HW answers  Set 2­1 days 1­4.notebook

1

September 12, 2014

p73 # 56b)Iron III Bromide c) cobalt II sulfide# 61c) sulfur dioxide d) diphosphorus pentasulfide#64) acetic acid b) ammonium nitritee) lead II phosphate h) strontium nitridek) sodium chromate66a) CrO3 b) S2Cl2 d)K2HPO4 h)Na2Cr2O7 i)(NH4)2SO3

67l) HNO2

68g) HBr h) HBrO2 i) HBrO4

p118 #58 a) 165.4 g/molb) 500.0 g CH x 1mol CH = 3.023 mol CH 165.4gc) 2.0 x 10-2 mol CH x 165.4 gCH = 3.3 g CH

1 mol CHd) 5.0g CH x 1mol CH x 3 mol Cl x 6.022 x 10 23 Cl atoms 165.4g CH 1 mol CH 1 mol Cl atoms= Cl atomse) 1.0 g Cl x 1mol Cl x 1 mol CH x 165.4 gCH = 1.6 g CH 35.5 g Cl 3 mol Cl 1 mol CHf) 500 molecules CH x 1 mole CH x 165.4g CH = 1.373 x 10-19g 6.022 x 1023 1 mole CH

5.5 x 1022

Day 1

#70)63.68g C x 1mol C = 5.302 mole C/.8837 = 6 12.01 g C12.38g N x 1 mol N =.8837 mol N/.8837 = 1 14.01 g N9.80g H x 1 mol H =9.72 mol H/.8837 = 11 1.008 g H14.14g O x 1 mole O =.8838 mol O/.8837 = 1 16.00 g OC6NH11O

Day 2

HW answers  Set 2­1 days 1­4.notebook

2

September 12, 2014

p 120 #7549.31g C x 1 mole C = 4.106 mol C/2.737 = 1.5 x 2 = 3 12.01 g C43.79g O x 1 mol O = 2.737 mol O/2.737 = 1 x 2 = 2 16.00 g O6.900 g H x 1 mol H = 6.845 mol H/2.737 =2.5 x 2 = 5 1.008 g H

C3O2H5 = empirical formulaMolar mass = 146.1 g/mol Empirical mass = 73g/mol

146.1/73 = 2 Molecular formula = C 6O4H10

HW answers  Set 2­1 days 1­4.notebook

3

September 12, 2014

p 125 80) CxHy Oz

10.68mg -> 16.01mg CO2 and 4.37mg H20 molar mass= 176.1 g/mol

.01601g C02 x 1 mole C02 x 1 mol C x 12 g C = .00437g C 44gC02 1mole C02 1 mol C

.00437g H20 x 1 mole H20 x 2 mol H 1.01g H = .000490gH 18.02 g H20 1 mole H20 1 mol H

.01068g-(.00437g C + .000490g H) = .00582 g O

.00437 g C x 1 mol C = .000364/.000364 = 1 x 3 =3 12.0

.000490g H x 1 mol H =.000485/.000364 = 1.33 x 3 =4 1.01 g H

C3H4O3

.00582g O x 1mol O = .000364/.000364 = 1 x 3=3 16.0g O176 = 288

C3H4O3 = empirical formula C6H8O6 = molecular formual

HW answers  Set 2­1 days 1­4.notebook

4

September 12, 2014

92. Ba(OH)2*8 H2O + 2 NH4SCN ---> Ba(SCN) 2 + 10H20 + 2NH3

b) 6.5g Ba(OH)2*8H20 x 1 mol x 2 mol NH4SCN x 76.13g NH4SCN 315.4g 1 mol Ba(OH)2*8H2O 1 mol = 3.2g NH4SCN