Lecture 23. Final Exam Review
This is a modal window.
The media could not be loaded, either because the server or network failed or because the format is not supported.
Formal Metadata
| Title |
| |
| Title of Series | ||
| Part Number | 23 | |
| Number of Parts | 23 | |
| Author | ||
| License | CC Attribution - ShareAlike 3.0 Unported: You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this | |
| Identifiers | 10.5446/18988 (DOI) | |
| Publisher | ||
| Release Date | ||
| Language |
Content Metadata
| Subject Area | ||
| Genre | ||
| Abstract |
|
Chem 1A: General Chemistry23 / 23
4
5
6
8
11
17
18
19
23
00:00
Man pageMagmaMagnetometerManganeseMashingDielectric spectroscopyMixtureAnomalie <Medizin>AuxinVancomycinGrading (tumors)Initiation (chemistry)Rydberg atomIce sheetRiver deltaWalkingHydrogenWine tasting descriptorsSetzen <Verfahrenstechnik>SingulettzustandWursthülleGesundheitsstörungFiningsSet (abstract data type)LightningGas cylinderLibrary (computing)Electronic cigaretteSulfurSea levelOctane ratingBeerFireComputer animation
09:19
Man pageVancomycinGraukäseHydroxybuttersäure <gamma->Nitrogen fixationAuxinMashingEthanolamineInsulinMagnetometerDominican RepublicSunscreenData conversionFoodWine tasting descriptorsPotato chipChlorineGesundheitsstörungColourantDrainage divideWursthülleMolecular beamHydrogen chlorideManganese dioxidePressureValence (chemistry)Meat analogueChemical reactionGemstoneWaterUltraschallschweißenBoyle-Mariotte-GesetzStuffingMan pageLibrary (computing)LightningBisacodylActive siteElectronic cigaretteOrganische ChemieComputer animationLecture/Conference
18:25
FormaldehydeMan pagePan (magazine)NitrosamineHydroxybuttersäure <gamma->MagnetometerTau proteinArginineDigital elevation modelElectronic cigaretteChlorineChemistryCalculus (medicine)HydrogenChemical formulaSample (material)Data conversionSunscreenMolar volumeMolekulargewichtsbestimmungCarbon (fiber)Amount of substanceWursthülleChemical compoundSea levelSense DistrictAtomic numberAzo couplingFormaldehydeCupcakeGemstoneExplosionGas cylinderAreaStockfishGrowth mediumComputer animationLecture/Conference
27:32
MagnetometerMan pageAtomFormaldehydeBrown adipose tissueBohriumEthylene-vinyl acetateDigital elevation modelHydroxybuttersäure <gamma->Secondary ion mass spectrometryMagmaPropanVolumetric flow rateQuantum chemistrySea levelAtomic orbitalCarbon dioxideCarbon monoxideCatalytic converterCobaltoxideElectronic cigaretteRedoxLakeHydrogenElectronChemical formulaQuantum chemistryChemical reactionWursthülleChemical propertyStuffingErdölraffinationCarbon (fiber)Common landReaktionsgleichungBranntweinThermoformingHalogenPropionaldehydStickstoffatomWaterProteinSea levelLevomethadonAtomic orbitalPropanAtomic numberEthylchloridDrainage divideWalkingChlorineComputer animationLecture/Conference
36:38
Digital elevation modelQuantum chemistryMan pageSolid phase extractionNitrosamineMagnetometerVolumetric flow rateMixtureSetzen <Verfahrenstechnik>LithiumCalculus (medicine)Noble gasAtomic numberElectronAzo couplingAtomic orbitalMolekulargewichtsbestimmungThermoformingCHARGE syndromeSilverSunscreenRiver deltaAtomData conversionAdamantaneAusgangsgesteinValence (chemistry)AgeingMarch (territory)Electronic cigaretteMedical historyCatalytic converterTrauma (medicine)Food additiveGemstoneCadmiumMatchWursthülleComputer animationLecture/Conference
45:44
Computer animationLecture/Conference
Transcript: English(auto-generated)
00:08
OK. So now, to get started with our final exam review. So here's the worksheet that I sent out. Remember final exam review, while interruptions are always allowed, they're kind of extra
00:23
encouraged here. So feel free to interrupt me. It's a big class so you may have to, you know, wave your hands around, get your neighbors to wave their hands, whatever. OK. What is the energy frequency and wavelength of the light associated with the emitted
00:42
photon in a hydrogen transitioning from an N5 to an N equals 1 level? So which equation are we going to be starting with? Rydberg equation. Now, there's the way that me and your book does it and there's the way that Sapling does it. Either way, I'll get you the right answer. I prefer this way.
01:02
And in this case, I kind of force you into it because I'm requiring you to give me the energy too. OK. So we start there. We fill in our values for the Rydberg constant.
01:26
We fill in our values for N, being really careful to make sure that we put the right ones in the right places.
01:42
Now, what's Z? 1, so I just left it out for this one. So remember this is the place that—so he asked is it initial minus final. So remember this is the place where it's written two different ways. Sometimes it'll be written as initial minus final and then there won't be a negative
02:02
sign out here. Sometimes there will be a negative sign out here and it'll be written as final minus initial. You don't have to memorize the equation so you can just steal it off the equation sheet. But that flip-flops—because what happens is when you derive it, you end up with a negative sign out here. But you can just take and distribute that negative N and end up with this equation and then it flip-flops it.
02:23
OK. So we fill in our energy and we get this. Now, are you going to put that on the line for energy in this case and walk away?
02:41
No. Why? Yeah, we have to change the sign. Why? I asked for the emitted photon, right? So if I asked for delta E, then it would be negative. But if I asked for the energy of the photon, it's the positive version of it. The energy of a photon is always positive.
03:04
So that's your energy value. Now I ask for lambda and frequency, so which equations are we going to use? Yeah. E equals hc over lambda and E equals h nu. And in each case, solve for what we need.
03:25
I want to make a quick note on these about sig figs because it's not all that obvious what sig figs you need to use if I ask the question to you this way. So basically the Rydberg constant here sets up how many sig figs that you need to
03:43
use. So you would just fill in the amount of sig figs that you—however many you cared about if you were doing this experiment. If you needed this value for something, you would fill out the Rydberg constant to the right number of sig figs. And that's not really something that I can test. So what I do on the—there's three or four different types of problems where
04:01
this comes up, and next to it I tell you how many sig figs to do it. And really that's just for consistency's sake so that like the TAs don't see you hold out like a huge amount of sig figs and mark it wrong not really knowing what the question was and not knowing that it really doesn't matter. So on any of the ones that in parentheses I say round to this many sig figs, just do
04:21
it. The idea there being that you could pick somewhere between like two and four and be okay, and I just want to make sure that there's no grading mistakes. So I also have a note not to grade on those, but just to be extra sure I figured I would just tell you how many sig figs to round to. So we do this and we get that.
04:50
If I don't tell you what units to put it in, you're more than welcome to leave it in meters. You don't need to switch it into nanometers or anything like that. Yeah?
05:00
Can you talk louder? I can't hear you. I still can't hear you. Much louder. Oh, you're talking about one where we actually are going to grade on sig figs?
05:22
You have to have the right number of sig figs. So if on a problem where sig figs matter, if you end up with, you know, if you say if you need to have three sig figs and you have three sig figs, you're good. If you have four sig figs, you're wrong. Two sig figs, you're wrong. Now, if there's a rounding error, like my answer says 2.18 and your answer says 2.19,
05:43
no that's fine. That's just a rounding error. That's right. And those are always pretty easy to tell when they're rounding errors. Okay. So now we have this one.
06:02
And all of these values are positive as well, right? Because you're filling in the positive value of e. What's our units on frequency? Yeah. So you can write it a few different ways. You can write it one over seconds, or you can write it seconds inverse, or you can write
06:20
it Hertz. All of those are fine. So any way you want. So remember, delta e is positive or negative, and whenever you see something where it says you're emitting a photon, what do you want to think? You want to think that that's going to be negative, right?
06:42
Because your final is lower than your initial. This direction, there was a, the only one that you would be too hurt on if you forgot that would be this, right? You would get that one wrong. But these you would know to make positive, so you'd be okay. When you go the other direction, what happens if you, I give you a lambda or a frequency, and you solve for e, and then you fill e into here without making it negative?
07:06
You won't get it right. And more than that, most likely, statistics say you won't even get an integer. So if you are doing your exam and you get, you know, 1.57 for instance, is that going to be the right answer? No.
07:21
And are you going to round that to 2? No. So this is, the reason I came up with that number is that's what happened in the first midterm, right? And you know, you should go back and the first place you should check is, is that e negative? Did I forget to make that e negative? That's the place where most of the mistakes happen. The other place where most of the mistakes happen is me giving you lambda in wavelengths,
07:42
or excuse me, in nanometers or some other thing like that, and you forgetting to convert it over. Yeah. When e, when a photon is emitted, that's usually your q that this is going to be negative. So the other thing is whenever it's going from a high energy to a low energy, because
08:04
that means your final is lower than your initial. So if your final is lower than initial, and you always have final minus initial being an energy, you have a low number subtracted from a high number, or a low number subtract a high number from that and it's going to be negative.
08:20
Okay. Anything around this subject? So her question was, if you do the other parts correct but you get e wrong, yes, except if you make this and this negative, you're still going to get them wrong. Because you should know that wavelength can never be negative, and you should know
08:43
that frequency can never be negative. So if you get e wrong and you fill in a negative there, and you go down, you'll be fine. As far as if you just have the wrong number completely, yes and no. It depends on if you show the right work. If you show the work and the graders can look at it and say, okay, they knew what they were doing, I'll give you the point.
09:01
They knew what they were doing, I'll give you the point. Yes. If you just put down a number, they're not going to go through and calculate and make sure that you did it right. So you have to show really good work. And of course on short answers, there's not these sort of multiple step things. Anything else in quantum? Yeah.
09:26
Don't round. Never round. Never round until the end. You only ever round when you're giving a final answer. You hold all your numbers until then. Next one.
09:41
So we have chlorine can be prepared in a laboratory using this reaction. How much MnO2 is should be added to excess HCl to get that much chlorine? And I gave you all the parameters for it. So this would be like if you needed to make a certain amount of chlorine for some sort of industrial application and you needed to know how much you were going to get.
10:02
So when we go to do this, we start where? What do we need to do if we need to go from, we have a certain amount of chlorine and we need to get MnO2? We need to find moles. Now we've been doing that through conversion factors when we were in the fundamentals chapter, but we can't do that in conversion factors here.
10:22
Why? We're not at STP. So in the first midterm, you couldn't have done this problem, but now we can. Because how can we find moles when we aren't at STP? Tv equals R2. We're solving for N, and so we can fill all of this in.
10:42
Should I fill in Torr for pressure? We need to convert that. And you can always convert it separately. I tend to like to just convert it on top of the problem.
11:04
And in a case like this, do you need to worry about all your units? Yes. How do you know that? You have R. And if you forget what units everything needs to be in, since I give you R, what can you do?
11:24
Yeah, look at your units. R has all these units right here, so if you forget which volume unit, you just look here. If you forget which pressure unit, you just look here. And we always have to worry about temperature.
11:40
We always have to put that in Kelvin, now and for all the other problems. So now we have moles of chlorine, and what are we trying to solve for again?
12:02
Oh, okay. Okay. The answer is still fine. I just, it was a handwriting typo. Okay. So now that we have moles of chlorine, what do we do?
12:21
So we go from moles of chlorine to moles of what? And we do that using mole ratio.
12:45
And we are looking for, okay, so I didn't actually tell you a unit of MNO2. I would be more careful on an exam. I'd be asking for grams. Mostly because you can weigh it. Right? It's hard to weigh out moles of something. So I'd be asking for grams on an exam.
13:02
So how do we get this mole ratio? Yep. We come up here and we look. And then how do we get our conversion factor here? Periodic table. Good.
13:24
And don't forget, of course, all these questions can be asked backwards. So when you're doing these problems for practice, you want to be able to solve them the other way too. So you can always take all of these problems and just reverse them. You can take your answer and say, okay, well for this given frequency, this given frequency, what's the wavelength and energy?
13:42
For this many moles of MNO2, what is, how much CL can I get? Both in moles and, this one would be, you'd have to give yourself some of these parameters. But you can always work backwards. It's the easiest way to make lots and lots of questions for exams, right? You just flip them. So make sure you can do these both the way I did them here and backwards.
14:05
Okay. Questions on this sort of thing? Yeah. So if I want you to use something like the Van der Waals equation, I'll make it really
14:21
obvious by saying, can't we treat it as ideal? Also, I'd have to give you A and B for it. Otherwise you're sort of stuck assuming ideal gas. Technically, you can do everything that we've done through this entire chapter using the Van der Waals equation. The algebra just gets a little bit hairy.
14:42
Can you guys quiet the chatter please? I can't hear what people are saying. Yeah, I'll tell you it's not an ideal gas. I'll say, you know, such and such a gas cannot be treated ideally and then give you
15:02
a bunch of parameters and ask you to solve for something. Yeah. I knew it was from Cl2 because these conditions were set up for Cl2.
15:30
So we have a monochromatic beam of light. What does monochromatic mean? All one color. In other words, all one.
15:41
What decides color? Wavelength, right? Wavelength frequency energy, that decides color. So monochromatic means one color. So all one wavelength. OK. So it has a total energy of 2.5 joules. It contains this many photons. What is the wavelength of the beam?
16:02
OK. So what do we have to do to find a wavelength in general? If we have energy and we need to find wavelength, what do we fill into? E equals hc over lambda. So can I just fill in this for E? No.
16:21
So what is that E of in hc over lambda? It's energy of one photon, right. So this energy is of this many photons. So we need to figure out the energy of one photon. So my analogy for how to remember how to do this is always food. Because most people can think about it with calories and food and things like that.
16:43
So if I tell you that a bag of chips is 100 calories and I tell you that there are 10 chips in the bag, how many calories is each chip? 10, right? A hundred calories per bag, 10 chips, so that means you have to have 10 calories per chip. How do you do that?
17:00
Total divided by the number. It's no different here. It's the total energy and the number of things that have that. So we'll take the energy total, divide by the number to get the energy of the photon.
17:25
So 2.5 joules. Now can I just fill in the 8.56? No that would be joules per mole, right? And we don't want joules per mole, we want joules per photon.
17:41
So, again, I'm going to do my conversion factor inside here. You don't need to do it that way, you can do it outside. And fill in. And I'm going to fill in the little shorthand notation for Avogadro's number.
18:16
Now how many sig figs on this one? Three.
18:20
Wait, no, not three. How many? Two. So we're going to go around there. Um, hang on, I was mildly number dyslexic there and threw in a one that didn't need to be there. There we go. It's 4.8514.
18:46
Ok, so now that we have our energy per photon, what do we do? We plug in E equals hc over lambda.
19:03
So we have our energy now, we're solving for our lambda.
19:28
Should I just fill in 4.8 here? No, right? We hold out some sig figs. We underline the last significant one so that we make sure that we can keep track of it quickly without having to go all the way back through.
19:46
And then we write out our number. So we have 4.1 times 10 to the negative 5th meters.
20:00
And again, since I didn't say give it in nanometers, you're fine to go ahead and just leave it in meters. You don't have to convert into any particular thing. Yeah. So equals hc over lambda, you mean times by the number of moles up here?
20:30
Oh, so you basically want to find wavelengths per mole. It may work mathematically, but it's a dangerous way of doing it.
20:41
Mostly because, especially in an exam where I'm probably going to ask you for the energy per photon anyways, mostly because I have to for partial credit purposes. So I would do it this way for that. Doing it your way would work because mathematically that's how they would cancel. But you lose the sense of joules per photon.
21:01
So it would work, but it works because of the mathematics of it. So unless you just happen to be really good at the math and the algebra and you know what you're doing that way, I would stick with that way. I would have to draw it all out, which I don't want to do because I don't want to confuse people the day before the exam. But I can do it afterwards if you want and I can make sure it'll work.
21:24
Yeah. Say that a little bit louder. So usually the rule is you hold out one or two past the last significant one you're safe.
21:44
In this case I went to three. In all honesty, when I'm doing this in real life, I'm just doing second answer for everything and filling the whole string of numbers in. That level of rounding error on an exam, we know. It would be the difference between getting four, well this one, since there's only
22:00
two sig figs, it would probably just come out right. If I held out a few more sig figs, it would be the difference between getting like 4.12 and 4.13 and we know that that's a rounding thing.
22:21
So here we have a compound made up of CH and CL, contains 55 percent chlorine by mass, we have five grams of the compound containing this many hydrogen atoms, what's the empirical formula? OK. So a lot of you didn't like the empirical formula one. So we'll spend some extra time on this, make sure you interrupt me if I do something
22:42
you're not sure why. So at this point, we need to go through and we need to solve for each one. We're going to need to solve for the moles of carbon, the moles of hydrogen and the moles of chlorine. We want to solve for the ones that we know first. So we can say that it's 55 percent chlorine by mass and we can say that there's this
23:00
many hydrogen atoms. So we're going to start with chlorine and hydrogen. If at this point you're saying, I don't know how I'm going to get carbon, we'll get there at the end. One of the big things with doing chem problems is being able to start it, not being 100 percent sure where you may be going. Just go with what you know. We're going to have to have moles of chlorine, we're going to have to have moles of hydrogen and we can see how to get those.
23:21
So we're going to start there. So with chlorine, I tell you that it's 55 percent by mass. So we've done a lot of these where we've just started with assuming 100 gram sample. We don't want to do that here because of the way that the hydrogen was written in. It was written in as the number of atoms.
23:42
So we want to just go ahead and go with the sample that we have. So we have a 9 gram sample. 55 percent of it by mass is going to be chlorine. How can we get moles of chlorine? Yeah, we multiply by .55.
24:02
That gives us how many grams of chlorine we have and then we can get moles of chlorine.
24:22
Now you may notice on empirical formula problems I get pretty lazy about sig figs and I don't talk about them much. It's because eventually you're going to come out with whole numbers and so if you don't do sig figs completely correct and if there's not enough sig figs here you're just going to be off by like less than a tenth so it's not a big deal. So I just left it as three sig figs and didn't worry about holding out any extras.
24:45
Now let's do hydrogen. Hydrogen we know how many atoms we have in this sample. So how do we get moles? Yeah, I got his number and I'm going to do the shorthand notation.
25:14
Now we need to figure out carbon which is a little bit tougher.
25:23
So we know how much chlorine we have, we know how much hydrogen we have. How would we know how much carbon we have then? Whatever's left over, right? So we're going to subtract. So we take the nine grams and this is one of the many places in these problems there's
25:41
a couple of different ways you can do it. I think the way that makes the most sense is to start with the nine grams and then subtract your grams off of each one. So how many grams of hydrogen do we have? We have 0.96, right? Why? What's hydrogen's molar mass? One. So we don't really have to do a conversion factor there.
26:07
And we need to look at how much chlorine we have so you may, it depends on how far ahead you are looking whether you maybe just left that as grams to begin with and then converted it.
26:21
In that case I didn't so I just will put that little half conversion factor in there. So this is our hydrogen and this whole section is our chlorine. So we have our total mass, subtract from both our hydrogen, and then subtract our hydrogen and our chlorine mass from that.
26:47
Right here? Yeah. So you can take this and you can multiply by 35.45. The reason I would suggest doing it this way is because you're less, if you messed up your calculator button pushing right here then you're going to have two things wrong.
27:02
If you go back and you just multiply these two it's the same number of button pushes but you're less, you know, you're not using something that you already used. Yeah. Yep. Because it's one gram per mole. So you can do a conversion factor, you can go ahead and write out, but you basically
27:36
just have one over one and so it's the same thing.
27:43
So we get this number and then to compare it what do we have to do? Convert to moles. And so now we have all our mole values, so how do we find the empirical formula?
28:16
Divide by a smallest number, so we get that.
28:38
So when we write this out, we write C2H5Cl.
28:47
I ask that you know that you really should be writing carbon and then hydrogen. Ask for what else to write the ordering, like normal to the oxygen, nitrogen, or halogen. Don't worry about what order to write them in, just write them down. But you should know by now that we pretty much always write the carbons and then the
29:03
hydrogens and then the rest of them. Okay, so that's our empirical and molecular formula actually.
29:41
Okay, actually you don't know that. You would just do empirical. I didn't give you the molecular mass, so you'd have to actually do the molecular mass. So because I didn't give you any molecular mass, you don't actually know for sure if
30:04
it is. You don't know if maybe this is actually C4H10Cl2. I'd have to give you the molecular mass to be able to figure it out.
30:22
So the difference between molecular and empirical is a good thing to talk about. It's the idea between lowest number and then being able to multiply by 2 or 3 or 4. And with the 9 gram thing, you would get the same number of carbons, hydrogens, and chlorines if actually it was a higher amount.
30:41
Because all we're doing is comparing moles, and we're comparing just moles of the actual atoms together, we don't technically know whether or not this is the empirical or molecular. If I gave you a molecular mass, now what would you do to check? Well, you'd find the mass of this, and you would compare it to the molecular mass.
31:03
If it was the same, what would be the molecular formula? It would be this. If it was times 2, it would be 2 times that. So yeah, you can't know whether it's the molecular formula unless I give you the mass.
31:27
It burns an oxygen to produce carbon dioxide and water. Write a balanced equation and then calculate the liters of carbon dioxide if we measure it at 3 atm in 60 atmospheres that we could get from a certain amount of propane. So first step with this, you have to have a formula, right?
31:43
You have to know what you're doing so that you have mole ratios and things of that sort. So propane, on an exam I would give you the formula for propane too. I would go ahead and I'd say C3H8. Technically we covered that in fundamentals, but I did tell you not to worry too much about it. So I would give you that.
32:05
The rest of it I wouldn't necessarily give you. So whenever you combust something or you burn something, what's the reaction with? And then I tell you that it makes carbon dioxide and water.
32:24
Okay, so now we have to balance this thing. So we look at our carbons and we say okay we're going to need 3. We look at our hydrogens and we say okay we're going to need 4. So now our carbons and our hydrogens are good, let's look at our oxygens. So we have 6 and 4 is 10 and so we need 5.
32:47
Okay, so now we have 7.45 grams of propane and we want to know how much carbon dioxide we can make at this. So it starts out, this is sort of the reverse problem of the one I did earlier today.
33:01
So we're going to start with the 7.45 grams of propane. We convert to moles of propane and we use the mole ratio which is why we had to
33:25
have the equation and then at that point we're done.
33:44
Well we're not done done, but we're done here. How many sig figs? Three so far. Okay, so now what do we do?
34:03
PV equals nRT and you fill everything in.
34:27
Not forgetting to convert into Kelvin. Okay, technically that would only be one sig fig. We're going to add a few more.
34:43
So it's not three sig figs. I try to avoid writing questions with one sig fig in them. Okay, so that's the reverse of the question we did earlier in the day.
35:06
Excellent. Some quantum number review. Okay, this has been a while, right? So, quantum numbers.
35:21
If n equals four, what quantum numbers are possible? How many electrons are in this energy level? Alright, so when we go through this, what values for L are allowed? n minus one, so in this case that's three
35:41
two one and zero. Now what about m sub L? Plus or minus three, plus or minus two, plus or minus one or zero, right? Those are kind of allowed. This depends on exactly which L we're talking about.
36:06
So really we can do this. Now on an exam if I just asked you which ones are possible, you don't have to write all these out. The reason I'm writing all these out is because it makes counting easier. So now when I say how many electrons
36:20
and for the record how many orbitals are in this level, it's a little bit easier to make sure we count right. So we have two, four, six, eight, ten, twelve, fourteen, sixteen. So we have sixteen orbitals.
36:40
So how many electrons? Thirty-two electrons. So you don't have to write out the quantum numbers for each one, but it is helpful so that when you go through and count you can just go two, four, six, eight, ten, twelve, fourteen, sixteen.
37:05
Next ones. For n equals five, m sub s equals one-half. So we've really only added one thing to this, right? We've added in L equals four.
37:22
And then all the rest is the same. Three, two, one, and zero. So if L is equal to four, what does m sub L allow to equal? Plus or minus four, plus or minus three, plus or minus two, plus or minus one, and zero.
37:40
Okay, so when we go to count orbitals we can go through and we just add on two, four, six, eight, nine to our original sixteen that we had here. Because for the L equals four we're also allowed to have all of these. So you can write them all out again if you want. I'm just saving some time.
38:01
So we have sixteen plus the two, four, six, eight, nine. So we have twenty-five orbitals. How many electrons are present? How many? What's the question say for m sub s?
38:23
We're only allowed to equal plus one-half now. So how many electrons do we have? Twenty-five. Right? Because the reason we're allowed to have double is because we have spin up and spin down or plus one-half minus one-half. So now I'm saying we're only allowed to have one of them. And so we only get to have twenty-five.
38:42
Okay. We're going to go ahead and skip the last one for the sake of time.
39:05
We have a lithium atom with an average speed of that and there's a little decimal point to indicate we have extra sig figs. It has an uncertainty of velocity within seven hundred nanometers per second. What is the uncertainty in position? Okay.
39:21
So which equation do we use for this? So, and what is P here? Momentum, which is equal to
39:42
m times v or delta v. And we want to know uncertainty in position which is delta x
40:08
to this. So some things to be careful of. This is now effectively filling in. So h is just h. We don't have to worry about that one too much. 4 pi is just 4 pi. We don't have to worry about that too much.
40:20
What is mass? Do we just look on the periodic table and see what the mass of lithium is? No. We see that it's 6.94 but that is 6 grams per mole. And what do we want? Kilograms. That's good. Yes, we definitely want kilograms. What else do we want?
40:44
Do we want it to be per mole? What do we see here? Lithium atom. So we also need to get rid of the moles and put atoms on the bottom instead.
41:05
And that's what we fill into our mass. Okay. Now for velocity. What is our delta v? Is it 2000? What is 2000?
41:23
It's our velocity, right? Is that velocity? No, it's uncertainty in velocity. So why is that 2000 there? Is it to trick you? Wait, where are you people who went to my review session last night? Same joke, right? It's not to trick you, it's to make sure you know what you're doing.
41:41
We don't do things to trick you. We do it to make sure you know what you're doing. Okay. So, we're going to use this. Are we filling it into here? Like, how it is right now? No.
42:08
We fill in our conversion factor. And that's what we're going to fill into here.
42:24
I tend to just write it out like that. Yeah. Because this is the actual speed, this is the uncertainty. And what we're always looking at here is the uncertainties. We don't really care how fast it's going. We care how much we're sure or not sure of how fast it's going.
42:41
So it's kind of like a plus or minus. And then this is what we fill in. To here. So when we do all that, delta x has to be greater than or equal to some number that you can type into your calculator.
43:05
Okay. Yeah. It's just there to make sure that you know the difference between velocity and uncertainty in velocity. And you know that this is the uncertainty in velocity, not the velocity. Because in real life you have lots of data, right? So we want to give you lots of data just like you have in real life
43:21
to make sure that you can solve for what you really need. Okay. So, yeah.
43:41
So we have to convert it to kilograms because of h. What unit is h in? Joules times seconds. And what does joules have in it? It has kilograms. So most of the time when you're dealing with mass, if you ever see any sort of joule unit and you're dealing with mass, you usually have to convert it to kilograms. Okay.
44:01
How are we doing on time? Oh, we have two minutes. Okay. So we can do a couple of these. Alright. So some going through and doing a couple of electron configurations. Well, unless you don't want me to, then I don't have to.
44:24
We'll do something. So let's do cadmium. Actually, let's not. Let's do silver. So if we look at the periodic table, we go to silver and we see that we have krypton as our last noble gas. Please write the noble gas configuration. While it's not technically wrong,
44:41
it does make the graders crabby and you don't want to make graders crabby, right? So, and then you look and you would, if you were just kind of going through blindly on the periodic table, you would write this. And what would happen if you wrote that? You'd get it wrong. And these are usually short answer questions,
45:02
so would you get partial credit? You would want to promote that up. Why do you want to promote it up? To make it more stable, right? That filled energy shell lowers the total energy. And so we want to do it like that.
45:20
So what charge does silver normally form because of that? Plus one, right? Because you just want to take away that so that it's closer to a noble gas configuration. And so you get 4d10. So this is also, silver is an example of one where I could ask you
45:40
to explain to me why it forms a plus one charge, too. Okay, so we didn't really make it through the rest, but that's okay. I kind of figured we wouldn't. So good luck. Study a lot.