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AP Chemistry review

AP Test review project
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Amy Cantrell

on 8 May 2013

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Transcript of AP Chemistry review

Amy Cantrell AP TEST REVIEW Atomic Structure & Periodicity
1997 D
Explain each of the following observations using principles of atomic structure and/or bonding.
(a) Potassium has a lower first-ionization energy than lithium.
(b) The ionic radius of N3- is larger than that of O2-.
(c) A calcium atom is larger than a zinc atom.
(d) Boron has a lower first-ionization energy than beryllium. Bonds Chemical Reactions & Stoichiometry Rutherford's Gold Foil Experiment The unexpected results of the experiment demonstrated for the first time the existence of the atomic nucleus, leading to the downfall of the plum pudding model of the atom, and the development of the Rutherford (or planetary) model. Top: Expected results: alpha particles passing through the plum pudding model of the atom undisturbed.
Bottom: Observed results: a small portion of the particles were deflected, indicating a small, concentrated positive charge. AP Problems 1984 D
Discuss some differences in physical and chemical properties of metals and nonmetals. What characteristic of the electronic configuration of atoms distinguishes metals from nonmetals. On the basis of this characteristic explain why there are many more metals than nonmetals. O2- has one more proton than N3- does so it has a stronger attraction for electrons and it causes the radius to be smaller than N3-. Zinc has a larger nuclear charge. Boron has an electron tin the 2p orbital
field, so that one is less attracted to the
nucleus than Beryllium's electrons.
The outermost electrons of a potassium atom are farther from its nucleus than those of lithium. So, removing an electron from a potassium atom is easier than removing one from a lithium atom. Metals' valence electrons in s or d orbital of their atoms. Nonmetals have valence electrons in the s and p orbitals.
There are more metals than nonmetals because filling d orbitals in that level involves the atoms of ten elements and filling the p orbitals involves only 6. Answers Periodic Trends
Protons are positively charged and so would be deflected on a curving path towards the negative plate.

Electrons are negatively charged and so would be deflected on a curving path towards the positive plate.

Neutrons don't have a charge, and so would continue on in a straight line. When passed between negative and positive plates: "... the atoms of the elements consist of a number of negatively electrified corpuscles enclosed in a sphere of uniform positive electrification...[" -J.J. Thompson Plum Pudding Model Cathode Ray Tube The cathode ray tube (CRT) is a vacuum tube containing an electron gun and a fluorescent screen used to view images. It has a means to accelerate and deflect the electron beams onto the fluorescent screen to create the images. The image may represent electrical waveforms, pictures (television, computer monitor), radar targets and others. #1 \ Schrodinger's Model Electronegativity 1 mol #2 Positively charged ions are called cations.
Negatively charged ions are called anions.
The cation is always named first. Naming Ionic Compounds Covalent molecules share electron pairs in the bonds that attach atoms to one another. Ionic molecules' electrons change places. At least one electron moves from one of the atoms to the other atom, leaving each ion with a positive or negative charge. Naming Ions, Molecules, and Acids Bases always have OH-1 as the anion with some metal as the cation. Acids & Bases Naming Acids always have hydrogen as the cation and some other anion. Covalent vs. Ionic Bonding Metallic Bonding Lewis Dot Structures Formal Charge VSEPR Theory Electron geometry depends on the electronic structure of the central atom in a molecule, while molecular geometry depends on whether there are other atoms bonded to the central atom or free electron pairs. Molecular Geometry vs. Molecular Shape Solubility Rules Common Polyatomic Ions 1973 D
Discuss briefly the relationship between the dipole moment of a molecule and the polar character of the bonds within it. With this as the basis, account for the difference between the dipole moments of CH2F2 and CF4. In CH2F2, the bonds are polar, and the molecule is non-symmetrical, so the molecule would have a dipole moment.
In CF4, the bonds are polar, but the molecule is symmetrical, so it would not have a dipole moment. 1988 D
Using principles of chemical bonding and/or intermolecular forces, explain each of the following.
(a) Xenon has a higher boiling point than neon has.
(b) Solid copper is an excellent conductor of electricity, but solid copper chloride is not.
(c) SiO2 melts at a very high temperature, while CO2 is a gas at room temperature, even though Si and C are in the same chemical family.
(d) Molecules of NF3 are polar, but those of BF3 are not. a.) Xenon, and Neon are both held together by London Disperson Forces; however, Xenon has more electrons, thus requiring more LDFs to bond it together.
b.) Because Copper metal has open valence shells, it is an excellent conductor of electricity, and Cl is an ionic solid, so it does not conduct electricity.
c.) SiO2 has strong covalent bonds, and is polar. CO2 is nonpolar, and only has LDF bonds. SiO2's bonds are much harder to break, requiring a greater temperature.
d.) Nf's molecular structure is polar because of a lone pair. Bf3's is not, becasue it has no lone pair on its molecular structure. AP Problems: Hydrogen Bonding London Dispersion Forces #3 Electrolytes An electrolyte is a compound that ionises when dissolved in suitable ionising solvents such as water. This includes most soluble salts, acids, and bases. NaCl(s) --> Na+(aq) + Cl−(aq) Example: Acids/Bases Weak electrolytes do not conduct electricity well. Strong electrolytes conduct electricity well. HCl + H2O--> H+(aq) + Cl-(aq) Acidic reaction: Basic reaction: NaOH + H2O--> Na+(aq) + OH-(aq) Acid Base Water Salt
HCl + NaOH --> H2O + NaCl
HBr + KOH --> H2O + KBr Neutralization Neutralization : acid + base= water + salt Empirical & Molecular Formula Empirical Formula: lowest possible whole number ratio
Molecular Formula: actual ratio present Percent Composition Solution:

Step 1 - Find the atomic weights

Step 2 - Find mole ratio between product and reactant

For this reaction, two moles of AgNO3 is needed to produce one mole of Ag2S.

The mole ratio then is 1 mol Ag2S
2 mol AgNO3

Step 3- Find the amount of product produced.

The excess of Na2S means all of the 3.94 g of AgNO3 will be used to complete the reaction.

Note the units cancel out, leaving only grams Ag2S

Answer:

2.87 g of Ag2S will be produced from 3.94 g of AgNO3. Problem:

Given the reaction

Na2S(aq) + 2 AgNO3(aq) Ag2S(s) + 2 NaNO3(aq)

How many grams of Ag2S will form when 3.94 g of AgNO3 and an excess of Na2S are reacted together? From the periodic table:
Atomic weight of Ag = 107.87 g
Atomic weight of N = 14 g
Atomic weight of O = 16 g
Atomic weight of S = 32.01 g

Atomic weight of AgNO3 = (107.87 g) + (14.01 g) + 3(16.00 g)
Atomic weight of AgNO3 = 107.87 g + 14.01 g + 48.00 g
Atomic weight of AgNO3 = 169.88 g

Atomic weight of Ag2S = 2(107.87 g) + 32.01 g
Atomic weight of Ag2S = 215.74 g + 32.01 g
Atomic weight of Ag2S = 247.75 g Theoretical Yield Percent Yield Stoichiometry 2003 D Required
For each of the following, use appropriate chemical principles to explain the observations. Include chemical equations as appropriate.
(a) In areas affected by acid rain, statues and structures made of limestone (calcium carbonate) often show signs of considerable deterioration.


(b) When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed that
(i) both solutions have higher boiling points than pure water, and

(ii) the boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq).

(c) Methane gas does not behave as an ideal gas at low temperatures and high pressures.


(d) Water droplets form on the outside of a beaker containing an ice bath. AP Problems The nonvolatile solute lowers the vapor pressure, so it also elevates the boiling point. NaCl dissociates into 2 moles, whereas C12H22O11 does not dissasoscate because it is a covalently bonded. At these conditions, kinetic enery is dropped, At low temperatures, the molecules are slower, and closer together. At high pressure, the molecules get closer together, and this leaves less room for the methane atoms. The water vapor outside, in the air finds the cool glass, and the vapor molecules attach to each other, and collect. 1971
Molarity and molality are two ways of expressing concentration.
(a) Clearly distinguish between them


(b) Indicate an experimental situation where expressing concentrations as molarity is particularly appropriate.

(c) Indicate an experimental situation where expressing concentration as molality is particularly appropriate. Molarity is moles per liters solute of solution, and it shows the concentration.
Molality is the solution's concentration written as moles of solute per kilogram of solvent. Acid-base titrations Calculating the change in freezing point depression. CaCO3 + H2o + HCl --> H+ + CL- + Ca+ + CO3-
Calcium Carbonate deteriorates into Calcium and Carbonate, so strongly effecting the decomposition of the statue. #4 Equilibrium Solution Dilution Beer's Law Absorbance Concentration Keq Q vs. Keq Q: ratio of concentration not at equilibrium
Keq: concentration at equilibrium Q > K: Reaction shifts left towards the products
K > Q: Reaction shifts right towards the reactants Ksp When you stress the reaction, the reaction will move to the side that relieves the stress. Le Chatelier's Principle Solubility Temperature
Concentration
Pressure Kp Equilibrium pressure Ka (acid) Ka= [H3O+][A-]
[HA] ____________ HA+ H2O <-> H3O+ + A- Acid Conjugate base Kb (base) Kb= [BH+][OH-]
[B] ____________ B+ H2O <-> OH- + BH+ Base Conjugate Acid Strong Acids:
HCl
HI
HBr
H2SO4
HNO3
HCLO4 NO EQUILIBRIUM!! Kc Concentration KaKb=Kw 1.00 x 10-14 Kw #5 Acids & Bases Classifications of Acids/Bases Equivalence point: [H3O+] = [OH-] Prefix p: log [thing] Kw: 1.00 x 10-14 KaKb=Kw pH vs. pOH ICE Table Example #6 Thermodynamics 1st Law of Thermodynamics Energy of the universe is constant 2nd Law of Thermodynamics Heat can never flow from a cold body to a warmer one 3rd Law of Thermodynamics Entropy of the universe is always increasing ΔGibb's Free Energy G= reaction's ability to do work -Delta G: spontaneous
+Delta G: non-spontaneous Entropy (S) "Disorder," "dispersion of energy and matter" -S = less disorder
+S = more disorder #7 Electrochemistry REDOX Reaction LEO the lion says GER Losing electrons:
oxidation Gaining electrons:
reduction Oxidation agent: reduction
Reduction agent: oxidation e- from cathode to anode Salt Bridge deposits electrons Galvantic Cell Standard Reduction Potential for Hydrogen is 0 Line Notation -ANE -ENE -YNE Meth
Eth
Prop
But
Pent
Hex
Oct
Non
Dec 1
2
3
4
5
6
7
8
9 3 of Carbons: Cyclo- Electrodes vs. Electrolytes
Electrodes are the charged plates that transmit the electrons or electricity to the electrolyte.

An electrolyte is the solution in which the electricity is passed through the electrodes. Cell potential E = E - E anode cathode cell + value: reaction occurs spontaneously 1) LEO goes GER
(Losing Electrons Oxidation, Gaining Electrons Reduction)

2) Step 1: Write the full equations
The two half cells you have given are:
1) Ni/Ni2+
so: Ni --> Ni2+ + 2electrons (e-)
2) Cu/Cu2+
so: Cu --> Cu2+ + 2e-
Step 2: Find standard REDUCTION potentials in a table

Ni2+ + 2e- --> Ni Eo = -0.257V
Cu2+ + 2e- --> Cu Eo = 0.340V

Step 3: Calculate the standard cell potential

The formula for calculating this is:
E cell = E cathode - E anode

Using the shortforms I mentioned earlier:
Cathode= reduction
Anode = oxidation

So we have:
Anode: Ni --> Ni2+ + 2e- Eo = -0.257V
Cathode: Cu2+ + 2e- --> Cu Eo= 0.340V

Ecell = Ecath - Eanode
= 0.340V - (-0.257V)
= 0.597V Cell potential example: Cell potential Free energy Nernst Equation Electrolytic Cell #8 Kinetics First order Second Order Zero Order Arrhenius Equation Graphing the Arrhenius Equation e- #9 Nuclear & Organic Alpha Decay 238
92 U 4
2 He + 234
90 Th Alpha particle Beta Decay protons decrease by 2 neutrons decrease by 2 mass decreases by 4 alpha particle He Proton changes into a neutron E P P + N P +1 since protons gain one, but neutrons lose one, the mass
number stays the same P + N 1 E 2 Gamma Decay
1970
H3PO2,H3PO3, and H3PO4 are monoprotic, diprotic and triprotic acids, respectively, and they are about equal strong acids.
HClO2, HClO3, and HClO4 are all monoprotic acids, but HClO2 is a weaker acid than HClO3 which is weaker than HClO4. Account for:
(a)The fact that the molecules of the three phosphorus acids can provide different numbers of protons.
(b)The fact that the three chlorine acids differ in strengths. a.) The hydrogen atom is bonded directly to the phosphorus atom, and is not acidic in aqueous solution because only those hydrogen atoms bonded to the oxygen atoms can be released as protons. (b) The acid strength is greater because the number of oxygen atoms increases as the really electronegative oxygen atoms are able to draw electrons away from the chlorine atom and the O-H bond. This results in the number of attached oxygen atoms increases. This means that a proton is most easily produced by the molecule with the largest number of attached oxygen atoms. 1972
Given a solution of ammonium chloride. What additional reagent or reagents are needed to prepare a buffer from the ammonium chloride solution?
Explain how this buffer solution resists a change in pH when:
(a) Moderate amounts of strong acid are added.
(b) Moderate amounts of strong base are added.
(c) A portion of the buffer solution is diluted with an equal volume of water. a.) When small amounts of a strong acid, H+ are added, the ammonia reacts with it. The concentration of the hydrogen ion remains essentially the same and so results in only a very small change in pH. b.) When moderate amounts of a strong base, OH- is added, the ammonium ion reacts with it. The concentration of the hydrogen ion remains essentially the same and so only a very small change in pH. c.) By diluting with water the concentration ratio of [NH4+]/[NH3] does not change, so there should be no change in pH. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q = mc∆T.
Assume the following.
• Both solutions are at the same temperature before they are combined.
• The densities of all the solutions are the same as that of water.
• Any heat lost to the calorimeter or to the air is neg¬ligible.
• The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mc∆T.
q has units of joules, m in grams, C in J/g°C or J/gK, T in °C or K
(b) List the measurements that must be made in order to obtain the value of q.
volume or mass of the HCl/NaOH solutions
initial temperature of HCl/ NaOH before mixing
highest temperature of solution after mixing
(c) Explain how to calculate each of the following.
(i) The number of moles of water formed during the experiment
since mixing equal volumes of same concentration and reaction has 1:1 stoichiometry, moles HCl =
moles NaOH = moles H2O.
(ii) The value of the molar enthalpy of neutraliza¬tion, ∆Hneut, for the reaction between HCl(aq) and NaOH(aq)
Determine the quantity of the heat produced, q, from q = mcT, where m = total mass of solution; divide q by mol H2O determined in part ci.) to determine Hneut 1970
Account for the observation that silver dissolves in 1 molar hydroiodic acid despite the fact that the standard electrode potential for the change, Ag  Ag+ + e- is –0.80 volt. The standard reduction potential is based on 1.0 M [Ag+], but Ag is insoluable, the the equilibrium is shifted in favor of the products. 1987 D

A dilute solution of sodium sulfate, Na2SO4, was elec-trolyzed using inert platinum electrodes. In a separate ex-periment, a concentrated solution of sodium chloride, NaCl, was electrolyzed also using inert platinum elec-trodes. In each experiment, gas formation was observed at both electrodes.
(a) Explain why metallic sodium is not formed in either experiment.

(b) Write balanced equations for the half–reactions that occur at the electrodes during electrolysis of the dilute sodium sulfate solution. Clearly indicate which half–reaction occurs at each electrode.

(c) Write balanced equations for the half–reactions that occur at the electrodes during electrolysis of the concentrated sodium chloride solution. Clearly in-dicate which half–reaction occurs at each electrode. (a) Na+ is not reduced as easily as H2O (or H+ or OH–) (b) Anode: 2 H2O + O2 + 4 H+ + 4e- cathode: 2 H2O + 2e- ® H2 + 2 OH- (c) anode: 2 Cl- ® Cl2 + 2e- cathode: 2 H2O + 2e- ® H2 + 2 OH- 1971
Ethyl iodide reacts with a solution of sodium hydroxide to give ethyl alcohol according to the equation.
CH3CH2I + OH-  CH3CH2OH + I-
The reaction is first order with respect to both ethyl iodide and hydroxide ion, and the overall-rate expression for the reaction is as follows: rate = k[CH3CH2I][OH-]
What would you do in the laboratory to obtain data to confirm the order in the rate expression for either of the reactants. Measure the pH of [OH] over time, and plot it on a graph. Plot ln[OH] vs. time. If there is a straight line, then there is a first order reaction. 1986 D
The overall order of a reaction may not be predictable from the stoichiometry of the reaction.
(a) Explain how this statement can be true.
(b) 2 XY  X2 + Y2
1. For the hypothetical reaction above, give a rate law that shows that the reaction is first order in the reactant XY.

2. Give the units for the specific rate constant for this rate law. Order of reaction is determined by the exponents in the rate law. Rate= k[XY] 1991 D
Explain each of the following in terms of nuclear models.
(a) The mass of an atom of 4He is less than the sum of the masses of 2 protons, 2 neutrons, and 2 electrons.


(b) Alpha radiation penetrates a much shorter distance into a piece of material than does beta radiation of the same energy.


(c) Products from a nuclear fission of a uranium atom such as 90Sr and 137Ce are highly radioactive and decay by emission of beta particles.


(d) Nuclear fusion requires large amounts of energy and to get started, whereas nuclear fission can occur spon-taneously, although both processes release energy. When nucleons are combined in nuclei, some of their mass is converted into energy which is released and stabilizes the nucleus. Alpha particles have a greater mass than beta parti-cles. So, their speed is less. The neutron/proton ratio in Sr-90 and Cs-137 is too large, so they emit beta particles to lower this ratio. Large amounts of energy are needed to start fusion reactions in order to overcome the repulsive forces between the positively charged nuclei. Large amounts of energy are not required to cause large unstable nuclei to break apart. 1989 D

The carbon isotope of mass 12 is stable. The carbon isotopes of mass 11 and mass 14 are unstable. However, the type of radioactivity decay is different for these two isotopes. Carbon-12 is not produced in either case.
(a) Identify a type of decay expected for carbon-11.

(b) Identify the type of decay expected for carbon-14.

(c) Gamma rays are observed during the radioactive de-cay of carbon-11. Why is it unnecessary to include the gamma rays in the radioactive decay equation of (a)?


(d) Explain how the amount of carbon-14 in a piece of wood can be used to determine when the tree died. (a) Positron decay (b) Beta decay Gamma rays have no mass or charge so they don't need to be shown in nuclear equations. Measure the amount of C-14 in the dead wood. Compare with the amount of C-14 in a similar living object.
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