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CHEMISTRY

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  1. CHEMISTRY The Study of Matter and the Changes that Matter Undergoes and The Energy Associated with The Changes

  2. What are the branches or subject areas of chemistry? Traditional 1. Inorganic a. Covers the chemistry all the elements and their compounds except the compounds of carbon 2. Organic a. Covers the chemistry of carbon compounds (except CO, CO2, metal carbonates, hydrogen carbonates, cyanides, cyanates, and thiocyanates) 3. Physical a. Uses physical or instrumental methods to measures physical properties of chemicals 4. Analytical a. Uses analytical methods to determine what constituents, and what percentages of constituents, are present in an unknown material Interdisciplinary 1. Biochemistry 2. Chemical Biology 3. Geochemistry 4. Pharmaceutical Chemistry/Medicinal Chemistry 5. Forensic Chemistry 6. Astrochemistry

  3. NUCLEOSYNTHESIS Big Bang Theory 1. Singularity exploded, and enormous energy was transformed into matter. 2. First matter formed as electrons and quarks. 3. In one millisecond, universe had expanded and cooled to 1012 K, and neutrons and protons formed. 4. By 4 minutes, the universe had expanded and cooled to about 109K. a. Protons outnumbered neutrons by about 7:1. b. Nucleosynthesis (fusing of fundamental and subatomic particles to create atomic nuclei) began c. Neutrons and protons fused together to make deuterons 5. After 5 minutes, universe was 75% by mass protons and 25% alpha particles

  4. Nucleosynthesis 6. Temperature dropped below 108 K, which was too low to sustain nuclear fusion. 7. After about 1 billion years, stars and galaxies started to form. 8. Compression heating inside stars increased temperatures above 108K and formation of heavier elements began. 9. Fusion reactions have continued for past 13 billion years. 10. Once 56Fe is formed in a giant star’s core, additional fusion reactions consumeenergy to make 60Ni, so star cools and collapses because of gravity produced by its enormous mass 11. As star collapses, compression reheats core above 109 K. 12. Repeated neutron capture and beta decay events in the cores of collapsing stars produce nuclei of the heaviest elements

  5. What produces the sizzling and aroma when a steak is fried or • barbecued? • (Answer: Glycerol [C3H5(OH)3] in the fat decomposes to water and acrolein [CH2=CHCHO], which: • a. Is a colorless liquid • b. Has a strong odor • c. Has a b. pt. of 52.50C

  6. What happens to green vegetables when they are cooked? • (Answer: Mg2+ ion in bright-green chlorophylls is replaced during heating by H1+ ion, producing dull, olive-green pheophytins: • C32H30ON4Mg COOCH3 + 2 H1+ COOC20H39 COOCH3 + Mg2+ C32H30ON4H2 COOC20H39

  7. Matter is anything that has mass and occupies space (has volume). 1. Can be visible: your book, your chair, salt, sugar 2. Can be invisible: clean air A substance is a type of matter that has a defined, fixed composition throughout. Physical property: a property that can be observed without altering the composition of the substance e.g. Color, melting point, boiling point, density, electrical and heat conductivity, hardness, ductility, malleability, state

  8. Chemical property: a property observed when a substance changes into or interacts with another substance e.g. Flammability, corrosiveness, reactivity with acids, etc.

  9. Chemical & Physical Properties of Copper Physical Properties 1. Reddish-brown 2. Metallic luster 3. Malleable 4. Ductile 5. Good heat & electrical conductor 6. Density = 8.95 g/cm3 7. Melting point = 1083oC 8. Boiling point = 2570oC Chemical Properties 1. Slowly forms a green carbonate in moist air 2. Reacts w/ HNO3 & H2SO4 3. Slowly forms deep-blue solution in aqueous ammonia

  10. Chemical change: a change occurring when a substance (or substances) is converted into a new substance (or substances) having new properties and new composition e.g. Octane + oxygen => carbon dioxide + water vapor 2 C8H18(l) + 25 O2(g) = > 16 CO2(g) + 18 H2O(g) Magnesium + oxygen => magnesium oxide 2 Mg + O2 => 2 MgO Silvery metal + colorless gas => white powder

  11. Physical Change: a change observed when a substances changes its state or condition (but not its composition) e.g. Change in condition: red HgO <=> yellow HgO e.g. Change of state: H2O(s) <=> H2O(l) <=> H2O(g)

  12. STATES OF MATTER -and the World Around US • SOLID - The Earth • LIQUID - Water • GAS - The Atmosphere

  13. Fig. 1.2

  14. Characteristics of Solid State • 1. Most solids are crystalline • 2. Some solids are amorphous • a. Glasses • b. Gels • c. Plastics • 3. Particles are very close together and well-organized • 4. Very small free volume • 5. Very strong interparticle forces • 6. Exhibit vibrational motion • 7. Are lower-energy materials than liquids or gases • 8 A condensed phase of matter; solids retain their own shape

  15. Characterisics of Liquids • 1. Particles are further apart than in solids but closer than in gases. • 2. Free volume = 3% • 3. Exhibit weak to moderately strong interparticle forces. • 4. Exhibit vibrational and rotational motion. • 5. Assume shape of portion of container occupied. • 6. Negligibly compressible. • 7. A condensed phase of matter.

  16. Characteristics of Gases • 1. Assume volume of container. • 2. Particles are very far apart. • 3. Have very weak interparticle forces. • 4. Are very compressible. • Free volume = 99%, even at P = 10 atm • Expanded state of matter • Exhibit rotational and translational motion, and vibrational motion if they are molecular gases

  17. gas + heat + heat vaporization sublimation - heat - heat condensation deposition + heat fusion solid liquid - heat solidification

  18. Phase Diagram P liquid solid gas T

  19. According to the First Law of Thermodynamics, energy is conserved, i.e., it cannot be destroyed. However, it can: 1. Change forms 2. Move from 1 part of the universe to another. Forms of energy: 1. Potential (Stored E or E of position) P.E. = mgh 2. Kinetic (E of motion) K.E. = 1/2 mv2 3. Chemical 4. Heat 5. Light 6. Electrical 7. Nuclear

  20. Fig. 1.3

  21. When the weight in figure A of the earlier slide fell to earth, there was a change in its potential energy of: P.E. change = P.E.final - P.E.init = 0 - mgh = -mgh But, according to the First Law of Thermodynamics, this energy cannot be destroyed. The potential energy lost by the weight as it falls must be converted to some other kind of energy, in this case, kinetic energy. It is natural for lower-energy states or conditions to be favored compared to higher-energy states. A lower-energy state is more stable, while a higher-energy state is less stable.

  22. SCIENTIFIC METHOD 1. Make observation; collect facts 2. Analyze data 3. State hypothesis (tentative proposal that explains observations) 4. Perform further experimentation 5. Revise or confirm hypothesis 6. Theory 7. Law

  23. Making Quantitative Measurements A quantitative measurement reports two things: 1. A number 2. A scale (or unit)

  24. SCIENTIFIC (EXPONENTIAL) NOTATION 1000 = 1 x 103 So, to write any number in 100 = 1 x 102 scientific notation, we: 10 = 1 x 101 1. Locate the decimal 1 = 1 x 100 2. Move decimal so there is 1 0.1 = 1 x 10-1 non-zero digit to its left 0.01 = 1 x 10-2 3. Rewrite number x 10 0.001 = 1 x 10-3 4. Power of 10 is number of places decimal was moved and has + sign if decimal moved left and minus sign if decimal moved right

  25. For example: 5,789,000 = 5.789 x 106 0.00000343 = 3.43 x 10-6 (11)2 = 121 = 1.21 x 102 1/4 = 0.25 = 2.5 x 10-1 Converting from scientific notation to standard notation reverses this process: 4.98x 105 = 498,000 6.76 x 10-3 = 0.00676

  26. MATHEMATICAL OPERATIONS USING SCIENTIFIC NOTATION 1. Multiplication (add exponents) (2 x 103)(3 x 102) = 6 x 105 (3.45 x 10-7)(2 x 104) = 6.90 x 10-3 (3.33 x 10-3)(4 x 10-9) = 13.32 x 10-12 = 1.332 x 10-? 2. Division (subtract exponents) 6.2 x 105 3.1 x 103 8.72 x 10-6 4 x 10-19

  27. 3. Raising to a power(multiplication of exponents) 23 = 2 x 2 x 2 = 8 so (2 x 103)3 = (2 x 103)(2 x 103)(2 x 103) = (2 x 2 x 2) x (103 x 103 x 103) = 8 x 109 i.e.. (2 x 103)3 = (2)3 x (103)3 = 8 x 109 4. Taking a root (division of exponents) 25 = 5 so 25 x 10-10 = 5 x 10-5 6.4 x 10-7 = ?

  28. 5. Addition In one of the numbers, the exponent of 10 must be changed to the same power as the exponent of 10 in the other number; also the decimal in the first factor of the number whose power of 10 was changed must be moved accordingly 5.23 x 10-3 + 3.09 x 10-4 = 52.3 x 10-4 + 3.09 x 10-4 = 55.39 x 10-4 = 5.539 x 10-3

  29. Subtraction Procedure is same as for addition, except numbers are subtracted instead of added

  30. Table 1. 2 (p. 17) SI - Base Units Physical Quantity Unit Name Abbreviation Mass Kilogram kg Length meter m Time second s Temperature Kelvin K Electric current ampere A Amount of substance mole mol Luminous intensity candela cd

  31. Table 1.3Common Decimal Prefixes Used with SI Units. Prefix Prefix Number Word Exponential Symbol Notation tera T 1,000,000,000,000 trillion 1012 giga G 1,000,000,000 billion 109 Mega M 1,000,000 million 106 Kilo k 1,000 thousand 103 hecto h 100 hundred 102 deka da 10 ten 101 ----- ---- 1 one 100 deci d 0.1 tenth 10-1 centi c 0.01 hundredth 10-2 milli m 0.001 thousandth 10-3 micro millionth 10-6 nano n 0.000000001 billionth 10-9 pico p 0.000000000001 trillionth 10-12 femto f 0.000000000000001 quadrillionth 10-15

  32. Solving Conversion Problems 1. Put what is given on left 2. Put units desired on right 3. Place x_______________ to right of units on left 4. Place units at left in denominator of conversion factor and units at right in numerator of conversion factor 5. From table of multiples of ten, determine relationship between numerator and denominator (i.e., how many multiples of 10 do they differ) 6. Place “1” in front of larger unit in conversion factor and multiple of 10 relationship in front of smaller unit and solve.

  33. Metric-Metric Conversions 1. Convert 4.2 km to m 4.2 km x 103m = 4.2 x 103 m 1 km 2. Convert 1.9 pm to cm 1.9 pm x 1 cm = 1.9 x 10-10 cm 1010 pm

  34. English-English Conversions 1. Convert 3.5 lb to oz 3.5 lb x 16 oz = 56 oz = 5.6 x 101 oz 1 lb 2. Convert 7 gal to qts 7 gal x 4 qts = 28 qts = 2.8 x 101 qts 1 gal

  35. Metric-English Conversions 1. Convert 4.5 ft to cm 4.5 ft x 12 in x 2.54 cm = 137.16 cm = 1.37 x 102 cm 1 ft 1 in 2. Convert 10 lb to g 10 lb x 453.6 g = 4536 g = 4.536 x 103 g 1 lb

  36. Convert 10 mi to mm 10.0 mi x 5.28 x 103 ft x 12 in x 2.54 cm x 101 mm 1 mi 1 ft 1 in 1 cm = 1.61 x 107 mm • Convert 6.1lb/in3 to g/mm3 6.1 lb x 453.6 g x 1 in3 x (1 cm)3 in3 1 lb (2.54 cm)3 (101 mm)3 = 1.7 x 10-1 g/mm3

  37. Table 1.4 Cont’dCommon SI-English Equivalent Quantities Quantity English to SI Equivalent Length 1 mile = 1.61 km 1 yard = 0.9144 m 1 foot (ft) = 0.3048 m 1 inch = 2.54 cm (exactly!) Volume 1 cubic foot = 0.0283 m3 1 gallon = 3.785 dm3 1 quart = 0.9464 dm3 1 quart = 946.4 cm3 1 fluid ounce = 29.6 cm3 Mass 1 pound (lb) = 0.4536 kg 1 pound (lb) = 453.6 g 1 ounce = 28.35 g

  38. Densities of Some Common Substances Substance Physical State Density (g/cm3) Hydrogen Gas 0.000089 Oxygen Gas 0.0014 Grain alcohol Liquid 0.789 Water Liquid 1.0 Table salt Solid 2.16 Aluminum Solid 2.70 Lead Solid 11.3 Gold Solid 19.3 Table 1.5 (p. 23)

  39. Density = mass of an object or substance own volume 1. Regularly-shaped object Calculate the density of a rectangular bar of lithium weighing 1.49 x 103 mg and measuring 20.9 mm x 11.1 mm x 11.9 mm d = m/v d = m/l x w x h d = 1.49 x 103 mg/20.9 mm x 11.1 mm x 11.9 mm d = 1.49 x 103mg/2.76 x 103mm3 d = 1.49 x 103 mg x 1g/103mg / 2.76 x 103mm3 x (1 cm)3/(101mm)3 d = 5.40 x 10-1 g/cm3

  40. An empty vial weighs 55.32 g. (a) If the vial weighs 185.56 g when filled with liquid mercury, which has a density of 13.53 g/cm3, what is the vial’s volume? (b) How much would the vial weigh if it were filled with water (d = 0.997 g/cm3 at 25oC)? (a) d = m/v v=m/d v vial = vHg filling vial = mHg/dHg = (185.56 – 55.32)g 13.53 g/cm3 Vvial = 9.626 cm3 (b) mvial filled = mvial empty + mwater mvial filled = mvial empty + (vd)water = 55.32 g + 9.626 cm3 x 0.997 g/cm3 = 55.32 g + 9.597 g mvial filled = 64.92 g

  41. 2. Irregularly shaped object Use displacement-of-fluid method Calculate the density of a rock weighing 4.56 g that raises the volume of water in a graduated cylinder from 44.6 mL to 48.4 mL d = m/v d = m/vf - vi d = 4.56 g/48.4 mL - 44.6 mL d = 1.20 g/mL

  42. Temperature = measurement of how hot or cold a substance is relative to another substance Heat = energy that flows between objects at different temperatures Temperature is an intensive property but heat (energy) is an extensive property. Extensive property = property that depends on the amount of substance present e.g. Energy (heat energy, potential energy, etc.), mass, volume Intensive property = property that doesn’t depend on the amount of substance present e.g. Temperature, density, melting point, boiling point

  43. Fig 1.13

  44. Temperature Conversions F = 180/100 C + 32 C = 5/9 (F - 32) F = 9/5 C + 32 K = C + 273.15 F = 1.8 C + 32 C = K - 273.15 Convert 250C to 0F Mercury melts at 234 K, lower than F = 9/5 C + 32 any other pure metal. Calculate its F = 9/5(25) + 32 melting point in 0C and 0F. F = 45 + 32 C = K - 273.15 F = 77 C = 234 - 273.15 = -39.150C F = 1.8 C + 32 F = 1.8(-39.15) + 32 = -70.47 + 32 F= -38.47

  45. If you wanted to calculate the melting point of mercury in 0F directly from K, instead of two steps, you could do the calculation like this: F = 1.8C + 32 F = 1.8[K - 273.15] + 32 F = 1.8(234 - 273.15) + 32 F = 1.8(-39.15) + 32 F = -70.47 + 32 F = -38.47

  46. Calculate the temperature that is the same on both the Fahrenheit and Celsius (Centigrade) scales. Assume C = F Now F = 1.8 C + 32 So F = 1.8(F) + 32 -0.8F = 32 F = 32/-0.8 F = -40 So C = F = -40

  47. Suppose that a new temperature scale has been devised on which the melting point of ethanol (-117.30C) and the boiling point of ethanol (78.30C) are taken as 00S and 1000S respectively, where S is the symbol for the new temperature scale. Derive an equation relating a reading on the S scale to a reading on the Celsius scale. C S b.pt. 78.3 100 S = 100/195.6(C + 117.3) 195.6 100 m.pt. -117.3 0

  48. Uncertainty of Measurements Every experimental measurement has error and is limited by: 1. Reliability of measuring device 2. Skill of operator Customarily report all the significant figures when making a measurement. Significant figures = all the certain digits (which are found using calibration marks) + 1 estimated digit (estimate to nearest 1/5th of last interval read precisely) So, when we report a measured mass as 2.0376 g, we know the estimated digit is the rightmost digit (the 6). We could write that the mass is between 2.0376 + 0.0001 and 2.0376 - 0.0001 g

  49. Rules for Finding Significant Figures in Measurements 1. All nonzero digits are significant 2. Zeros # sig. figs. (a) leading never significant 0.00387 3 (b) captive always significant 20.0895 6 (c) trailing may or may not be 500.0 4 significant 500. 3 500 at least 1