Problem set 6 Crib 2011

Problem set 6 Crib 2011 - orbital instead of having the...

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CHM 241 Spring 2011 Problem Set 6 15.7 (a) PCl 4 + ? tetrahedron. (b) PCl 4 ? see-saw. (c) AsCl 5 ? trigonal bipyramid. 15.8 (a) P 4 + 5O 2 → P 4 O 10 (b) P 4 O 10 + 6H 2 O → 4H 3 PO 4 (c) 2H 3 PO 4 (l) + 3CaCl 2 (aq) → Ca 3 (PO 4 ) 2 (s) + 6HCl(aq) 15.18 (a) A = NO 2 (b) B = HNO 3 ; C = NO (c) D = N 2 O 4 16.1 CO 2 , SO 3 , P 2 O 5 , are acidic Al 2 O 3 is amphoteric CO is neutral MgO and K 2 O are basic 16.5 SO 3 2– > SO 4 2– > S 2 O 8 2– The best reducing agent is sulfite and best oxidizing agent is peroxydisulfate. 16.10 Cu + , Fe 3+ , and Co 3+ will be reduced.
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A. Singlet oxygen is the excited state of dioxygen containing electrons paired into one of the π* orbital’s. Triplet oxygen is the ground state of dioxygen, containing two unpaired electrons in the π* orbitals. The higher energy is due to electrostatic repulsions associated with pairing the electrons in one
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Unformatted text preview: orbital instead of having the electrons in two separate degenerate orbitals. B. 2 S(s) + 3 O 2 (g) →2 SO 3 (g) (with catalyst) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq) SO 3 (g) + NH 3 (g) → NH 2 SO 2 OH(l) NH 2 SO 2 OH (l) + NH 3 (l) → (NH 4 ) + (NH 2 SO 3 )-(s) A: SO 3 (g) B: H 2 SO 4 (aq) C: H 2 NSO 2 OH (l) D: (NH 4 ) + (NH 2 SO 3 )-(s) C. a.) N 2 O 5 b.)S 2 O 8-2 c.) ClPF 4 Cl Can be in or above/below equatorial plane for isomers d.) H 4 N 4 S 4 D. O 2 O 2-H 2 O 2 H 2 O-.33 1 .77 1 .69 1 .23 .68 O 2 + 4e- + 4H + → 2H 2 O 1.23 – (0.0592/4) log (1/[10-7 ] 4 ) =0.816 V...
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This note was uploaded on 03/20/2012 for the course CHM 24100 taught by Professor Professordavidr.mcmillin during the Spring '10 term at Purdue.

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Problem set 6 Crib 2011 - orbital instead of having the...

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