Davis

1. Some zinc sulphate crystallizations were heated to constant hole with the following result crowd together of crucible=20.00g plenty of crucible& group Aere; crystal=25.74gm muckle of crucible& ampere; residure=23.22gm Calculate X in a formula ZnSO4XH2O from the entropy given above Solution host of a crystal=Mass of crucible& crystal-Mass of acrusible = 25.74gm-20.00gm =5.74gm The down of ZnSO4XH2O =5.74gm Mass of residure=Mass of crucible& residure-Massof crusible =23.22gm-20.00gm 20.00gm Mass of residure=3.22gm demand the response downstairs ZnSO4XH2O ZnSO4 + XH2O Let the molar mass of ZnSO4XH2O to be Y Y of ZnSO4XH2O = 161gmol-1 But the mass of ZnSO4 = 3.22 NowY of ZnSO4XH2O =161gmol-1 of ZnSO4 =3.22gm Y= 287gmol-1 ZnSO4XH2O = 287gmol-1 Zn + S + O4 +X H2O = 287gmol-1 = (65+32+64+18X)gmol-1 Qn.1 18 Xgmol-1 = (287- 161)gmol-1 X = 7 The numeral of water molecule is 7 Qn.2 What volume of Oxygen cross up at s.t.p will be obtained by bunkum of 4.32g of Mercury(II)oxide(H O) by heat Solution Data given:Mass of Mercury(II)oxide=4.32gm Volume of Oxygen(O2)=? date the reaction below Hgos 2Hg + O2g 2moles of HgO produces 1mole of O2g 1mole of any blow out at s.t.p = 22.

4dm3 Molar mass of HgO =201gmol-1 +16gmol-1 =217gmol-1 2(217gmol-1)of HgO = 22.4dm3 4.32gm of HgO = ? = 0.22229774dm3 or 222.96774cm3 The volume of group O at s.t.p is 2 22.96774cm3 or 0.22296774dm3 Qn.3 W! hat volume in cm3 of Hydrogen at s.t.p would be required to hydrogenate 6.5gm of ethyne to Ethene? (H=1.0,C=12.0&gaseous molar volume at s.t.p =22.4) solution data given Mass of ethyne = 6.5gm Volume of ethane = ? pass the reaction below CHCH + H2 CH2CH2 26gmol-1 of CHCH = 22400cm3mol-1 of H2g 6.5gm of CHCH = ? =...If you want to get a commensurate essay, order it on our website: BestEssayCheap.com

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