen and helium which havetwo). Since nitrogen is in Group VA, it has five valence electrons illus-tratedas

:N

Diatomic nitrogen must have the structure

:N : : :N : (or :N" N :)

to completely fill the valence shells of both atoms.

26. (A)Dissolving sodium chloride in water is an example of a physical

change. A physical change alters the physical properties of a substancewhile maintaining its composition. If the water solution of NaCI were tobe evaporated we would once again have solid sodium chloride. Chemicalchanges involve altering the composition and structure of a substance andare always associated with changes in energy. Wood and oxygen arechanged to CO

2, H

20 and nitrogen oxides while ozone is changed to

diatomic oxygen and sodium and water are changed to sodium hydroxideand hydrogen gas.

27. (E)A beta particle is a fast electron of mass 9.11 x lO....28 g while a proton

and a neutron both have a mass of 1.67 x 1O-24 g. A hydrogen nucleus is aproton, and an alpha particle is a helium nucleus (two protons and twoneutrons). Thus, the electron (beta particle) has the smallest mass of thechoices given.

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SA T Chemistry

28 (0)One mole of NaCI weighs 58.5 g as obtained by: the atomic weightof

Na plus the atomic weight of CI from the periodic table. Thus, 58.5 g ofNaCI in one liter of solution is 1 molar and 58.5 g of NaCI in one kilogramof solvent is I molal. By simple proportions, 58.5 g of NaCI in 2 kilo-grams of solvent is 0.5 molal.

29. (0)Molecules in the gaseous state have the greatest kinetic energy. The

difference in energy between the liquid and gas phases is greater than thedifference in energy between the solid and liquid phases. This may be readilyseen by the energy changes occurring in water; the heat of fusion of wateris80 calories/gram, while the heat of vaporization is 540 calories/gram.

30. (E)Cr(NH)sS04Br represents 27 atoms. They are: I x Cr; 5 x N:

15 x H; I x S; 4 x 0; and I x BI.

3l. (El. The element of atomic weight 197 is gold (Au-atomic number 79).

Since the atomic weight is equal to the number of protons and neutrons mthe nucleus and the atomic number is equal to the number of protons in thenucleus, the number of neutrons in the nucleus is 197 _ 79 or 118.

32. (A)

Diamond, composed solely of carbon cannot have ionic bonds orhydrogen bonds. Van der Waals attraction between the nucleus of oneatom and the electrons of an adjacent atom are relatively weak cornpar~dt~ the covalent bonding network (sp3 hybrid) between the carbon atoms IndIamond. On the other hand, graphite (another allotropic form of carbon)IS sp2 hybrid and not strongly bonded as compared to diamond.

33. (A)

. Atomic radius decreases as one goes from left to right across a p~-n~d,. so the atomic radius of carbon is greater than that of oxygen. ThiSehmmates choices (B) and (D). Now we must determine whether the CO,molecule is linear Or bent. Linearity means the O-C-O bond angle is 180.Recall that the nu I' f eriencec er 0 a molecule anent themselves so as to exp .the smallest repUlsions of the positive nuclei. Thus a triatomic molecule ISexpected to he Ii . C . . The lone. near as IS 02' However this IS not always true.electron pairs 0" that the

n oxygen III a water molecule bend the molecule so

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aDetailed Explanations of Answers / Test 1

hydrogen nuclei and the two electron pairs occupy the comers of a tetrahe-dron. Thus, the water molecule is bent. The same effect occurs in ammo-nia, NH3, where a lone pair of electrons on nitrogen distorts the expectedtrigonal planar geometry. Shape of a CO2 molecule is

:0= C =0:

~180)34. (Al

All three phases (solid, liquid and gas) may coexist at a single pres-sure/temperature combination known as the triple point. This point occursat the intersection of the solid-liquid, solid-gas and liquid-gas equilibriumCurves as illustrated by point C.

35. (A)Examining a labeled phase diagram we see that the solid phase can

only exist at point A.

G liquid H

D BPressure

solid E

A C F

gas

Temperature

36. (El .The critical point is the point above which a gas cannot change mto a

liquid. This means that a liquid cannot exist above this poi~t~ but a~ an.dbelow this point a liquid can exist. The temperature at the cntl~~l point IScalled the critical temperature and the pressure is called the c.f1tIcalp,res-Sure,The critical point in the phase diagram shown is the pomt H, smceabove it a gas cannot be liquefied.

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SA T Chemistry

37. (B)Referring to the phase diagram previously given we see that the solid

and liquid phases coexist on the line upon which point D is located.

38. (0)An electrolyte is a substance which, when melted or dissolved in a

suitable medium, conducts electricity. Therefore (D), a solution of sodiumchloride dissolved in water, is electrolytic since the sodium and chlorideions are free to move. Neither (A) mercury in water, (B) a benzene solu-tion of ethanol, nor (C) sucrose in water, are electrolytic.

39. (E)The oxidation state of sulfur in sodium bisulfate may be determined

by recalling that the oxidation states of sodium, hydrogen, and oxygen areusually +1, + I, and -2, respectively. Since the sum of the oxidation stalesfor the atoms of a neutral compound are zero we have:

oxidation state of S + 1 + 1 + 4(-2) = 0; thereforeoxidation state of S :;::+6

So, the oxidation state of sulfur in NaHS04 is +6.

40. (elThe solution in question had been supersaturated as is seen by the

precipitation of more solute than what had been added. The same amountof solute would have precipitated if the solution was saturated and noprecipitation would have occurred if the solution was unsaturated. Theterms dilute and concentrated cannot be used in this context since a dilutesolution may be saturated if the solute is only slightly soluble while aconcentrated solution may be unsaturated if the solute is exceptionallysoluble.

41. (0)This is an exam If cid andap e 0 a neutralization reaction where an a

base ~eac~to produce water and a salt. It must be known that barium hasan oxidation number of +2 and that perchloric acid is HCI04.

2HCI04 + Ba(OH)2 --'> 2H,O + Ba(CI04),

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Detailed Explanations of Answers / Test 1

42. (B)An endothermic reaction is one in which heat may be considered one

ofthe "reactants." An exothermic reaction releases heat upon formation oftheproducts. An equilibrium reaction may be either exothermic or endo-thermic.The same holds true for spontaneity; spontaneity can only bedeterminedif one also knows the entropy change (~) for the reaction.

43 (e)The transition metals have highest energy electrons in d subshells

(3d, 4d, and 5d). Lanthanides and actinides are characterized by highestenergyelectrons in the 4jand 5jsubshells, respectively.

44. (E)Neutral fluorine atoms have 9 electrons as determined by their atomic

number.Magnesium atoms have 12 electrons so Mg3+ has 9 electrons.Boronhas 5 electrons so B3- has 8 electrons (the same as oxygen). Nitro-genhas 7 electrons so N+ has 6 electrons (the same as carbon). Neon has10electrons so Ne- has 11 electrons (the same as sodium). Sodium has] 1electronsso Na- has 12 electrons (the same as magnesium).

45. (0)Le Chatelier' s Principle may be used to predict equilibrium reactions.

If a stress is placed on a system in equilibrium, the equilibrium shifts so asto counteract that stress. Hence, increasing the reactant concentration fa-vorsformation of the products while decreasing the reactant concentrationfavors formation of the reactants. The same holds true for altering theproductconcentrations. Increasing the temperature favors the reaction thatabsorbs heat while decreasing the temperature favors the reaction thatreleasesheat. Increasing the pressure favors the reaction that decreases the~olumeof a closed system while decreasing the pressure favors the reac-tion resulting in an increased volume (moles of gaseous product prod.ucedare the only things counted since liquids and solids occupy a relativelysmall volume in comparison). However, temperature and pressure depen-denciescannot be inferred from this question. The addition of a catalystalters the reaction rate but not the position of equilibrium. The only wayCompletioncan be obtained is that we remove the products as they aref . .armed. Now the state of the reaction becomes nonequilibrium, but It tnesto come into an equilibrium state once again. This lea~s to fo~ation ofmoreproducts which in turn leads to completion of the given reaction.

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46. (e)An acid H3X is classified as triprotic since it may "give up" three

protons to a base. An example of a triprotic acid is phosphoric acid,H3P04 Examples of monoprotic and diprotic acids are hydrochloric, Heland sulfuric, H2S04, respectively. The term bidentate, rather than referringto acids, is associated with ligands. Bidentate ligands have two atoms thatmay coordinate to a metal ion.

47. (B)The salts of strong bases and weak acids hydrolyze to form a basic

solution while the salts of weak bases and strong acids hydrolyze to forman acidic solution.

Ha(N03), + 2H,o -> 2HN03 + Ba(OH)z

A neutral solution is produced since both nitric acid and barium hydroxideare completely dissociated, and each is present in the same concentration(barium hydroxide has 2 hydroxy groups)

Na,S + 2H,O -> H,S + 2NaOH

A ~asic solution is produced since hydrosulfuric acid is a weak acid andsodium hydroxide is a strong base.

AI,(S04h + 6H,O -> 3H,S04 + 2Al(OH)3

An ac!dic ~ol.ution is produced since aluminum hydroxide is insoluble andsulfunc acid IS a strong acid.

Pb3(P04), + 6H,O -> 2HJPO, + 3Pb(OHjzAn acidic solution is produced since phosphoric acid is a weak acid andlead (II) hydroxide is insoluble.

NaCI + H20 -> HCI +NaOH

A n.eutral solution is produced since hydrochloric acid is a strong acid andsodium hydroxide is a strong base.

48. (A)

The reaction of the Haber process is:

N, (g) + 3H, (g) 2NH3 (g) + heatWe see that th 4 . and2I ere are moles of gas on the left side of the reactionmo es of gas th . . . creas-. h On e fight. According to Le Chatelier's PrinCiple, In

mg t e. pressure would force the rea on to the right to produce mo~ammon,. (Note Ch L, lbrium IIIhi h' . ange III pressun ffects only those equutw IC a zas or gas110 0 es are reactants orI cts.)

Detailed Explanations of Answers I Test 1

....

49. (B)The ionization energy is defined as the energy required to remove the

most loosely bound electron from an element in the gaseous state. Theenergy released by an element in forming an ionic solid with anotherelement is the lattice energy of that ionic compound. The electronegativityof an element gives the relative strength with which the atoms of thatelement attract valence electrons.

50. (B)Plutonium-238 has a mass of 238 and an atomic number of 94. The

atomic mass tells us the number of protons and neutrons in the nucleuswhile the atomic number tells us the number of protons. An alpha particle(a) is a helium nucleus j: He composed of 2 neutrons and 2 protons(atomic mass of 4). Hence, upon emitting an alpha particle, the atomicnumber decreases by 2 and the atomic mass decreases by 4. This gives us2~~X. Examining the periodic table we find that element 92 is uranium.Thus~our new atom is 2j: U. 2j: Pu and 2j:em are impossible since theatomic number of plutonium is 94 and that of curium is 96. 2~PU and2~~Cmare impossible since these nuclei could only be produced by fusionof 2lPuwith an alpha particle. In addition, ~~ Pu is incorrectly named.

The reaction (decay) is 238Pu~ 234U~ 4He94 92 2

51. (B)First we determine the number of moles present in solution taking the

molecular weight of ethylene glycol to be 62 g. Thus,

6.20 g x = 0.1 mole of ethylene glycolWemust also know the molality-the ratio of moles of solute to kilogramsof solvent. The number of kilograms of solvent is

0.2 L x I kg = 0.2 kgI L

since the density of water is I g/rnL. The molality of the solution is

Q.I mole0.2 kg = 0.5 molal.

For Ii0 . 1.86 'CimolaI.2 , the molal freezing point depression constant IS

Thus, the freezing point depression is

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SAT Chemistry

0.5 molal X 1.86 'C ~ 0.93 'Cmolal

Thus, the melting point would be

o 'C - 0.93 -c ~ -0. 93 "C

52. (E)A 100 g sample of this gas contains 25.9 g of nitrogen and 74.1.g of

oxygen. Dividing each of these weights by their respective atomic weighsgives us the molar ratio of N to for the gas. This gives

N 25.9074.1 = N1.8504.63- -14 16

Dividing both subscripts by the smallest subscript gives

N1.8504.63 =N102.51.85 1.85

Doubling both subscripts so as to have whole numbers gives us N20s.

4.48 Lx 16 g ~3.2 g22.4L

53. (C) .A molecular weight of 16 g tells us that a volnme of 22.4 hte"

(molar volume) of that gas weighs 16 g. To determine the weight of a4.48 L sample we multiply

54. (C)Th I hi . volving'ere ations rp M1 VI = M2V2 in neutralization problems III

strong acid and strong base. We have

(3 M HCI)(2,000 mL) ~ (5 M NaOH)(V2

mL)

~ ~ (3M)(2,000 mL) _ I 202 (5M) -, OmL

55. (B)A 1M f sulrate. . sodium sulfate (Na2S04) solution contains one mole a I of

IOn per .hter of solution. Thus 0.2 L of aIM solution contains 0.2 molertesulfate IOn 02 L f 2' . Ie ofsu a

. '. 0 a M solutIOn would then con tam 0.4 rooIon.

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Detailed Explanations of Answers I Test 1

56. (B)The atomic weight of lead is 207.2 g/mole from the periodic table.

The number of moles present in 103.6 g of lead is given by

1 mole103.6 g x = 0.5 mole

207.2 g

Since lead is in the +2 oxidation state, two moles of electrons are requiredfor every mole of lead to reduce it to the metals. However only one moleof electrons is required to reduce 0.5 mole of Pb2+ to Pb".

57. (B)Copper is being reduced from Cu2+ to the metal according to

Cu2+ + 2e- ---7 Cu

The amount of electricity that allows one mole of electrons to undergoreaction is the faraday (F) which is equal to 96,500 coulombs. Thus, twofaradays of charge are required to reduce one mole of Cu2+ to the metal.Now we must calculate the number of coulombs provided by the appliedCurrent

3.0 amps x 2 hOUISx 3,600 sec x I coulomb = 21 ,600 coulombsI hour I amp sec

Calculating the number of faradays donated to the copper we obtain

21,600 coulombs x 1 F =0.22 F96,500 coulombs

Now we may compute the amount of copper deposited by this amount ofcharge since we know that 2F of charge reduces one mole of Cu

2+ to CUD.

Thus, we have

0.22 Fx 1 moleCu x 63.5 g Cu =7.1 g2 F 1 moleCu

7.0 g of Copper deposited.

58. (E). The hydrogen electrode has been chosen as the standard electrode

WIth an assigned value of EO = 0.00 V.

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59. (A)The anode of any electrochemical cell is defined to be the siteof

oxidation. Thus, since Zn is being oxidized to 202+ in this cell it is deter-mined to be the anode. The cathode, the site of reduction, is Cr in thiscell.The solutions of the metal ions are not the anode or the cathode but ratherthe electrolytic medium.

60. (0)An sp2 configuration is represented by the trigonal planar orientation,

which looks like

61. (E)

. An sp3cfl configuration is represented by the square planar orienta-tron, which looks like the following. Note: two hybridizal pairs of elec-trons are unbounded.

--~"""

62. (E)

E~amining the balanced reaction equation of 2HzO ----7 2Hz + 02 yieldsthe ratio 2:2: 1.

63. (e)

Balancing the reaction equation gives

2H,O --? 2H, + 0,As may be seen 1 h . duce one. rom t e equation two units of water react to prounit 01 oxygen Th 10 L ' f oxygen. us of water are required to produce 5 L 0

114

Detailed Explanations of Answers I Test 1

4

64. (0)The molecular weight of a compound is the sum of its constituents'

atomic weights. Elements or groups followed by a subscript have theiratomicweight multiplied by that subscript. Thus, the molecular weight ofperchloricacid (HCI04) is

atomic weight of H + atomic weight of Cl+ 4 x atomic weight of 0

or I g/mol + 355 g/mol + 4(16 g1mol)= 1005g/mol

65. (C)The first member of the alkyne series is acetylene (or ethyne), whose

structure is

HC"CH

Thesecond is propyne: HC" C - CH3

Thethird is butyne: HC" C - CH, - CH3Note that there are no analogous compounds in the alkene or alkyne seriesfor the first member of the alkane series (methane - CH4)

66. (C)Alcohols are named by replacing the -e of the corresponding hydro-

carbon name by the suffix -ol. The position of the hydroxy substituent ISnumbered from the shorter end of the chain. Thus, the structure is named3-hexanol. It is a hexanol because the parent hydrocarbon has six carbonsand the prefix 3- (not 1-) is used to indicate the location of the hydroxygroup on the third carbon.

67. (E)The metals are found on the left side of the periodic table, with

~etallic character increasing as one goes down a group. All the choicesgiven are in Group IA, so the one farthest down in the group has thegreatest metallic character. This is francium (Fr).

68. (E)The oxidation states of the element comprising a neutral compound ~ust

~avea sum of zero. Thus, nitrogen in HN03 has an oxidation sta~eof +5, sincerdrogen and oxygen have oxidation states of +1 and -2, respectively.

i.e., (I) + (n) + (-{j) = 0; therefore

n=5115

SA T Chemistry

(025 L)(22.4 Limo!) = 56 L

69. (A)A mole is defined to be 6.02 x 1023 atoms, molecules, ions, particles,

etc. Thus, 1.5 x 1023 atoms represents 0.25 mole. Recalling that the molarvolume of any ideal gas at STP is 22.4 liters, we may calculate the volumeof 1.5 x 1023 atoms to be

116

---.---PRACTICETEST 2

a

1

SAT ChemistryPractice Test 1

(Answer sheets appear in the back of this book.)

PART A

TIME: 1 Hour85 Questions

DIRECTIONS: Each set of questions below consists of five letteredchoices followed by a list of numbered statements or questions. For eachstatement or question, select the answer choice that is most closelyrelated to it. Each answer choice may be used once, more than once, ornot at all.

Vote: For all questions involving solutions, assume that the solvent iswater unless otherwise noted.

Questions 1-6 refer to the following groups.(A) Alkali metals (D) Halogens

(B) Alkaline earth metals (E) Rare earths

(C) Metalloids

l. Used primarily in semiconductors

2. Some occur as diatomic molecules

3. Give oxides with die formula X20

4. Produce acid salts with the formula XS04

5. Have large electronegativity values

6. Have small ionization energies83

SAT Chemistry

Practice Test 1

(Answer sheets appear in the back of this book.)

PART A

TIME: 1Hour85 Questions

DIRECTIONS: Each set of questions below consists of five letteredchoices followed by a list of numbered statements or questions. For eachstatement or question, select the answer choice that is most closelyrelated to it. Each answer choice may be used once, mere than once, or

not at all.

Note: For all questions involving solutions, assume that the solvent is

water unless otherwise noted.

Questions 1-6 refer to the following groups.

(A) Alkali metals

(B) Alkaline earth metals

(e) Metalloids

(D) Halogens

(E) Rare earths

1. Used primarily in semiconductors

2. Some occur as diatomic molecules

3. Give oxides with the fannuia X20

4. Produce acid salts with the formula XSO 4

5. Have large electronegativity values

6. Have small ionization energies 83

SA T Chemistry

Questions 7-10 refer to the following species.

(A) A Brensted acid (D) A weak base

(B) A Brensted base (E) A buffer

(C) A strong acid

7. Is a solution made by the combination of a weak acid and the salt ofits conjugatebase

8. Always dissociates nearly completely in aqueous solution

9, Has a very high K,

10. Always accepts a proton

Questions 11-14 refer to the following species.

(A) A strong acid (D) A weak base

(B) A strong base (E) A salt

(C) A weak acid

12, CI-

13. NaHC03

14, NaOH

Questions 15-17 f 're er to the followmg values.(A) 0

(8) -t

(C) +\

(D) -2

(E) +2

15 The idan. ox! ation number of Na in NaCl

84

Test 1

16. The oxidation number of Cl in Cb

17. The oxidation number of Sin NazS

Questions 18-20 refer to the following colors.

(A) Yellow (D) White

(B) Purple (E) Colorless

(C) Green

18. S(s)

19. HCI(aq)

20. NaOH(aq)

Questions 21-23 refer to the following reaction and the given values.

4NH,(g) + 502(g) -> 4NO(g) + 6H,O(g)(A) 2.294 (D) 25.3

(B) 36.49 (E) 2.513

(C) 1.409

21. If you begin with 16.00 grams of ammonia, and an excess of oxygen,how many grams of water will be obtained?

22. If you begin with 66.00 grams of ammonia, and 54.00 grams ofoxygen, how many grams of water will be obtained?

23. How many moles of NH, are needed to produce 2.513 moles of NO?

85 --

SA T Chemistry

PARTS

DIRECTIONS: Each question below consists of two statements. Deter-mine if Statement I is true or false.a.r::K!.jf Statement II is true or false an?fill in the corresponding ovals on your answer sheet. Fill in oval CE IfStatement II is a correct explanation of the true Statement I.

(I) (II)

101. Acid rain isdestructive to limestone

because acid rain containsHC!.

102. The net ionic becauseequation for the mixingof aqueous solutionsof CaCl, and Na,S04is: Na+(aq) + C1-(aq) -> NaCI(s)

NaCI precipitatesout of the reaction.

103. Given that the Ksp

of AgBr and BaCO]are 5 X 10-13 and2 X 10-9 respectively,AgBr is less solublethan BaCO]

because a larger Ksp indicatesa larger conversionto products.

104. In any solution(H+] = 1 x ]()-7 because Kw = [OH-j (W]

= 1 X 10-14

105. HF is considereda strong acid because strong acids ionize

completely in anaqueous solution.

106. N~+ is consideredto be a strong acid because it is the conjugate

acid of a strong base.

107. By the Lewistheory of acids andbases, P- is consideredto be a base

because it can acceptprotons.

86

L ,Test 1

108.A 1N solution of because normality and

H2S04 is the samemolarity are both

as a 1M solution units of concentration

ofH2S04and refer to thesame thing.

109. A solution of NaOH because it contains OH-

and Ba(OHh would as a common Ion.

be a good buffer

110.The oxidation because Cl is missing onenumber of Cl in Ch electron to fill its

is +1 shell.

111. Water has a high boiling because of polar covalent

point for its molecular bonding.

weight

111. ~enzene is a poor electrolyte because it does not ionize.

In water solution

113.The reaction of zinc with because hydrogen gas is

hydrochloric acid does not go not evolved.

to completion in an opencontainer

114 At because they have the same oms of the same elementcombine covalently rather

electronegati viti es.

than by ionic attraction.

115 M the positive nuclei etals such as gold are becausemalleable

are surrounded by a "sea"of free electrons.

116 Ac ti " because it ionizes completely e lC acid IS a strong acid in water solution.

87

24. The element with atomic number 32 describes(A) a metal

(B) a nonmetal

(C) a metalloid

(D) a halogen

(E) a noble gas

PARTe

SA T Chemistry

. '. . I t the best answerDIRECTIONS: For each question In this section, se ec . thefrom among the given choices and fill in the corresponding oval onanswer sheet.

25. The Lewis dot structure of N 2 is best represented as(A) :N:N:

(D) :N:::N:. .(B) :N::N

(E) N::: N(C) 'N:: N

26. All of the follOwing are chemical changes EXCEPT(A) dissOlving NaCI in Water

(B) burning a piece of wood

(e) OZOneabsorbing ultraviolet light

(D) disSOlving Na metal in water(E) rusling of iron

27. Which of the following has the smallest mass?(A) a hydrogen nucleus

(B) an alpha panicle

(C) a neUlron

(D) a helium nucleus

(E) a beta particle

88

Test 1

28. 585 g of NaCI in

(A) 1 liter of solution is 2 molar

(B) 2 liters of solution is 0.75 molar

(C) I liter of solution is I molal

(D) 2 kilograms of solvent is 0.5 molal

(E) 1 kilogram of solvent is I molar

29. The greatest reduction of kinetic energy of water molecules occurswhenwater is

(A) cooled as a solid

(B) cooled as a liquid

(C) converted from a liquid to a gas

(D) converted from a gas to a liquid

(E) converted from a liquid to a solid

30. One formula unit of Cr(NH3),S04Br represents

(A) 4 atoms (D) 23 atoms

(B) 8 atoms (E) 27 atoms

(C) 12 atoms

31. How many neutrons are there in the nucleus of an element of atomicweight 197?

(A) 43

(B) 79

(C) 83

(D) 100

(E) 118

32 b ex-. The extremely high melting point of diamond (carbon) may eplamed by large numbers of

(A) covalent bonds (D) Van der Waals forces

(B) ionic bonds (E) polar bonds

(C) hydrogen bonds

89d

SA T Chemistry

.. tr nd atomic radii33. Which of the followmg best represents the geome y ain a carbon dioxide molecule?(A) 0-0-0

(B) 0-0-0

(C) 0/,,-o 0

(D) o/,,-o 0

(E) 0-0--0

Answer questions 34-37 using the phase diagram below.

G H

D BPressure E

A C F

Temperature34. At which point can all three phases coexist at equilibrium?

(A) C

(B) D

(C) E

(D) G

(E) H

35. At which point can only the solid phase exist?(A) A

(B) B

(C) C

(D) E

(E) F

36. Which is the criticnl point?(A) B

(B) E

(C) F

(D) G

(E) H

90

38. Whichof the following is electrolytic?

(A) mercury in water

(B) a benzene solution of ethanol

(C) sucrose dissolved in water

(D) sodium chloride dissolved in water

(E) vinegar in water

Test 1

(A) C

(B) D

(C) E

(D) A

(E) B

37. Atwhich point can only the solid and liquid phases coexist?

39. The oxidation number of sulfur in NaHS04 is

(A) O.

(B) +2.

(C) -2.

(D) +4.

(E) +6.

40. A small crystal of NaCl is added to a sodium chloride solution resul~-ing in the precipitation of more than I gram of sodium chloride. Thissolution had been

(A) unsaturated

(B) saturated

(C) supersaturated

(D) dilute

(E) concentrated

41, The salt produced by the reaction of perchloric acid with bariumhydroxide is

(A) BaCIa,

(B) BaCia.

(C) Ba(OH)2

(D) Ba(CIO.),

(E) BaCl2

91

d

SA T Chemistry

42. A reaction that occurs only when heat is added is best described as(A) exothermic (D) spontaneous(B) endothermic (E) non-spontaneous(C) an equilibrium process

43. The transition metals are characterized by

(A) highest energy electrons in s subshells

CB) highest energy electrons in p subshells

(C) highest energy electrons in d subshells

CD) highest energy electrons injsubshells

(E) stable electron configurations

44. Neutral atoms of F (fluorine) have the same number of electrons as

(A) B3~ (D) Na-

(B) N+ (E) Mg3+(C) Ne-

45. An equilibrium reaction may be forced to completion by(A) adding a catalyst

(8) increasing the pressure

(C) increasing the temperature

(D) remOVing the products from the reaction mixture as they areformed

(E) decreasing the reactant concentration

46. An acid H3X is classified as

(A) monoprotic

(B) diprotic

(C) triprotic

(D) bidentate

(E) polar covalent

92

Test 1

(A) Ba(NO,),

(B) Na,S

(e) Al,(S04h

(D) Pb,(P04)2

(E) NaCl

47. Which of the following salts will result in a basic solution whendissolvedin water?

48. TheHaber process is used for producing ammonia from nitrogen andhydrogen. This reaction could be forced to produce more ammoniaby

(A) increasing the reaction pressure

(B) decreasing the reaction pressure

(C) adding a catalyst

(D) adding a salt

(E) adding water

49. Theionization energy of an element is

(A) a measure of its mass(B) the energy required to remove an electron from the element in

its gaseous state(C) the energy released by the element in forming an ionic bond

(D) the energy released by the element upon receiving an additionalelectron

(E) the amount of ions in a molecule

so. The.radioactive decay of plutonium ~238 ( 2~Pu)produces an alphaparticleand a new atom. That new atom is

(A)

(B)

(C)

234pU92

234U92

234Cm92

(D)

(E)

242pU96242Cm96

93

SA T Chemistry

5]. What is the approximate melting point of 0.2 liters of water contain-ing 6.20 g of ethylene glycol (C,H,O,)?(A) -186"C

(B) -D.93 -c(C) a -c

(D) 0.93"C

(E) 1.86"C

52. What is the empirical formula of a compound composed of 25.9%nitrogen and 74.]% oxygen?(A) NO

(B) NO,

(C) N,O

(D) N,04

(E) N,O,

53. The molecular weight of a gas is 16. At STP, 4.48 liters of this gasweighs

(A) 2.3 g

(B) 2.7 g

(C) 3.2 g

(D) 4.1g

(E) 4.9 g

54. How many milliliters of 5 M NaOH are required to completely neu-tralize 2 liters of 3 M Hel?(A) 600 mL

(B) 900ml,

(C) 1,200mL

(D) 1,500 mL

(E) 1,800 mL

55. How many moles of SUlfateion are in 200 mL of a 2M sodium sulfateSOlution?

(A) 0.2 mole

(B) 0.4 mole

(C) 0.6 mole

(D) 0.8 mole

(E) 1.0mole

94

Test 1

56. How many moles of electrons are required to reduce 103.6 g of leadfromPb2+ to the metal?

(A) 0.5 mole

(B) I mole

(C) 2 moles

(D) 4moles

(E) 8 moles

57. How many grams of copper will be deposited from a solution ofCUS04by a current of 3 amperes in 2 hours?

(A) 5 g

(B) 7 g

(C) 8 g

(D) 11g

(E) 15g

58. The standard electrode in electrochemistry is composed of

(A) gold (D) magnesium

(B) platinum (E) hydrogen(C) copper

59. Zn ~ Zn2+ + 2e- ; Eg~:;;:+0.76 V

Cr3++ 3e- ~ Cr ; E~d:;;:~0.74 V

The anode in this cell is

(A) Zn (D) Cr3+

(B) Cr (E) 3e-(C) Z02+

60 A 2 . . .?. n sp configuration is represented by which onentatIon.

(A) tetrahedral (D) trigonal planar

(B) planar (E) square

(C) linear

95

SAT Chemistry

61. A sp3d2configuration is represented by which orientation?

(A) tetrahedral (D) trigona! hipyramidal

(B) trigonal planar (E) square planar(C) linear

62. The balanced molar relationship of the dissociation of water intohydrogen and oxygen is(A) I : I:!

(D) 2:1:2(B) 2: 1: I

(E) 2:2:1(C) I :2: I

63. What volume of water is required to produce 5 liters of oxygen by theprocess below?

H,O(g) --? H,(g) + O,(g)(A) 3 liters

(D) 14 liters(8) 5 liters(E) 16 liters(C) 10 liters

64.What is the molecular weight of HCJ0

4?

(A) 52.5 g/mol(D) 100.5 g/rnol(B) 73.5 g/mo](E) 116.5 g/mo!(C) 96.5 g/mol

65. The structure of the third member of the alkyne series is(A) H-C"C_H

(B) H-C"C-CH,

(C) H-C" C-CH,-CH,

(D) H-C" C- C" C_H

(E) H-C"C-CH~CH2

96

97

Test 1

66. The systematic (lUPAC) name of this structure is

(A) hexanol

HI I I I I I

H-C-C-C-C-C-C-HI I I I I IH H H H H H

(D) l-hexanol

(E) isohexanol

H H OH H H

(B) 3-hydroxyhexane

(C) 3-hexanol

67. Which of the elements in Group IA of the periodic table has thegreatest metallic character?

(A) Li

(B) Na

(C) K

(D) Rh

(E) Fr

68. The oxidation state of nitrogen in nitric acid (HN03) is

(A) +1

(B) +2

(C) +3

(D) +4

(E) +5

69. What volume does a sample of 1.50 x 1023 atoms of helium at STPrepresent?

(A) 5.6 liters

(B) 11.2 liters

(C) 17.8liters

(D) 22.4 liters

(E) 1.5 liters