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Chapter 1

Chapter 1. mass. density =. volume. m. d =. V. Density – SI derived unit for density is kg/m 3. 1 g/cm 3 = 1 g/mL = 1000 kg/m 3. 0 F = x 0 C + 32. 5. 0 C = x ( 0 F – 32). 9. 9. 5. K = 0 C + 273.15. 273 K = 0 0 C 373 K = 100 0 C. 32 0 F = 0 0 C

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Chapter 1

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  1. Chapter 1

  2. mass density = volume m d = V Density – SI derived unit for density is kg/m3 1 g/cm3 = 1 g/mL = 1000 kg/m3

  3. 0F = x 0C + 32 5 0C = x (0F – 32) 9 9 5 K = 0C + 273.15 273 K = 0 0C 373 K = 100 0C 32 0F = 0 0C 212 0F = 100 0C Kelvin is the SI Unit of temperature: absolute temperature scale. 0K is the lowest temperature that can be achieved theoretically. 1.7

  4. Significant Figures - The meaningful digits in a measured or calculated quantity - The last digit is uncertain; 6.0±0.1ml • Any digit that is not zero is significant • 1.234 kg 4 significant figures • Zeros between nonzero digits are significant • 606 m 3 significant figures • Zeros to the left of the first nonzero digit are not significant • 0.08 L 1 significant figure • If a number is greater than 1, then all zeros to the right of the decimal point are significant • 2.0 mg 2 significant figures • If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant • 0.00420 g 3 significant figures 1.8

  5. 89.332 + 1.1 two significant figures after decimal point one significant figure after decimal point 90.432 round off to 0.79 round off to 90.4 3.70 -2.9133 0.7867 Significant Figures Addition or Subtraction The answer cannot have more digits to the right of the decimal point than any of the original numbers. 3DP 1DP 1DP 2DP 4DP 2DP Answer can have only as many decimal places (DP)as the number in the operation with the least number of decimal places: 1.8

  6. 3 sig figs round to 3 sig figs 2 sig figs round to 2 sig figs Significant Figures Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.536366 = 16.5 3 sig figs 6.8 ÷ 112.04 = 0.0606926 = 0.061 2 sig figs 1.8

  7. Chapter 2

  8. A X Mass Number Element Symbol Z Atomic Number Atomic number, Mass number and Isotopes Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons number of protons = number of electrons 2.3

  9. 235 238 U U 92 92 The Isotopes of Hydrogen Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei 2.3

  10. Steps of naming ionic and binary molecular compounds Naming acids and base 2.7

  11. Chapter 3

  12. 7.42 x 6.015 + 92.58 x 7.016 100 Average atomic mass- average mass of naturally occurring mixture of isotopes. Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: = 6.941 amu 3.1

  13. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molecular mass(or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3

  14. 1 12C atom 12.00 g 1.66 x 10-24 g = 12.00 amu 6.022 x 102312C atoms 1 amu = molar mass in g/mol M Relationship between amu and grams x 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu NA= Avogadro’s number 3.2

  15. Percent composition of compounds Percent composition by mass is the percent of each element in a compound. 3.5

  16. Percent Composition and Empirical Formulas 3.5

  17. IS NOT How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO 2 grams Mg + 1 gram O2 makes 2 g MgO 2 Mg(s) + O2 (g) 2 MgO(s) Indicate physical state 2 HgO(s) O2 (g) + 2Hg(l) 3.7

  18. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Balancing Chemical Equations • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3.7

  19. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations • Start by balancing those elements that appear only once on each side of the equation and balance them. start with C or H but not O multiply CO2 by 2 multiply H2O by 3 3.7

  20. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2 3.7

  21. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7

  22. Amounts of Reactants and Products • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units 3.8

  23. 2NO + 2O2 2NO2 Limiting Reagents The reactant used up first in a reaction is called limiting reagent. 7 mole NO and 8 mole O2 NO is the limiting reagent O2 is the excess reagent The maximum amount of products formed depends on the amount of the limiting reagent. 3.9

  24. Chapter 4

  25. precipitate Pb(NO3)2(aq) + 2NaI (aq) PbI2(s) + 2NaNO3(aq) Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3- Pb2+ + 2I- PbI2 (s) Molecular equation, Ionic equation and net Ionic equation molecular equation ionic equation net ionic equation Na+ and NO3- are spectator ions: ions that are not involved in the overall reaction 4.2

  26. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Write the net ionic equation for the reaction of silver nitrate with sodium chloride. Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3- Ag+ + Cl- AgCl (s) Writing Net Ionic Equations • Write the balanced molecular equation. • Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.Weak and non electrolytes are written as molecules • Cancel the spectator ions on both sides of the ionic equation • Check that charges and number of atoms are balanced in the net ionic equation 4.2

  27. Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen isusually–2. In H2O2 and O22- it is –1. 4.4

  28. Oxidation numbers of all the elements in HCO3- ? • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds (e.g. LiH, CaH2). In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. • The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. For example,NH4+ , the sum of oxidation numbers is • -3+4(+1)=+1 7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2-, is -½. O = -2 H = +1 3x(-2) + 1 + ? = -1 HCO3- C = +4 4.4

  29. 5. Disproportionation reaction

  30. The Activity Series for Metals The Activity Series * Arranges metals according to their ease of oxidation *The higher the metal on the Activity Series, the more active that metal (the easier it is oxidized.)Any metal can be oxidized by the metal ions below it. *Any metal above hydrogen will displace it from water or from an acid.

  31. 3. Halogen Displacement F2 > Cl2 > Br2 > I2 F2 is the greatest oxidizing halogen I2 is the least oxidizing halogen Example: -1 0 -1 0 2 Br- + Cl2 --> 2 Cl- + Br2 Br2 + Cl- --> no reaction (NR)

  32. moles of solute mole n liters of solution liters V Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = = = • Read 2.5 M NaCl as 2.5 molar sodium chloride Often use square bracket [ ] to indicate the concentration Example: What is the molarity of a solution made from 4.00 g of NaOH diluted to a final volume of 250 mL? First find the number of moles of NaOH. Then divide by the volume in Liters. 4.5

  33. Chapter 5

  34. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation. 5.4

  35. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) Molar volume of gas • 1 mole of gas at STP = 22.4 Liters • Example: 2 moles of gas at STP = 44.8 L PV = nRT R = 0.082057 L • atm / (mol • K)=8.314J/(K·mol) = 8.314 L·kPa/(K·mol) 5.4 In calculation, the units of R must match those for P,V,T and n.

  36. The combined gas law P1V1 n1T1 P2V2 n2T2 R= R= P1V1 n1T1 P1V1 T1 P2V2 n2T2 P2V2 T2 = = If n1=n2

  37. Dalton’s Law of Partial Pressures Partial pressure is the pressure of the individual gas in the mixture. V and T are constant P1 P2 Ptotal= P1 + P2 5.6

  38. PA = nART nBRT V V PB = nA nB nA + nB nA + nB XB = XA = ni mole fraction (Xi) = nT Consider a case in which two gases, A and B, are in a container of volume V. nA is the number of moles of A nB is the number of moles of B PT = PA + PB PA = XAPT PB = XBPT Pi = XiPT 5.6

  39. Chapter 6

  40. Another form of the first law for DEsystem DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDVwhen a gas expands against a constant external pressure 6.3

  41. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ/mol x H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 x DH = 6.01 kJ/mol DH = 44.0 kJ/mol 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. = 6470 kJ 266 g P4 6.4

  42. Consider the reaction below:  2CH3OH(l) + 3O2(g)  4H2O(l) + 2CO2(g) ∆H = -1452.8 kJ/mol  What is the value of ∆H for the reaction of  8H2O(l) + 4CO2(g)  4CH3OH (l) + 6O2 (g)? If you reverse a reaction, the sign of ΔH changes. If you multiply both sides of the equation by a factor n, then ∆H must change by the same factor n. ∆H = 2*(+1452.8 kJ/mol) = +2905.6 KJ/mol Practice problem of chap6:Q2

  43. Calorimetry- the measurement of heat. The specific heat(s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity(C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial 6.5

  44. Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. “0”-standard state at 1atm; “f”-formation f DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Standard enthalpy of formation and reaction Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero. 6.6

  45. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn mDH0 (reactants) dDH0 (D) nDH0 (products) cDH0 (C) aDH0 (A) bDH0 (B) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 6.6

  46. Chapter 7

  47. Maxwell (1873), proposed that visible light consists of electromagnetic waves.– provides a mathematical description of the general behavior of light. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves, which has an electrical field component and a magnetic field component . Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c 7.1

  48. Energy of light: Quantum = packet of energy Photon = packet of light • E = h n Planck’s constant (h) h = 6.63 x 10-34 J•s

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