Elementary - 3.6(a V =(b 195.5 kg H 2SO 4 1 kg solution L 0.35kg H 2SO 4 12563 1000 kg 195.5 kg H 2 SO 4 = 445 L Videal = L 18255 1.00 kg 195.5 kg

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3- 3 3.6 (a) V = × = 1955 1 0 35 12563 1000 445 . .. . kg H SO kg solution L kg H SO kg L 24 (b) V ideal 2 2 kg H SO L kg kg H O L L = × += 18255 100 065 035 1000 470 4 . %. error = ×= 445 100% 56% 3.7 Buoyant force up Weight of block down bg b g = E Mass of oil displaced + Mass of water displaced = Mass of block ρρ ρ oil HO c 2 0542 1 0542 b g VV V +−= From Table B.1: g / cm , g / cm g / cm 33 o 3 cw i l == = 2 26 100 3325 . mV oil oil g / cm cm g = × = 353 117 4 . m oil + flask g g g =+ = 117 4 124 8 242 3.8 Buoyant force up = Weight of block down b g ⇒= = WW V g V g displaced liquid block disp. Liq block () Expt. 1 : ρρρ wBB w Ag 15 2 2 . . b g =⇒ = × w B B SG = = 1 075 .00 g/cm 3 3 g / cm Expt. 2 : soln soln 3 soln SG BB b g = = = 2 2 3.9 W + W hs AB hb h 1 Before object is jettisoned 1 1 Let w = density of water. Note: Aw > (object sinks) Volume displaced : VA h A h h db s i b p b 11 1 di (1) Archimedes = + wd A B Vg W W 1 weight of displaced water
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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