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Lecture Comments (5)

1 answer

Last reply by: Professor Hovasapian
Sun Feb 22, 2015 7:37 PM

Post by Shih-Kuan Chen on February 21, 2015

Professor, where can I learn more about flame color and colors of ions and compounds?

0 answers

Post by Mwongera Mwarania on October 27, 2013

Sorry, on re-listening the recording, you have explained that e0 is not like Enthalpy: you don't multiply with. You only reverse the sign if you reverse the reaction. Please ignore my earlier question.

0 answers

Post by Mwongera Mwarania on October 27, 2013

Q.57: Why did you not multiply the e0 by 3 for the second step in accordance with the Hess' law?

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Post by Professor Hovasapian on July 18, 2012

Link to the AP Practice Exam:

Take good Care


AP Practice Exam: Multiple Choice, Part II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Multiple Choice 0:12
    • Multiple Choice 42
    • Multiple Choice 43
    • Multiple Choice 44
    • Multiple Choice 45
    • Multiple Choice 46
    • Multiple Choice 47
    • Multiple Choice 48
    • Multiple Choice 49
    • Multiple Choice 50
    • Multiple Choice 51
    • Multiple Choice 52
    • Multiple Choice 53
    • Multiple Choice 54
    • Multiple Choice 55
    • Multiple Choice 56
    • Multiple Choice 57
    • Multiple Choice 58
    • Multiple Choice 59
    • Multiple Choice 60
    • Multiple Choice 61

Transcription: AP Practice Exam: Multiple Choice, Part II

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

We are going to continue our multiple-choice practice, and we left off at problem #41, so we are going to start with problem #42: let's just jump right on in.0004

OK, when the equation above is balanced and all coefficients reduced to lowest whole-number terms, the coefficient of OH- is going to be...0013

All right, so 42: this is just a simple balancing issue, so when you balance this out, the coefficient of the OH- is going to be 3, so 42 is C.0021

OK, #43: A sample of 61.8 grams of H3BO3 (a weak acid) is dissolved in 1,000 grams of water to make a 1 molal solution.0034

Which of the following would be the best procedure to determine the molarity of the solution? Assume no additional information is available.0046

OK, of these choices, the best way to check the molarity of the situation...because molarity is moles per liter, you need the total volume, which is in the denominator of molarity; it actually is going to be D.0055

You need to take a measurement of the total volume of the solution.0068

That is it--nice and straightforward.0072

OK, #44: A rigid metal tank contains oxygen gas (rigid, so there is no volume change).0076

Which of the following applies to the gas in the tank when additional oxygen is added at a constant temperature?0084

That is the important part: additional oxygen is added at a constant temperature.0091

"The volume of the gas increases"--no, it's rigid.0097

"The pressure of the gas decreases"--no, you are actually injecting gas into the system, so the pressure actually increases.0100

"The average speed of the gas molecules remains the same"--this is where it is a little bit tricky.0107

Now, a constant temperature: that is the secret here--a temperature constant implies that the average kinetic energy is unchanged.0114

Remember, temperature is a measure of the energy of the actual molecules: if the temperature doesn't change, the average energy of the molecules doesn't change.0125

So, that is unchanged.0134

Well, that means one-half the mass times the average velocity squared is unchanged; this means that the velocity, the average velocity, is unchanged.0140

So, yes, C is the answer for #44.0151

OK, #27 What is the hydrogen ion concentration in a .05 Molar HCn?--the Ka for HCn is 5.0x10-10.0156

OK, so again, this is a weak acid problem; we do have to use an ICE chart, but again, the numbers are going to be really, really simple, so there doesn't have to be any complicated calculations.0168

Let's go ahead and run through what we normally run through.0181

The number of species in solution: well, we have HCn in solution, and we have H2O in solution.0186

Of those two things that can dissociate, the 5.0x10-10 Ka for HCn...this is the dominant equilibrium, so what we are going to have is the following.0192

We are going to have: HCn is in equilibrium with H+ + Cn-.0203

Well, they tell us that we are going to have an Initial; we are going to have a Change; and we are going to have an Equilibrium.0210

They tell us we have a .05 Molar, so before anything happens, there is none of this, and there is none of that.0215

I'm sorry, I shouldn't say they don't matter; I'll use 0; we use straight lines for things that actually are irrelevant.0221

A certain amount is going to dissociate; a certain amount is going to show up; and a certain amount is going to show up.0228

We are looking for x, because that is going to be the hydrogen ion concentration.0232

0.05-x, x, and x; so they tell us that the Ka, which is 5.0x10-10, equals this times that, divided by that.0236

We get x2 over 0.05-x, which is just about equal to x2 over 0.05.0253

OK, and when I do that calculation, I end up with (so now we go over here; we end up with) x2=0.25x10-10; that means x is equal to the hydrogen ion concentration (right?--x is hydrogen ion concentration); it is equal to 0.5x10-5.0264

Square root: just basically take the square root of .25 and the square root of 10-10, and then, when you convert this to scientific notation, you end up with (oh, these stray lines are crazy! OK) 5.0x10-6.0292

That is D, so our answer for 45 is D.0311

Again, set up your ICE chart; the numbers are really easy to work with--nice and straightforward.0316

OK, so #46 says: Which of the following occurs when excess concentrated NH3 is mixed thoroughly with .1 Molar copper nitrate, aqueous?0322

OK, so we have a copper nitrate solution; a copper nitrate solution is going to be a colored solution.0336

It is a transition metal: most transition metals are colored--they absorb visible light.0343

And again, like I mentioned earlier, the copper nitrate dissolves; the water molecules surround the copper (2 of them, 4 of them, 6 of them...however many); there is a certain color that that complex ion absorbs.0349

When you put NH3 into this solution, well, the NH3...a couple of molecules of ammonia may actually replace a couple of molecules of water surrounding the copper ion.0363

What happens is: the color actually gets darker or changes or gets lighter; so, in this case, the best choice would be: The color of the solution turns from a light blue to a dark blue; so the answer would be C.0374

OK, #47: When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is found to contain 62.2% hafnium by mass and 37.4% chlorine by mass; what is the empirical formula of the compound?0390

OK, so this is a straight empirical formula problem where you convert percentages to grams; you can go to moles and divide by the least number, and you should be able to get some whole numbers.0410

Let's go ahead and do this really quickly: Hafnium--we have 62.2%, is 62.2 grams, times 1 mole of hafnium is roughly 179 grams; so I end up with; so let's go ahead and put .34 (I'll round it slightly up, because this is a little bit more than that).0419

When I do chlorine, that is going to be...they say 37.4%, which is 37.4 grams times 1 mole/35.45 grams, so you are looking at about 1.0448

So now, we go ahead and divide by that to get 1; get 3; there is your answer, HFCl3; it looks like C is your final answer for #47.0464

OK, #48 says: if 87.5% of a sample of pure iodine-131 decays in 24 days, what is the half-life of iodine-131?0482

Well, we know what the half-life of something is: that is the amount of time it takes for half of a sample to vanish.0498

OK, this 87.5%...again, on these AP exams, you are going to be given very easy numbers to work with.0505

Here is what happens: I start with a sample; OK, half of it decays; now I am left with 50%.0512

Well, now half of that sample decays; well, half of 50% is 25%, so now, I am at 75% of the original sample (is gone).0522

Now, half now I have 25% of the sample left; well, half of that 25% is 12 and a half percent; so after three cycles of half-life, half-life, half-life, 75 plus another 12.5 puts me at 87.5.0533

So, after 3 cycles...the original sample--half goes away; what is left over--another half goes away; what is left over--another half goes away; at that point, I have 87.5% that has decayed.0552

Well, I have three cycles; 24 days to go through three cycles; that means each cycle is 8 days long, so the half-life of iodine-131 decay is 8 days.0568

Take a look at the percentage, and chances are it is either going to be 1, 2, 3, or 4; and then, just take the number of days or hours or minutes or seconds and divide by the number of cycles.0582

We went through one cycle, two cycles, three cycles; 24 days; 3 into 24 is 8 days; so 48 is B.0592

I hope that makes sense.0604

OK, #49: Which of the following techniques is most appropriate for the recovery of solid potassium nitrate from an aqueous solution of potassium nitrate?0606

If you have an aqueous solution, and you need to recover the solid, it's very simple: just boil off the water--so "evaporation to dryness"--49 is E.0615

OK, #50--let's see: #50 says: In the periodic table, as the atomic number increases from 11 to 17, what happens to the atomic radius?0628

As you move from left to right on the periodic table, atomic radius (not ionic radius, atomic radius) decreases.0641

As the positive charge gets bigger, even though you are adding one electron, you are actually going to be sucking in the electrons a little bit more.0649

So, as you move from left to right (from your perspective, left to right on the periodic table), the atomic radius gets smaller.0656

The answer is D: it decreases only.0662

OK, #51--let's see what we have: Which of the following is a correct interpretation of the results of Rutherford's experiment, in which gold atoms were bombarded with alpha particles?0667

Well, you remember the Rutherford scattering experiment: he shot alpha particles (which are helium nuclei) at a thin piece of gold--through a thin piece of gold--and very few of the alpha particles actually ended up bouncing backward.0680

Most of the alpha particles just passed right on through.0697

What that means is that the positive charge of the atom is concentrated in a very small region.0701

An atom like this--a positive charge is actually concentrated in a very tiny part of it; most of the alpha particles just pass through the atom.0706

So, 51 is E.0715

OK, #52: Under which of the following sets of conditions could most O2 be dissolved in H2O?0722

OK, if you are trying to dissolve a gas in a liquid, what you need is high pressure, low temperature.0730

So, here: high pressure--let me see: we look at 5 and 5 atmospheres (that is A and B); low temperature--between 80 and 20, we are going to take the 20 degrees, so we will take B.0737

52 is B; so, again, when you need to dissolve a gas in a liquid, you need conditions of high pressure, but you need conditions of low temperature.0751

If the liquid actually is a high temperature, it is going to be bubbling, bubbling, bubbling; it is going to be all excited; high temperature is actually going to force the gas out.0761

Low temperature means the gas can actually occupy a space in that liquid.0769

All right, let's see--#53: OK, gases W and X react in a closed, rigid vessel to form gases Y and Z according to the equation above; the initial pressure of W is 1.2 atm, and that of X is 1.6 atm.0775

No Y or Z is initially present; the experiment is carried out at constant temperature; what is the partial pressure of Z when the partial pressure of W has decreased to 1 atmosphere?0796

OK, so let's work this one out; it is actually very, very simple.0809

W gas, plus X gas, goes to Y gas plus Z gas.0814

They tell me this is initially at 1.2 atm, and this is at 1.6 atm.0825

It is 0 and 0 here; so at the end, when this is dropped down to 1 atmosphere, they want to know what is the concentration of...yes, what is the partial pressure of Z.0832

OK, now, notice what they said...equation...container...carried out...OK, here is what is important: the experiment is carried out at constant temperature.0851

Because it is carried out at constant temperature, that means nothing actually changes in terms of pressure.0860

So, the beginning total pressure, which is 1.2 plus 1.6 (which is a total of 2.8)--that is going to actually end up being the final pressure: 2.8.0865

Now, the final pressure consists of the pressure of the W, the X, the Y, and the Z.0877

Now, pressure is the same as concentration (right?--pressure and concentration are just different sides of the same coin): now, this pressure went from 1.2 to 1.0.0884

Well, this is a 1:1 mole ratio; so, if .2 moles of this was used up, I can think of pressure as just moles.0895

.2 moles of this was used up: that means this pressure dropped down to 1.4.0903

Well, this is 1:1; if .2 moles was used up, that means .2 moles of this was formed, and .2 moles of this was formed.0908

So now, we want the total pressure to be 2.8.0919

Well, 1, 2.4, 2.6, 2.8; that is how you work out this problem; the idea is constant temperature--that means pressure doesn't change: the initial pressure equals the final pressure.0924

You use the mole ratios, and you use the fact that pressure is like concentration: if this pressure drops from 1.2 to 1.0, since this is 1:1, this one drops, also, from 1.6 to 1.40.0937

Well, if this loses .2 atmospheres, there is gain of .2 atmospheres here, gain here; as long as the sum of those equals the sum of this, everything is fine.0951

So, 1.2 is the answer: so we get: 53 is equal to A.0963

OK, #54: Let's see, which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above?0972

OK, so they say that we have: 2 NO gas, plus O2 gas, is in equilibrium with 2 NO2; and they tell me that this ΔH is less than 0, which means that it is exothermic.0989

Exothermic means that it releases heat, so I'm going to write +heat; heat is one of the products.1006

Now, which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above?1012

Let's see: if you decrease the, if you decrease the temperature, it is going to want to increase the temperature, so the reaction is going to go to the right.1021

Decreasing the value of Keq means pushing the equilibrium to the left, backward towards the reactants: the only thing here that does that--if you increase the temperature, which means increasing the heat, that means you are going to increase one of the products, which is heat.1031

Well, the system is going to offset; it is going to offset by wanting to move to the left.1049

A reaction that moves to the left: that means the equilibrium constant is actually--all things being equal, the equilibrium constant will decrease.1055

That is what you have recognize: a decrease in the value of Keq means that more of the reactant is going to be favored.1067

In this particular case, because it's exothermic, by increasing the heat--increasing the temperature--you are forcing the reaction to move to the left.1074

I hope that makes sense; so here, 54--the best answer is B.1086

OK, #55: According to the balanced equation above, how many moles of HI would be necessary to produce 2.5 moles of I2, starting with 4 moles of KMnO4 and 3 moles of H2SO4?1092

OK, this is basically just a limiting reactant problem; so we need to find out...we need to just look at the equation to see...which is which.1110

Let me actually write this equation know what, that is fine; we can just work right from the page; we don't really need to write anything out--there is nothing too complicated going on here.1119

We want to produce 2.5 moles of I2, starting with 4 moles of KMnO4.1130

OK, so we have 4 moles of KMnO4; how many moles of HI?...and 3 moles of H2SO4.1136

OK, so we have 3 moles of H2SO4; for that, 2 moles of KMnO4 are necessary; well, we have 4 moles of KMnO4, so that is not a problem.1145

3 moles of H2SO4 requires 10 moles of HI (right?--because there is a 10 in front of the HI).1155

10 moles of HI gives 5 moles of I2; in order to produce 2.5 moles of I2, which is half of 5, I need half of 10.1162

Therefore, I need 5 moles of HI; so the answer is D.1171

That is it: just look at the mole ratio; see which one is the limiting reactant (in this case, the H2SO4); the rest is just mole ratios.1175

And again, the numbers are very, very simple.1183

OK, #56: A yellow precipitate forms when .5 Molar NaI is added to a .5 Molar solution of which of the following ions?1186

OK, so you have sodium iodide: sodium is not going to do anything--it is going to be soluble with anything; iodide ion is not going to mix with chromate, sulfate, or hydroxide, and it is going to be soluble with zinc; so A is your final answer.1196

Lead iodide: that is the one that is the yellow precipitate.1210

OK, #57: Let's see, according to the information above, what is the standard reduction potential for the half-reaction M3+ + 3 electrons goes to M, solid?1215

OK, we need to look at what they gave us; we need to rearrange it in such a way that, when we add it, we get what we want, which is: M3+ + 3 electrons goes to M(s).1229

All right, now notice what they want: the M3+ is on the left; what they give us--the M3+ is on the right; so I'm going to flip that first equation.1240

I'm going to write that first equation as (oops, let me write--this is #57)...I'm going to write: 3 Ag, solid, plus M3+, goes to M, solid, plus 3 Ag+.1249

Now, the potential for that is +2.46, but since I flipped the equation, now the electric potential is going to be -2.46 volts.1273

Now, I have an Ag+ on the right, and the other equation they give me has an Ag+ on the left; so I'm going to leave that as written.1283

Ag+ + an electron goes to Ag, solid.1291

Let's see: the electrostatic potential for that is positive 0.80, because I haven't done anything to it.1308

However, I am going to go ahead and put a 3 in front of the Ag, 3 in front of the electron, and 3 in front of this Ag, because the final equation that I am looking for--I need the Ag's to cancel.1317

This cancels with that; that cancels with that; and I am left with: M3+ + 3 electrons goes to M, solid--which is exactly what I wanted.1327

Now, I just add these two, and I end up with -1.66 volts, which is A.1339

That is it; I just need to arrange this in such a, the only thing you have to remember here is: notice, I change this second equation by actually multiplying everything by a 3.1350

However, this is not like enthalpy: reduction potential is an intrinsic property--it doesn't depend on how much of something you have; it is just a property.1361

So, even though I put 3 here, I didn't multiply the electrostatic potential by 3; the only thing that I change when I change electrostatic potentials is the sign, if I flip the equation.1373

Flip the equation--just go positive/negative, negative/positive; but by adding coefficients in order to make things cancel, I do not change the electrostatic potential.1385

It is not like enthalpy or entropy or free energy; OK, so very, very important: electrostatic potential is an intrinsic property--does not depend on the amount that you have.1394

If this were enthalpy and we were doing Hess's Law, yes, you would have to multiply this by 3; OK.1403

Let's see, where are we?--we are at #58.1411

OK, #58: On a mountaintop, it is observed that water boils at 90 degrees Celsius, not at 100 degrees Celsius, as at sea level.1415

This phenomenon occurs, because on the mountaintop...of our choices, C is the best choice.1425

Equilibrium water vapor pressure equals the atmospheric pressure at a lower temperature.1433

Remember what boiling point is: the boiling point is the temperature at which the vapor pressure is equal to the atmospheric pressure.1439

As you go higher in elevation, the atmospheric pressure drops; there is less atmosphere, so there is less pressure.1448

Well, because there is less pressure, you reach vapor pressure at a lower temperature.1454

That is why boiling point decreases as you go higher in elevation; so C is the best answer here.1460

OK, #59: Let's see, a 40-milliliter sample of a .25 Molar KOH is added to 60 milliliters of a .15 Molar BaOH2.1468

What is the molar concentration of OH- in the resulting solution?--assume that the volumes are additive.1479

OK, so we need to calculate the total moles of OH- and divide by the total volume.1485

All right, so we know that potassium hydroxide dissociates as 1:1: one mole of potassium hydroxide produces one mole of hydroxide ion.1491

Let's go ahead and do that one: that is going to be 40 milliliters, times 0.25 millimoles per milliliter (moles per liter--I can do millimoles per milliliter, or centimoles per centiliter, as long as I do it consistently); so I end up with 10 millimoles of OH-.1501

Now, barium hydroxide (BaOH2): when it dissociates, it produces 1 mole of barium 2+, but it produces 2 mol of hydroxide.1524

Here, I have 60 milliliters times 0.15 millimole per milliliter, and then I multiply that by 2 moles of OH- per one mole of barium hydroxide; I end up with 18 millimoles.1534

Or, I could have just done this first calculation and gotten 9 and just multiplied by 2, because 1 mole of this produces 2 moles.1556

So now, I have a total of 28 millimoles of OH-, and I divide by my total volume, which is 40 milliliters plus 60 milliliters, which is 100 milliliters; so I get a molarity of 0.28 M.1563

Now, you notice: if you remember, my m has a line on top of it; the molarity symbol that is used nowadays is a capital M; so you will see it as .28 M.1583

There you go; that is your answer, and .28 is C.1594

#59 is C; all right.1599

Just a couple more to go here for this lesson: let's see, #60 says: A .03 mole sample of NH4NO3 is placed in a 1-liter evacuated flask, which is then sealed and heated.1603

The NH4NO3 decomposes completely according to the balanced equation above.1621

The total pressure in the flask is measured at 400...I'm sorry, the total pressure of the flask is measured at 400 Kelvin; it is closest to which of the following?--and they give you the value of R; OK.1627

Here, we need to find the total number of moles of gas that is produced; we have a solid, and it is evacuated, so the pressure is 0 to begin with, but you are producing N2 gas and H2O gas.1644

OK, well, let's see: we have .03 let's go ahead and write our equation here.1657

We have: NH4NO3--that is a solid--and it is going to form 1 mole of N2 gas and 2 moles of H2O gas--2 moles of that.1665

Well, if I have 0.03 moles of this to start off with, 1:1 ratio, it is going to end up forming .03 moles of this; 1:2--it is going to end up forming .06 moles of that; the total number of gas that is going to be floating around in that container is going to be 0.09.1683

So now, we use our PV=nRT; the total pressure is equal to nRT over V; number of moles is 0.09; 0.082 is what they give us for R; the temperature is going to be 400; and then, it is in a one-liter flask.1705

When you multiply all of this out (which you should be able to do just on sight): .09, .08, times are looking at about 2.95; the answer is going to be 3 atmospheres.1730

So the answer is A; so 60 is A--that is it.1745

You want to find the total moles of gas, and then just use the ideal gas law to find the pressure.1750

OK, #61: For the reaction of ethylene represented above, ΔH is -1323 kilojoules.1756

What is the value of ΔH if the combustion produced liquid water, rather than water vapor?1766

ΔH for the phase change of H2O gas to H2O liquid is -44 kilojoules per mole.1773

OK, so we have this reaction: C2H4 + 3 O2 gas forms 2 moles of CO2 and 2 moles of H2O.1781

They tell me that the ΔH for this is equal to -1323 kilojoules.1794

Notice, it is not kilojoules per mole; it is kilojoules.1800

OK, now, this H2O is as a gas; well, they are saying, "What would happen to the ΔH if, instead of forming water vapor, it actually formed liquid water?"1803

Well, they are telling me that the ΔH for the phase change of H2O gas to H2O liquid...they are telling me that the ΔH for this process is equal to -44 kilojoules per mole.1815

Here we have to watch a couple of things: we have to watch the direction, and we have to watch the units.1838

This one is in kilojoules; this is in kilojoules per mole.1843

Well, we have 2 moles of water that are going to go from gas to liquid; OK.1846

Here is what this looks like: the final ΔH is going to be the sum of the ΔH's, so we are going to start with -1323 kilojoules, and we are going to add...1852

Now, in going from a gas to a liquid, you are actually giving off heat, which is what this negative 44 says; so, we are going to add the heat given off in going from gas to liquid.1865

But notice: we have 2 moles of water (let's get rid of these crazy lines that are showing up all over again), so we have -44 kilojoules per mole, times 2 moles, because we have 2 moles of water that is going from gas to liquid.1879

That does that; when we actually do this, 2 times -44 is -88; -1323 plus a -88...our final answer is -1411 kilojoules, and that is E; so 61 is E.1904

OK, thank you for joining us for the second part of the multiple choice; we have one more section to go.1926

So, thank you for joining us here at

We'll see you for the next lesson; goodbye.1933