CHEM Cheat Sheet (2)

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Chemistry

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Feb 20, 2024

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CHEM 113 Cheat Sheet Temperature Conversions: SI/Metric Prefix Chart Key Points  Any two things that are equal can be used as a conversion factor  “x” is used to denote any base unit. Replace “ x” with the base unit utilized in your problem. Prefix Symbol Relationship to Base Unit ( x ) x = grams (g), liters(L), meters (m) Scientific Notation Standard Notation Tera- T 1 T x = 10 12 x 1 T x = 1,000,000,000,000x Giga G 1 G x = 10 9 x 1 G x = 1,000,000,000x Mega- M 1 M x = 10 6 x 1 M x = 1,000,000x Kilo- k 1 k g = 10 3 x 1 k g = 1,000x Hecto- h 1 h x = 10 2 x 1 H x = 100x Deca- da 1 da x = 10 1 x 1 da x = 10x Base Units x = Base Units x = grams (g), liters(L), meters (m) Deci- d 1 d x = 10 -1 x 1 d x = 0.1x Centi- c 1 c x = 10 -2 x 1 c x = 0.01x Milli- m 1 m x = 10 -3 x 1 m x = 0.001x Micro- μ 1 μ x = 10 -6 x 1 μ x = 0.000001x Nano- n 1 n x = 10 -9 x 1 n x = 0.000000001x Pico- p 1 p x = 10 -12 x 1 p x = 0.000000000001x

CHEM 113 Cheat Sheet SI/Metric Prefix Chart Usage Help ( values below are in standard notation, but the same applies to scientific notation ) Length (meters – m)  “ x ” would be replaced with m  Example for kilo (k): 1 k x = 1,000x would be become 1 k m = 1,000m  1km = 1,000m can be turned into two conversion factors, 1 km 1000 m ∨ 1000 m 1 km Mass (grams – g)  “ x ” would be replaced with g  Example for milli (m): 1 m g = 0.001g would be become 1 m g = 0.001g  1 m g = 0.001g can be turned into two conversion factors, 1 mg 0.001 g ∨ 0.001 g 1 m g Volume (liter – L)  “ x ” would be replaced with L  Example for micro (μ): 1 μ L = 0.000001L would be become 1 μ L = 0.000001L  1 μ L = 0.000001L can be turned into two conversion factors, 1 µ L 0.000001 L ∨ 0.000001 L 1 µL Inter-system Conversion Factors Length 1 kilometer (km) = 0.6214 mile (mi) 1 meter (m) = 39.37 inches (in.) 1 meter (m) = 1.094 yards (yd) 1 foot (ft) = 30.48 centimeters (cm) 1 inch (in.) = 2.54 centimeters (cm) Mass 1 kilogram (kg) = 2.205 pounds (lbs) 1 pound (lb) = 453.59 grams (g) 1 ounce (oz) = 28.35 (g) Volume 1 milliliter (mL) = 1 cubic centimeter (cm3) 1 liter (L) = 1.057 quarts (qts) 1 U.S. gallon (gal) = 3.785 liters (L)

CHEM 113 Cheat Sheet Electron Filling ( Aufbau Principle ) Chart

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CHEM 113 Cheat Sheet Electronegativity Polyatomic Ions

CHEM 113 Cheat Sheet VESPR Chart Electron Density (bonds or lone pairs) 0 Lone Pair 1 Lone Pair 2 Lone Pairs 3 Lone Pairs 2 Geometry Electron Linear Linear Molecular Linear Linear 3 Geometry Electron Trigonal Planar Trigonal Planar Trigonal Planar Molecular Trigonal Planar Bent Linear 4 Geometry Electron Tetrahedral Tetrahedral Tetrahedral Tetrahedral Molecular Tetrahedral Trigonal pyramidal Bent Linear 5 Geometry Electron Trigonal Bipyramidal Trigonal Bipyramidal Trigonal Bipyramidal Trigonal Bipyramidal Molecular Trigonal Bipyramidal See Saw T Shape Linear 6 Geometry Electron Octahedral Octahedral Octahedral Molecular Octahedral Square Pyramidal Square Planar Stoichiometry: The Molar Key Method

CHEM 113 Cheat Sheet This Stoichiometry Method has been developed to aid students in their ability to conduct basic stoichiometric calculations in a single step conversion. This method relies on developing a stoichiometric “Key” Steps of Method: 1. Determine what amount you are given . 2. Determine what unknown amount you are trying to find. 3. Determine the stoichiometric key s for the given amount and the unknown amount . ( The stoichiometric key is simply a visualization of the basic stoichiometric equivalences for any molecule as represented by its molar mass.) Example: The stoichiometric “Key” for H 2 O would be, D ŽůĂƌ <ĞǇ ĨŽƌ H 2 O D ĂƐƐ;ŐͿ Atomic Weight (element) Molecular Weight (molecules) с ŵŽů с ƚŽŵƐ;ĞůĞŵĞŶƚƐͿ ŽƌD ŽůĞĐƵůĞƐ ϭϴϬϭϱ ͘° Ő H 2 O с ϭŵŽů H 2 O с ϬϮϮǆϭϬ ϲ͘ Ϯϯ ƚŽŵƐ ͬ"D ŽůĞĐƵůĞƐ H 2 O 4. Create a coefficient conversion template using coefficients from the chemical equation . Place the given on the bottom and the unknown on the top. Conversion Template: Givenamount x unknowncoefficient ( Key ) givencoefficient ( Key ) = unknown amount 5. Find the given key that matches the given units and the unknown key that matches the unknown units and enter them into their key positions. Perform the calculations. Example 1 : ( Moles to Moles ): How many mols of O 2 are needed to react with 5.21 mols of CH 4 CH 4 ( g ) + 2 O 2 ( g )  CO 2 ( g ) + 2H 2 O ( l ) 1. Given amount (Value you are given) – 5.21 moles CH 4 2. Unknown amount - moles of O 2 3. Stoichiometric Keys 4. Coefficient conversion template . (Coefficients from chemical equation are shown in red.) 5.21 mole CH 4 x 2 ( Unknown Key ) O 2 1 ( GivenKey ) CH 4 = ? moles O 2 5. Find the given key that matches the given units and the unknown key that matches the unknown units and enter them into their key positions. Perform the calculations. 5.21 mole CH 4 x 2 ( mole ) O 2 1 ( mole ) CH 4 = ? moles O 2 5.21 mole CH 4 x 2 ( mole ) O 2 1 ( mole ) CH 4 = 10.42 moles O 2 *The coefficient in front of CH 4 is an implied “ 1 ”

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CHEM 113 Cheat Sheet Example 2 : ( grams to grams ): What mass (g) of NaOH , would be required to produce 14.2 g of magnesia [Mg(OH) 2 ] by the following reaction? MgCl 2 ( aq ) + 2 NaOH ( aq ) ⟶ Mg(OH) 2 ( s ) + 2NaCl( aq ) 1. Given amount (Value you are given) – 14.2 g Mg(OH) 2 2. Unknown amount - ? g NaOH 3. Stoichiometric Keys 4. Coefficient conversion template . (Coefficients from chemical equation are shown in red.) 14.2 g Mg(OH) 2 x 2 ( Unknown Key ) NaOH 1 ( Given Key ) Mg ( OH ) 2 = ? g NaOH 5. Find the given key that matches the given units and the unknown key that matches the unknown units and enter them into their key positions. Perform the calculations. 14.2 g Mg(OH) 2 x 2 ( 40.0 g ) NaOH 1 ( 53.3 g ) Mg ( OH ) 2 = ? g NaOH ; 14.2 g Mg(OH) 2 x 2 ( 40.0 g ) NaOH 1 ( 53.3 g ) Mg ( OH ) 2 = 21.3 g NaOH Example 3 : ( moles to grams ): What mass of oxygen gas, O 2 , from the air is consumed in the combustion of 0.532 moles of octane, C 8 H 18 , one of the principal components of gasoline? 2 C 8 H 18 + 25 O 2  16CO 2 + 18H 2 O 1. Given amount (Value you are given) – 0.532 moles C 8 H 18 2. Unknown amount - ? g O 2 3. Stoichiometric Keys 4. Coefficient conversion template . (Coefficients from chemical equation are shown in red.) 0.532 moles C 8 H 18 x 25 ( Unknown Key ) O 2 2 ( Given Key ) C 8 H 18 = ? g O 2 5. Find the given key that matches the given units and the unknown key that matches the unknown units and enter them into their key positions. Perform the calculations.

CHEM 113 Cheat Sheet 0.532 moles C 8 H 18 x 25 ( 32.00 g ) O 2 2 ( mole ) C 8 H 18 = ? g O 2 ; 0.532 moles C 8 H 18 x 25 ( 32.00 g ) O 2 2 ( mole ) C 8 H 18 = 213 g O 2 Example 4 : ( grams to molecules ): How many molecules of AgNO 3 would be needed to react with 8.50 g of copper metal, Cu, in the following reaction? Cu( s ) + 2 AgNO 3 ( aq )  2Ag( s ) + Cu(NO 3 ) 2 ( aq ) 1. Given amount (Value you are given) – 8.50 g Cu 2. Unknown amount - ? molecules AgNO 3 3. Stoichiometric Keys 4. Coefficient conversion template . (Coefficients from chemical equation are shown in red.) 8.50 g Cu = 2 ( Unknown Key ) AgNO 3 1 ( Given Key ) Cu = ? molecules AgNO 3 5. Find the given key that matches the given units and the unknown key that matches the unknown units and enter them into their key positions. Perform the calculations. 8.50 g Cu x 2 ( 6.022 x 10 23 molecules ) AgNO 3 1 ( 63.55 g ) Cu = ? molecules AgNO 3 8.50 g Cu x 2 ( 6.022 x 10 23 molecules ) AgNO 3 1 ( 63.55 g ) Cu = 1.02 x 10 25 molecules AgNO 3

CHEM 113 Cheat Sheet Units of Pressure Unit Name and Abbreviation Definition or Relation to Other Unit pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit kilopascal (kPa) 1 kPa = 1000 Pa pounds per square inch (psi) 14.7 psi = 1atm atmosphere (atm) 1 atm = 101,325 Pa = 760 torr bar (bar, or b) 1.01 bar = 1 atm = 101 kPa torr 760 torr = 1 atm millimeters of mercury (mm Hg) 1 mm Hg ~1 torr Gas Laws V 1 T 1 = V 2 T 2 P 1 V 1 =P 2 V 2 P 1 V 1 T 1 = P 2 V 2 T 2 P Total = P A + P B + P C + … PV = nRT Universal Gas Constants (Common):  R = 0.08206 L atm mol -1 K -1  R = 8.314 kPa L mol -1 K -1 pH pH = −log [H + ]

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