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AU / Chemistry / CHEM 1030 / Can a measurement be both precise and accurate explain?

Can a measurement be both precise and accurate explain?

Can a measurement be both precise and accurate explain?


School: Auburn University
Department: Chemistry
Course: Fundamentals Chemistry I
Professor: John gorden
Term: Fall 2015
Cost: 50
Name: CHEM1030 Final Exam StudyGuide
Description: This is a cumulative study guide for the final exam on Dec. 7.
Uploaded: 12/04/2016
22 Pages 89 Views 4 Unlocks

CHEM 1030  

Can a measurement be both precise and accurate explain?

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

Chapter 1

∙ Expressing Scientific Quantities

o Measurements can be accurate and or precise.

 Accurate: multiple measurements are close to the actual value.  Precise: multiple measurements are close in proximity to each  other.

o Significant Figures are used to express the precision of a measuring  instrument

∙ Scientific Measurement

o SI System is the universal system of measurement. If you want to learn more check out In 2011 a country had a real gdp of $13.89 trillion and gdp deflator of 110. in 2012 it had a nominal gdp of $17.8 trillion and real gdp of 14.24 trillion. what is the rate of inflation in 2012?

o This first test will only deal with SI units for length (m), mass (kg)  and time (s).

When a measurement is close to the true value?

o Derived Units: combinations of base units.

 Volume is a combination of meter base units cubed (m3)

∙ 1 mL = cm3 We also discuss several other topics like What are the nutritional requirements for different age groups?

∙ 1 L = dm3 We also discuss several other topics like What is the social science concerned with how individuals?

 Density is a combination of (kg/m3)

o Dimensional Analysis: in problems involving conversion factors,  dimensions must be tracked throughout the calculation. If you want to learn more check out Externalities can be either what?

 Dimensional Analysis can NOT tell you if an equation is correct  though, so don’t use it for this purpose.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

CHEM 1030  

Exam 1 Study Guide

How close a group of measurements are to each other?

Rik Blumenthal Written By: Caroline Smith









































Don't forget about the age old question of When did jesus start his public ministry?

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

∙ Classification of Matter

o Matter is a substance or mixture of substances.

 Substances have a definite composition.

∙ Ex. Sodium (Na) will always have the properties of  


o 3 different states of matter:

 Solid: fixed neighbor distances between molecules and fixed  geometric shape (ice).

 Liquid: fixed neighbor distances between molecules but NO  fixed geometric shape (water).

 Gas: no fixed neighbor distances between molecules but fixed  geometric shape

o Mixture: combination of substances NOT chemically bonded.  Homogeneous mixture: concentration of solute in solvent (like salt in water) is consistent throughout the mixture.

 Heterogeneous mixture: concentration of solute in solvent is  not consistent throughout the mixture (like rocks in mud).

∙ Properties of Matter:

o Physical Properties: can be observed or measured WITHOUT  changing the chemical composition (color, density). We also discuss several other topics like What is meant by the sensitivity of a dependent measure? what are ceiling and floor effects?

o Chemical Properties: a chemical change must occur to observe this  change.

o Extensive Properties: depend on the amount of matter there is. o Intensive Properties: independent of the amount of matter there is. ∙ Changes In Matter

o Physical Changes: occur without changing the chemical composition  of a substance (freezing, melting).

o Chemical Changes: change the chemical composition of a substance  (burning, rusting).

Chapter 2

∙ Discovery of Electrons

o J. J. Thomson used cathode ray tube, a contraption with a  

negatively charged plate and a positively charged plate  

enclosed in a glass tube and connected to voltage source. The  

negatively charged plate emitted cathode ray, a stream of negatively  charged particles, and the ray was deflected by the negative electric  field and attracted to the positive electric field.

 CONCLUSION: atoms must contain negatively charged  


 Charge-to-Mass Ration = e- = 1.76 x 108 C/g

o Robert Millikan made contraption that suspended oil droplets in an  electric field (in an enclosed area). He used the magnitude of the  electric field necessary to keep a drop suspended in air to calculate the charge of an electron

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith  CONCLUSION: Charge of e- = -1.6022 x 10-19 C  

 Mass of e- = 9.10 x 10-28 g

∙ Discovering Atomic Structure

o J. J. Thomson came up with the “Plum Pudding Model” of the atom  (after his experiment with the cathode ray tube), which suggested an  atom was made of positively charged particles with electrons  

embedded throughout.

o Ernest Rutherford performed the gold foil experiment by aiming a  stream of positively charged alpha particles at a thin sheet of gold foil.  Most of the particles went through the sheet, but a very small  

percentage of particles were completely deflected. He proposed that  most of the atom is “empty space” (filled with electrons) and that a  small mass of positively charged particles called protons are in the  center of the structure (the small cluster of protons and neutrons is  called the nucleus).

∙ Atomic Number and Mass Number

o Atomic number: number of protons in an atom (also the number of  electrons in an atom)

o Mass Number: number of protons + number of neutrons


∙ Isotopes: atoms of the same element with different atomic masses. o Isotopes were discovered when two different nuclei of the same  element were deflected at different amounts/angles by an electric  field.

∙ Average Atomic Mass: measures in atomic mass units (amu) o Formula for calculating average atomic mass:  

 Avg. Atomic Mass = F1M1+ F2M2

∙ F = fraction representing natural abundance of isotope  

(usually a percentage)

∙ M = mass number of isotope

∙ The Mole and Molar Mass

o Mole: unit that tells how many atoms are in a sample of something   1 mole = 6.022 x 1023 atoms

o Molar Mass tells the mass in grams in 1 mole of substance

 1 amu = g/mol

Chapter 3

∙ Forms of Energy

o Kinetic Energy (KE): energy caused by motion

 KE = (1/2)mv2 

o Potential Energy (PE): energy of position of object

 Electrostatic Potential Energy: energy resulting from  

interactions between charged particles

1. E = (q1 q2)/r

o q = charge of particle

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith o r = distance between two particles

o Kinetic and potential energy can be converted into one another  (pushing a rock up a hill)

∙ Light Energy

o Light has properties of waves and particles

 Wave properties

1. Wavelength (lambda): distance from peak-to-peak

2. Frequency (nu): number of waves per second

3. Amplitude (meters): distance between midline and top  

of peak (or bottom of wave)  

 Speed of light (c) = 3.0 x 108 m/s

1. Wavelength, frequency and speed are related by this  


o C = wavelength (lambda) x frequency (nu)  

 Light energy adds like waves

1. Coherent interference is when waves add when aligned


2. Incoherent interference is when waves add aligned  


o A completely incoherent wave adds to zero

o In case Blumenthal decides to put this as a bonus  

or something, remember that noise-cancelling  

headphones work by using incoherent interference

o Double Slit Experiment shows the wave properties of light  A light source was passed through a slit. The bright light that  was generated from this was then passed through two closely  

placed slits. The result was a series of alternating light and dark  lines called an interference pattern.

∙ Quantum Theory proposes that energy travels in small quantities (this is  called quantization)

o Max Planck suggested this theory that energy can only travel in  discrete quantities called a quantum.

 Planck’s Constant (h) = 6.63 x 10-34 J-s

 The formula for a quantum of energy (E)  

1. E = h(nu)

o Albert Einstein used Planck’s theory to explain the phototoelectric  effect which is the emission of electrons when light hits a metal  surface

 The particle-like properties of light could NOT be explained by  the wave theory of light. So Einstein suggested that light also  

behaves as a particle.

1. Ephoton = h(nu)

o CONCLUSION of quantum theory: Light behaves as a wave and a  particle

∙ Hydrogen Spectra

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith o Radiating hydrogen gas produced a distinct pattern of light called an  atomic line spectra.

o Johannes Rydberg created an equation that could calculate all  wavelengths of radiation that hydrogen was emitting

 1/(lambda) = Rinfinity symbol (1/n12 – 1/n22)

1. n = an integer number

2. Rinfinity (Rydberg’s Constant) = 1.097 x 107 m-1 

o Neils Bohr suggested electrons can only occupy certain orbitals of  specific energies to explain the hydrogen spectra. He also suggested  that electrons can move from higher orbitals (excited state) the  lowest orbital of an atom (ground state) to cause energy emission.  

o De Broglie reasoned if waves can act like particles, then particles can  also behave like waves

 Neither of these theories explained the spectra for atoms with  more than one electron (in other words, any element other than  hydrogen), so they are collectively called “Old Quantum  


o Schrodinger’s equation explained what Bohr’s and Broglie’s theories  couldn’t explain

 Schrodinger’s equation used three quantum numbers instead of  one.

1. n = principle quantum number

o tells SIZE of an orbital

o has integer values (1,2,3…)

2. L = angular momentum quantum number

o Describes SHAPE of orbital and amount of total  


o Range from 0 to n-1

3. mL = magnetic quantum number



o mL = 2L + 1

o from the negative of L to the positive of L

o Atomic Orbitals  

Use the different quantum numbers to describe shape of orbitals  s-orbitals

o n = 1, so L = 0 (no nodes)

Spherical shape is possible (no nodes)

 p-orbitals

o n = 2, so L = 0, 1 (one node)

Spherical shape AND two-lobed shape (like figure ‘8’) are  


o P-orbitals have 3 orientations: Px Py Pz

 d-orbitals

o n = 3, so L = 0, 1, 2 (two nodes)

Spherical shape AND two-lobed shape AND d-orbital  

shape are possible

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

∙ Electron Configurations

Show how electrons are distributed among different orbitals.

o Rules for writing electron configurations:

 The Pauli Exclusion Principle: no two electrons can have the  same magnetic spin (ms) number. Notate this by drawing one  

arrow up and one arrow down.

 The Aufbau Principle: You must make configurations based on increasing orbital energies. So fill in order of 1s, 2s, 2p, 3s, 3p,  

4s, 3d, 4p, etc.

 Hund’s Rule: Fill empty orbitals with SINGLE electrons before  pairing electrons.

Paired electrons are called diamagnetic, and single electrons  

are called paramagnetic.

Chapter 4

∙ The Periodic Table  

The periodic table is organized in a way that certain trends are followed. o Classification of Elements

 Group 1A: Alkali Metals

 Group 2A: Alkaline Earth Metals

 Group 3B-12B: Transition Metals

 Group 3A-6A: Mix of metals, semimetals and nonmetals

 Group 7A: Halogens

 Group 8A: Noble Gases

o Periodic Trends

 Atomic Radius: the radius of the orbital decreases as you go  UP and ACROSS the periodic tables.

 Ionization Energy: the energy it takes to remove an electron  from an atom. Ionization energy increases UP and ACROSS the  

periodic table.  

 Electron Affinity: measure of how “willingly” an atom accepts  an electron. This increases UP and ACROSS the periodic table. In  other words, it increases with ionization energy.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith  Electronegativity: how strongly an atom attracts electrons.  Electronegativity increases UP and ACROSS the periodic table. In other words, it increases with ionization energy and electron  


∙ Chapter 5

o Ionic Compounds and Ionic Bonding

 Formed by ionization (complete transfer of electron from one  atom to another atom)

 Ionization causes ions to form a crystalin structure called a  


∙ Lattice energy is the energy it takes to convert one  

mole of a solid ionic compound into a gas.

 When naming anions (negatively charged atoms), add the  

ending "-ide" to the element name.

∙ Ex. Cl- is chloride

 When naming cations (negatively charged particles), add the  word "ion" after the element name

∙ Ex. Na+ is sodium ion

 Remember when making ionic compounds, the resulting charge  must be neutral.

∙ Ex. Write the formula for magnesium fluoride

Magnesium has +2 charge and fluoride has -1 charge.

Make the charges equal zero to make a neutral compound. (2) + (-1)(2) = 0

As seen by the formula, fluorine needs a +2 charge to make the compound neutral

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith MgF2

o Polyatomic Ions

 Ions that have more than one charged atom bonded together.  Below is a chart of polyatomic anions to memorize

Common Polyatomic Ions








mercury (I)

















dihydrogen phosphate


hydrogen carbonate or bicarbonate


hydrogen phosphate


hydrogen sulfate




























 Rules to help with memorization (DON"T FORGET TO MEMORIZE  CHARGES):

1. The "-ate" form of the ion has one more oxygen atom  

than the "-ite" form.

2. The "hyper – ite" form of the ion has one less oxygen than

the "-ite" form.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith 3. The "per – ate" form of the ion has one more charge than  

the "-ate" form.

 When asked to balance an ionic compound, you must balance  the charges.

 Use the stock system (uses roman numerals to indicate  

charge) if the metal is a trasition metal ('d' orbital element)

o Organic Compounds

∙ Compunds containing a carbon atom

∙ Name of compound depends on number of carbon atoms in molecule

 Below is a chart of the alkanes (simple hydrocarbons) you must memorize

 There is a chart on page 163 of organic compounds you need to  memorize.

∙ A friend told me they memorize the compounds by  

thinking "Me Eat Peanut Butter" for the first 4 (methane,  

ethane, propane, butane). After those, the next 6 use the  

common prefix names that tell how many carbon atoms  

there are with the name "-ane" added to the end of the  

word (pentane, hexane, heptane, octane, nonane,  


∙ Also, the formula for knowing the formula of an organic  

compound is CnH(2n + 2)

o Ex. If asked to write the molecular formula for  


 There are 4 carbons, and if you plug 4 into the subscript  

 formula you get 10 hydrogens.  SO  C4H10 is the formula for butane. ∙ Molecular Formulas and Formula Masses

 To find molecular mass of a compound, add the molar masses  of each elemental ratio together.

 Molecular Mass = Σ(number of atoms of  element) x (atomic mass of element)

∙ Ex. Find molecular mass of H2O.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith  MM = 2(1.001) + 15.999 = 18.001 g

 To find percent composition of one element in a compound,  use a ratio of the elemental mass and the total mass

 % Composition = ((mass of element) / (mass of  total compound)) x 100

∙ Ex. Find the percent composition of oxygen in a water  

molecule (H2O).

 % O = (15.999) / ((15.999) + (2 x 1.001)) =  5.926%

 To determine the empirical formula from the percent  


1. Find grams of each element

2. Convert grams of each element to moles

3. Divide the molar amounts of by the smallest molar  

amount o all elements involved (this will give you the  

whole number ratio of the empirical formula

∙ Ex. Determine the empirical formula of a compound that  

is 30.45% N and 69.55% O.  

 First, use the percent composition as  the masses of each element (we

 can do this if we assume the molecular mass is 100g).

 30.45% N = 30.45 g N

 69.55% O = 69.55 g O

 Now, convert molar mass to moles.  30.45 g N x (1 mol / 14.009 g N) = 2.173 mol N  69.55 g O x 1 mol/15.999 g O = 4.347 mol O  

 Now divide each molar amounr by the  smallest of the molar amounts.

 2.173/2.173 = 1 mol

 4.347/2.173 = 2.0004 = 2 mol

 Write the completed epirical formula  with whole number subscripts.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith  Empirical Formula: NO2

∙ If you are given the molar mass of the compound, you can

then find the molecular formula (which includes  

subscript ratios of elements as they actually appear).  

 So, say the question tells you that the molar  mass of the compound is  

 92 g/mol.  

 Divide the empirial formula's mass by  the molecular formula's mass to  

 get a whole number.

 Empirical Formula Mass = 14.001 + (2 x  15.999) = 46.01.  

 92/46.01 = about 2.  

 Multiply each subscript in empirical  formula by this whole number  


 Molecular Formula: N1 x 2O2 x 2 = N2O4

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

Chapter 6

o Lewis Dot Structure Theory says atoms will lose or gain electrons in order to obtain a noble gas configruation (more stablility).

o Rules for drawing Lewis Structures for 2nd period elements:

1. Count the number of total number of valence electrons in the  compound.

2. Place double bonds and surrounding elements around the  

central atom of a molecule (the central atom is usually the least  electronegative atom UNLESS hydrogen is involved).

3. Subtract the number of electrons you've already drawn from the  total number of valence electrons. This difference tells how  

many electrons you still need in your structure.

4. Place remaining electrons around the surrounding elements.  Make sure to add double bonds where needed to make atoms  

fuflill the octet rule.  

Atoms in the 3s orbital and higher can hold more than 8 electrons because the 'd'  orbitals can hold electrons.

o If the compound contains an added postive charge, make sure you  accommodate one (or more) less electron into your structure. If the  compound contains an added negative charge, make sure you  

accommodate one (or more) more electron into the structure.

 If the compound contains any added charge, make sure you  indicate this with brackets around the structure with the added  


o Sometimes there are multiple ways to draw the same Lewis Structure.  This results in resonance structures.

Chapter 7

o Valence-Shell Electron-Pair Repulsion (VSEPR) Model of atomic  structure uses the fact that electron domains will get as afar away from each other as possible to get into a more stable structure.

1. Electron Domain Geometries  

∙ Geometry depends on number of electron domains  

(including lone paris of electrons)

∙ 5 Electron Domain Geometries

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith 1. Linear - 180°

2. Trigonal Planar - 120°

3. Tetrahedral – 109.5°

4. Trigonal Bipyrimidal - 90°, 120°

5. Octahedral - 90°

 Molecular Geometries

∙ Geometry depends on number of electron domains AND if  

lone pairs of electrons or double bonds are present.

∙ The chart on page 226 in the book lists all of the  

molecular geometries based on the electron domain  

geometries. (Dr. Blumenthal didn't go in depth about any  

of these. Actually, he only went over ONE of them, so I  

wasn't sure if he expected us to know all of them or not.  

Just in case, I have at least mentioned them).

∙ Molecular geometries that differ from electron domain  

geometries are caused by deviation from ideal bond  


o Lone pair of electrons and double bonds between  

atoms repel surrounding atoms more effectively.

 This causes angles that differ slightly  from the previously listed  

 ideal angles.

o Polarity results from an unequal sharing of electrons due to a  difference in electronegativity or atoms.

 Polarity occurs when a dipole moment (when an atom has a  partial negative or positive charge due to electronegativity)  


∙ Atoms that are more electronegative are more negative  

(because they are more attracted to electrons).

∙ Rules for determining polarity of a molecule:

1. Draw the Lewis Dot structure

2. Determine which atoms will have a partial negative  

charge and which ones will have a partial positive  


3. Draw corresponding arrow directions. On a graph.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith 4. Use vector sum rules to add the vectors.

5. If overall vector sum is zero, there is NO dipole  

moment, which means the molecule is nonpolar.

If the overall vector sum is NOT zero, there IS a dipole moment, which means the  molecule is polar.

o Valence Bond Theory uses quantum mechanic description of orbitals (meaning it uses the actual shapes of different orbitals) to show a 3-D  structure of a molecule.

 Utilizes fact overlapping orbitals can hold single electron to bond

 Hybridization is the mixing of atomic orbitals to show how  atoms can make double bonds. (It is a continuation of the  

valence bond theory).

 It is almost impossible for me to show this process in depth  

without pictures, although it is in the uploaded written chapter 7 notes.  

 However, here is an overview of the rules:

1. Draw the Lewis Dot Structure of the given molecule

2. Observe the geometry using the VSEPR model. The  

number of electron domains in the Lewis structure is the  

number of hybrid orbitals you should end up with.

3. Write out the electron configuration of the central atom. If

there are any paired electrons, move one of the electrons  

in each pair to a higher orbital.

4. Combine the orbitals as necessary.

5. Once hybridized central atom is drawn, add the  

surrounding atoms in their respective orbital shape.

Once the structure is drawn, it will contain sigma bonds, bonds that occur where  probability of electron-sharing is high (usually directly between nuclei), and pi  bonds, bonds that occur where there is a low probability of electron-sharing (not  directly between nuclei).

 Molecular Orbital Theory uses diagrams to describe if a  

diatomic molecule is paramagnetic (has unpaired electrons) or  diamagnetic (has all paired electrons).

∙ States that orbitals combine to make NEW orbitals called  

bonding molecular orbitals and antibonding  

molecular orbitals

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith ∙ When filling out the diagram, remember to follow Hund's  


∙ Bond Order can be determined from the resulting  


BO = ½(# of electrons in bonding molecular orbitals – # of electrons in  antibonding molecular orbitals).

Overview of Models

 VSEPR Model: uses number of electron domains around central atom to determine structure and angles of structure. Provides  

relative structure of  

 Valence Bond Theory: explains why covalent bonds form: to  reduce the overall potential energy of the isolated atoms (in  

other words, to become more stable)

 Hybridization: continuation of valence bond theory. Adds  

explanation of why double bonds can form in molecules (double  bonds form when one sigma bond is combined with one pi  


 Molecular Orbital Theory: uses diagrams that can describe  important properties of diatomic molecules (paramagnetism  

and diamagnetism).

Chapter 8

∙ Rules for writing chemical equations

o Reactants are on the left of the arrow

o Products are on the right of the arrow

o Chemical equations must be balanced to follow the Law of  

Conservation of Mass

o When writing chemical equations, remember to notate the physical  state of the substance with ‘s’ for solid, ‘l’ for liquid, ‘g’ for gas, and  ‘aq’ for aqeous.

∙ Patterns of Reactivity

o Combination: two substances combine to make one product. o Decomposition: one substance breaks into 2 or more products. o Combustion: reactants are burned in the presence of oxygen. ∙ Combustion Analysis

Combustion reactions can be analyzed to find the empirical formula.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith o SECTION 8.2 IN THE BOOK EXPLAINS HOW TO DO THE MATH

∙ Remember when doing calculations involving chemical equations, the  equation must be balanced first.

o Stoichiometric coefficients are the numbers in from of chemical  formulas that will be used in calculations.

o When finding the limiting reactants in a reaction, you must convert the  given mass of the sample to its molar amount. JUST BECAUSE ONE  REACTANT’S STOICHIOMETRIC COEFFICIENT MIGHT BE SMALLER THAN  ANOTHER DOESN’T MAKE IT THE LIMITING REACTANT.

∙ Remember reactivity trends

o Group 1A: alkali metals are reactive because they can easily lose their  one valence electron

o Group 2A: alkali earth metals reactive because they easily lose their  two valence electrons.

o Group 7A: halogens are reactive because they easily gain one electron o Group 8A: noble gases are NOT REACTIVE, because their valence shells are full  

Chapter 9

∙ Soluble substances (dissolve in water) fit into one of two categories o Electrolytes: dissolve in water to yield solution that conducts  electricity. This is due to the dissociation of chemicals into ions

 When a base dissolves in water, it produces hydroxide ions  


 When a strong electrolyte dissolves in water, it dissociates  completely into separate ions. A LIST OF STRONG ACIDS ARE IN  CHAPTER 9.1 OF THE BOOK.

 When a weak electrolyte dissolves in water, some molecules  will dissociate, but most will stay chemically bonded as whole  


(In a chemical equation, weak electrolytes are notated with  

double arrows to show the substance goes between being  

separated and being bonded together.)

o Nonelectrolytes: dissolve in water to yield solution that doesn’t  conduct electricity.

∙ Insoluble substances

o Precipitates are compounds that are separated as solids when  dissolved in water.


o Ionic Equations

 It is more realistic to represent aqueous species in their ionic  forms.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith Instead of writing a chemical equation with whole molecules,  

break the molecules into their separate ions

Ex. Na2 SO4(aq)  Na2+(aq) + SO42- (aq)

o Net Ionic equations

 Instead of representing every ion involved in the reaction, only  represent the ions that make the insoluble substance  


o Oxidation-reduction reactions (redox reactions)

 Occur when electrons are transferred from one reactant to  


 Oxidation is the loss of an electron

∙ The reactant that RECIEVES the lost proton is called the  

oxidizing agent

 Reduction is the gaining of an electron

∙ The reactant that PROVIDES the proton is called the  

reducing agent

 Oxidative State represents number of electrons lost or gained  by an atom

∙ When figuring oxidative states, remember the sum of the  

charges of the ions must add to be zero to be neutral. If  

there is an overall charge on the compound, then the sum

of the separate charges must add to equal the overall  


o Molarity is the concentration of a substance in water

 Units: mol/Liter

 Molarity will be important when doing calculations where  

conversions are involved.

Chapter 10

∙ Energy and Energy Changes

o Energy changes can occur in an open system or closed system o Exothermic Reaction: a reaction in which heat is released, therefore  energy is released.

 In energy calculations, a negative number indicates an  

exothermic reaction.

o Endothermic Reaction: a reaction in which heat is absorbed,  therefore energy is absorbed.

 In energy calculations, a positive number indicates an  

endothermic reaction.

∙ Thermodynamics: study of interconversion of heat with other types if  energy.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith o First Law of Thermodynamics: energy can be converted from one  form to another but cannot be created or destroyed.

 This means that energy released or absorbed in a system must  be equal on both sides of the equation.

 ∆U = Uproducts – Ureactants

o Work and Heat Equations

 Work: W = -P∆V

 Change in Energy: ∆U = q + w OR ∆U = q - P∆V

 Heat at constant volume: q = ∆U

 Heat at constant pressure: q = ∆U - w OR q = ∆U + P∆V

o Enthalpy Equations

 Enthalpy: H = U + PV

 Constant pressure: q = ∆H

o Thermochemical Equations

 When given the change in enthalpy (∆H) of a chemical equation: 1. Find the molar amount of mass of product you are asked  

to find

2. Multiply the ∆H by that molar amount to get the energy  

absorbed or released

∙ Calorimetry

o Specific Heat: amount of heat required to raise the temperature of 1  g of substance by 1 degree celcius

o Heat Capacity: amount of heat required to raise temperature of  object by one degree celcius.

o Calorimetry Equations

 Heat absorbed or released: q = sm∆T

‘s’ is the specific heat of solution, which would be given to you if  asked on a test.

o Constant Pressure Calorimeter: includes two Styrofoam cups and  thermometer. Is used to measure the heat exchanged between the  system and surroundings

∙ Hess’s Law

o States the change in enthalpy of a reaction is the same whether  reaction takes place in one or multiple steps (net change in enthalpy  will always be the same).

o In a multiple-stepped reaction, you would need to add the enthalpy  changes of all reactions to get the net enthalpy change.

∙ Standard Enthalpies of Formation  

o Equation: ∑∆Hproducts - ∑∆HReactants

o If asked questions about this, you would be given the enthalpy of the  respective element or compound.

Remember from ALEKS that some equations will require you to  

manipulate the given equations, such as multiplying by certain  

integers to get a desired stoichiometric coefficient.

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith ∙ Bond Enthalpies

o Equation: ∑BEproducts - ∑BEreactants

o You would be given the bond enthalpies of particular bonds if asked  questions about this on the test.

You will need to know how to draw lewis structures to do these  

problems (if you recall from ALEKS).

Chapter 11

∙ Gas Laws

All gases can be described by these four things: pressure, volume,  temperature and molar amount

o Boyle’s Law: describes pressure-volume relationship (constant temp)  Volume is inversely related to pressure  

V = 1/P (the equals sign means ‘proportional to’)

∙ Ex. If volume is doubled, pressure is halved

 If pressure is doubled, volume is halved

 V1P1 = V2P2

This equation says a sample of gas under different conditions  

with constant pressure and constant amount of gas will have the same product of VP in both conditions

o Charles’s and Gay-Lussac’s Law: describes temperature-volume  relationship (constant pressure)

 Volume is directly proportional to temperature  

V = T (the equals sign means ‘proportional to’)

∙ Ex. If volume is doubled, pressure is doubled

 If volume is tripled, pressure is tripled

 V1/T1 = V2/T2

o Avagadro’s Law: describes molar amount-volume ratio

 Volume is directly proportional to number of moles

V = n (the equals sign means ‘proportional to’)

 V1/n1 = V2/n2

o Combined Gas Law:  

 V1P1/T1n1 = V2P2/T2n2

o Ideal Gas Law: describes behavior of gas under ideal conditions  PV = nRT

CHEM 1030  

Exam 1 Study Guide

Rik Blumenthal Written By: Caroline Smith

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