Ebook Instant notes Inorganic chemistry (2nd edition) Part 2

... hydroxylamine (NH2OH), semicarbazide (NH2NHCONH2) and 2, 4-dinitrophenylhydrazine takes place by the same mechanism described for primary amines to give oximes, semicarbazones, and 2, 4-dinitrophenylhydrazones, ... H2NNH NH2 R R R' O C R' R NH2 C N C C R' H N C O R' Oxime Semicarbazone NO2 NO2 c) O + C R H N H N N H2N NO2 Fig 2, 4-Dinitrophenylhydrazone C R' R R' Synthesis of oximes, semicarbazones, and 2, 4-dinitrophenylhydrazones ... synthesized by using diols rather than alcohols (Fig 6) H O CH2CH3 H R C -H2O O O H R H H H H C R OCH2CH3 H C H R O O O CH2CH3 OCH2CH3 -H H R H C O O CH2CH3 Oxonium ion CH2CH3 C CH2CH3 CH2CH3 Acetal...

... solutions 7 12 24 28 32 35 44 48 50 71 71 72 73 75 80 91 102 10 7 11 4 14 1 14 1 14 4 14 5 14 7 15 1 15 3 16 0 viii The Finite Element Method Further topics in the finite element method 5 .1 The variational ... Consider the equation (2 .13 ) −u = f, < x < 1, 21 Weighted residual and variational methods Table 2.3 The integrals for the Galerkin method i, j vi Lvj dx and 10 15 21 15 10 10 5 84 10 10 35 14 84 15 10 5 ... 24 c0 c1 = , giving c0 = 12 , c1 = 16 The approximate solution is then u 1 (x) = x (1 − x) (1 + 2x) 12 Example 2.6 The final method to be considered is the Galerkin method In this method the integral...

... + 32 (1 + 2U2 + U3 )U2 − 32 (U2 + U3 )U3 = − , − 32 (U2 + U3 ) + 32 (U2 + 2U3 )U3 = − Solving these yields U 22 = 49 , U 32 = 19 21 2 The Finite Element Method Just as in Example 5.5, the positive ... after the conditions U11 = 0, U21 = 29 , U31 = 89 , U41 = 2, U 12 = 1, U 42 = are enforced, becomes ⎤ ⎥ ⎢ ⎢ U 22 ⎥ + ⎣ U 32 ⎦ ⎡ which yields U 22 = 11 , U 32 = 17 29 46 29 29 46 29 ⎡ ⎤ ⎡ ⎤ 0 2 ... Performing the integrations, ⎤ ⎡ 156 22 h 54 −13h kh ⎢ 4h2 13h −3h2 ⎥ ⎥ + EI ⎢ ke = ⎣ 156 22 h ⎦ h3 420 sym 4h2 fe = fh [−6 12 −h h dξ, 4EI N N h2 i j ⎡ ⎢ ⎢ ⎣ 12 − 12 −6h 12 6h 4h2 sym ⎤ 6h 2h2 ⎥ ⎥,...