Chemistry Review Mr. Halfen June 2011. Substances Pure substance - a substance composed on only one...
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Transcript of Chemistry Review Mr. Halfen June 2011. Substances Pure substance - a substance composed on only one...
![Page 1: Chemistry Review Mr. Halfen June 2011. Substances Pure substance - a substance composed on only one type of element or molecule Examples - diamond, oxygen,](https://reader031.fdocuments.in/reader031/viewer/2022013011/56649e555503460f94b4cf2d/html5/thumbnails/1.jpg)
Chemistry ReviewMr. HalfenJune 2011
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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
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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
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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)
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Elements
• mass number = #protons + #neutrons
• 41Ca has 20 p+ and 21 n0
• isotopes - same #p+ & different #n0
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Periodic Table
• Families, natural phase, etc.
• www.chemicool.com
• Trends - pp. 132-141 in Holt
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Periodic Table
• Using the text and your periodic table, list at least 10 trends or patterns in the periodic table.
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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
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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
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Mass Ratios
• mass of 1 element:mass of 2nd element
• CO - carbon monoxide
• 12.0107:15.9994
• 1:1.332
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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
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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
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Formulas - Practice
Structural Empirical
Acetic Acid - C2H402(HC2H302)
CH2O
Formaldehyde - CH2O CH2O
Glucose - C6H12O6 CH2O
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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
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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%
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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.
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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
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• 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.
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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
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Chemical Reactions - Types
• Combustion
• Ex: C3H8 + 5O2 --> 3CO2 + 4H2O
• Synthesis
• Ex: 2H2 + O2 --> 2H2O
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• Decomposition
• Ex.: CH4 --> C + 2H2
• Displacement
• Ex.: Cu + FeO --> CuO + Fe
• Double Displacement
• Ex.: H2SO4 + Ca(OH)2 --> 2H2O + CaSO4
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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.
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Balancing Chemical Reactions
• worksheet
• work thru the first few together
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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)
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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
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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
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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
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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
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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
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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
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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
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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