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Chemistry-Chem04-01_150930_01
1.
© Art Traynor
2011 Chemistry Solution Stoichiometry Solutions Solvent Referring to a substance possessing the propensity to dissolve another ( i.e. the Solute ) into a Solution of which the Solvent is characterized as forming the majority constituent. Wiki: “Solvent” Section 4.3, (Pg. 140) Zumdahl Dissolution / Dilution An interaction between two substances whereby one species – the Solvent – acts to effect a uniform disassociation of another – the Solute – thus maximizing both Disorder and Potential Energy in the Solute ( i.e. Solvation ) to the degree that the Solvent can affect and forming a homogenous Solvent-Solute Complex. Wiki: “Dissolution” Solute Referring to a substance possessing the propensity to become dissolute in combination with another of which the Solute is characterized as forming the minority constituent. Wiki: “Solvent” Section 4.1, (Pg. 127) Zumdahl Section 4.2, (Pg. 130) Zumdahl Section 4.2, (Pg. 130) Zumdahl
2.
© Art Traynor
2011 Chemistry Solution Stoichiometry Solutions Solution A Solvation Mixture forming a homogenous, single phase, Solvent-Solute complex ( in contra-distinction to other multi-phase Mixtures such as colloids, suspensions, or emulsions ) which does not precipitate a chemical reaction or change of chemical configuration in either of the constituent species , but which nevertheless can be accompanied by changes in composition energetics and characterized by the Concentration of its Solute. Wiki: “Solvent” Concentration A characteristic of a Solution expressing the relative abundance of its Solute – canonically denoted as a mass-volumetric quantity by a variety of disparate measures ( e.g. Molarity , Mass Percent, Mole Fraction, Molality, Density ) . Wiki: “Concentration” Section 11.1, (Pg. 500) Zumdahl Density is the mass per unit volume of any substance ⍴ = → m V mass volume
3.
© Art Traynor
2011 Chemistry Solution Stoichiometry Aqueous Solutions Aqueous Solution A solution in which the solvent is water Wiki: “ Aqueous Solution” Section 4.1, (Pg. 127) Zumdahl
4.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula A Dilution can be considered as a mathematical “scaling” of a fixed ( i.e. Constant ) concentration of a substance – in this interpretation, concentration can be thought of as “scaling” a Volume: [ X ]i · Vi , as a linear combination Section 4.3, (Pg. 142) Zumdahl Vf[ X ] i [ X ] i Vi Vi For some [ X ]¬ i < [ X ]i the product [ X ]i · Vi may be equated with a similar product Vf · [ X ]¬ i where Vf is an additive mixture of Vi and a quantity Vsv of a shared Solvent Solvent = Sv Vsv Vf [ X ] ¬ i Vf = Vi + Vsv Vi Vsv = Vf – Vi The two product terms equated thus describe a proportional relationship of the form: [ X ]i · Vi = Vf · [ X ]¬ i M1 · V1 = M2 · V2 or ab = cd or or = a c b d = a· d – b · c a b c d a b c dA = det ( A ) = a· d – b · c
5.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula ( PST ) Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H2 SO4 solution ? Section 4.3, (Pg. 141) Sample Exercise 4.9 Zumdahl Problem Solving Technique (PST) Variables Initial Volume Vi [ A ]i Final Concentration Vf [ A ]f Units Molarity M = Moles of Solute Liters of Solution Initial ConcentrationInitial Final [ X ]k Vk A = H2 SO4 (l ) [ A ]i = 16 M = 16 mol 1 L [ A ]f = 0.10 M = 0.10 mol 1 L Vi = UNKOWN Vf = 1.5 L Final Volume Formulae Dilution [ A ]i Vi = [ A ]f Vf16 mol 1 L Vi 1 = 0.10 mol 1 L 1.5 L 1 [ A ] i V i [ A ] f V f
6.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula ( PST ) Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H2 SO4 solution ? Section 4.3, (Pg. 141) Sample Exercise 4.9 Zumdahl Problem Solving Technique (PST) 16 mol 1 L Vi 1 = 0.10 mol 1 L 1.5 L 1 [ A ] i V i [ A ] f V f 16 mol 1 L Vi 1 = 0.10 mol 1 L 1.5 L 1 Solve for Vi 1 L 16 mol [ A ] i – 1 16 mol 1 L Vi 1 = 0.10 mol 1 L 1.5 L 1 1 L 16 mol 1 L 16 mol Vi = 0.10 1 1.5 1 1 L 16 [ A ] i – 1
7.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula ( PST ) Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H2 SO4 solution ? Section 4.3, (Pg. 141) Sample Exercise 4.9 Zumdahl Problem Solving Technique (PST) V i [ A ] f V f Vi = 0.10 1 1.5 1 1 L 16 [ A ] i – 1 Vi = ( 0.10 ) ( 1.5 ) L 16 Variables Initial Volume Vi [ A ]i Final Concentration Vf [ A ]f Units Molarity M = Moles of Solute Liters of Solution Initial Concentration Final Volume Formulae Dilution [ A ]i Vi = [ A ]f Vf Vi = ( 0.15 ) L 16 Vi = 0.009375 L 0.0.0.9.375 L = 9.375 x 10 – 3 L Moving 3 positions in the “+” directionVi = ①② ③
8.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula ( PST ) Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H2 SO4 solution ? Section 4.3, (Pg. 141) Sample Exercise 4.9 Zumdahl Problem Solving Technique (PST) V i Variables Initial Volume Vi [ A ]i Final Concentration Vf [ A ]f Units Molarity M = Moles of Solute Liters of Solution Initial Concentration Final Volume Formulae Dilution [ A ]i V1 = [ A ]f V2 Vi = 0.0.0.9.375 L = 9.375 x 10 – 3 LVi = 9.375 m LVi = 0.0.0.9.375 L = 9.375 x 10 – 3 L Moving 3 positions in the “+” direction ①② ③
9.
© Art Traynor
2011 Chemistry Solution Stoichiometry Dilution / Dissolution Dilution Formula ( PST ) Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H2 SO4 solution ? Section 4.3, (Pg. 141) Sample Exercise 4.9 Zumdahl Variables Initial Volume Vi = 9.375 m L [ A ]i = 16 M Final Concentration Vf = 1.5 L [ A ]f = 0.10 M Units Molarity M = Moles of Solute Liters of Solution Initial Concentration Final Volume Formulae Final Volume Vf = Vi + Vsv Volume of Solvent Vsv = Vf – Vi Vi = 9.375 mL Vf [ A ] = 16M [ A ] ¬ i = 0.10 M Vsv = 1.490625 L Vf = Vi + Vsv Vsv = 1.5 mL – 0.009375 Final Volume V f Initial Volume V i Volume of Solvent V sv Vf = Vi + Vsv 1.5 L = 9.375 mL + Vsv Volume of Solvent V sv Vsv = Vf – Vi Vsv = 1.5 L – 0.009375 L Vsv = Vf – Vi Solvent Vsv = 1.490625 L Vf = 1.490625 L + 0.009375 L Vf = 1.5 L
10.
© Art Traynor
2011 Chemistry Pressure Definitions Pressure – n. 1. The exertion of a force upon a surface by an object, fluid, etc., in contact with it. Physics. 2. Force per unit area ( symbol “ P ” ). ( Pressureless – adj.; Pressure, Pressured, Pressuring – v. ). Webster’s Encyclopedic Unabridged Dictionary of the English Language ( Pg. 1532 ) A measure of Force, normal to a surface, expressed per unit area of the referent surface over which the Force is distributed. Variables Pressure (Normal) Force F P P = F A Area of Contact Surface A A scalar quantity relating a vector surface element (normal to the referent surface) with the normal force which acts upon it As a scalar quantity pressure has no direction
11.
© Art Traynor
2011 Chemistry Pressure Units Pascal – SI Unit One Newton per square meter. Webster’s Encyclopedic Unabridged Dictionary of the English Language ( Pg. 1532 ) N m2 Newton - Defined the amount of Net Force One Newton ( N ) is that gives an acceleration of one-meter per second squared 1m s 2 to a body with a mass of one kilogram 1 kg 1 Newton = ( 1 kilogram ) ( 1 meter per second squared) 1 N = 1kgm s 2 kg m · s2 = Wiki: “ Pressure”
12.
© Art Traynor
2011 Chemistry Pressure Units Standard Atmosphere ( atm ) – Legacy Unit One Standard Atmosphere is defined as: Webster’s Encyclopedic Unabridged Dictionary of the English Language ( Pg. 1532 ) 1 atm = 101,325 Pa = 1.01325 x 105 Pa = 101.325 kPa Torricelli ( torr ) – Legacy Unit One torr is defined as: Pa = 1.33 Pa = atm 101,325 760 1 760 Wiki: “ Torr ” Wiki: “ Pressure”
13.
© Art Traynor
2011 Chemistry Gas Laws Boyle’s Law Boyle’s Law An expression positing a constant relationship between the product of the volume of a confined gaseous substance and its characteristic pressure Section 5.2, (Pg. 187) Zumdahl PV = k Stated in the alternative, Boyles’ Law posits an inverse relationship between the Volume of a gaseous substance and its characteristic Pressure (at a particular temperature) V = = k k P 1 P This form of Boyle’s Law reveals its consonance with the linear form y = mx + b ( with b = 0 ) A gas strictly conforming to Boyle’s Law is referred to as an Ideal Gas
14.
© Art Traynor
2011 Chemistry Gas Laws Charles’s Law Charles’s Law An expression positing a proportional relationship between the characteristic Volume of a Gas and its Temperature Section 5.2, (Pg. 187) Zumdahl V = bT Gaseous Volume is a function of Temperature, scaled by a proportionality constant
15.
© Art Traynor
2011 Chemistry Gas Laws Avogadro’s Law Avogadro’s Law An expression positing a proportional relationship between the characteristic Volume of a Gas and the profusion of its constituent particles Section 5.2, (Pg. 191) Zumdahl V = an Gaseous Volume is a function of particle profusion scaled by a proportionality constant
16.
© Art Traynor
2011 Chemistry Gas Laws Ideal Gas Law Ideal Gas Law ( IGL ) An expression amalgamating three of the principal relationships characterizing gaseous medium behavior Section 5.2, (Pg. 191) Zumdahl Boyle’s Law : Gaseous Volume is a function of particle profusion scaled by a proportionality constant Charles’s Law : Avogadro’s Law : V = k P V = bT V = an Ideal Gas Law : V = R T n P V = R T n P P 1 P · V = R · T n P P 1 → PV = RTn PV = nRT
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