# Chemistry Spring Final Exam Study Guide Richard Molina
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Synthesis/Combination

2 or more reactants combine to form a single product 2Mg (s) + O2 (g) –> 2MgO (s)
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Decomposition

A single compound breaks down into 2 or more simpler products 2HgO (s) –> 2Hg (l) + O2 (g)
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Single-Replacement

1 element replaces a second element in a compound; both the reactants and the products consist of an element and a compound 2K (s) + 2H2O (l) –> 2KOH (aq) + H2
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Double-Replacement

Two ionic compounds are the reactants and 2 new compounds are the products (the metals switch places) 2Kl (aq) + Pb(NO3)2 (aq) –> PbI2 (s) + 2KNO3 (aq)
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Combustion

Reactants include a hydrocarbon ( a compound composed of hydrogen and carbon) and oxygen; products include carbon dioxide and water CH4 (g) + 2O2 (g) —–CO2 (g) + 2H2O (g)
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Calculate percent error

Error = experimental value – accepted value [(l error l) / ( accepted value )] x100
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Mass of element / compound =

1 mole
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22.4 L =

1 mole
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6.02 x 10^23 particles (atoms / molecules) =

1 mole
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1000 mLs =

1 L
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1 mL =

1 g
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Percent composition of a compound

( Parts / total ) x 100 = the percent
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Calculate empirical formula from percent composition of a compound

1. Percent composition of a compound = mass of compound 2. Take the masses and figure out the number of moles by dimensional analysis 3. Take the lesser of the 2 amounts and divide both of the numbers of moles by it 4. If numbers come out nice then that is the empirical formula. If they don’t, multiply them both by a number that will get them to come out nice.
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Calculate the molecular formula

( Given molar mass / empirical formula mass ) = a number Take that number and multiply the entire empirical formula by it
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Percent yield =

Actual yield (given) / theoretical yield
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Determine limiting reagent

Determine the moles of reactants and products; smaller one is answer because you can only make as much product as the smallest reactant amount
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As ionic solids and molecular solids dissolive in water they

Break up into their components (or dissociate)
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Factors that determine rate at which substance dissolves

Agitation, temperature, and particle size of the solute
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Conditions that determine the amount of solute that will dissolve a given solvent

Temperature affects the solubility of solid, liquid, and gaseous solutes; both temperature and pressure affect the solubility of gaseous solutes
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Freezing points are different from the pure solvent in that

The freezing point of a solution is lower than the freezing point of the pure solvent
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Boiling points are different from the pure solvent in that

The boiling point of a solution is higher than the boiling point of the pure solvent
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Percent by mass (%(m/m)) =

( Mass of solute / mass of solution ) x 100%
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Percent by volume (%(v/v)) =

( Volume of solute / volume of solution ) x 100%
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Molarity (M) =

Moles of solute / liters of solution
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Molality (m) =

Moles of solute / kilograms of solution
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Boiling point elevation

Triangle T = molality of particles x Kb
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Kb for water solutions =

+ .512 degrees Celsius per molal particles
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Freezing point depression

Triangle T = molality of particles x Kf
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Kf for water solutions =

– 1.86 degrees Celsius per molal particles
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Heat flows in this way

Spontaneously fom a warmer object to a cooler object ( H –> C )
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Endothermic processes

Heat is absorbed bby the system from the surroundings; heat flowing into a system from its surrounds is defined as positive; q has a positve value
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Exothermic processes

A process that releases heat from the system to its surroundings; heat flowing out of a system into its surroundings is defined as negative; q has a negative value
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Units heat flow is measured in

1. Joule (J) 2. Calorie (cal) 4.184 J = 1 cal
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Calculate heat involved in a process using calorimetry

(Triangle H) = (- qwater) = (Triangle Twater) x (masswater) x (Cwater of water)
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Cwater of water =

4.184 J/g degrees Celsius
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Write a thermochemical equation

Includes the balance equation of the reaction and the calculated experimental molar heat of combustion
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4 facotrs that influence the rate of a chemical reaction

Temperature, concentration, particle size, and the use of a catalyst; (also pressure on gaseous reactants)
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3 main parts of the collision theory

1. Reactions happen because of collisions between reactant particles 2. The more collisions the faster the rate 3. Raising the temperature increases the number of effective collisions and increasing the concentration increases the number of collisions
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The amount of reactants and products change in a chemical system at equilibrium in this way

The rate of the forward and reverse reactions is equal in an equilibrium system so the amounts of reactants and products become constant but not necessarily equal.
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Vapor-liquid equilibrium equation

H2O (g) H2O (l)
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3 stresses that can causes a change in the equilibrium position of a chemical system

Cooling, Heating
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LeChatelier’s Principle of stresses on equilibrium

If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress
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2 characteristics of spontaneous reactions

Exothermic, decreasing in energy content and they increase the entropy, randomness of the system
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2 factors that determine the spontaneity of a reaction

Entroy and enthalpy
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Arrhenius

Acid = H+ producer Base = OH- producer
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Bronsted-Lowry

Acid = Proton: H+ donor Base = Proton: H+ acceptor
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Lewis

Acid = electron-pair acceptor Base = electron-pair donor
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Acids and bases

Neutralize each other
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Kw =

[H+] x [OH-] = (1.0 x 10^-14)
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Acids on pH scale =

1 – 6
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Bases on pH scale =

8 – 14
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Write equations for neutralization and identify the salts formed

> Involves writing and balancing an equation > Then identifying the proton donor and proton acceptor: acid = always donor and base = always acceptor
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Calculate the unknown concentration of acid or base using titration lab results

(Molarity H+ acid) x (Liters of acid) = (Moles of OH- base) x (Liters of base)
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Boyle’s Law

(P1 / P2) = (V2 / V1) Pressure and volume are inversely proportional
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Charles’s Law

(T1 / T2) = (V1 / V2) Temperature and volume are directly proportional
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Gay-Lussac’s Law

(T1 / T2) = (P1 / P2) Temperature and pressure are directly proportional
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Ideal Gas Law

PV = nRT
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P =

Pressure in atms or kilopascals
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V =

Volume in Liters
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N =

Numbers of moles
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T =

Temperature in Kelvins
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Kelvins =

Degrees Celsius + 273
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R =