Study Guide For midterm 2
Study Guide For midterm 2 Chem 141
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This 6 page Study Guide was uploaded by Cassidy Zirko on Saturday October 10, 2015. The Study Guide belongs to Chem 141 at University of Montana taught by Mark Cracolice (P) in Fall 2015. Since its upload, it has received 47 views. For similar materials see College Chemistry 1 in Chemistry at University of Montana.
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Date Created: 10/10/15
Midterm 2 Chem 141 Prof Cracolice Study Guide Midterm 2 Chapter 1017 Key Points Chapter 10 Number of oxygens Prefix su ix Example formula Example Name related to ic acids One more Per ic H Cl04 Perchloric Acid Same ic H C l03 Chloric Acid One Less ous H Cl02 Chlorous Acid Two Less Hypo Ous H ClO Hypochlorous Acid No oxygen Hydro ic HCl Hydrochloric acid 0 Acids ending in ous form anions ending in ite hypochlorous acid I hypochlorite ion 0 Acids ending in ic form anions ending in ate chloric acid I chlorate ion 0 When naming acids think of acids you already know the names for 0 Polyprotic acids acids With more than 1 hydrogen ion 0 Total Dissociation an acid that has all of the H removed 0 Remember Ammonium NH4 and hydroxide OH39 0 Ionic Compound compound made up of a cation and an ion that have no net charge 0 Hydrate Ionic Compound that contains water molecules 0 Anhydrous compound Without water Midterm 2 Chem 141 Prof Cracolice Key Points Chapter 11 0 Formula Mass Based on the chemical formula of the compound 0 Molecular mass average mass of molecules or formula units comparted with the mass of an atom of carbon12 which is exactly 12 atomic mass units 0 Mole the amount of any substance that contains the same number of units in the number atoms in exactly 12 grams of a carbon12 atom 23 0 Avogadro s number Na 602 10 atoms molecules or formula un1ts 0 Molar mass mass in grams of one mole of a substance 0 Percent the amount of one part of a mixture per 100 total parts in the mixture 0 Percent composition the percentage by mass of each element in the compound Changes you make Conversion factors used M016 DD units 602 1023 Molecules atoms or formula units Mole II grams Molar mass 0 Formula mass sum of the atomic mass in the formula unit 0 Molar mass gmol Key Points Chapter 12 0 Combustion Analysis a compound burned for the purpose of analyzing its composition 0 Empirical formula simplest formula of a compound 0 Procedure 1 Find masses of different elements in a sample of a compound 2 Convert the masses into moles 3 Express the moles of atoms as the lowest possible ratio of whole number integers 4 Write the empirical formula using the number of each atom in the integer ratios as the subscript 0 Use this type of table to solve empirical formula problems Element grams Moles Mole ratio Formula ratio 0 Procedure for Molecular Formula 1 Determine the empirical formula 2 Calculate the molar mas of the empirical formula 3 Determine the molar mass of the compound usually given 4 Divide the molar mas of the compound by the molar mass of the empirical formula to get the number of empirical formula units per the molecular given 5 Multiply the empirical formula by the factor from step four and write the molecular formula Key Points Chapter 13 Midterm 2 Chem 141 Prof Cracolice 0 Five indicators of a chemical change color change formation of a solid formation of a gas expulsionabsorption of heat production of light 0 State symbols indicate the state of each compound and is sometimes omitted 0 4 state symbols always written as a subscript 0 Aqueous solution a substance that is dissolved into water 0 Qualitative Equation an chemical equation that is not balance 0 Balanced placing a coefficient in front of one or more of the formulas indicating it is used more than once 0 Procedure of Balancing an Equation 1 Write a qualitative description of the reaction In this step write the formulas based on the reactants and products 2 Quantify the description by balancing the equation Done by adding coefficients 0 Combustion reaction 0 Reactants Any combination of elements andor compounds 0 Reaction type combination 0 Equation type A X I AX 0 Products one compound 0 Decomposition reaction 0 Reactants one compound 0 Reaction type decomposition 0 Equation type AX I A X 0 Products any combination of elements or compounds 0 Single Replacement Reaction 0 Reactants element A plus a solution of either an acid or ionic compound BX 0 Reaction type Single replacement 0 Equation type A BX I AX B 0 Products an ionic compound usually in solution AX plus and element B 0 Burning Oxidation Reaction 0 Reactants CXHy or CXHyOZ compound and 02g 0 Reaction type burning or oxidation reaction 0 Equation type CXHyOZ 02 I C02 H20 0 Products C02 and H20 0 Double Replacement reactions 0 Reactants solution of two compounds each with positive and negative ions AX BY Reaction type Double replacement Equation type AX BY I AY BX 0 Products two new compounds AY BX which may be a solid water or an aqueous ionic compound 00 Midterm 2 Chem 141 Prof Cracolice Key Points Chapter 14 Stoichiometry Given a macroscopic quantity of one species before and after a chemical change you can find the quantity of a second species You can use Avogadro s number to convert to particles Use grams to convert to mass Mass I molar mass is grams I Mole Mass of given I moles of given I Moles of wanted I mass of wanted Procedure 0 1 Change mass of given to moles of given 0 2 Change moles of given to moles of wanted 0 3 Change moles of wanted to mass Key Points Chapter 15 Limiting reactant the reactant that is completely used up by the reaction Excess reaction the amount of the reaction which is still left over after the reaction is complete Procedure 0 1 A the reactant that yields the smaller amount of product is the limiting reactant 0 B the smaller amount of product is the amount that will be formed when all of the limiting reactant is used up 0 2 Calculate the amount of excess reactant that is used by using that limiting reactant and converting it to the excess reactant O 3 Subtract the amount of excess reactant that was given in the problem initially form the amount calculated in step two This difference will give you the amount of the excess reactant that is unreacted Key Points Chapter 16 Ideal yield amount of product formed from the complete calculation Actual yield measured quantity determined by experiment Percentage yield actual yield expressed as a percentage of the ideal yield actual product yield 100 yield ideal product yield x parts of A x grams actual yield 2 100 parts of mixture I x 100 gramsideal product yield Key Points Chapter 17 Midterm 2 Chem 141 Prof Cracolice 0 Law of Combining Volumes reacting volumes are always in the ratio of small whole number if the volumes are measured at the same temperature and pressure 0 Avogadro s law equal volumes of all gases at the same temp and pressure contain the same number of molecules 0 Adding gas constant R 0821 atmL molK 0 Ideal gas equation PVZRRT 0 Ideal Gas Law volume is directly proportional to the number of moles and the temperature while being indirectly proportional to pressure m 0 Ideal Gas equation with molar mass PV W RT 0 Characteristics of a gas 0 Gases can be compressed O Gases can expand O Gases have low densities due to the large amount of space between each of the individual molecules 0 Gases may be fixed in volume 0 Gases exert constant pressure on the walls of its container uniformly in all directions 0 Ideal gas model allows us to be able to visualize the nature of gas by comparing it with a physical system we can see or imagine 0 Properties of an Ideal gas 0 1 Gases consist of particulates moving at any given instant in a straight line 0 2 Molecules collide with each other and with the container walls without a loss of total kinetic energy Loss in energy I decrease in pressure Drop in temperature I energy drops I decrease in pressure 3 Gas Molecules are very widely spread out 4 The actual volume of molecules is negligible compared to the space they occupy O 5 Gas molecules are independent attractive forces between them are negligible 0 Ideal gas a model of a gas constructed of identical particles that occupy no sufficient 0000 volume and exert no forces on one another 0 Dalton s Law of Partial Pressure total pressure exerted by a mixture of gases is the sum of the partial pressures of the gases in the mixture 0 Pp1p2p3 O P total pressure 0 pm partial pressure of each of the gasses 2 Pa V nbnRT 0 van de Waals Equation Midterm 2 Chem 141 Prof Cracolice 0 a constant adjusts for the attractive forces among the particles 0 b constant corrects for the volume of the molecules the closer a and b are to zero the more the gas Will behave like an ideal gas
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