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43YS58bxWuU-029|So we can do that.
43YS58bxWuU-030|We have a mini neon tube.
43YS58bxWuU-031|And it's called neon generically, of course.
43YS58bxWuU-032|In our tubes, there's several different gases.
43YS58bxWuU-033|We have, in fact, all the gas of the atmosphere.
43YS58bxWuU-034|We have nitrogen, oxygen, neon, helium, hydrogen. All the gases that are found, to a certain extent, in the atmosphere.
43YS58bxWuU-035|When I excite them with an electric charge, the electrons will be promoted, and they will emit specific energies.
43YS58bxWuU-036|They'll emit them all at once, so the colors will be a mixture of these, they'll also be emitting that ultraviolet radiation.
43YS58bxWuU-037|We won't be able to see that, but it's emitted at the same time.
43YS58bxWuU-038|So let's look at the emission spectrum of atoms and molecules.
43YS58bxWuU-040|Let's look at that.
43YS58bxWuU-042|You can see the various colors, a lot of them bluish, where there is high energy photons.
43YS58bxWuU-043|And there's even photons, as we said, in the ultraviolet being emitted.
43YS58bxWuU-044|So we're doing kind of a spectroscopy experiment here.
43YS58bxWuU-046|If we know exactly the wavelength, we can calculate exactly the energy difference of the atomic orbitals in these atoms.
43YS58bxWuU-047|So you can see how powerful spectroscopy is to understand the properties and the structure of atoms.
SUkBTHfKqf4-000|Titration is the process of reacting acid and base together.
SUkBTHfKqf4-001|And I do that in a controlled way.
SUkBTHfKqf4-002|So I can take an acid solution and a base.
SUkBTHfKqf4-003|The acid solution will originally have low pH, a high acid concentration.
SUkBTHfKqf4-004|As you add base, that base will react with the acid and lower that acid concentration.
SUkBTHfKqf4-005|Lowering the acid concentration raises the pH.
SUkBTHfKqf4-006|The solution becomes more basic, which makes sense because you're adding a base.
SUkBTHfKqf4-007|Eventually, you'll use up all of the acid that's there.
SUkBTHfKqf4-008|And if you continue to add base, then that base will dominate the pH.
SUkBTHfKqf4-009|And you'll have a basic solution.
SUkBTHfKqf4-010|You can plot that in something that's called a titration curve.
SUkBTHfKqf4-013|This totally dissociates, forms 0.1 or 10 to the minus 1 moles of H3O+ moles per liter.
SUkBTHfKqf4-014|The pH is minus log of 10 to the minus 1 or 1.
SUkBTHfKqf4-015|So I start with pH 1.
SUkBTHfKqf4-016|I add some base.
SUkBTHfKqf4-017|And as I add moles of base, those react with my moles of acid to form water.
SUkBTHfKqf4-018|And that continues to happen until I use up all the original acid that is there.
SUkBTHfKqf4-019|When I've added 1 mole of base for every mole of acid in my original solution, I've used up all my acid, and I've used up all the base that I've added.
SUkBTHfKqf4-020|They've all completely reacted together.
SUkBTHfKqf4-022|As you continue beyond that, then the base that you're adding dominates the pH, and the pH rapidly goes to a basic level.
SUkBTHfKqf4-023|You can do the reverse.
SUkBTHfKqf4-024|You can start with a strong base solution or any base solution and add an acid solution.
SUkBTHfKqf4-025|Here the pH will start high because the H3O+ concentration is low.
SUkBTHfKqf4-026|And as you add acid, that pH will drop.
SUkBTHfKqf4-028|But a sodium chloride solution has pH 7.
SUkBTHfKqf4-029|The ions sodium and chlorine don't contribute to the pH.
SUkBTHfKqf4-030|We'll talk about that more later.
SUkBTHfKqf4-031|So you have a pH of 7, a neutral salt water solution.
SUkBTHfKqf4-032|As I continue to add acid, the acid will dominate the pH.
SUkBTHfKqf4-033|And you'll go to a low pH solution.
SUkBTHfKqf4-034|So this is a strong acid or strong base titration.
qvaEyckBI0g-000|So let's look at electrons in their various orbitals in forms of a ChemQuiz.
qvaEyckBI0g-001|When an electron spin flips, for which of the following is the energy not altered?
qvaEyckBI0g-009|So we're talking about nitrogen, oxygen, and fluorine, and we're trying to decide can we flip an electron spin and not change the overall energy of the atom?
qvaEyckBI0g-010|Nitrogen has the electronic configuration 1s2 2s2 2p3, three unpaired electrons in those 2p orbitals.
qvaEyckBI0g-011|Oxygen, 2p4, fluorine, 2p5.
qvaEyckBI0g-014|In oxygen and nitrogen, there's other electrons.
qvaEyckBI0g-015|So these electrons will try to be parallel.
qvaEyckBI0g-016|So the energy of this electron depends on the spin of this electron.
qvaEyckBI0g-017|But fluorine, that's not the case, everybody is paired up except for this unpaired electron, so fluorine can either have spin up or spin down.
7TZXu6s0KA0-000|Let's look at the Maxwell Boltzmann distribution for a pair of gasses.
7TZXu6s0KA0-001|I've plotted them here, and I've indicated where the root mean square velocity would fall for gas X at 400 Kelvin, and gas Y at 800 Kelvin.
7TZXu6s0KA0-002|The question is what are gases X and Y?
7TZXu6s0KA0-009|We're looking at the Maxwell Boltzmann distribution plots for two different gases at two different temperatures.
7TZXu6s0KA0-012|The temperature is doubling, but that wouldn't give us a factor of 2.
7TZXu6s0KA0-013|That would give us a factor of square root 2 in the rms velocity.
X5KLAdXoDxc-000|Let's look at the effect of temperature on a chemical reaction.
X5KLAdXoDxc-001|Now, we know intuitively if you raise the temperature, chemical reactions proceed more rapidly.
X5KLAdXoDxc-002|The reason is that at higher temperature, there are more molecules that have energies in excess of the activation energy for the chemical reaction.
X5KLAdXoDxc-003|So they can get over the barrier that separates products and reactants.
X5KLAdXoDxc-006|Now, that doesn't mean all those molecules will react.
X5KLAdXoDxc-007|They still have to collide, and collide with the appropriate orientation.
X5KLAdXoDxc-010|So now I have a higher percentage of molecules with energies in excess of the activation energy.
X5KLAdXoDxc-011|So why do reactions go faster with the increasing temperature?
X5KLAdXoDxc-012|More molecules, more particles with higher energies.
X5KLAdXoDxc-013|Now if you think about kinetics, our kinetic rate laws look like the rate is the rate constant times some function of the concentrations.
X5KLAdXoDxc-014|Now, the concentrations are not a function of temperature.
X5KLAdXoDxc-015|So for rates to increase with temperature, our rate constants must be a function of temperature.
X5KLAdXoDxc-018|So rate constants are a function of temperature.
X5KLAdXoDxc-023|k, the rate constant for the chemical reaction, is a function of temperature.
vhlYDz16s0w-000|Let's look at some examples of oxidation reduction reactions.
vhlYDz16s0w-001|Here's a common one-- the rusting of iron.
vhlYDz16s0w-002|Now, it's interesting to note, simply putting a nail made of iron in pure water, that won't rust.
vhlYDz16s0w-003|You actually need the presence of dissolved oxygen.
vhlYDz16s0w-004|So here's the actual reaction.
vhlYDz16s0w-007|Iron is in its 0 oxidation state, it's oxidation number increases.
vhlYDz16s0w-008|That's an oxidation to plus 2.
vhlYDz16s0w-009|So to help us balance this, let's go to our table of standard production potentials.
vhlYDz16s0w-012|We can use those to help balance this reaction, so let's do that.
vhlYDz16s0w-016|And I'll change the sign of my reduction potential.
vhlYDz16s0w-021|The cell potential for the oxidation of iron is 0.81 volts.
vhlYDz16s0w-023|These are farther down the table.
vhlYDz16s0w-026|They'll protect the iron from oxidation.
vhlYDz16s0w-027|And you know this.
NLGrUdD5uDw-000|To describe molecules quantum mechanically, we can take the atomic orbitals from the atoms and recombine them to form molecular orbitals.
NLGrUdD5uDw-001|And that works well for diatomics.
NLGrUdD5uDw-002|We can take s and p orbitals and overlap them and form molecular orbitals.
NLGrUdD5uDw-003|The s and p orbitals have the appropriate arrangement and symmetry for a diatomic molecule.
NLGrUdD5uDw-004|But what if we get more complicated?
NLGrUdD5uDw-005|Even if you go to a very simple triatomic or if you go to something with steric number 2, how do we accommodate steric number 2?
NLGrUdD5uDw-006|How do we have a carbon bond to something on this side and something on this side?
NLGrUdD5uDw-007|If I use my p orbital over here to make a molecular orbital with this, I don't have a p orbital available to make a molecular orbital with this.
NLGrUdD5uDw-008|I can't reuse my orbitals to make more bonds.
NLGrUdD5uDw-009|So I have to have bonds and orbitals that have the right orientation.