AP Chemistry AP Practice Exam: Multiple Choice, Part III

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

We are going to continue our discussion of the AP practice exam: this is the third part of the multiple choice.0004

We left off on problem 61, so let's continue on with problem #62.0011

OK, so #62 says: The reaction represented above has an equilibrium constant equal to 3.7x10⁴; which of the following can be concluded from this information?0017

OK, now, let's see...so we take a look at this reaction: we have acetic acid reacting with the cyanide ion, and it forms the hydrocyanic acid, the HCn, plus the acetate ion.0030

So basically, the H has jumped from the acetic acid to the cyanide ion.0047

Or, another way of looking at it: the cyanide ion has taken the H away from the acetic acid.0052

The best information that is given here is the equilibrium constant, 3.7x10⁴; that is a huge equilibrium constant.0057

Now, a high equilibrium constant means that, at equilibrium, the reaction is all the way to the right, meaning that there is virtually no reactant left; it is all product.0069

So, as such, basically what this says...these choices, A, B, C, D, and E: A is the best choice--it says that the cyanide ion is a stronger base than the acetate, because the cyanide has taken the H.0079

A stronger base: it takes the H.0095

And because it's such a high equilibrium constant, that means it is virtually all HCn and all acetate, and no acetic acid and Cn^-.0099

So, for #62, A is our best choice: 62, A.0109

OK, #63: Let's see, the graph above shows the results of a study of the reaction of X, with a large excess of Y, to yield Z.0116

The concentration of X and Y were measured over a period of time; according to the results, which of the following can be concluded about the rate law for the reaction under the conditions studied?0128

OK, so Y actually doesn't matter altogether this much, because they said it is mostly excess; and you notice the Y is just sort of a steady concentration.0139

That doesn't really tell us anything; and because they tell us that it is in excess, it is not really going to show up; so we are going to be concerned mostly with what is happening with X.0150

OK, so this is a plot of concentration versus time.0159

Now, remember, with rate laws, when we have time and concentration data, what we do is: we create three plots.0164

We create a plot of time versus concentration, time versus logarithm of concentration, and one over concentration against time; so we have three plots.0172

Depending on which one is a straight line, it will tell us what the order of the reaction is.0184

Here, unfortunately, all we have is just the concentration and the time; well, it's not a straight line--as you can see, it drops down that way; it's curved; so we definitely know it's not a 0 order reaction.0189

Well, here what you want to look at is concentration and time; and what you notice--we start with a .2 Molar concentration at time 0.0207

Well, at the first time increment (which is 1), it has dropped down to .1 molarity; so it has dropped by half.0216

At the next equal time increment (2), now the concentration is .5; at the next equal time increment (which is 3), it has dropped down to .5; so, for equal time increments, the concentration always drops by half.0225

This says that the half-life is constant; in other words, the half-life doesn't depend on concentration.0243

The only time that the half-life doesn't depend on concentration is for a first-order reaction: the half-life is 0.693, divided by K, which is the rate constant.0253

That means that every equal time interval, half the reaction is gone; half the concentration vanishes; half the concentration vanishes.0269

And, based on these numbers (.2 to .1 to .05, from 0 to 1 to 2 to 3), equal time intervals--half of the concentration at the beginning of that time interval vanishes.0280

So, because that is constant, I know that this is a first-order reaction; so in this particular case, B is our answer.0298

So again, when you see a graph of time versus concentration (or Y versus X, I guess, so concentration versus time; I never know which one goes first), take a look at the values themselves, and recognize: again, these are multiple choice questions, so they are not going to be altogether that strange.0308

If you were to be given a problem like this on the free response, then you might be given a couple of different graphs.0326

But here, since they only gave you one graph of time and concentration, chances are they are going to be asking you something about the half-life.0332

And, in this case, the numbers confirm that the half-life is constant: for every equal time interval, there is a 50% depletion of the concentration of reactant, since the beginning of that time interval.0339

The only reaction that has a constant half-life is the first-order reaction: the 0 order and the second-order are not like that; so 63 is B.0352

OK, let's see: #64 says: An equal number of moles of He, Ar, and Ne are placed in a glass vessel at room temperature; if the vessel has a pinhole-sized leak, which of the following will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the mixture has effused?0370

They put a bunch of gases in this vessel, and after a certain amount of time, they measure how much gas is in the vessel (that means some has escaped); and here is what they find.0397

Well, helium, argon, neon: so argon is a bigger atom than neon; neon is bigger than helium, which means that...the heavier something is, the slower it moves (the slower it effuses).0407

Therefore, helium is going to escape faster, because it is the smallest; neon will escape the next fastest, because it is next smallest; and argon will escape the slowest.0426

Well, at the end, because helium has escaped the fastest, there is less helium than there is neon; there is less neon than there is argon.0436

Therefore, the partial pressure of argon is going to be greater than the partial pressure of neon, which is going to be greater than the partial pressure of helium.0447

As far as the choices that they gave us: helium--less than neon, less than argon; so it is A.0461

That is how you work that out: a heavier molecule escapes slower; a lighter molecule escapes faster.0468

A heavier molecule moves slower--a heavier atom/a heavier molecule moves slower; a lighter molecule or atom moves faster.0475

OK, #65: Which of the following compounds is not appreciable soluble in water, but is soluble in dilute hydrochloric acid?0484

The answer is A, magnesium hydroxide, and here is why: magnesium hydroxide is a solid; magnesium 2+, plus 2 OH^-; it is not very soluble.0494

This equilibrium is mostly over here--it is mostly solid.0510

What happens, though, is: when you put it in acid, the acid reacts with the OH^- that has dissolved a little bit, and it forms water.0513

Well, when you add acid, it uses up this, so it depletes that; when this is depleted, this reaction needs to create more OH^-, needs to create more OH^-, needs to create more OH^-, as it is being produced.0527

As it is being produced, it is being eaten up by the hydrogen, so eventually, all of this dissolves.0541

So, in acidic solution, magnesium hydroxide is soluble, because the hydroxide ion that does dissolve a little bit is used up in this neutralization reaction.0547

OK, #66: When solid ammonium chloride is added to water at 25 degrees Celsius, it dissolves, and the temperature of the solution decreases.0559

Which of the following is true for the values of ΔH and ΔS for the dissolving process?0573

OK, so let's deal with ΔH first: they say that the temperature decreases--in other words, when I touch the solution, it gets colder; that means it is sucking heat from the environment.0578

That means that ΔH is greater than 0; it is endothermic.0588

Or...endothermic...that means it is absorbing heat; I'll write it this way: Endothermic--that means that the ΔH is greater than 0 (it is positive).0595

Now, we are going from a solid to aqueous: a solid is a highly ordered thing; aqueous, less ordered.0608

Something that is less ordered has higher entropy, so ΔS is greater than 0--it is positive.0618

A is ΔH positive, ΔS positive, so our answer is A.0626

That is it--nice and straightforward; just reason it out; this means this, this means this, this means this.0632

OK, #67: Wait, that is number...yes, #67--it says: What is the molar solubility in water of Ag₂CrO₄?0640

OK, #67: so, Ag₂CrO₄ dissociates as 2 Ag⁺ + CrO₄^2-, and they tell me that the K_sp is equal to 8x10^-12.0657

Well, I know the K_sp for this: it is equal to the silver ion, raised to the 2 power because of that 2, times the CrO₄^2- ion raised to the 1 power.0677

Well, for every chromate produced, twice as much silver is produced.0689

So, if I call the chromate x, well, that means 2x of silver is produced (right?--when this dissociates, it produces 1 mole of chromate and 2 moles of silver ion, so if I produce x amount of chromate, I produce 2x amount of silver ion).0703

Now, I put these into here; so I write the K_sp, which they gave me as 8x10^-12, equals 2x squared times x, which is equal to 4x squared times x, which is equal to 4x cubed.0725

That gives me x³; when I divide both sides by 4, I get 2x10^-12; and x is equal to the cube root of 2x10^-12.0746

That is my answer: 67--that would be E; there you go.0762

We spent a fair amount of time on solubility.0770

OK, #68: Now, in this particular course in AP Chemistry, I did not discuss the bonding of solids in networks (for example, metallic bonding, ionic solids, things like that); that was one of the things that I left to you, because it was reasonably straightforward.0774

It was mostly informational--just stuff that you sort of have to know; there is nothing to explain, so much--nothing to sort of describe what is going on--no real calculations.0797

It was just one of those things that I felt that, if I had covered it, it would have just been sort of going through a laundry list of information; so that is the reason I didn't cover it, but it will show up in a couple of questions here, like this one.0808

In which of the following processes are covalent bonds broken?0822

OK, iodine solid to iodine gas--those are not covalent bonds that hold the iodine molecules together; those are London dispersion forces.0826

CO₂ solid to CO₂ gas--no; NaCl solid to NaCl liquid--no; diamond to carbon gas--yes.0838

Diamond is a covalent network solid; diamond is a bunch of carbons that are covalently bonded; gas--gaseous carbon--that means all of the bonds have been broken, and there are free carbon atoms just bouncing around all over the place--very high temperatures, but the answer is D.0848

The rest are essentially intermolecular forces; OK.0869

69: What is the final concentration of barium ions in a solution, when 100 milliliters of a .1 Molar barium chloride is mixed with 100 milliliters of a .05 Molar H₂SO₄?0873

OK, well, here is where we have to be careful: so let's do this--barium chloride dissolves into barium 2+, plus 2 Cl^-; OK.0888

We have 100 milliliters, times 0.1 millimoles per milliliter; that gives us 10 millimoles of barium +...barium 2+; I'm sorry.0903

OK, so now, they say they mix it with 100 milliliters of a .05 Molar H₂SO₄.0920

What I have to recognize here is that this 10 millimoles of barium that is floating around in solution is now going to be bound up with sulfate, because barium sulfate is insoluble.0928

In other words, this barium ion that is floating around is going to react with the sulfate to form this solid barium sulfate (and remember, an arrow going down means that it is a...).0938

This is a 1:1 relationship, so we need to find out how many moles of sulfate there is.0951

Well, 100 milliliters, times 0.05 millimole per milliliter: that is going to give me, if I'm not mistaken, 5 millimoles of sulfate.0956

Well, if I have 5 millimoles of sulfate and 10 millimoles of barium, that means that 5 millimoles of the sulfate is going to react with 5 millimoles of the barium.0975

So, if 5 millimoles of the barium ends up reacting, I am going to be left with 5 millimoles left over, of barium floating around.0987

What is it that they want?--they want to know the final concentration; so millimoles over the total volume: I had 100 milliliters and 100 milliliters, so now I have a total volume of 200 milliliters; and that, when I do that, is going to equal...so 5/200 is 0.025 Molar.0997

That is it: so here, the crux of the problem is recognizing that barium sulfate is insoluble.1021

When you have some barium ions, there is a certain number of moles of that; you have some sulfate ion, a certain number of moles of that; you know that it is a 1:1 ratio.1028

Again, this is chemistry; chemistry is always going to involve some kind of a reaction, some kind of an equation; so when you write this equation, it's 1:1; 5 moles reacts with 5 moles; that leaves 5 moles left over.1036

Now, you have a total volume of 200 milliliters, so the molarity is .025 Molar.1049

OK, #70--we are almost done: we have 5 more questions to go.1054

OK, when 100 milliliters of a 1 Molar sodium phosphate is mixed with 100 milliliters of a 1 Molar AgNO₃, a yellow precipitate forms, and silver ion becomes negligibly small.1060

Which of the following is a correct listing of the ions remaining in solution, in order of increasing concentration (lowest to highest)?1074

OK, so let's see: so let's do Na₃PO₄; 3 Na⁺ + PO₄^3-.1083

OK, so 100 milliliters times 1 mole per liter...1 millimole per milliliter...100 millimoles...OK; here is where you are basically going to have to look at mole ratios.1101

1 mole of this produces 3 moles of that and 1 mole of that (right? 1 mole of sodium phosphate produces 3 moles of sodium ion and 1 mole of phosphate ion).1117

Well, now let's do AgNO₃: AgNO₃--one mole of this produces 1 mole of silver and 1 mole of nitrate.1131

This is just a mole ratio problem; well, the silver ion and the phosphate ion are going to mix: silver--three silvers--plus one phosphate is going to go to Ag₃PO₄ solid.1146

So, 3 moles of this, plus 1 mole of this, is going to form 1 mole of this.1167

So now, let's see what we have: here, I have 1 mole of the silver...100 milliliters...I actually don't have to use the 100 milliliters and the molarity--I can just use these numbers.1174

OK, I don't have to actually find out the number of moles; I can just work with mole ratios.1191

All of the silver is going to react with all of the phosphate.1195

Well, three moles of silver per one mole of phosphate...well, if I have 1 mole of silver, I need 3 moles to react with 1 mole of phosphate.1200

Therefore, all of the silver will be used up: so the silver ion...there is going to be nothing left.1213

The phosphate ion--well, 1 mole is produced; 1 mole is used; but if I have 1 mole of silver, I need 3 moles of silver for 1 mole of phosphate; therefore, since I only have 1 mole of silver, I'm only going to end up using 1/3 of the amount of phosphate.1222

Phosphate--I have 2/3 of that left over; and now, sodium--I have 3 moles left over, because it produced 3 moles of sodium; and the nitrate is 1 mole NO₃^-.1245

So, in terms of increasing concentration, I'm going to end up with PO₄, and then the next highest concentration is going to be nitrate, and then the next highest concentration is going to be sodium.1271

That means A is our best answer.1292

Now, notice: I didn't have to use the 100 milliliters times the 1 Molar, 100 milliliters times the 1 Molar; I mean, I certainly recognize that this is 100 times 1 Molar of the sodium phosphate and 100 milliliters times 1 Molar of the silver nitrate.1297

I looked at those numbers; but because they were the same--the 100, 1; 100, 1; I realized that I could just go ahead and use 1 mole, straight forward.1313

100 milliliters times 1 millimole per milliliter gives me 100 millimoles; so I could use 100; I could use the numbers directly; but because it was the same, I could just pick a standard, and a standard is just 1--I could just use the mole ratios as is.1324

If those numbers were different--if it were something like 100 milliliters of a 1 Molar sodium phosphate and, let's say, 50 milliliters of a .5; then, I would actually have to use the specific number of millimoles, and I would have to do more conversions.1340

But here, because they were the same, I could just use 1, 3; 1, 1; 1, 1; 1, 1; 3 to 1.1353

I could just use them straight; I hope that makes sense.1360

OK, #71: Let's see, in a qualitative analysis for the presence of lead, iron, and copper ions in aqueous solution, which of the following will allow the separation of lead from the other ions at room temperature?1364

OK, let's see: we'll call this x; OK, so qualitative analysis: we discussed qualitative analysis briefly when we were talking about complex ions and ions in solution.1388

We talked about how to separate a collection of ions in a solution; so let's say you have lead, mercury, zinc, iron, copper, all floating around in a solution: you could add different things to the solution to precipitate out different ions.1414

In the case of lead, what we added was the first step in that qualitative analysis; if you look back in one of the lessons, I actually showed very, very carefully what the analysis was.1430

The first step was adding dilute hydrochloric acid to any collection of ions: that is sort of the first step in classical qualitative analysis.1441

You add dilute hydrochloric acid, and that automatically pulls out the lead ion and the mercury; in this case, because we have lead 2+, iron, and copper, you would add dilute hydrochloric acid.1451

So, in this case, B would be the best answer.1462

OK, #72: After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of 38%.1467

The correct value for the percentage of water in a hydrate is 51%: which of the following is the most likely explanation for this difference?1479

OK, the best way to think about this is: pick a sample size--I chose 10 grams--and then deal with that; so let's say I have a 10-gram sample.1487

If I heat up the sample to drive off the water, I'm going to end up with a certain mass--let's call it x grams.1500

OK, now, the percentage of water is going to be...so I have this 10-gram sample, which is a hydrate (it contains the sample and water attached to it); I drive off the water by heating it; I weigh it again; now it is going to weigh x grams.1508

Well, so the sample, in and of itself, would be the x grams: that is how much the sample weighs without any water attached to it.1523

Well, the water weight would be 10-x, because I had 10 grams to begin with, and I weighed out the dried sample; so 10-x is the water weight; over 10, times 100--that gives me the percentage of water.1532

OK, so now, they are telling me that this...well, you know what, I don't even need to multiply it by 100; I can just go ahead and just use decimals.1550

When they say 38%, that means this thing is equal to 0.38.1558

Now, they are saying what it should be is (oops, let's get rid of these stray lines)...they are telling me that what it should be is 51%.1564

What I should have is: 10-x, over 10, equals 0.51.1573

When I solve this for x, x equals 6.2; what I should have is x = 4.9.1583

In other words, when I measured the mass (or when this person doing the experiment measured the mass), they measured a mass that was higher than what should have been.1592

She got 38%; it should be 51%; this is the equation for how she got the 38%; this is the equation for the 51%.1602

Well, that means she measured a mass, after driving off the water, of 6.2 grams; the mass should have been 4.9 grams--it is heavier than it should have been.1613

Heavier than it should have been--that means it has more water; that means she didn't drive off all of the water.1623

Or, the other possibility: if she did drive off all of the water, after drying it, she didn't weigh it fast enough, and it ended up actually recollecting moisture.1630

There is actually still water attached to this, so B is the best answer for this.1641

The dehydrated sample absorbed moisture after heating: so she may have driven off the water, but before she got to weighing it, it actually reabsorbed, and therefore the mass that she measured was higher than it should have been.1647

It should have been 4.9; it ended up being 6.2, which gave her a lower percentage (38 instead of 51).1659

OK, 73 (we are almost done--it's kind of exciting!): OK, let's see: The volume of distilled water that should be added to 10 milliliters of a 6 Molar HCl in order to prepare a .5 Molar HCl solution is approximately what?1667

OK, this is a dilution problem; I'm going to use...OK, molarity 1 times volume 1 (or molarity initial...actually, let me use initial and final; I think it's better)...initial molarity times initial volume equals final molarity times final volume.1687

Well, the initial molarity is 6; the initial volume is 10 milliliters; and I can just leave it: 6, 10 milliliters, 6 millimoles per milliliter; so, I can just deal in milliliters; it's not a problem--I don't have to change to liters or anything like that.1717

The final molarity is 0.5; now, the final volume is 10+x: x is what we want--they want to know how much distilled water I have to add to this 10 milliliters, so it's 10+x.1739

There you go: you end up with 60=5+0.5x; you end up with 0.5x=55; x=55/.5, which is 110: your answer is 110 milliliters of distilled water, which is D; that is it.1755

OK, #74: Which of the following gases deviates most from ideal behavior?1779

The gas that will deviate most from ideal behavior is the gas that is the most polar.1785

SO₂ happens to be the most polar; neon, no; CH₄, no; N₂, no; H₂, no; so SO₂, by process of elimination.1791

By process of elimination, you can get to SO₂; however, just so you know, the Lewis structure is this.1801

There is a dipole moment here--the net dipole is this way: it is a polar molecule; it will deviate from ideal behavior.1811

So, the answer is A for 74.1819

OK, and #75, the last question of the multiple choice: All right, which of the following pairs of liquid forms the solution that is most ideal (most closely follows Raoult's Law)?1823

An ideal solution (in other words, a mixture of two different liquids) is when the liquids that you mix are the most alike.1835

In this list, A: C₈H₁₈: it is a hydrocarbon, nonpolar; H₂O: hydrogen bonding and polar; no way.1847

You have the propanol, and you have water; they are kind of cool, but you have this long carbon chain on here; not really.1858

The propanol and the C₈H₁₈...not really.1866

Of the choices, C₆H₁₄, C₈H₁₈...they are practically identical; they are only off by two carbons; they are pure hydrocarbons; the answer is D.1870

75--the answer is D: an ideal solution--the one that obeys Raoult's Law, one that doesn't deviate positively or negatively (remember?--that is one of the last lessons that we discussed--Raoult's Law and deviation)--you want the ideal solution to be where the two solutions are the most alike.1881

So, for example, if I had something like water and methanol, water and methanol are very much alike: there is hydrogen bonding...even though there is that one carbon, they are actually quite alike.1900

Water and acetone: a great mixture--you have hydrogen bonding...I'm sorry, no: water and acetone--that will actually end up giving you deviation.1911

In this case, you want an ideal solution, so that they are alike, so that the interactions between particles of A among themselves and particles of B among themselves are the same as the interactions between the particles of A and B.1921

In this particular case, the C₆H₁₄ and the C₈H₁₈...the interactions among these and the interactions among these are virtually identical with the interactions among those.1938

That will give you the most ideal situation.1954

OK, that takes care of the multiple choice for this practice exam.1959

Thank you for joining us here at Educator.com, and for AP Chemistry.1963

Next lesson, we will start our discussion of the free response questions; take care; goodbye.1966