AP Chemistry Exam Review - SUPERTALLTEACHER

AP Chemistry Exam Review - SUPERTALLTEACHER

+Properties of a Gas - Factors Source Dont worry about individual gas law names, but do worry about the effect of changing moles, pressure and temperature on a sample of gas Virtua l Lab LO 2.5: Refine multiple representations of a sample of matter in the gas phase to accurately represent the effect of changes in macroscopic properties +The Ideal Gas Law Source Video Click reveals answer and explanation. LO 2.6: The student can apply mathematical relationships or estimation to

determine macroscopic variables for ideal gases +Chromatography Source Video LO 2.7: The student is able to explain how solutes can be separated by chromatography based on intermolecular interactions. + Dissolving/Dissociation: Solute and Solvent Source When drawing solute ions: Video 1.

pay attention to size (Na+ is smaller than Cl-) 2. Draw charges on ion, but not on water 3. draw at least 3 water molecules around each 4. the negative dipole (oxygen side) points toward cation and the positive dipoles (H side) points towards the anion LO 2.8: The student can draw and/or interpret representations of solutions that show the interactions between the solute and solvent. Molarity and Particle Views

+ Source Video QUESTION: Rank the six solutions above in order of increasing molarity. Pay attention to volume, and some have equal concentration C,D, and E (tied); Click A andreveals F (tied);answer most concentrated is B LO 2.9: The student is able to create or interpret representations that link the concept of molarity with particle views of solutions +Distillation to Separate Solutions Source Video

In the diagram above, ethanol has lower IMFs and a resulting lower boiling point than water, so it can be heated, vaporized and condensed easily. Ethanol hydrogen bonds as water does and is polar, but part of the ethanol has only weaker LDFs because its nonpolar resulting in a lower boiling point LO 2.10: Design/interpret the results of filtration, paper/column chromatography, or distillation in terms of the relative strength of +London Dispersion Forces and Noble and Nonpolar Gases This answer is VITAL! Remember with increased number of ELECTRONS a particle becomes more polarizable, not with increased mass! Source Video

Click reveals answer and explanation. LO 2.11: The student is able to explain the trends in properties/predict properties of samples consisting of particles with no permanent dipole on the +Deviations from Ideal Gas Behavior Source Video When watching the video, dont concern yourself with Van der Waals AP Exam focuses on LDFs instead Click reveals answer and explanation. LO 2.12: The student can qualitatively analyze data regarding real gases to identify deviations from ideal behavior and relate these to molecular

Hydrogen Bonding + Hydrogen bonding is seen in the following molecules: water, DNA, Source ammonia, HF, and alcohols. H-bonding is an attraction or force not a true intramolecular bond. Hydrogen bonds are like a sandwich with N, O, and/or F as the Video bread. H will be in a intramolecular (same molecule) bond with one N, O, and/or F and have an intermolecular attraction (different molecule) with the other. Remember this tip: hydrogen bonds just wanna have FON LO 2.13: The student is able to describe the relationships between the structural features of polar molecules and the forces of attraction between

+Coulombs Law and Solubility Source Ionic compounds can dissolve in polar liquids like water because the ions are attracted to either the positive or negative part of the molecule. There is a sort of tug-of-war involved with species dissolved in water. The water pulls individual ions away from the solid. The solid is pulling individual Video ions back out of the water. There exists an equilibrium based on how strongly the water attracts the ions, versus how strong the ionic solid attracts the ions. We can predict the degree of solubility in water for different ionic compounds using Coulomb's law. The smaller the ions, the closer together they are, and the harder it is for the water molecules to pull the ions away from each other. The greater the charge of the ions, the harder it is for the water to pull them away as well.

QUESTION: Predict which of the following pairs should be more soluble in water, based on Coulombic attraction. LiF or NaF NaF or KF BeO or LiF LO 2.14: Apply Coulombs law to describe the interactions of ions, & the attractions of ions/solvents to explain the factors that contribute to solubility + Entropy in Solutions Source Video Generally

speaking. There are exceptions Do NOT say like dissolve s like. Youll, like... get no points. DO refer to LDFs, hydrogen bonding and dipole-dipole interactions LO 2.15: Explain observations of the solubility of ionic solids/molecules in water and other solvents on the basis of particle views that include IMFs and + Physical Properties and IMFs Source Video

Hg Click reveals answer and explanation. LO 2.16: Explain the properties (phase, vapor pressure, viscosity, etc.) of small and large molecular compounds in terms of the strengths and types of +Bonding and Electronegativity Source Differences in electronegativities lead to different types of bonding*: 0.0 0.4: Bond is generally considered nonpolar 0.5 1.7: Bond is generally considered polar > 1.7: Bond is generally considered ionic Video Electronegativities are assigned values and are relative to fluorine. Electronegativity is a function of shielding / effective nuclear charge. *Values presented are one possibility other scales exist. LO 2.17: The student can predict the type of bonding present between two atoms in a binary compound based on position in the periodic table and the electronegativity

+Ranking Bond Polarity Source Video LO 2.18: The student is able to rank and justify the on the ranking of bond polarity on the basis of the locations of the bonded atoms in the periodic table. +Ionic Substances and their compounds are brittle. As the Properties Ionic crystal structure is struck, the ions become displaced. The displaced ions will repel like charges and fracture. Source Video LO 2.19: The student can create visual representations of ionic substances that connect the microscopic structure to macroscopic properties and/or use representations to connect microscopic structure to macroscopic properties (e.g., boiling point, solubility, hardness, brittleness, low volatility, lack of malleability, ductility, or conductivity). Metallic Properties Sea of

+ Electrons Source Video The metallic bond is not the easiest type of bond to understand, so an analogy may help. Imagine filling your bathtub with golf balls. Fill it right up to the top. The golf balls will arrange themselves in an orderly fashion as they fill the space in the tub. Do you see any spaces between the balls? If you turn on the faucet and plug the drain, the water will fill up those spaces. What you now have is something like metallic bonding. The golf balls are the metal kernals, and the water represents the valence electrons shared by all of the LO 2.20: The student is able to explain how a bonding model involving delocalized electrons is atoms.properties of metals (e.g., conductivity, malleability, ductility, and low consistent with macroscopic volatility) and the shell model of the atom. +Lewis Diagrams / VSEPR

Source Video LO 2.21: The student is able to use Lewis diagrams and VSEPR to predict the geometry of molecules, identify hybridization, and make predictions about polarity. +Ionic or Covalent? Bonding Tests As the type of particles and forced of attraction in ionic and covalent compounds differ, their properties also differ! Great Lab Examp le Properties Ionic Compounds Covalent Compounds

Melting/Boiling Points High Low except for some giant covalent molecules Electrical Conductivity Conduct electricity in molten and in aqueous solution Does not conduct electricity in any state when pure, may conduct in aqueous solution (i.e., acids) Solubility in water and organic solvents Soluble in water Insoluble in organic solvent Insoluble in water, except for some simple molecule

Soluble in organic solvent Volatility Not volatile Highly volatile Click here to do a virtual lab on bonding type (chart pictured below) Video Source Use properties of compounds to differentiate them from one another. Other tests may be performed to positively identify the compound, but are not necessary to observe types of bonds present.

LO 2.22: The student is able to design or evaluate a plan to collect and/or interpret data needed to deduce the type of bonding in a sample of a solid. +Crystal Structure of Ionic Compounds Source Video LO 2.23: The student can create a representation of an ionic solid that shows essential characteristics of the structure and interactions present in the substance. +Crystal Structure of Ionic Compounds Source Video The +2 and -2 ions attract each other more strongly than +1 attracts -1. The ions Mg+2 and O-2 are smaller than Na+1 and Cl-1, therefore the ions can get closer together, increasing their electrostatic attractions.

LO 2.24: The student is able to explain a representation that connects properties of an ionic solid to its structural attributes and to the interactions present at the atomic level. Alloys and their Properties + Source Video LO 2.25: The student is able to compare the properties of metal alloys with their constituent elements to determine if an alloy has formed, identify the type of alloy formed, and explain the differences in properties using particulate level reasoning. +Alloys! Source Video LO 2.26: Students can use the electron sea model of metallic bonding to predict or make claims about macroscopic properties of metals or alloys. +Metallic Solids - Characteristics

Source Video LO 2.27: The student can create a representation of a metallic solid that shows essential characteristics of the structure and interactions present in the substance. +Properties of Metallic Solids Source Video LO 2.28: The student is able to explain a representation that connects properties of a metallic solid to its structural attributes and to the interactions present at the atomic level. +Covalent Compounds Graphite are Interactions sheets of carbon atoms bonded together and stacked on top of one another. The interactions between sheets is weak, much

like the substance itself. Diamonds carbon atoms are more connected in a three dimensional structure, adding strength to the LO 2.29: The student can create a representation of a covalent solid that shows essential network. characteristics of the structure and interactions present in the substance. Source Video +Covalent Solids Source Video LO 2.30: The student is able to explain a representation that connects properties of a covalent solid to its structural attributes and to the interactions present at the atomic level.

Molecular Compounds + Interactions Water (H2O) Source Iodine (I2) Video Polar Covalent compounds align according to dipole-dipole interactions. Non-Polar Covalent compounds align according to LDFs as a solid. LO 2.31: The student can create a representation of a molecular solid that shows essential characteristics of the structure and interactions in the substance. + Molecular Compound Interactions a. Covalent bonds b. Hydrogen bonds c. Dipole-dipole interactions d. London Dispersion Forces Source

Video LO 2.32: The student is able to explain a representation that connects properties of a molecular solid to its structural attributes and to the interactions present at the atomic level. + Big Idea #3 Chemical Reactions + Changes in matter involve the rearrangement and/or reorganizations of atoms and/or the transfer of electrons. +Types of Chemical Reactions Synthesis A + B AB Source Decomposition

AB A + B Video Single Displacement A + BC AC + B Double Displacement AB + CD AD + CB Images from: Wilbraham, Antony C. Pearson Chemistry. Boston, MA: Pearson, 2012 Print. LO 3.1: Students can translate among macroscopic observations of change, chemical equations, and particle views. +Types of Chemical Reactions C. Pearson Chemistry. Boston, MA: Pearson, 2012. Print. Combustion CxHx + O2 CO2 + H2O Oxidation-Reduction A + + e- A B B + e-

Source Acid-Base (Neutralization) HA + BOH H2O + BA Video Precipitation AB (aq) + CD (aq) AD (aq) + CB (s) LO 3.1: Students can translate among macroscopic observations of change, chemical equations, and particle views. Source +Balanced Equations Complete Molecular: Complete Ionic : Net Ionic : AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq) Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) AgCl (s) + K+(aq) + NO3(aq) Video Quizlet

Ag+(aq) + Cl-(aq) AgCl (s) Spectator ions should not be included in your balanced equations. Remember, the point of a Net Ionic Reaction is to show only those ions that are involved in the reaction. Chemists are able to substitute reactants containing the same species to create the intended product. You only need to memorize that compounds with nitrate, ammonium, halides and alkali LO 3.2: The student can translate an observed chemical change into a metals are soluble. balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. +Making Predictions

Source Solid copper carbonate is heated strongly: CuCO3 (s) CuOClick (s) + reveals CO2 (g) answer and explanation. What evidence of a chemical change would be observed with this reaction? Video One would observe a color changeanswer and evolution of a gas Click reveals and explanation. What is the percent yield of CO2 if you had originally heated 10.0g CuCO 3 and captured 3.2g CO2 ? Step 1: Find the Theoretical Yeild Click reveals answer and

explanation. 10.0g CuCO3 x(1mol/123.555g) x (1mol CO2 /1 molCuCO 3 ) X 44.01gCO2/mol = 3.562 gCO2 Step 2: Find Percent Yield (3.2 g / 3.562 g) * 100 = 89.8 % 90% with correct sig figs How could you improve your percent yield? and explanation. -reheat the solid, Click to see reveals if there is answer any further mass loss -make sure you have pure CuCO3 LO 3.3: The student is able to use stoichiometric calculations to predict the results of performing a reaction in the laboratory and/or to analyze deviations from the expected results. +Limiting Reactants D.A. Source

Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S 15.00 g aluminum sulfide and 10.00 g water react a) Identify the Limiting Reactant 15.00g Al2S3 x (1mol/ 150.158 g) x (6mol H2O/1mol Al2S3) x (18g/mol H20 ) = 10.782 g H20 needed Click reveals answer and explanation. Video Sim pHet 10g H20 x (1mol/ 18.015 g) x (1 mol Al2S3 / 6mol H2O) x (150.158 g/mol) = 13.892g Al2S3 needed b) What the maximum mass ofmore H 2S than which formed from these reagents? H20 isislimiting, because we

need wecan werebe given Theoretical Yield 10.00 g H20 x (1mol/ 18.015 g) x (3/6) x (34.0809 g/mol ) = 9.459 g H2S produced Click reveals answer and explanation. c) How much excess reactant is left in the container? 15.00 g 13.892 g = 1.11g Al2S3 Click reveals answer and explanation. **Dimensional Analysis is not the only way to solve these problems. You can also use BCA tables (modified ICE charts), which may save time on the exam LO 3.4: The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion. Source +Limiting Reactants BCA Table 15.00 g aluminum sulfide and 10.00 g water react according to the following equation: Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S

a) Video Identify the Limiting Reactant 15.00g Al2S3 x (1mol/ 150.158 g) = .100mol 10g H20 x (1mol/ 18.015 g) = .555 Before Complete the table using the molar relationships Al2S3 .0999 6 H2O 2Al(OH)3 .5551 0 Click reveals Change answer and explanation.

-.0925 -.5551 + .1850 Water is the limiting reactant. After .0074 0 .1850 3 H2S 0 + .277 5 .2775 b) What is the maximum mass of H2S which can be formed from these reagents? 0.2775 mol H2S x (34.0809 g/mol ) = 9.459 g H2S produced Click reveals answer and explanation. c) How much excess reactant is left in the container? .0074mol Al2S3 x 150.158 g/mol = 1.11g Al2S3

Click reveals answer and explanation. LO 3.4: The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion. +Experimental Design Source Synthesis A sample of pure Cu is heated in excess pure oxygen. Design an experiment to determine quantitatively whether the product is CuO or Cu2O. Video Find the mass of the copper. Heat in oxygen to a constant new mass. Subtract to find the mass of oxygen that combined with the copper. Click reveals basic steps Compare the moles of oxygen atoms to the moles of original copper atoms to determine the formula.

Decomposition CaCO3(s) CaO(s) + CO2(g) Design a plan to prove experimentally that this reaction illustrates conservation of mass. Find the mass of calcium carbonate and seal it in a rigid container. Evacuate the container of remaining gas. Heat the container and take pressure readings (this will be the pressure exerted by the CO2). Using PV=nRT, calculate the moles of carbon dioxide Click reveals basic steps gas present in the container and compare it to the molar relationships afforded by the balanced chemical equation. LO3.5: The student is able to design a plan in order to collect data on the synthesis or decomposition of a compound to confirm the conservation of matter and the law of definite proportions. Source +Data Analysis When tin is treated with concentrated nitric acid, and the resulting mixture is strongly heated, the only remaining product is an oxide of tin. A student wishes to find out whether it is SnO or SnO2. Mass Mass Mass Mass Mass of pure tin 5.200 grams.

of dry crucible 18.650 g of crucible + oxide after first heating after second heating 25. 253 g after third heating 25. 252 g Video 25.500 g How can you use this data, and the law of conservation of mass, to determine the formula of the product? 1) Determine the number of moles of tin. 5.200/118.7 = 0.0438 moles. Sn 2) Subtract the mass of the crucible from the mass after the third heating. 25.252-18.650 = 6.602 g SnOx 3) Subtract the mass of tin from the mass of oxide to get the mass of oxygen. 1.402 grams of oxygen. Click reveals answer and explanation. 6.602-5.200 = 4) Calculate the moles of oxygen atoms, and divide by the moles of tin atoms to get the formula ratio. 1.402 g/16.00 g/mol of atoms = 0.0876 moles. 0.0876/0.0438 = 2.00 The formula must be SnO .

2 LO 3.6: The student is able to use data from synthesis or decomposition of a compound to confirm the conservation of matter and the law of definite +Bronsted-Lowery Acids & Bases Source According to Bronsted-Lowery (B.L.) an acid is a "proton donor" and a base is a "proton acceptor. The proton here is shown as a hydrogen. Video The acids conjugate base is the anion. The bases conjugate acid now has the proton (hydrogen ion). Quizlet Amphoteric nature of water Water acts as both an acid & a base.

H2O H+ + OH2H2O H3O+ + OH- LO 3.7: The student is able to identify compounds as Bronsted-Lowry acids, bases and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification. Redox Reactions + When an electron is transferred, it is called a redox reaction. When something is reduced, the RED part of redox, it gains electrons. You may have a difficult time with this definition because when something is reduced, it usually means that it is losing something. In this case, it is a reduction in charge. Remember, electrons are negatively charged so if something is being reduced, it's getting more negatively charged by receiving more electrons. The other reaction that is coupled with this is called oxidation--the "OX" part of redox. Whenever something is reduced, the electron it gains has to come from somewhere. The oxidation is the loss of an electron, so if an atom is oxidized it loses its electron to another atom. And these are always coupled reactions. If one molecule is oxidized, another molecule must be reduced and vice versa: the electron must go somewhere. OILRIG

LO 3.8: The student is able to identify redox reactions and justify the identification in terms of electron transfer Source Video +Redox Titrations Source A redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. A common example is the redox titration of a standardized solution of potassium permanganate (KMnO4) against an analyte containing an unknown concentration of iron (II) ions (Fe2+). The balanced reaction in acidic solution is as follows: MnO4- + 5Fe2+ + 8H+ 5Fe3+ + Mn2+ + 4H2O In this case, the use of KMnO4 as a titrant is particularly useful, because it can act as its own indicator; this is due to the fact that the KMnO4 solution is bright purple, while the Fe2+ solution

is colorless. It is therefore possible to see when the titration has reached its endpoint, because LO 3.9: The student is able to design and/or the solution will remain slightly purple from the interpret the results of an experiment a redox titration unreactedinvolving KMnO Video Evidence of Chemical Change + Source Note: it is a common misconception that boiling water makes O2

Video and H22 gas. Notice that Video the water molecule stays Video intact as the water boils. Chemical Changes: Physical Changes: Production of a gas:Covalent bonds are not may produce similar visible evidence (i.e. this physical 2KClO3 (s) + heat broken 2KCl (s) +with 3O2 (g) boiling water creates bubbles, but bonds are not broken and reformed. No new change- only substances are made. Formation of a precipitate: intermolecular AgNO3) (aq) + KCl (aq) attractions AgCl (s)+ 2KNO 3 (aq) (hydrogen bonds) between water Change in color: molecules.

Two white solids react to produce a mixture of a yellow and a white solid when shaken forcefully! Pb(NO3)2 (s) + 2KI (s) PbI2 (s)+ 2KNO3 (s) Production of heat*: 2 Mg (s) + O2 (s) 2MgO (s) + heat *can also Evaluate include thethe absorption of heat of a process as a physical, chemical, or LO 3.10: classification ambiguous change based on both macroscopic observations and the distinction between rearrangement of covalent interactions and noncovalent +Energy Changes Chemical reactions involve the formation of new products Bonds between atoms or ions in the reactants must be BROKEN (the enthalpy of the system is increasing ENDOTHERMIC process) Bonds are then FORMED between atoms or ions to make the producsts of the reaction. (the enthalpy of hte system is decreasing...EXOTHERMIC process)

LO 3.11: Source Video Video The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process to generate a relevant symbolic and/or graphical representation of the energy changes. Source + Galvani c Cell Potentia l Video Video Click reveals answer and explanation. LO 3.12: Make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/or

+Redox Reactions and Half Cells Source Video Video LO 3.13: The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions + Big Idea #4 Kinetics +Factors Affecting Reaction Rate Factors that Affect Reaction Rate Collision theory states that reactants must collide in the correct orientation and with enough energy for the molecules to react; changing the number of collisions will affect

the reaction rate Rate is the change in concentration over time [A] / tA] / t Source State of reactants Rate increases as state changes from Video solid gas as increased molecular movement allows for more opportunity for collision Greater surface area of solids will increase rate as more reactant is exposed and able participate in collisions Temperature - more kinetic energy leads to more successful collisions between

molecules Concentration more reactants more collisions Use of a catalyst affect the mechanism of reaction leading to faster rate LO 4.1: The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. Source +Determining Rate Order Rate law for a reaction has the form: rate = k [A]m[B]n (only reactants are part of the rate law) Exponents (m, n, etc. ) are determined from examining data, not coefficients: forWhen

A + [A] B isC When [A] is doubled, the rate do not change, so the reaction is zero order with respect to A Trial Initial [A] (mol/L) Initial [B] (mol/L) Initial Rate (mol/

(Ls) 1 0.100 0.100 0.002 2 0.200 0.100 0.002 3 0.200 0.200 0.004 When [B] is

doubled, the rate doubles, so the reaction is first order with respect to B Video The overall rate expression for the reaction is rate = k [B] k is the rate constant and is determined Plot to create a straightexperimentally line graph: by plugging in data into the rate expression Zeroth Order [A] / Time First Order ln[A] / Time Second Order The first and second 1/[A] / time

order integrated rate laws can be found on the Kinetics section of the AP Equations Sheet LO 4.2: The student is able to analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or second-order reaction. Source +Half-life (First Order) Time needed for the concentration of reactant to reach half its initial value The first order half life equation is derived from the first order integrated rate

law Time to reach half concentration is dependent on k, not initial concentration Half life remains constant in a first order reaction Video Example: when t1/2= 30 sec, the concentration is halved each 30 seconds Initial Conditions seconds (12 molecules) After 30 seconds (6 molecules) After 60 (3 molecules) LO 4.3: The student is able to connect the half-life of a reaction to the rate constant of a firstorder reaction and justify the use of this relation in terms of the reaction being a first-order

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