Fundamentals%20CDE%20Solns

Fundamentals%20CDE%20Solns - C.3 C.7 The chemical formula...

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Unformatted text preview: C.3 C.7 The chemical formula of xanthophyll is C40H56O2. (a)Cesium is a metal in Group 1; it will form Cs + ions.(b)Iodine is a nonmetal in Group 17/VII and will form I - ions.(c)Selenium is a Group 16/VI nonmetal and will form Se 2- ions.(d)Calcium is a Group will form Ca 2+ ions. C.9 2 metal and (a)10Be2+ has 4 protons, 6 neutrons, and 2 electrons.(b) 17O2- has 8 protons, 9 neutrons, and 10 electrons.(c) 80Br- has 35 protons, 45 neutrons, and 36 electrons.(d) 75As3- has 33 protons, 42 neutrons, and 36 electrons. C.13 (a)Aluminum forms Al3+ ions; tellurium forms Te 2- ions. Two aluminum atoms produce a charge of 2 h +3 = +6. Three tellurium atoms produce a charge of 3 h - 2 = -6. The formula for aluminum telluride is Al2 Te3 . (b)Magnesium forms Mg 2+ ions and oxygen forms O 2- ions. A magnesium ion produces a charge of +2, which is required to balance the charge on one O 2- ion. The formula for magnesium oxide is MgO. sodium (c)Sodium forms +1 ions; sulfur forms 2 ions. The formula for sulfide is Na 2 S. (d)Rubidium forms +1 ions and iodine forms 1 ions. One iodide ion is required to balance the charge of one rubidium ion, so the formula is RbI. C.15 (a) HCl, molecular compound (in the gas phase); (b) S8, element (molecular CS2, molecular (f) SrBr2, ionic compound substance); (c) CoS, ionic compound; (d) Ar, element; (e) compound; D.1 (a)MnCl2. Mn forms 2+ ions and chlorine forms -1 ions. (b)Ca3(PO4)2. Calcium forms +2 ions and the phosphate ion is PO43-. (c)Al2(SO3)3. Aluminum forms +3 ions and the sulfite ion is SO32-. (d)Mg3N2. Magnesium forms 2+ ions and the nitride ion is N3- . D.3 (a)calcium phosphate;(b)tin(IV) sulfide, stannic sulfide;(c)vanadium(V) oxide;(d)copper(I) oxide, cuprous oxide D.5 (a) TiO 2 ;(b) SiCl 4 ; (c) CS2 ; (d) SF4 ; (e) Li 2 S; (f) SbF5 ; (g) N 2 O5 ; (h) IF7 D.7 D.9 (a)sulfur hexafluoride;(b)dinitrogen pentoxide;(c)nitrogen triiodide; chlorine dioxide acid; (a)hydrochloric acid;(b)sulfuric acid;(c)nitric acid;(d)acetic (d)xenon tetrafluoride;(e)arsenic tribromide;(f) (e)sulfurous acid;(f)phosphoric acid D.11 D.13 (a)ZnF2 (b)Ba(NO3)2 (c)AgI (d)Li3N (e)Cr2S3 (a)sodium sulfite;(b)iron(III) oxide or ferric oxide;(c)iron(II) sulfate oxide or ferrous oxide;(d)magnesium hydroxide;(e)nickel(II) hexahydrate;(f)phosphorus pentachloride; (g)chromium(III) dihydrogen phosphate;(h)diarsenic trioxide;(i) ruthenium(II) chloride D.15 (a) CuCO3 Copper (II) carbonate; (b) K2SO3 potassium sulfite; (c) LiCl, lithium chloride D.17 D.25 (a)heptane;(b)propane;(c)pentane;(d)butane (a)selenic acid;(b)sodium arsenate;(c)calcium tellurite;(d) barium arsenate;(e)antimonic acid;(f)nickel(III) selenate E.7 (a)mass of average Li atom 7.42 92.58 = (9.988 10-24 g) + (1.165 10-23 g) 100 100 -23 -1 = 1.153 10 g atom molar mass = 10-23 g atom -1 ) (6.022 1023 atoms mol -1 ) (1.153 = 6.94 g mol -1 100 5.67 - 5.67 = (9.988 10-24 g) + (1.165 10-23 g) 100 100 -23 -1 = 1.1556 10 g atom molar mass = 10-23 g atom -1 ) (6.022 1023 atoms mol -1 ) (1.1556 = 6.96 g mol-1 (b)mass of average Li atom E.9 (a) MgSO 4 h7 H 2 O formula mass = 246.48 g mol -1 5.15 g 11 mol O atoms atoms of O = -1 7 246.48 g mol mol MgSO 4 H 2 O (6.022 23 atoms -1 ) = 1.38 23 10 mol 10 5.15 g = (6.022 1023 atoms mol-1 ) -1 246.48 g mol (b)formula units = 1.26 1022 5.15 g = 0.146 mol (c)moles of H 2 O = 7 -1 246.48 g mol 10 E.11 The percentage B = 100 - percentage 11 B 10 10 B % molar mass = (mass 100% 11 B % B) + (mass 11 B) 100% 10 100% - % 11 B = (mass 100% Rearranging gives 11 B % B) + (mass 11 B) 100% 100 molar mass - 100 mass % B= 11 10 mass B - mass B 11 10 B 100(10.81 g -1 ) - 100(10.013 g -1 ) mol mol -1 11.093 g mol - 10.013 g -1 mol = 73.8 % = % E.17 10 B = 26.2 % -1 (a)molar mass of Al2 O3 = 101.96 g mol nAl2O3 = 10.0 g = 0.0981 mol 101.96 g mol-1 N Al2O3 = (0.0981 mol)(6.022 23 atoms -1 ) = 5.91 22 molecules 10 mol 10 (b)molar mass of HF = 20.01 g mol-1 nHF = 25.92 10-3 g = 1.30 10-3 mol -1 20.01 g mol N HF = (1.30 10-3 mol)(6.022 1023 atoms mol -1 ) = 7.83 1020 molecules (c)molar mass of hydrogen peroxide = 34.02 g mol-1 nH2 O2 = 1.55 10-3 g = 4.56 10-5 mol 34.02 g mol-1 N H2 O2 = (4.56 10-5 mol)(6.022 1023 atoms mol-1 ) = 2.75 1019 molecules (d)molar mass of glucose = 180.15 g mol-1 nglucose = 1250 g = 6.94 mol 180.15 g mol -1 N glucose = (6.94 mol)(6.022 23 atoms -1 ) = 4.18 24 molecules 10 mol 10 (e)molar mass of N atoms = 14.01 g mol-1 nN = 4.37 g = 0.312 mol 14.01 g mol-1 N N = (0.312 mol)(6.022 23 atoms -1 ) = 1.88 23 atoms 10 mol 10 molar mass of N 2 molecules = 28.02 g mol-1 nN2 = 4.37 g = 0.156 mol 28.02 g mol -1 N N2 = (0.156 mol)(6.022 23 atoms -1 ) = 9.39 22 molecules 10 mol 10 (a)molar mass of CuBr2 = 223.35 g mol-1 3.00 g CuBr2 mol Cu 2+ 1 2+ nCu 2+ = = 0.0134 mol Cu -1 223.35 g mol CuBr2 mol CuBr2 1 (b)molar mass of SO3 = 80.06 g mol-1 0.700 g SO3 -3 nSO3 = = 8.74 10 mol SO3 -1 80.06 g mol SO3 (c)molar mass of UF6 = 352.03 g mol-1 1000 g 1 mol UF6 mol F- 6 - nF- = 25.2 kg = 430. mol F 1 kg 352.03 g UF6 mol UF6 1 (d)molar mass of Na2CO3 10H2O = 286.21 gmol-1 2.00 g Na 2 CO3 2 O 10H 10 mol H 2O nH2 O = 286.21 g -1 Na 2CO3 2 O mol Na 2 CO3 2 O mol 10H 1 10H = 0.0699 mol H 2 O E.25 (a)molar mass of CuCl24H2O = 206.53 g.mol-1 8.61 g CuCl2 2 O 4H n= = 4H 0.0417 mol CuCl 2 2 O -1 206.53 g mol CuCl2 2 O 4H (b)Because there are 2 mol Cl- per mole of compound, the number of will be twice the amount in part (a), 0.0834 mol Cl- (c)There are 4 mole of H2O for each mole compound. Therefore, number of water molecules = (0.0417 4) mole H2O (6.022 v 1023 molecules mol-1) = 1.00 1023 H2O molecules 4(16.00 g mol -1 ) (d)fraction of mass due to O = = 0.3099 206.53 g mol -1 moles E.19 ...
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This note was uploaded on 04/24/2008 for the course CHEM 6A taught by Professor Pomeroy during the Fall '08 term at UCSD.

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