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Get Full Access to University Physics, Volume 3 - 17 Edition - Chapter 4 - Problem 107
Get Full Access to University Physics, Volume 3 - 17 Edition - Chapter 4 - Problem 107

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ISBN: 9781938168185 2032

## Solution for problem 107 Chapter 4

University Physics, Volume 3 | 17th Edition

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University Physics, Volume 3 | 17th Edition

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Problem 107

How far would you place a screen from the slit of the previous problem so that the second minimum is a distance of 2.5 mm from the center of the diffraction pattern?

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Pg – 1  Benzene (looks like 1,3,5 – cyclohexatriene) o Uniform C-C bond length (1.39 Angstrom)  2 resonance structures  extra stable o M.O. Diagram  6 p orbitals combine and make 6 MO orbitals  3 bonding and 2 anti-bonding  Pi molecular Orbitals of Benzene o When comparing energy of electrons in benzene to 6 electrons in other alkenes, benzene is the most stable. o Heats of hydrogenation can be used to compare stability of nonconjugative, conjugated and aromatic systems. Heats of Hydrogenation (~29 kcal/mol for each double bond  slightly lower than expected)  Resonance stabilizes benzene  Resonance energy or delocalization energy lowers ΔH  Actual ΔH for benzene = -49.3 kcal/mol  Calc valve = - 82.2 kcal/mol 1 In order for  the ΔH to show conjugation, that the rotation orbitals must isn’t free line up Heats of Hydrogenation  Benzene b/c of its resonance is 33.3 kcal more stable!  There’s a BIG benefit to bring aromatic Requirements of Aromaticity  Structure must be cyclic w/ conjugated pi bonds  Each atom in the ring must have and unhybridized p orbital  P orbitals must overlap continuously around the ring Ex)  Must obey Hüchel’s Rule o Count the electrons in Heat in the pi system (not all electrons)

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