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CHAPTER 10 AND 11 AP CHEMISTRY.

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Presentation on theme: "CHAPTER 10 AND 11 AP CHEMISTRY."— Presentation transcript:

1 CHAPTER 10 AND 11 AP CHEMISTRY

2 INTERMOLECULAR FORCES
Molecules tend to be Nonconductors of electricity when pure substances Most molecules when placed in water won’t conduct EXCEPTION- highly polar molecules i.e. HCl Insoluble in water but soluble in nonpolar solvents (iodine dissolves in alcohol) i.e. toluene, CCl4 A few are very soluble in water i.e. ethyl alcohol(C2H5OH) Low melting and boiling points Many are gases at room temperature Many boiling points are directly related to the strength of the intermolecular forces

3 DIPOLE FORCES Ion-dipole Dipole-dipole What are dipole forces?
Ions attracted to a dipole substance i.e. NaCl in water Dipole-dipole What are dipole forces? Attraction between polar molecules As the dipole moment increases the boiling point increases London dispersion forces (Van der Waal) What is a dispersion force? Electrical temporary dipoles (POLARIZABILITY) All molecules have dispersion forces, the strength depend on two factors Number of electrons in the molecule Ease with which electrons are dispersed to form temporary dipoles As the molar mass increases the strength will increase

4 HYDROGEN BONDS Polarity has a small effect on boiling point
Compounds have to have hydrogen attached to nitrogen, oxygen or fluorine Hydrogen in one compound is attracted to a lone pair on the N, O, or F of a different compound Two reasons hydrogen bonding is stronger than other dipole forces Electronegative difference is large between hydrogen(2.2) and nitrogen(3.0), oxygen(3.5), or fluorine(4.0) Small size of hydrogen allows the unpaired electron pair of F, O, or N atom to approach the H atom closely The molar mass of H2S is double that of water, both are polar molecules, but the boiling point of water is 167 °C higher than H2S. Why? Hydrogen bonding What types of intermolecular forces are present in N2, SiCl4, OCCl2, N2H4

5 PROPERTIES OF LIQUIDS Viscosity Resistance to flow Surface tension
Energy required to increase surface area of liquid by a unit amount Water increases surface tension because of hydrogen bonding

6 PHASE CHANGES Vapor pressure Pvapor = Patm - PHg column page 460-461
Heat of fusion Enthalpy change associated with melting Ice = 6.01 kJ/mol Heat of vaporization Heat required to vaporize a liquid Water = kJ/mol Calculate the enthalpy change associated with converting 1.00 mol of ice at - 25 oC to water vapor at 125 oC at 1 atm pressure. Heat capacities of ice, water, and steam are 2.09 J/g·C, 4.18 J/g·C, 1.84 J/g·C. The heat of fusion of ice is 6.01 kJ/mol, heat of vaporization is kJ/mol. 1 mol of water has 18 g/mol, use the specific heat formula ΔH = (18 g/mol)(2.09 J/g·C)( ) = 940J = .94kJ ΔH = (18 g/mol)(4.18 J/g·C)( ) = 7524J = 7.52 kJ ΔH = (18 g/mol)(1.84 J/g·C)( ) = 828J = .83kJ 55.97 kJ/mol

7 PROBLEM A 428 mL sample of hydrogen gas was collected by displacement of water in Salt Lake City a 25 oC at a pressure of 647 mm Hg. The vapor pressure of water at 25 oC is 23.8 mm Hg. Calculate the partial pressure of hydrogen gas and the number of moles of hydrogen gas formed. PH2 = Patm -PH2O 623 mm Hg PV = nRT 1.43 X 102- mol H2

8 CRITICAL TEMPERATURE Highest temperature at which a substance can exist as a liquid Critical pressure Pressure required to bring about liquidefaction Sublimation When a substance goes from the solid phase to the gaseous phase without going through the liquid phase Deposition When a substance goes from the gaseous phase to the solid phase without going through the liquid phase

9 VAPOR PRESSURE Dynamic equilibrium Volatile Phase diagram
Opposing processes are occurring at the same rate Evaporation and condensation Volatile Liquids with high vapor pressure evaporate quickly Phase diagram On the overhead

10 STRUCTURE OF SOLIDS Crystalline solid Amorphous solid Crystal Lattice
Geometry shaped to their arrangement of ions Amorphous solid No orderly arrangement They do not melt at a specific temperature Glass Crystal Lattice Unit cell repeated, page 447 If the unit cell of NaCl is 6.54 Å on an edge, calculate the density of NaCl (6.54 Å)3(10-8cm/1Å)3 2.78 X cm3 8 atoms X 58.5 g/mol X 1mol/6.02 X 1023 atoms 7.78 X g/ 2.78 X cm3 2.76 g/cm3

11 CONTINUE Close packing of ions cause three types of unit cells
Simple Body centered

12 X-RAY DIFFRACTION Scattering of x-rays by a regular arrangement of atoms or ions Constructive interference Higher intensity if waves are the same Destructive interference Cancels each other out Bonding in solids Table 10.3 page 451 Graphite and diamonds are allotropes of carbon (same number of carbons (60) but in different forms You can convert graphite into diamonds Read pages 454 to 470 take notes

13 PROBLEM Silver crystallizes in a cubic closest packed structure. The radius of a silver atom is 1.44 Å (1 Å = 10-8cm). Calculate the density of solid silver. This is face-centered so you would have one half (2r) and two fourths (2r). Using Pythagorean theorem d2 + d2 = (4r)2 d= r8 d= (1.44 Å)( 8) (4.07 Å)3│ (10-8 cm)3 │ ( 1 Å ) 6.74 X cm3 (4 atoms)(107.9 g ) (1 mol ) (1 mol )(6.02 X 1023 atom)(6.74 X cm3) 10.6 g/cm3

14 SOLUTION COMPOSTION Molarity = moles of solute L of solution
Percent mass = mass of solute X 100 Mass of solution Mole fraction = A = XA = nA nA + nB nA = moles of solute nB = moles of solution or other substance Molality = mole of solute kg of solvent

15 CONTINUE A solution is prepared by mixing 1.00 g ethanol(C2H5OH) with g water to give a final volume of 101 mL. Calculate molarity, mass percent, mole fraction and molality of ethanol. 1.00g/ 1 mol 46 g 0.0217mol 0.0217mol/.101L .215M 1.00 g X 100 100.0g water g 0.990% 0.0217mol mol .0217mol/.1000kg .217m

16 NORMALITY Equivalents/L solution Acid-base Oxidation-reduction 5
Equivalent mass of acid or base that can get 1 mole of protons Equivalent mass of sulfuric acid is ½ it’s molar mass because you get 2 protons Oxidation-reduction Equivalent amount of oxidizing or reducing agent to accept or furnish 1 mole of electrons KMnO4 MnO4- + 5e- + 8H+ ----> Mn2+ + H2O, so MMKMnO4 to find normality 5

17 ENERGY SOLUTION FORMATION
Enthalpy Hsol’n = H1 + H2 + H3 Fig 11.2 page 505 and table 11.3 page 507 Read and take notes Vapor pressure Nonvolatile solute lowers the vapor pressure of a solvent Raoult’s law Psol’n =Xsolvent·Posolvent Po = vapor pressure of solvent X = mole fraction of solvent Calculate the expected vapor pressure of a solution that contains 215 g urea, NH2CONH2, dissolved in 435g of water (Vp pure water = mmHg) P = 24.2mole water/( mol urea) X 23.76 20.7mmHg

18 CONTINUE Ptotal = XAPoA + XBPoB
Liquid-liquid solution that obey Raoult’s law is an ideal solution Table 11.4 page 520

19 COLLIGATIVE PROPERTIES
Nonvolatile solute raises the boiling point and lowers the freezing point ΔTf = kfm m = mol solute/kg solvent ΔTb = kbm ΔT = change in temperature kf or kb = boiling point or freezing pt. constant Calculate the boiling point of a solution that contains 50.0g of glucose (MM = g/mol) in 400. g of water. The molal constant is 0.52 °C/m ΔTb = (0.52)(50.0g/ 1 mol / ) ( / g / .400 kg ) .36 °C °C

20 CONTINUE A student dissolves 1.00 g of a new compound in 25.0g cyclohexane. The solution has a freezing point of 2.0 °C and that of cyclohexane is 6.5 °C. Cyclohexane has a kf of 20.2 °C/m. Using this data, calculate the molar mass of the compound. =20.2°C/m (Xmol of unknown solute) (0.0250kg of cyclohexane) mol of unknown solute 1.00g/0.0056mol 180 g/mol unknown

21 OSMOTIC PRESSURE Semipermeable membrane
Solvents can pass through but not the solute molecule The volume of solution increases and the solvent decreases, this flow is called OSMOSIS Osmotic pressure When you have an excess pressure on the solution in comparison to the pure solvent

22 CONTINUE Reverse osmosis Colligative properties of electrolytes
If an external pressure greater than the osmotic pressure is applied the solvent particles will go from the solution to the pure solvent Colligative properties of electrolytes Van’t Hoff factor i: i = moles of particles in solution Moles of solute dissolved Colloids Suspension of tiny particles in some medium Blood Coagulation - destroy a colloid - heating or adding an electrolyte (ions)


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