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Chemistry Review

Chemistry Review. Mr. Halfen June 2011. Substances. Pure substance - a substance composed on only one type of element or molecule Examples - diamond, oxygen, water Mixture - a physical combination of more than one pure substance Examples - salad, steel, soda

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Chemistry Review

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  1. Chemistry Review • Mr. Halfen • June 2011

  2. Substances • Pure substance - a substance composed on only one type of element or molecule • Examples - diamond, oxygen, water • Mixture - a physical combination of more than one pure substance • Examples - salad, steel, soda • Compound - a chemical combination of two or more pure substances

  3. Mixtures • Heterogeneous Mixture - substances are not uniformly mixed together • Examples - salad, granite, chocolate chip cookie • Homogeneous mixture - substances are evenly mixed; each part is identical to all others, even under a light microscope • Examples - steel, soda, air

  4. Atomic Structure • 3 main particles - protons, neutrons & electrons • nucleus - center of atom; contains neutrons & protons • electron cloud - electrons found in region outside of nucleus in defined energy levels • valence electrons - electrons in outer shell (highest energy level)

  5. Elements • mass number = #protons + #neutrons • 41Ca has 20 p+ and 21 n0 • isotopes - same #p+ & different #n0

  6. Periodic Table • Families, natural phase, etc. • www.chemicool.com • Trends - pp. 132-141 in Holt

  7. Periodic Table • Using the text and your periodic table, list at least 10 trends or patterns in the periodic table.

  8. Mass Calculations • molar mass = atomic weight in grams/mole • Avogadro’s number = 6.022 x 1023 atoms/mole • mass of atom = molar mass/A.N. • #moles = measured mass/molar mass

  9. Molar Mass • Glucose - C6H12O6 • C - 6 x 12.0107 = 72.0642 g • H - 12 x 1.00794 = 12.09528 g • O - 6 x 15.9994 = 95.9964 g • 72.0642 + 12.09528 + 95.9964 = 180.1559g • Glucose molecule = 180.1559g/6.022 x 1023

  10. Mass Ratios • mass of 1 element:mass of 2nd element • CO - carbon monoxide • 12.0107:15.9994 • 1:1.332

  11. Mass Ratio - 2 • Glucose - C6H12O6 • C - 6 x 12.0107 = 72.0642 • H - 12 x 1.00794 = 12.09528 • O - 6 x 15.9994 = 95.9964 • 72.0642 : 12.09528 : 95.9964 • 5.96 : 1 : 7.94

  12. Formulas • Structural formula - chemical formula showing the ratios of different elements in a chemical and the arrangement of the atoms • Empirical formula - chemical formula showing the relative ratios of the elements in a molecule

  13. Formulas - Practice

  14. Calculation of Percent Composition from Chemical Formulas • Calculate the percent composition for Mg(CN)2. • 1 - Find the molar mass. • Mg - 1 x 24.3050g = 24.3050g • C - 2 x 12.0107g = 24.0214g • N - 2 x 14.00674g = 28.01348g • molar mass = 76.33988g

  15. Percent composition - 2 • 2 - Find the % of each element • % Mg = 24.3050/76.33988 = 31.84% • % C = 24.0214/76.33988 = 31.47% • % N = 28.01348/76.33988 = 36.70%

  16. Calculation of Empirical Formulas from Percent Compositions • What is the empirical formula for a liquid with a composition of 18.0% C, 2.26% H and 79.7% Cl? • Assume 100 g of substance. • Convert the % to grams. • So there is 18.0 g of C, etc.

  17. Empirical formula -2 • Calculate moles of each substance. • mole-C = 18.0g/(12.01g/mole) = 1.50 • mole-H = 2.26g/(1.01g/mole) = 2.24 • mole-Cl = 79.7g/(35.4527g/mole) = 2.24

  18. Divide moles of each by the smallest subscript. • mole-C = 1.50/1.50 = 1 • mole-H = 2.2/1.50 = 1.5 • mole-Cl = 2.24/1.50 = 1.5 • By inspection, C2H3Cl3 • Structurally, it could be ClH2C2HCl2 or H3C2Cl3.

  19. Chemical Reactions - Evidence • release or absorption of heat • production of light (or flames) • production of sound • release or absorption of electricity • formation of a gas • formation of a precipitate • change in color • change in odor

  20. Chemical Reactions - Types • Combustion • Ex: C3H8 + 5O2 --> 3CO2 + 4H2O • Synthesis • Ex: 2H2 + O2 --> 2H2O

  21. Decomposition • Ex.: CH4 --> C + 2H2 • Displacement • Ex.: Cu + FeO --> CuO + Fe • Double Displacement • Ex.: H2SO4 + Ca(OH)2 --> 2H2O + CaSO4

  22. Balancing Chemical Reactions • In the examples of types of chemical reactions, all of the equations are balanced. • Balancing chemical reactions is simply applying the Law of the Conservation of Mass. • That is, if you start with 2 atoms (or moles) of Hydrogen, you have to end up with 2 atoms of Hydrogen.

  23. Balancing Chemical Reactions • worksheet • work thru the first few together

  24. Gas Laws - Ideal Gas Law • PV = nRT • Others derived this by combining the other gas laws. Einstein derived it from a mathematical description of the kinetic theory of gases. I suspect you would rather not see this derivation right now.... • R is the “gas constant.” • R = 8.314 L kPa/(mol K), or • R = 0.0821 L atm/(mol K)

  25. Gas Laws - Boyle’s Law • PV = nRT • If the amount (i.e., moles) of the gas are constant (k1) and the temperature is held constant (k2), then the equation becomes... • PV = k1 R k2 • Let k1 R k2 = kB and PV = kB

  26. Boyle’s Law - 2 • If we start with conditions 1, then • P1V1 = kB • If we change to conditions 2, then • P2V2 = kB • Since the kB’s are equal, P1V1 = P2V2

  27. Gas Laws - Charles’ Law • PV = nRT • If the amount (i.e., moles) of the gas are constant (k1) and the pressure is held constant (k3), then the equation becomes... • k3 V = k1 R T and • let k1 R/k3 = kC and we get V = kC T or V/T = kC

  28. Charles’ Law - 2 • If we start with conditions 1, then • V1/T1 = kC • If we change to conditions 2, then • V2/T2 = kC • Since the kC’s are equal, V1/T1 = V2/T2

  29. Gas Laws - Gay-Lussac’s Law • PV = nRT • If the amount (i.e., moles) of the gas are constant (k1) and the volume is held constant (k4), then the equation becomes... • P k4 = k1 R T • Let k1 R /k4 = kG and P = kGT

  30. Gay-Lussac’s Law - 2 • If we start with conditions 1, then • P1/T1 = kC • If we change to conditions 2, then • P2/T2 = kC • Since the kC’s are equal, P1/T1 = P2/T2

  31. Gas Laws - Avogadro’s Law • PV = nRT • If the pressure of the gas is constant (k5) and the temperature is held constant (k6), then the equation becomes... • k5 V= n R k6 • Let k6 R /k5 = kA and V = kA n

  32. Gas Law Problems • Boyle’s: p.425 #1 & #3 • Charles’: p.428 #1 & #3 • G-L’s: p. 431 #1 & #3 • Section Review: #5, #7 & #9 • Ideal Gas Law: p.435 #1 & #3

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