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CHEMISTRY THE CENTRAL SCIENCE

CHAPTER 10: GASSES


#5a
Pressure = [density] X [height] X [acceleration due to gravity]
Pressure(water) = Pressure(mercury)
1000 kg/m3 X [h] X [9.8 m/s2] = 13600 kg/m3 X [0.760 m] X [9.8 m/s2]
[h] = [13600 kg/m3 X 0.760 m X 9.8 m/s2] / [1000 kg/m3 X 9.8 m/s2]
[h] = 10.3 meters

#5b
Total Pressure = Atmospheric Pressure + H2O Pressure
Water pressure = [mm of H2O at below surface] / [mm of H2O at surface]
There are 12 inches in one foot --> inches of H2O = 36 X 12 = 432
There are 2.52 cm in one inch --> cm of H2O = 432 X 2.54 = 1097.28 cm of H2O
There are 10 mm in 1 cm --> mm of H2O = 10972.8 mm of H2O
Answer from part a in mm --> 10.3 m = 10.3 X 103 mm
H2O Pressure = [10972.8 mm] / [10.3 X 103 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 = CH2 --> 12 + [2 X 1.0] = 14.0, since 14.0 fits three
times into 42.0 then the molecular formula is C3H6

#100b
Ar is a monatomic gas, whereas C3H6 is a complex and relatively large molecule and is expected to deviate from ideal behavior. C3H6 also has rotational and vibrational degrees of freedom not available to Ar atoms, which only possess 3 translational degrees of freedom.

#101
Mol of CH4 = [PV] / [RT]
Mol of CH4 = [1.0 atm X 2.7 X 1012 L] / [R X 273 K]
Mol of CH4 = 1.205 X 1011 mol
From the equation:
CH4 + 2O2 --> CO2 + 2H2O + 890.4 kJ
From stoichiometry CH4: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 1011 mol = -1.073 X 1014 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 S2F2

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 104 cm3
2.83 X 104 cm3 = 2.83 X 104 mL = 28.3 L/ft3
Volume of methane = 10.7 X 109 ft3 X 28.3 L/ft3 = 3.03 X 1011 L
Mol of methane = n = [PV] / [RT]
n = [1.00 atm X 3.03 X 1011 L] / [R X 298 K] = 1.24 X 1010 mol
From mole ratio metane:methanol is 1:1
Grams of methanol = 1.24 X 1010 mol X 32.0 g/mol = 3.97 X 1011 grams
Volume of methanol = 3.97 X 1011 grams / 0.791 g/L = 5.01 X 1011 L

104b
From the equation:
CH4 + 2O2 --> CO2 + 2H2O + 890.4 kJ
From stoichiometry CH4: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 1010 mol = -1.10 X 1013 kJ
From the equation:
CH3OH + 3/2 O2 --> CO2 + 2H2O + 726.6 kJ
From stoichiometry CH3OH: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 1010 mol = -9.01 X 1012 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 104 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 104 kJ/L
Methane is more efficient per unit volume.

105a
From the equation:
I2 + 5 F2 --> 2 IF5
Mol of F2 = 10.0 g / 38.0 g/mol = 2.63 X 10-1 mol
Mol of I2 = 10.0 g / 254 g/mol = 3.94 X 10-2 mol
From mole ratio I2:F2 is 1:5 then I2 is limiting
Remaining F2 gas = 2.63 X 10-1 mol / [5 X 3.94 X 10-2 mol] = 6.60 X 10-2 mol
Mol of IF5 = 2 X mol of I2 = 3.94 X 10-2 mol X 2 = 7.88 X 10-2 mol

105b
PV = nRT
P(IF5) = nRT / V = [7.88 X 10-2 mol X R X 398 K] / 5.00 L
P(IF5) = 0.515 atm
Mole fraction of IF5 = mol IF5 / [mol IF5 + F2]
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 IF5 = 0.544

106a
MgCO3 + 2 HCl --> MgCl2 + H2O + CO2
CaCO3 + 2 HCl --> CaCl2 + H2O + CO2

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 MgCO3 = Ψ grams/ [84.32 grams/mol]
Mol of CaCO3 = [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 MgCO3
% MgCO3 by mass = [1.52 g X 100] / 6.53 g = 23.3 % MgCO3 by mass





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