CHEM 1441 1st Edition Exam 1 Study Guide Lectures 1 7 Lecture 1 January 23 Chapter 1 Keys to Study of Chemistry Significant Figures SIG FIG PROBLEMS WILL BE ON TEST What is an exact number and what is an inexact number Exact Numbers include counting numbers average number of terms and defined measurements Have infinite number of sig figs For Example 3 pens 1 in 2 54 cm 1000g 1kg Inexact Numbers include numbers that are not exact and have uncertainty such as measurements taken in a lab using equipment like a beaker graduated cylinder etc For Example 1kg 2 205lb 1 mile 1 609km Counting Sig Figs All non zero digits are significant For Example 5 32 3 sig figs 4 4 2 sig figs 35 52 4 sig figs 0 65 2 sig figs All zeros between other significant figures are significant For Example 5 002 4 sig figs Leading zeros are not significant For Example 0 045 2 sig figs 0 306 3 sig figs Trailing zeros are always included if decimal is present For Example 0 003040 4 sig figs 150 2 sig figs 150 3 sig figs 150 10 5 sig figs Scientific notation helps remove ambiguity by using a decimal point For Example 150 turns into 1 50 x 103 3 sig figs 1 5 x 103 2 sig figs Calculations using Significant Figures Addition and Subtraction look for least precise term find the common significant decimal place for all terms For Example 29 52 3 001 7219 5 Line them up by the decimal point 29 52 3 001 Answer is 7251 0 5 sig figs or Least precise 7219 5 7 2510 x 10 3 7251 021 25 57 g 25 49 g 25 57 25 49 00 08 5 68 8 3 5 68 Answer is 00 08 g 1 sig fig or 8 x 10 2 g 8 3 13 98 14 0 need to round off to least precise term Answer is 14 0 3 sig figs or 1 40 x 10 1 Multiplication and Division use smallest number amount of sig figs For Example 2 99 x 7 3 Count number of sig figs in each term 3 sig figs x 2 sig figs answer with 2 sig figs 2 99 x 7 3 21 827 round off to 2 sig figs Answer is 22 Or 2 2 x 10 1 5 982 x 0 00201 4 sig figs x 3 sig figs answer with 3 sig figs 5 982 x 0 00201 12 02382 Answer is 12 0 or 1 2 x 10 1 Mixed Operations For Example 12 3 8 72 295 767 Use PEMDAS do mul div first then add sub Don t round off until the end keep track of sig figs by marking them mark the sig fig 1 4105505 295 767 1 4105505 use least precise term 295 767 297 1775504 5 sig figs in answer Answer is 297 18 or 2 9718 x 10 2 Lecture 2 January 26 Chapter 1 Keys to Study of Chemistry cont d Significant Figure Concept Check 31 49 12 x 434 049 432 1 19 49 x 1 949 37 98601 2 sig fig 2 sig fig ans with 2 sig fig Answer is 38 or 3 8 x 101 Doing Dimensional Analysis Step 1 Write down what you know Step 2 Start with the given measured quantity Step 3 Apply conversion factors MEMORIZE THESE 1in 2 54cm 1kg 2 205 1ml 1cm3 12in 1ft 1000g 1kg For Example How many feet are in 6 0 yds Step 1 1yd 3 ft this is an exact measurement so it has infinite sig figs Step 2 3 6 0 yds x 3 ft 18 ft 1 yd Know the SI Unit System CGS System and Metric Prefixes SI Unit System Mass kg kilograms Volume volume is a combined unit l h w m3 meters cubed or area m2 meters Length m meters Temperature K Kelvin Time s seconds squared Chemistry CGS System Mass g grams Volume L liters Length cm m centimeters meters Temp K Time s Metric Prefix Table MEMORIZE Power always belongs to the base unit g s L m deci 1 dg 1x10 1 g 1x101 g 1 Dg Deca 2 centi 1 cg 1x10 g 1x102 g 1 hg hecta milli 1 mg 1x10 3 g 1x103 g 1 kg kilo micro 1 g 1x10 6 g 1x106 g 1 Mg Mega 9 nano 1 ng 1x10 g 1x109 g 1 Gg Giga pico 1 pg 1x10 12 g 1x1012 g 1 Tg Tera femto 1 fg 1x10 15 g 1x1015 g 1 Pg Peta Conversion Problems Convert 0 00275 dL to L Step 1 1 dL 1 x 101 L 1 L 1 x 10 6 L Step 2 3 0 00275 dL x 1x101 L x 1 L 275 L 1 dL 1x10 6 L Express 65 miles hour in m s Step 1 1 mile 1 609km 1 hr 60 min 1 min 60 s 1 km 1x103 m Step 2 3 65 miles x 1 609 km x 1x103 m x 1 hr x 1 min 29 m s 1 hr 1 mile 1 km 60 min 60 s Express the density 13 6 g mL in lbs ft3 Hint this may be on test 3 Step 1 2 205 lb 1 kg 1 kg 1x10 g 1 mL 1 cm3 1 ft 12 in 1 in 2 54 cm Step 2 3 13 6 g x 1 mL x 2 54 cm 3 x 12 in 3 x 1 kg x 2 205 lbs 849 lbs 1 mL 1 cm3 1 in 1 ft 1g 1 kg Calculating Density d m V Density can be used as a conversion factor m Given that the density of ethanol is 0 789 g conversion factors can be expressed as so 0 789 g of Ethanol or 0 789 g of Ethanol 1 mL of solution 1 cm3 of solution 3 For Example Write the mass of 1 5 m of ethanol in lbs Step 1 1 kg 2 205 lbs 1 kg 1x103 g 1 cm 1x10 2 m 0 789g 1 cm3 d V Step 2 3 1 5 m3 x 1 cm 3 x 0 789 g x 1 kg x 2 205 lbs 2 6 x 103 lbs 1x10 2 m 1 cm3 1x103 g 1 kg Lecture 3 January 28 Dimensional Analysis Concept Check Professor Abayan weighs 220 50 lb What is this in grams 2 205 lb 1 kg 220 50 lb x 1 kg x 1x103 g 1 000 x 105 g 2 205 lb 1 kg Atomic Structure Abbreviation Mass Weight Charge Proton p H 1 672 x 10 27 kg 1 amu 1 Neutron n 1 675 x 10 27 kg 1 amu 0 electron e 9 109 x 10 31 kg 0 0005 amu 1 Atoms When referring to an atom such as a Hydrogen atom we refer to the ground state The ground state of an atom is where the of protons equal the of electrons and the …