AP Chemistry AP Practice Exam: Multiple Choice, Part I

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to begin actually going through an AP exam completely.0004

We are going to start with the multiple-choice section.0010

Before I launch into answering the questions on the multiple-choice section, I want to talk to you, just briefly, about it.0012

Now, you are going to have about 90 minutes to do 75 multiple-choice questions; you are going to get 1 point for every one that you get right, and you are going to be subtracted a quarter point for the ones that you miss.0020

Now, you should know that you could actually get a raw score (which is the number you get correct, minus the number you miss) of about 50; so you could literally skip one-third of the multiple choice questions, and still end up with a 5.0033

You can answer about 35 or 40 only of the multiple choice questions and still end up with a 4; so don't think like you have to answer absolutely every single one.0048

If you are not exactly sure about one, move on: chances are that your score will still be sufficiently high to keep you in the 4-5 range.0058

Having said that, let's just go ahead and get started.0067

We are going to be going through the 1999 AP practice exam; so hopefully, you have it in front of you; I have it in front of me here.0071

We are just going to run through it; OK.0079

The first question gives you a choice of activation energy, free energy, ionization energy, kinetic energy, and lattice energy: so those are your answer choices, and we are going to...the questions that follow actually ask you to identify which energy is being talked about.0082

#1: The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion: so, this one is the ionization energy, so this one is C.0102

If you remember the ionization energy, it is if you have some ground-state gaseous atom; you hit it with a certain amount of energy, and you convert it to a gaseous phase ion.0118

You knock off one electron, and it is written like this (this is the first ionization energy--the second ionization energy, and so on, will just be this thing when you knock off another electron).0129

That is it; #1 is C.0141

OK, #2: The energy change that occurs in the conversion of an ionic solid to a widely-separated gaseous ion; OK, now this is one of the things that, in our particular course, we did not discuss formally: this is called the lattice energy.0143

The lattice energy is basically the energy that is keeping an ionic compound together.0160

When you have, let's say, sodium and chlorine, they come together; the chlorines separate; they each steal an electron from the sodium, and then you have a chloride ion, which is negatively charged, and sodium, which is positively charged.0167

They combine, because positive and negative attract; when they combine, they actually are more energetically stable--remember, we said all systems seek the lowest energy.0180

Ionic compounds, because of the positive and negative charge, release energy; so that is the lattice energy--it's the amount of energy that is necessary to keep that together.0192

In order to break that apart--to separate them as far away as possible--to separate sodium ion from chloride ion--that is called the lattice energy, because these ionic compounds are composed of a crystal lattice (positive/negative, positive/negative, arranged in a particular arrangement).0201

OK, so #3: The energy in a chemical or a physical change that is available to do useful work: OK, that is the important word here, "useful work": that is free energy.0219

OK, that is B; and you remember free energy: the reason we called it free energy--it is energy available to actually do work, not random energy, which would be heat.0231

OK, so these are reasonably straightforward; and again, this is how it's going to be on the multiple choice exam.0243

I would say a good 75% of the multiple choice questions are things that you either know or don't know, and can answer pretty quickly.0248

There are some calculations; now, of course, you can't use a calculator on this part of the exam--all you are given is a periodic table--but that is all you need.0257

Any calculations that you do need to make are actually really, really simple numerical calculations--things like .5, .25, 2, 4--numbers that you can multiply pretty quickly.0264

OK, #4: The energy required to form the transition state in a chemical reaction: so you remember, when we go from reactants to products, we have to pass over a transition state: that is called the activation energy.0275

Activation energy is A, so 4 is A.0287

OK, let's see--let's move on.0292

#5 gives you a series of electron diagrams (you know, the ones with the arrows up and down); it gives you a choice of 5 of them, and now we are going to ask you some questions about which one is which.0299

We have to identify.0314

#5 says it represents an atom that is chemically unreactive; well, if you take a look at the choices..."unreactive"--generally, you are looking at a noble gas configuration; you are looking for the s and the p to be completely filled, so s², p⁶.0317

The only one in there that is s², p⁶, is D: 1s², 2s², 2p⁶.0332

The answer to #5 is D.0338

OK, "represents an atom in an excited state": so, in an excited state means that an atom...the normal electron configurations, when we do them--they all go into the lowest energy; we go from the lowest energy, and we work our way up.0342

An excited state means an atom has been kicked up from its ground state energy level (whatever that may be) to a higher level; so it is going to have orbitals lower than it that may not be filled.0355

Well, the only one that satisfies that one is A: you have the 1s, which is empty; you have the 2s with the one electron in it; that is an excited state atom.0366

#6 is A.0375

#7: "represents an atom that has 4 valence electrons": OK, 4 valence electrons--remember, the valence electrons are all of the electrons in the highest primary shell.0381

In here, we have...let's see: 2s, no; B is 2s², 2p⁴...no, I'm sorry, 2s², 2p²; there you go.0393

C is the answer here, and the reason is because the valence shell is the primary electronic level 2: 2s, 2p: there are 2 in the s orbitals; there are 2 in the p; that gives you a total of 4 valence electrons; so #7 is C.0405

OK, #8: "represents an atom of a transition metal": well, a transition metal is the one where the d orbitals are actually filled--in the periodic table, that middle section with the 10 columns.0423

Those are the transition metals; so here, you are looking for something that has filled d orbitals, and that is going to be E.0436

#8 is E; OK.0445

Let's see: so now, questions 9 through 12 refer to an aqueous solution containing a 1:1 mole ratio of the following pairs of substances; assume all concentrations are 1 Molar.0452

So, we have ammonia-ammonium chloride; phosphoric acid and sodium dihydrogen phosphate; hydrochloric acid and NaCl; NaOH and NH₃; and NH₃ and acetic acid.0465

Those are our pairs of...a solution with a pair of those species; those are our choices.0479

And again, they said they are in a 1:1 mole ratio.0485

So, #9: we are looking for the solution with the lowest pH; OK.0488

Now, lowest pH means most acidic.0494

OK, now we look over here: as far as the most acidic is concerned, the only thing that I see that is the strongest acid is HCl.0507

And then, we look at NaCl; well, the sodium chloride, when they actually break up in solution into sodium and chloride--they are going to be spectator ions; they don't actually do anything.0518

They are not going to act as bases; they are not going to affect the equilibrium concentration of the hydrogen ion; so, in this case, the answer is C.0529

Let me put a little box around that: so lowest pH=most acidic.0539

OK, #10: the most nearly neutral solution: OK, in this particular case, the most nearly neutral solution is going to be E, the NH₃ and the acetic acid.0544

All right, you might think that it...well, let's just sort of run through the possibilities.0561

It is not going to be the NaOH and the NH₃, because they are both bases; it is not going to be C, the HCl and the NaCl, because that is acidic; the H₃PO₄ is a weak acid, and the sodium dihydrogen phosphate--that is going to dissociate.0568

H₂PO₄^- is also a weak acid, so you are probably looking at an acidic solution there.0586

NH₃ and NH₄Cl: well, this one--you might think that it is A, but as it turns out, let's take a look at why it's not A.0593

Now, if I take NH₃ in solution, I'm going to get something like this: the equilibrium for this involves NH₄⁺ plus OH, the standard dissociation (or association) equilibrium of a weak base--in this case, NH₃.0609

Well, we know (or we should know) that the K_b for this is a lot less than 1.0628

Well, the K_b for this is equal to the NH₄⁺ concentration, times the OH^- concentration, divided by the NH₃ concentration.0637

But they said that the NH₃ and the NH₄Cl are in a 1:1 ratio, so even though there is going to be some dissociation, the K_b is really, really small, so it's not going to be that much.0654

So, for all practical purposes, the ammonium concentration and the ammonia concentration are going to be the same, so what you are left with is: the concentration of hydroxide is actually going to be equal to the K_b (which is going to be somewhere on the order of 10^-5 and 10^-6).0664

When you take the negative log of that, you are going to get the pOH.0681

Well, the pOH is not going to be anywhere near neutral; but E, where you have ammonia (which is a weak base) and acetic acid (which is a weak acid), they are in a 1:1 molar ratio, so that one actually will end up neutralizing completely and giving you a pH close to about 7.0685

So, E is the best answer for #10.0707

OK, let's see: #11: A buffer at a pH of greater than 8...so we are looking for a buffer that is basic.0711

Let's write down here (remember what a buffer is?): A buffer is a weak acid plus its conjugate base, or it is a weak base plus its conjugate acid.0730

Well, when you have a weak acid and its conjugate base, you are going to get a solution that is acidic.0760

That is going to be a pH less than 7; we don't know where--I mean, we can adjust the pH any way we want; that is the whole idea of a buffer--you choose the pH beforehand, and then you create the solution.0765

If this is going to be...but they want the buffer with a pH that is going to be greater than 8, so you are looking for a weak base and its conjugate acid.0775

As it turns out, A satisfies that; so 11 is A: the weak base is ammonia; its conjugate acid is ammonium; the chloride is irrelevant.0784

OK, #12: A buffer at a pH less than 6: so now, this is the one--this is the buffer--that is going to be acidic.0796

So here, we are looking for a weak acid, and we are looking for its conjugate base.0807

As it turns out, the only one that satisfies that is B.0814

Phosphoric acid is a weak acid; its conjugate base...remember how we get a conjugate base--we just pull off one of the hydrogen ions; it is going to be the H₂PO₄^-, and that is your conjugate base.0819

B is the combination.0831

OK, let's see: let's move on to questions 13 through 16.0833

OK, so questions 13 through 16 refer to the following descriptions of bonding in different types of solids.0844

The key word here is "solids."0850

OK, well, let's just run through it; so #13: cesium chloride--cesium chloride is an ionic compound; it is made of cesium cations and chloride anions; and that is going to be A.0854

It is a lattice of positive and negative ions held together by electrostatic forces.0875

So, it will always be that for ionic compounds: bonding in an ionic solid is actually pretty easy--it's just positive and negative charges stuck together--so no worries there.0880

#14: Gold--OK, so gold is a metal, and it consists of closely-packed lattice with delocalized electrons throughout.0892

#14 is going to be B; so what we mean when we say this: one of the things that makes metals such good conductors of electricity is precisely because the electrons are delocalized.0903

So, if I have a bunch of gold atoms--a piece of gold--the electrons aren't necessarily attached to any one gold atom; they are delocalized.0917

They are actually free to move: that is why metals conduct electricity, because the electrons can actually move through the sample (or over the surface of the sample, anyway).0928

When you see delocalized electrons, that represents metallic bonding.0938

OK, #09 carbon dioxide--all right, carbon dioxide--well, if you remember carbon dioxide...the Lewis structure for carbon dioxide is this.0944

Our choices would be: Strong multiple covalent bonds, including pi bonds (yes, we have pi bonds here; a double bond represents one pi bond), with weak intermolecular forces.0958

Well, CO₂ is a nonpolar molecule (right?--the dipole moments--they actually cancel out: it is nonpolar), so this one would be D; that is the best choice.0968

16: We have (let me see) strong single covalent bonds with weak intermolecular forces.0983

The same thing: when you look at methane, if you remember the Lewis structure for methane, you have something that looks like a little tripod here, and the dipole moments of each individual CH bond do cancel out, and when they cancel out, you get a nonpolar molecule.0989

These are single bonds, so the best answer would be C.1007

OK, so questions 17 and 18 refer to the following elements: we have lithium, nickel, bromine, uranium, and fluorine.1014

#17 says: Is a gas in its standard state at 298 Kelvin (298 Kelvin is room temperature, just about 25 degrees Celsius).1025

That one is going to be fluorine.1033

17 is E, fluorine.1038

Now, "reacts with water to form a strong base": lithium, nickel, bromine, uranium, fluorine--it reacts with water.1043

The alkali metals (you just know this), when you put them in water, react to form a strong base; they react violently, which is why lithium, sodium, things like that actually can't be stored in water, because they will react.1053

Let me go ahead and put 18 over here; 18 is going to be lithium, which is A, and just so you know what the reaction actually looks like...it's actually the reason why they call them alkali metals, because they form hydroxide in solution.1068

If you take lithium metal, and if you add it to water (which I am going to write as HOH), what ends up happening is...well, literally, what ends up happening is (I'm going to draw 2 of these and 2 lithiums): a hydrogen atom--one of these hydrogen atoms--steals an electron from lithium; another hydrogen atom steals another electron from lithium.1085

They form hydrogen gas (they come together to form hydrogen gas); it leaves 2 hydroxides free, and therefore, what you end up with--the lithium + and the hydroxide minus--they come together; you end up with lithium hydroxide in solution, which is basic; that is what happens.1114

OK, so let's see: Starting with #19, each of the questions or incomplete statements below is followed by 5 suggested answers or completions; select the one that is best in each case, and then fill in the corresponding oval on the answer sheet.1137

OK, so which of the following best describes the role of the spark from the spark plug in an automobile engine?1159

OK, of these choices, C is the best: the spark plug supplies some of the energy of activation for the combustion reaction.1167

Combustion reactions are highly exothermic--they want to move forward; they are thermodynamically favorable.1176

However, just because you put some gas, like some gasoline, in a container with oxygen doesn't mean that the reaction is going to happen.1185

They have a very high activation energy; the spark gets it the activation energy.1196

If I put oxygen and hydrogen in a container and don't do anything, well, they want to form water really badly, but they can't get over the activation energy.1203

When I ignite it--when I just do a little spark--that is what makes the reaction go forward.1212

The answer here is C; 19 is C.1218

OK, what mass of Au is produced when .05 moles of Au₂S₃ is reduced completely with excess H₂?1224

OK, so let's write out the reaction for this: so we have: Au₂S₃ is reduced completely with excess H₂.1235

The mass of Au...so they say that gold is actually formed; so our reaction is going to be...gold is formed, and then it's going to be H₂S, and then we are going to have to balance this equation out, so we have ourselves 3, 2, 3; I think it's going to be 3, 2, and 3 (3 H₂, 3 S, 3 S, 2 Au); yes, that is our balanced equation.1247

OK, so they say we have 0.0500: 0.0500 moles of the Au₂S is reduced completely, so we want to produce...let me see, so for every 1 mole of this, what mass of Au?1273

For every one mole of this, we produce 2 moles (right? for every 1 mole, we produce 2 moles) of gold.1295

.05 moles of this produces 0.100 moles of gold; well, now we do 0.100 mol (and again, these numbers are very, very easy), times 197.0 grams per mole (which, again--you will have a periodic table available for you, so you will have molar masses, and this is a really simple calculation): .1 times 197 is just 19.7 grams.1301

19.7 grams is the answer; that makes it B.1333

Good; OK, now, when a solution of sodium chloride is vaporized in a flame, the color of the flame is yellow.1339

This is just one of those things that you should know: when we do flame testing (we actually didn't discuss this; it is part of the qualitative analysis that we discussed briefly, when we talked about separating ions)...when you have a solution of ions, or if a solution that you have separated out--separated the separate ions--you need to sort of quickly test to make sure what ion is in there.1350

As it turns out, sodium ion burns yellow when you dip a piece of metal (a paper clip or something) in the solution and put that in a flame.1374

The flame will actually turn yellow, because the heat is exciting the electrons up to higher energy levels; they fall back down; and they emit light, yellow light.1386

Sodium burns yellow, lithium burns red, and potassium will burn blue; so just sort of know that.1396

And again, it is just one of those things that just sort of comes up.1404

OK, #22: Of the following reactions, which involve the largest decrease in entropy?1410

OK, so let's see: the largest decrease in entropy...so we are looking for something that goes to a smaller number of gas particles, or maybe from a gas to a liquid, a gas to a solid, liquid to solid...things like that--something that becomes more ordered.1420

A decrease in entropy means something that becomes more ordered.1441

Of these, it looks like E is the best choice, because you have three moles of oxygen gas that are all of a sudden contained in this lanthanum oxide, so you have a gas going to a solid; so yes, our best choice is E, for #22.1445

OK, let's see: #23: A hot air balloon, shown above, rises; which of the following is the best explanation for this observation?1467

OK, the best explanation for this observation: air in a balloon is surrounded on the outside by normal air; things rise when their density is less than the density of the surrounding medium.1484

So, for example, ice floats because ice is less dense than water.1504

Metal sinks in water because metal is more dense than the water.1507

So, in this case, E is the best choice: because the air density inside the balloon is less than that of the surrounding air.1511

That is it; so 23--we have E.1518

OK, #24: The safest and most effective emergency procedure to treat an acid splash on skin is to do which of the following immediately?1524

24 is going to be: Flush the affected area with water and then with dilute sodium bicarbonate solution.1533

That is going to be D; the reason that is, is: you flush it with water to get rid of as much of it as possible, but if there is any excess, you want to use bicarbonate, which is a very, very weak base, to compensate for the acid.1540

You don't want to use any strong base (like a sodium hydroxide solution or anything like that), because you run the risk of going over the top.1554

A strong base will also do damage the same way that an acid will; they actually will do the same damage, so you want to use a weak base like sodium bicarbonate (baking soda).1562

That is the best: wash it off; put a little baking soda on there--it will neutralize everything; then wash it off again.1573

So, D for #24.1578

OK, #15 The cooling curve for a pure substance as it changes from a liquid to a solid is shown above; the solid and the liquid coexist at...this one--the answer is C.1581

The answer is C because all points between Q and S...OK, this is a heating curve; so basically, what it is telling you is: anytime you see a straight line on a heating curve, that means it is making the transition.1601

By "a straight line," I mean a horizontal line, where the temperature is not changing; so you notice, the temperature is vertical; the time is the horizontal axis.1617

The temperature is dropping, dropping, dropping, dropping, dropping; all of a sudden it hits Q, jumps up to R, and then it stays horizontal from R to S.1625

That means there is a phase change; that means--and since we are doing a drop in temperature, that means the liquid is going to a solid.1633

And, of course, we know as much, because that is what the question asks.1642

Well, in the transition from a liquid to a solid, both the liquid and the solid coexist; until all of the liquid is converted to a solid, they coexist; so, certainly they coexist between R and S.1645

Well, this from-Q-to-R part--as it turns out, you can drop the temperature, drop the temperature, drop the temperature of a liquid; and if you drop it kind of quickly, the molecules don't have time to arrange themselves completely in the lattice necessary for the solid.1657

You can actually drop the temperature below its normal freezing point; it is called supercooling.1675

Well, as you drop to a supercooling temperature, at a certain point they start to rearrange; so between Q and R, you still have...that means the ice is starting to form.1680

As the ice starts to form, the temperature of that ice will actually start to rise back up until it reaches the temperature where it is the normal freezing point.1692

This little supercooling part--you will often see that in heating curves--at least experimentally, you will see it.1704

Under ideal conditions, you wouldn't see that supercooling, because we would go very slowly, very slowly; you would just see a drop--or from your perspective, a drop, a horizontal, and then another drop.1711

That horizontal, again, is where it is making the transition from liquid to solid, or from liquid to vapor, or from vapor to liquid.1721

OK, temperature doesn't change at that point--only when it is actually completely solid, completely liquid, or completely gas--that is when the temperature changes, depending on its heat capacity.1729

We did discuss heating curves.1739

So in this case, yes, C would be the best answer.1742

OK, let's see: When the equation above is balanced, and all the coefficients are reduced to their lowest whole-number terms, the coefficient of O₂ is...1745

All right, so let's do this one: 26.1754

We have: C₁₀H₁₂O₄S, plus O₂; OK, we are going to form CO₂; we are going to form H₂O; and, if you have sulfur, sulfur is also going to oxidize.1759

It is going react with the oxygen, so you are going to end up with SO₂.1777

OK, so let's go ahead and do the carbon first; so we'll stick a 10 in here for the carbon.1782

We'll do the hydrogen next, which is 12 hydrogen here; so we'll put a 6 there.1789

And then...now, let's see what we have: we have one S; we have one S; so now, let's count the oxygens: oxygens--we have 20 oxygens, plus 6 oxygens, plus 2 oxygens, for a total of 28.1794

We have 28 oxygens on the right; we have 2 here; we have 4 here.1812

We want 28; if we get 28-4, that is 24; how do we get 24 out of this?--we put a 12 here...and what do they actually ask for?1819

What is the coefficient of O₂?--the coefficient of O₂ is 12, so 26 is C.1830

There you go: just a straight balancing problem--nothing too strange.1838

OK, #27: let's see, appropriate uses of a visible spectra photometer include which of the following?1842

Determining the concentration of a solution of copper nitrate (copper (2) nitrate); measuring the conductivity of a solution of potassium permanganate; determining which ions are present in a solution that may contain sodium, magnesium, and aluminum ion...1855

A visible light spectra photometer is used for the determination of a concentration of a particular species, so it would be 1 only, so 27--it would be A.1871

OK, by the way, I should let you know so that...we didn't actually really talk about this, because I didn't discuss coordination chemistry in this particular AP Chemistry course.1886

Coordination chemistry usually comes at the end; some teachers talk about it; some don't.1899

If you see any questions about coordination chemistry, whether it is naming or something like this, it may be anywhere from maybe 1 to 3 questions...well, 1 question--really, you are not going to see more than 1 or 2.1904

So, the reason I didn't want to discuss it is because, again, it is one of those things that, running through a practice exam, you can actually get the little bit of information that you need regarding coordination chemistry.1915

I didn't want to spend an entire lesson or two lessons on it; it wasn't worth it, which is the same reason why I decided not to discuss organic chemistry.1927

Anything that comes up in organic chemistry is going to be 2 questions, at most; and again, if you miss them--if you don't know them--it is not going to hurt you in absolutely any way at all.1934

I didn't want to actually spend a whole bunch of lesson time on that; I wanted to concentrate on the things that were absolutely important and essential.1946

You should know: transition metals like copper--basically, what happens is: when you--if you--were to dissolve this copper nitrate in water, the copper nitrate dissolves, but what ends up happening is that the copper...the water molecules actually arrange themselves around the copper ion: maybe 4 of them, maybe 6 of them, maybe 2 of them.1955

Transition metals are notorious for absorbing visible light, which is why transition metal solutions actually are of different colors (green, blue, orange, pink, things like that--really, really beautiful colors; that is the reason why).1977

We can use a visible light spectra photometer to measure how much of the light they actually absorb, based on their color; and that tells us the concentration.1991

That is a little sideline explanation of what it is that is going on here.2000

OK, 27...let's see, 28: OK, the melting point of magnesium oxide is higher than that of sodium fluoride; explanations for this observation include which of the following?2005

Magnesium is more positively charged than sodium; O₂^- is more negatively charged than F^-; the O₂ ion is smaller than F^- ion.2020

Well, when we are talking about melting point, we are talking about the charges between; so, if I have magnesium oxide and I want to melt it, that means I am trying to break the magnesium and oxygen bond, which is ionic.2033

Well, the stronger the bond, the more energy I have to put into it to break that bond.2048

So, as it turns out, here it is going to be B: 1 and 2: magnesium is more positively charged; oxygen is more negatively charged.2052

Magnesium is more positively charged than sodium; oxygen is more negatively charged than fluoride; more positive charge, more electrostatic force, stronger bond, higher melting point; so this is going to be B.2061

OK, 29: let's see, the organic compound represented above is an example of...that is a keytone.2079

Again, this is...so this is going to be E, and just to let you know, a keytone consists of a carbonyl group (which is this thing) connected to two...this carbon is connected to 2 carbons.2088

There can be more carbons here, but it is connected to at least two carbons.2107

This is called a carbonyl group, and this whole thing...when this carbonyl group is attached to two carbons, at the very least, it is called a keytone.2111

Again, organic--it is not going to hurt you; besides, those of you that are going to continue on--you are going to discuss organic chemistry, for those of you that are biology majors and chemistry majors.2122

Not a problem--no worries about that.2135

OK, #18 Which of the following is true regarding the reaction represented above?2139

These things...it looks like they are asking about oxidation numbers, mostly; so let's go ahead and calculate some oxidation numbers here.2144

We have H₂Se + 4 O₂F₂ goes to SeF₆ + 2 HF + 4 O₂ (I hope I copied that correctly).2151

Let's just do some oxidation numbers here: hydrogen is going to be +1; selenium is going to be...there are 2 of them, so it's +2, so selenium here is -22170

O₂F₂...this is a very unusual compound; the only atom that is actually more electronegative than oxygen is fluorine, so in this case, oxygen will not have a -2 oxidation state.2181

The fluorine will have a -1 oxidation state; and because there are two of them, the total charge is going to be -2.2192

Well, it has to be balanced by a +2, because it's a neutral...that means oxygen has a charge of +1.2202

You don't see that too often.2211

Fluorine--again (oops, this is not sulfur; this is selenium; this is selenium fluoride), fluoride is -1; there are 6 of them, so a total charge of -6; that means selenium is a +6.2213

This is going to be -1, +1, and 0 on that; so in this particular case, it looks like it's going to be D: the oxidation number of selenium changes from -2 to +6, so our answer is D.2229

OK, let's see--31: If the temperature of an aqueous solution of sodium chloride is increased from 20 degrees to 90 degrees Celsius, which of the following statements is true?2247

The density of the solution remains unchanged--no; the molarity of the solution remains unchanged--no.2264

Density--as you heat something up, it's going to expand, so the density is going to decrease; the molarity of the solution remains...it's going to expand, which means the total volume is going to change, so molarity doesn't stay.2274

The molality of the solution remains unchanged--yes: the molality is moles of solute per kilograms of solvent; the mass of water doesn't change--its volume changes.2285

The mass of the solution (of the water) doesn't change; its volume changes, so C is the answer here.2295

Density changes, molarity changes...OK, that is 31.2305

Let's move on to #32: let's see, types of hybridization exhibited by the C atoms in propene...OK.2308

Let me go ahead and draw a propene atom: now again, this is one of those questions...so we didn't discuss organic chemistry; is it possible that a molecule like propene actually shows up in a Lewis structure problem that you have studied, maybe, in class, or in the problem sets?--maybe.2319

I mean, you have three carbons in there; each carbon is going to be attached to each other; yes, it is possible, but let me just go ahead and do it.2343

Propene is going to look like this.2351

Basically, what you have is a double bond, which is going to be sp² hybridized (any time you see a double bond, the atoms are sp² hybridized), and you have single bonds all the way around here (so sp³ hybridized).2361

In this case, it is going to be sp² and sp³ hybridized; so it looks like the correct answer is D.2375

That is it: single bonds--sp³ hybridization; double bonds--sp² hybridization; triple bonds--sp hybridization.2382

OK, #33: A 1-liter sample of an aqueous solution contains .1 moles of sodium chloride and .1 moles of calcium chloride.2391

What is the minimum number of moles of AgNO₃ that must be added to the solution in order to precipitate all of the chloride as AgCl?2403

Assume that AgCl is insoluble.2411

OK, so basically, we have this: NaCl will dissociate into Na⁺ + Cl^-, so that is going to be one source of the chloride.2414

Our other source is CaCl₂; it is going to dissociate into calcium 2+, plus 2 Cl^-.2425

So, for every mole of this, it is going to be 2 moles of chloride; so we need to calculate some moles here.2434

1 liter...so let's do this one...times 0.1 millimole per milliliter...actually, I don't need to do...this is 1 liter, so I can just leave it as mole per liter...so we end up with 0.1 moles of sodium chloride solution.2439

.1 moles of this produces .1 moles of chloride ion; so I have 0.1 mol of chloride ion that comes from the sodium chloride solution.2465

OK, 1-liter sample...let's see: so this one--again, they tell me I have...oh, the same thing; OK, 1 liter times 0.1 moles per liter equals 0.1 moles of the calcium chloride.2476

Well, one mole of this produces 2 moles of chloride; so what I end up with is 0.2 moles of chloride ion.2496

I end up with a total of 0.3 moles of chloride ion.2504

Now, AgCl: Ag + Cl^- goes to AgCl, solid; well, the question is: What is the minimum number of moles of AgNO₃?2510

Well, I need .3 moles of Ag, and AgNO₃...one mole of that produces 1 mole of Ag⁺ plus one mole of NO₃^-; so I need 0.3 moles of AgNO₃.2525

That is it; you could obviously have done this a little bit faster, maybe, but I want you to see where this comes from.2544

A certain amount of chloride from sodium chloride; a certain amount of chloride from magnesium chloride: you have to find the total number of moles of the chloride ion if you are trying to take all of that out of solution.2550

The relationship is 1:1 with silver and chloride, and the silver comes from silver nitrate, which is 1:1.2563

That is it; so the answer is C.2572

That is #33; OK.2576

#34: Questions 34 and 35 refer to an electrolytic cell that involves the following half-reaction.2579

Let's see: which of the following...so we have the AlF₆^3-, plus 3 electrons, goes to aluminum metal, plus 6 fluoride ions.2588

So, in this case, it looks like the aluminum has been reduced from aluminum 3+ to aluminum 0.2601

Which of the following occurs in the reaction: AlF₆^3- is reduced at the cathode--yes, oxidation occurs at the anode; reduction occurs at the cathode.2609

The equation as written, which is AlF₆^3- + 3 electrons goes to aluminum + 6 F^-: this is a reduction.2621

The oxidation state of fluorine doesn't change; aluminum is a +3; this is a 0; it is a reduction, and we know it is a reduction because the electrons are on the left-hand side.2639

This is a reduction; reduction happens at the cathode, so our answer for 34 is A.2651

OK, #21 A steady current of 10 amps is passed through an aluminum production cell for 15 minutes; which of the following is the correct expression for calculating the number of grams of aluminum produced?2659

1 faraday equals 96,500 coulombs: OK, this is an electrolytic cell; this is where we are going to pass the particular current through a cell for a certain amount of time to see how many grams we want.2673

OK, so let's start with the time: this is the conversion.2687

We have 15 minutes; we are going to multiply that by 60 seconds in 1 minute to cancel minutes; we are going to have 10 amps, so this is going to be 10 coulombs per second, times 1 mole of electrons is 96,500 coulombs.2692

We have 3 moles of electrons per 1 mole of aluminum, times 27.0 grams of aluminum per 1 mole.2722

Therefore, our answer is C, because C is 15 times 60 times 10, times 27 divided by 96,500, and 3 on the denominator; so 35 is C.2737

OK, now 36: the initial rate data in the table above were obtained for the reaction represented below; what is the experimental rate law for the reaction 2 NO + O₂ gas goes to NO₂?2753

OK, so we know that our rate law...rate is going to equal K times concentration of NO...2773

Well, actually, you know what, let me sort of run through it.2781

I take a look at...I start with the nitrogen oxide, and I notice experiment 1 is .1 Molar initial, and experiment 2 is .2 Molar; so this is the initial rate, so we are using the method of initial rates here.2786

It says that, for .1 Molar, the initial rate is 2.5x10^-4; and for .2 Molar, it's 5.0x10^-4; so this is...when I double the concentration, the rate doubles.2804

Well, that means that it is first order in NO.2819

Now, I go to the other reactant; I go to oxygen, which is in the second column; so now, I use experiment 2 and 3, because I'm going to hold the NO concentration fixed.2824

I notice, when I do .1, for a .1 concentration of O₂, the rate is 5x10^-4; and if I multiply that by 4 and go to .4 for an initial concentration, my rate goes to 8.0x10^-3.2836

So, when I multiply the concentration by 4, I actually end up multiplying the rate by 16, which means that it is 4 squared.2857

Therefore, it is second order in O₂; so that means our answer is B: the rate is K times NO times O₂ squared.2868

OK, #37: The ionization energies for element X are listed in the table above; on the basis of the data, element X is most likely to be _____.2883

OK, so take a look at these energies of ionization: I have 580, 1815, 2740; the jump from 580 to 1815 is actually pretty high; however, notice the fourth and fifth: the jump from 2740 to 11,600 is huge.2892

So, chances are, the best that this represents is probably aluminum; and the reason it is aluminum is because of the following.2916

The 1s², 2s², 2p¹...the first ionization energy that takes this electron is reasonably low.2928

Well, to go here, it is going to be a significant jump in order to start taking away these two electrons; and the 1815 and the 2740 represent a pretty significant jump.2940

And then, of course, the fourth and the fifth--the 1s² electrons are just way too high--it's way too much energy, so just based on these numbers, aluminum is the best choice.2952

#38: Let's see, a molecule or an ion is classified as a Lewis acid if it accepts a pair of electrons to form a bond; that one is just something that you should know.2966

A Lewis acid is something that accepts a pair of electrons to form a bond.2978

OK. #39: A phase diagram for a pure substance is shown above; which point on the diagram corresponds to the equilibrium between solid and liquid phases at normal melting point?2983

Normal melting point--the atmospheric pressure is 760 millimeters, 1 atmosphere; so we go up to point C: that is where the phase difference is between the solid and the liquid--solid on the left, liquid in the middle, gas on the right.2998

So, C is the answer.3013

#24: We have: Of the following molecules, which has the largest dipole moment?3017

D--that is the one that is ionic; CO does have a dipole moment...well, HF...some people consider it ionic; some people, not; the largest difference in electronegativity between H and F is a bigger difference in electronegativity between C and O.3022

CO₂ is nonpolar; O₂ is nonpolar; F₂ is nonpolar; so it is between A and D; D is actually bigger--fluorine is the most electronegative.3039

OK, and #41: After the equilibrium represented above is established, some pure O₂ gas is injected into the equilibrium vessel at constant temperature.3049

After equilibrium is established, which of the following has a lower value, compared to its value at the original equilibrium?3061

OK, I'm at equilibrium, constant pressure; I put some O₂ into this reaction vessel.3070

When I put some O₂ into this reaction vessel, Le Chatelier's principle--it is going to shift to the left.3079

So, what it is actually going to do: is the K_eq for the reaction going to change?--no, at a given temperature, the K_eq stays the same.3086

The equilibrium position might change; in other words, the SO₃, SO₂, and O₂...different concentrations; but the K_eq does not change.3095

The total pressure in the reaction vessel...let me see, what is it saying?--which of the following has a lower value?3104

No, the total pressure does not go down; it actually goes up.3110

The amount of SO₃ gas doesn't go down; the amount of O₂ in the reaction doesn't go down, because you are injecting O₂; the amount of SO₂ in the reaction vessel--yes.3114

Since the reaction is going to move to the left, the SO₂ is going to be used up--the excess that you use; the SO₂ concentration that was there before anything happened--that is going to deplete, because the reaction is going to move to the left.3125

So, 41 is E.3139

OK, thank you for joining us here at Educator.com for this particular discussion of the multiple choice questions.3144

The next lesson, we will continue on with question 42.3150

Take care; goodbye.3153