I. The Study of matter and changes it undergoesCHEM 1320 1st EditionFinal Exam Lecture 1 (January 23)I. The Study of matter and changes it undergoes .Matter is anything that has mass and volume. Composition is the types and amounts of simplersubstances within. Properties are characteristics identifying a substance. Properties can be physical or chemical. II. Composition of matterThe periodic table consists elements. These elements can be made into compounds which consist of two types of atoms, expressed by a chemical formula. There are also compounds that are called mixtures. These mixtures can’t be expressed by a formula. They consist of at least twoatoms. III. Pure SubstancesPure substances are compounds. They consist of two or more different elements. They have a constant composition and break down chemically. Elements consist of the same atoms. They can’t break down chemically. They are all the substances listed on the periodic table. IV. Mixtures Mixtures can be broken down physically. They can filter and evaporate. They consist of two or more substances and have a variable composition. Chemistry consists of mostly mixtures. Reactions deal with mostly pure substances. V. PropertiesThere are physical, extensive, and intensive. Physical properties are observed without doing chemical reactions. These physical properties can be color, melting point, and boiling point. Extensive properties depend on the amount of matter. These properties include mass and volume. Intensive properties do not depend on matter. There properties include density, temperature, and color. Lecture 2 (January 26) I. Nucleus and ElectronsElectrons orbit the nucleus in a constant motion. Protons are positive and neutrons are negative. Electrons have a negative charge. Mass is the number of protons and neutrons and is shown to the top left of every element on the periodic table. The atomic number is the number of protons and electrons. There are also ions and isotopes. There are two types of ions; cations and anions. Cations are positive while anions are negative. Lecture 3 (January 28)I. Periodic TableGroup one (the column all the way to the left of the periodic table) is the alkali metals. Group 2 (to the right of group one) is called the alkaline earth metals. Group eighteen (all the way to the right on the periodic table) is called the noble gasses. Group seventeen (the column to the right of the noble gasses) is called the halogens. The metals in the middle of the periodic table are called transition metals, made up of metals and nonmetals. II. Ionic CompoundsThese are metals and nonmetals that combine cations and anions to make neutral species. III. Molecular compoundsThese are combinations of nonmetals. We cannot predict them based on charge neutralization.Lecture 4 (January 30)I. Common Acids (H)HCl is hydrochloric acidHBr is hydrobromic acid HI is hydroiodic acidHNO3 is nitric acidH2SO4 is sulfuric acidC2SO4 is sulfuric acidH3PO4 is phosphoric acid II. Common Bases ( OH-) NaOH is sodium hydroxideKOH is potassium hydroxide Lecture 5 (February 2)I. Chemical EquationsThere are two parts to every equation; reactants and products. The law of conservation of mass states that matter cannot be lost in a chemical reaction. II. Balancing Chemical Equations Stoichiometry is the conservation of mass. There are three steps in balancing chemical reactions. 1). Translate the words in the reaction and product.2). Balance the elements that appear once and then the remaining elements.3). Check to make sure the numbers are equal on both sides.Lecture 6 (February 4)I. Translating Written EquationsSolid nickel (II) sulfide + oxygen  nickel (II) oxide + sulfur dioxide2NiS+3O2  2NiO + 2SO2Combustion reaction- when something burns it reacts with oxygen in the air II. Atomic MassesAtomic mass is one unit or an amu which equals 1/12 mass of 12 carbon. 1 amu = 1.661 x 10-24 g or 1.1661 x 10-27kg1g =6.022 x 1023amuNa= 6.022 x 1023mol-1 (avogadro’s number) III. Molecular Mass or Formula MassKOH: K= 39.098g/molO= 15.99g/molH= 1.008g/mol55.096 g/mol (add all the elements together) Lecture 7 (February 6)I. Conversion Problems Example 1: If there are 3.60 molecules of carbon monoxide, what is the number of moles in carbon dioxide? CO+ O2  CO2 BALANCED 2CO + O2  2CO23.6mol/ 2 Example 2: If there are 1.140 grams of KCN treated with HCL, how much HCN is formed? KCN + HCl  KCl + HCN 0.0140g KCN 1mol KCN 1mol HCN 27.03g 65.11g 1mol KCN 1mol = 0.058g HCNII. Limiting ReagentBefore the reaction starts all the molecules float around separately. After the reaction is complete some of the molecules attach to one other molecule. Lecture 8 (February 9)I. Key TermsA solution is a homogenous mixture of two or more substances. A solute is a substance present in a similar amount. Lastly, a solvent is a substance that dissolves causing the solutionII. Chemistry in SolutionA. Qualitative Ionic compounds that are soluble: - Common compounds of Group 1 (alkali metal) elements or of an aluminum ion. - Common nitrates (NO3-) and acetates (CH3OO-) - Common halides, chlorides, bromides, and iodides (EXCEPT: silver, mercury, and lead)- Common sulfates (EXCEPT: Ag+,Hg2+,Pb2+,Ca2+,Sr2+,Ba2+)- Common hydroxides (EXCEPT: most metal hydroxides are insoluble) - Common carbonates (EXCEPT: most metals)- Common sulfides (EXCEPT: metal sulfides) B. Quantitative Concentration is the amount of solute in a solvent. Molarity is the number of moles solute in a given volume of solvent. M= moles solute/ volume solvent (mol/L) Moles solute= M x VMol/L x L = molLecture 9 (February 11)I. Word ProblemIf there is a 2.324g sample of an unknown acid, and H2X is dissolved in water. The sample requires 68.83mL of a 0.7500M NaOH solution for a complete reaction. What is the molar mass of the acid?Molar mass = mass/molH2X (aq) + 2NaOH (aq)  Na2X (aq) + 2H2O (l) (BALANCED)Net Ionic Equation: 2H+ (aq) + X2- (aq) + 2Na+ (aq) + 2OH- (aq)  2Na+(aq) + X2-(aq) +2H20 (l)Cancel all the identical terms on each side: 2H+ (aq) + 2OH- (aq)  2H20 (l)Re-Balance Equation: H+ (aq) + OH- (aq)  H2O (l)M x V = mol/L x L(68.85mL) (1L) (0.7500mol) (1000 mL) (1L) = 5.162 x 10-2 mol OH-(0.05)(2 mol H2X/ 1 mol NaOH)  0.0258 mol H2XMolar mass = mass/mol 2.324g/0.025g = 90.04g/molLecture 10 (February 13)I. Dilution Dilution is when you add solvent to a solution. Example: Prepare 60mL