#5a Pressure = [density] X [height] X [acceleration due to gravity] Pressure(water) = Pressure(mercury) 1000 kg/m^{3} X [h] X [9.8 m/s^{2}] = 13600 kg/m^{3} X [0.760 m] X [9.8 m/s^{2}] [h] = [13600 kg/m^{3} X 0.760 m X 9.8 m/s^{2}] / [1000 kg/m^{3} X 9.8 m/s^{2}] [h] = 10.3 meters #5b Total Pressure = Atmospheric Pressure + H_{2}O Pressure Water pressure = [mm of H_{2}O at below surface] / [mm of H_{2}O at surface] There are 12 inches in one foot --> inches of H_{2}O = 36 X 12 = 432 There are 2.52 cm in one inch --> cm of H_{2}O = 432 X 2.54 = 1097.28 cm of H_{2}O There are 10 mm in 1 cm --> mm of H_{2}O = 10972.8 mm of H_{2}O Answer from part a in mm --> 10.3 m = 10.3 X 10^{3} mm H_{2}O Pressure = [10972.8 mm] / [10.3 X 10^{3} mm] = 1.065 atm Notice this is like when you devide mm of Hg by 760 to get the answer in atm. Total Pressure = 0.95 atm + 1.065 = 2.01 atm #100a PV = nRT --> n = mass / molar mass Molar mass = [mass X RT] / [PV] Molar mass = [1.56 g X R X 323 K] / [00.984 atm X 1.00 L] Molar mass = 42.0 g/mol Carbon --> 85.7 / 12 = 7.42, Hydrogen --> 14.3 / 1.0 = 14.3 14.3 / 7.42 = 1.93, 7.42 / 7.42 = 1.00 --> 1:2 Empirical formula = CH_{2} --> 12 + [2 X 1.0] = 14.0, since 14.0 fits three times into 42.0 then the molecular formula is C_{3}H_{6} #100b Ar is a monatomic gas, whereas C_{3}H_{6} is a complex and relatively large molecule and is expected to deviate from ideal behavior. C_{3}H_{6} also has rotational and vibrational degrees of freedom not available to Ar atoms, which only possess 3 translational degrees of freedom. #101 Mol of CH_{4} = [PV] / [RT] Mol of CH_{4} = [1.0 atm X 2.7 X 1012 L] / [R X 273 K] Mol of CH_{4} = 1.205 X 10^{11} mol From the equation: CH_{4} + 2O_{2} --> CO_{2} + 2H_{2}O + 890.4 kJ From stoichiometry CH_{4}:thermal energy is 1:1 and since the thermal energy is a product the enthalpy change is negative (the reacting particles loose energy). ΔH = [-890.4 kJ/mol] X 1.205 X 10^{11} mol = -1.073 X 10^{14} kJ #102a PV = nRT --> n = mass / molar mass --> density = mass / volume Molar mass = [mass X RT] / [PV] Molar mass = [density X RT] / P Molar mass = [0.803 g/L X R X 301 K] / 0.197 atm Molar mass = 101 g/mol Mol of gas reacted with water: PV = nRT --> n = [PV] / [RT] n = [0.166 atm X 480 X 10^{-3} L] / [R X 301] = 3.22 X 10^{-3} mol grams of gas reacted with water: 3.22 X 10^{-3} mol X 101 g/mol = 0.325 grams mol of F from the 0.081 M solution of HF Molarity X volume = 0.081 M X 80 X 10^{-3} L = 6.48 X 10^{-3} mol HF = mol of F Grams of F = # of mol X molar mass = 6.48 X 10^{-3} mol X 19.0 g/mol = 0.123 grams of F Since a silver containing compound is most likely to be solid, a compound in the gas phase would contain sulfur: Grams of S = grams of compound - grams of F Grams of S = [0.325 - 0.123] grams = 0.202 grams of S Mol of S = 0.202 / 32 = 6.31 X 10^{-3} mol S = mol of F (from 1:1 ratio) Empirical formula = SF --> 32 + [19] = 51.0, since 51.0 fits two times into 102.0 then the molecular formula is S_{2}F_{2} 102b Sulfur is like oxygen so it can bind at least two times, Fluorine can only bind once. However, the bond distance between the two sulfurs has been found to be sorter than the standard single bond and longer than a double bond, let's make it 3/2 bond. 104a One cubic foot in SI units (see above image) = 30.48 cm X 30.48 cm X 30.48 cm = 2.83 X 10^{4} cm^{3} 2.83 X 10^{4} cm^{3} = 2.83 X 10^{4} mL = 28.3 L/ft^{3} Volume of methane = 10.7 X 10^{9} ft^{3} X 28.3 L/ft^{3} = 3.03 X 10^{11} L Mol of methane = n = [PV] / [RT] n = [1.00 atm X 3.03 X 10^{11} L] / [R X 298 K] = 1.24 X 10^{10} mol From mole ratio metane:methanol is 1:1 Grams of methanol = 1.24 X 10^{10} mol X 32.0 g/mol = 3.97 X 10^{11} grams Volume of methanol = 3.97 X 10^{11} grams / 0.791 g/L = 5.01 X 10^{11} L 104b From the equation: CH_{4} + 2O_{2} --> CO_{2} + 2H_{2}O + 890.4 kJ From stoichiometry CH_{4}:thermal energy is 1:1 and since the thermal energy is a product the enthalpy change is negative (the reacting particles loose energy). ΔH = [-890.4 kJ/mol] X 1.24 X 10^{10} mol = -1.10 X 10^{13} kJ From the equation: CH_{3}OH + 3/2 O_{2} --> CO_{2} + 2H_{2}O + 726.6 kJ From stoichiometry CH_{3}OH:thermal energy is 1:1 and since the thermal energy is a product the enthalpy change is negative (the reacting particles loose energy). ΔH = [-726.6 kJ/mol] X 1.24 X 10^{10} mol = -9.01 X 10^{12} kJ 104c Volume of one mol of methane: V = 16.0 g / 466 g/L = 3.43 X 10^{-3} L This much methane generates 890.4 kJ of energy. Volume:enthalpy ratio = 890.4 kJ / 3.43 X 10^{-3} L = 2.60 X 10^{4} kJ/L Volume of one mol of methanol: V = 32.0 g / [786 g/L] = 4.07 X 10^{-2} L This much methanol generates 726.6 kJ of energy. Volume:enthalpy ratio = 726.6 kJ / 4.07 X 10^{-2} L = 1.79 X 10^{4} kJ/L Methane is more efficient per unit volume. 105a From the equation: I_{2} + 5 F_{2} --> 2 IF_{5} Mol of F2 = 10.0 g / 38.0 g/mol = 2.63 X 10-1 mol Mol of I_{2} = 10.0 g / 254 g/mol = 3.94 X 10^{-2} mol From mole ratio I_{2}:F_{2} is 1:5 then I_{2} is limiting Remaining F_{2} gas = 2.63 X 10-1 mol / [5 X 3.94 X 10^{-2} mol] = 6.60 X 10^{-2} mol Mol of IF_{5} = 2 X mol of I_{2} = 3.94 X 10^{-2} mol X 2 = 7.88 X 10^{-2} mol 105b PV = nRT P(IF_{5}) = nRT / V = [7.88 X 10^{-2} mol X R X 398 K] / 5.00 L P(IF_{5}) = 0.515 atm Mole fraction of IF_{5} = mol IF_{5} / [mol IF_{5} + F_{2}] Mole fraction of IF5 = 7.88 X 10^{-2} mol / [7.88 X 10^{-2} mol + 6.60 X 10-2 mol] Mole fraction of IF_{5} = 0.544 106a MgCO_{3} + 2 HCl --> MgCl_{2} + H_{2}O + CO_{2} CaCO_{3} + 2 HCl --> CaCl_{2} + H_{2}O + CO_{2} 106b PV = nRT n = PV / RT = [(743/760) atm X 1.72 L] / [R X 301 K] = 6.808 X 10^{-2} mol Mol of MgCO_{3} = Ψ grams/ [84.32 grams/mol] Mol of CaCO_{3} = [6.53 grams - Ψ grams] / [100.09 grams/mol] Let's remove the units to make the equation less fuzzy; and don't forget what Ψ is. 6.808 X 10^{-2} = [Ψ / 84.32] + {[6.53 - Ψ] /100.09} Remember Lowest Common Denominator? Here the LCD is [100.09 X 84.32] 6.808 X 10^{-2} X 100.09 X 84.32 = 100.09Ψ - 84.32Ψ + 84.32(6.53) 574.57 = 100.09Ψ - 84.32Ψ + 550.61 574.57 - 550.61 = 100.09Ψ - 84.32Ψ 23.96 = 15.77Ψ 1.52 g = Ψ = grams of MgCO_{3} % MgCO_{3} by mass = [1.52 g X 100] / 6.53 g = 23.3 % MgCO_{3} by mass |