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D4R7ho3YnSo-017|So I can calculate the pH directly from that.
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D4R7ho3YnSo-018|The pH is about 2.88.
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D4R7ho3YnSo-020|I haven't done any base.
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D4R7ho3YnSo-021|The pH can be calculated just as if it were a weak acid solution.
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D4R7ho3YnSo-022|Weak acid solution of acetic acid, the initial pH of 0.1 molar acetic acid has an initial pH of 2.88.
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-fHWcbYcLeE-000|When you react phosphorous with oxygen, you get a brilliant white light and the production of a phosphorus oxide.
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-fHWcbYcLeE-001|Let's look at that in terms of a ChemQuiz.
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-fHWcbYcLeE-002|31 grams of phosphorus will react with excess oxygen to form 71 grams of a phosphorus oxide.
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-fHWcbYcLeE-003|The question I have for you is, what's the empirical formula of that phosphorus oxide?
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-fHWcbYcLeE-011|When the phosphorus and oxygen react to form the phosphorus oxide, 31 grams of phosphorus in excess oxygen produce 71 grams of the phosphorus oxide.
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-fHWcbYcLeE-012|So that difference of 40 grams must have been the oxygen.
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-fHWcbYcLeE-013|So it's 31 grams of phosphorus react with 40 grams of oxygen for a mass ratio of 3 to 4.
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-fHWcbYcLeE-014|But when we write chemical formulas, we do not use mass ratios.
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-fHWcbYcLeE-015|We use mole ratios.
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-fHWcbYcLeE-019|We could denote that with the subscript 2.5, but we like to have integer subscripts in our molecular formulas.
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-fHWcbYcLeE-020|So let's multiply through by 2 and use the mole ratios 2 to 5.
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-fHWcbYcLeE-021|So the empirical formula of this phosphorus oxide is P2O5.
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-fHWcbYcLeE-022|The correct answer is A, P2O5.
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-fHWcbYcLeE-023|It turns out the actual molecule is P4O10.
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CNAU31Q7cJg-001|We can calculate the energy difference and calculate the wavelength of the photon that's emitted.
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CNAU31Q7cJg-002|So the energy levels go as minus z squared over n squared times the Rydberg constant.
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CNAU31Q7cJg-004|We can put in all those numbers.
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CNAU31Q7cJg-005|Here is z, we're talking about a hydrogen atom, so the charge on the nucleus z is 1.
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CNAU31Q7cJg-006|The two states from the n equal 2 to n equal 4 state.
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CNAU31Q7cJg-007|I'm doing a transition where it's emission.
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CNAU31Q7cJg-008|I'm starting out in 4 and going to the n equal 2 state.
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omeeRx3Zl84-000|Now let's write the electronic configuration of rubidium atoms.
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omeeRx3Zl84-001|And give all the quantum numbers of the outermost electron that was ejected in that photo electron spectroscopy experiment.
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omeeRx3Zl84-002|So when you're writing electronic configurations you need to know the total number of electrons.
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omeeRx3Zl84-003|And in a neutral atom, the total number of electrons is equal to the positive charges, the number of neutrons, the atomic number of the atom.
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omeeRx3Zl84-004|Rubidium atomic number 37, 37 protons in its nucleus.
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omeeRx3Zl84-005|It has 37 electrons in its neutral state.
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omeeRx3Zl84-006|Now, we're going to take all the possible orbitals.
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omeeRx3Zl84-007|And we're going to write them out in kind of a diagrammatic way here.
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omeeRx3Zl84-008|Remember, as you add electrons, the energy levels of the orbitals shift around a bit.
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omeeRx3Zl84-009|So there's a little device, you can use to remember the actual order, which energy level is the highest.
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omeeRx3Zl84-010|So you write them all out like this, the 1s, the 2s, all the threes, all the fours, all the fives, and you can keep going, sixes et cetera.
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omeeRx3Zl84-011|We won't need that many orbitals for rubidium.
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omeeRx3Zl84-012|And then you make diagonal lines.
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omeeRx3Zl84-013|This is the order that the electrons fill in.
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omeeRx3Zl84-014|It's the energy ordering of orbitals when there are many electrons present.
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omeeRx3Zl84-015|So 1s is the first orbital to fill, then 2s.
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omeeRx3Zl84-016|And you keep running diagonal lines.
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omeeRx3Zl84-017|Then the 2p, then the 3s.
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omeeRx3Zl84-019|That just goes by principle quantum level.
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omeeRx3Zl84-020|Then the 3p, then the 4s, then the 3d, then the 4p, then the 5s.
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omeeRx3Zl84-021|And that would continue.
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omeeRx3Zl84-022|You would do 4d, 5p, 6s.
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omeeRx3Zl84-024|So now we've to do is go through that and count up to 37 electrons.
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omeeRx3Zl84-025|So we can do that 1s is the first orbital we'll fill.
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omeeRx3Zl84-026|And the 1s will hold two electrons, so 1s, two.
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omeeRx3Zl84-027|We can keep total electron count, so we remember when we get to 37.
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omeeRx3Zl84-028|So far we have two electrons.
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omeeRx3Zl84-029|Next we'll fill the 2s.
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omeeRx3Zl84-030|It also holds two, bringing our count to four.
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omeeRx3Zl84-031|Then we fill the 2p and the 3s.
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omeeRx3Zl84-032|The 2p hold six electrons.
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omeeRx3Zl84-033|Remember, three equivalent p orbitals, each holding two electrons is six.
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omeeRx3Zl84-034|Brings our total count to 10.
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omeeRx3Zl84-035|Then the 3s we'll fill with it two bringing our count to 12.
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omeeRx3Zl84-036|The 3p and the 4s we'll fill.
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omeeRx3Zl84-037|3p holds six.
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omeeRx3Zl84-038|The 4s holds two.
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omeeRx3Zl84-039|20 total electrons.
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omeeRx3Zl84-040|Now, rather than fill the 4p, we're going to fill the 3d next.
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omeeRx3Zl84-043|Then the 4p accommodates six electrons.
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omeeRx3Zl84-044|Then the 5s can accommodate two, but we're already up to 36.
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omeeRx3Zl84-045|All we need is 37.
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omeeRx3Zl84-047|Quite a mouthful of electronic configuration.
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omeeRx3Zl84-048|What are the quantum numbers for that outermost electron.
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omeeRx3Zl84-049|Well, the quantum numbers we generally write as n, l, m sub l, and m sub s.
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omeeRx3Zl84-050|So n for the outermost electron is five.
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omeeRx3Zl84-051|It's in the fifth principle quantum level, so five.
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omeeRx3Zl84-052|L is zero because its an s orbital.
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omeeRx3Zl84-053|Remember, the designation for l equals zero is s.
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omeeRx3Zl84-054|So we'll change that s to a zero for l.
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omeeRx3Zl84-055|M sub l will be zero because that's the only possible value of m sub l for l equals zero.
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omeeRx3Zl84-059|It can either be in the plus one half or the minus one half state.
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omeeRx3Zl84-060|So there are our electronic configuration and our quantum numbers for the outermost electron for rubidium atoms.
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AlwmlHuZRUU-000|When we make molecular orbitals from atomic orbitals we want to make molecular orbitals that have the right geometry for the molecular problem we're solving.
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AlwmlHuZRUU-001|If we have a tetrahedral molecule, or a linear molecule.
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AlwmlHuZRUU-002|So what we need to do is look at our atomic orbitals and see if they satisfy that.
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AlwmlHuZRUU-005|Two orbitals 180 degrees from each other.
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AlwmlHuZRUU-006|How do we do that?
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AlwmlHuZRUU-007|Well, let's start by taking the s and the p orbital, the s and the pz for instance, from an atom.
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AlwmlHuZRUU-008|If I take the s and the pz, that's two atomic orbitals.
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AlwmlHuZRUU-012|Now again, I'm adding, so green areas are positive amplitude.
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AlwmlHuZRUU-013|They'll get bigger.
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AlwmlHuZRUU-014|Red areas are negative amplitude.
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AlwmlHuZRUU-015|So when I add red and green, I should see a decrease in amplitude in those areas.
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AlwmlHuZRUU-016|So adding those two together gives me an orbital that looks like this.
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AlwmlHuZRUU-017|The green and the green added, giving me higher amplitude.
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AlwmlHuZRUU-018|And the green and the red added, giving me lower amplitude.
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AlwmlHuZRUU-019|And that's a new orbital.
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AlwmlHuZRUU-020|It's a combination of an s and a p.
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AlwmlHuZRUU-021|We'll call it an sp atomic orbital.
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AlwmlHuZRUU-022|So that sp atomic orbital is going to have a pair that's only one.
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AlwmlHuZRUU-023|I started with two atomic orbitals.
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AlwmlHuZRUU-024|I need to create two new ones.
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AlwmlHuZRUU-025|So I can take the opposite.
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