Chem 111 Week 6 lecture notes
Chem 111 Week 6 lecture notes 111/40551
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This 11 page Class Notes was uploaded by firstname.lastname@example.org Notetaker on Friday October 16, 2015. The Class Notes belongs to 111/40551 at University of St. Thomas taught by Uzcategui-White in Summer 2015. Since its upload, it has received 15 views.
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Date Created: 10/16/15
Chemistry Week 6 Lecture Notes Chapter 4 0 Displacement Reaction atoms or ions in a compound are replaced by an atom or ion of another compound 0 A B 9 C D 0 Does not change the number of substances productreactant o 3 Kinds metal hydrogen halogen 1 Metal displacement formation of a metal Ex Zn 5 CuSO4 aq 9 ZnSO4 aq Cu 5 Not precipitation reaction because that has 2 aqueous reactants displacement has a solid 2 Hydrogen displacement formation of H2 g H20 l H20 g strong acid Most reactive metals 9 H2 g from H20 l Group 1A and 2A Ex 2Na 5 ZHZO l 9 H2 g 2NaOH aq The slightly less reactive metals 9 H2 g from H20 g Ex 2 Al 5 6H20 g 9 3H2 g 2Al OH3 aq Still less reactive metals 9 H2 g with strong acid Ex Ni 5 2HC aq 9 H2 g NiCl2 aq The least reactive metals 9 do not produce H2 3 Halogen displacement formation of halogen group 7A F2Br2 I2 Cl2 Ex C2 g 2KBr aq 9 KCI aq Br2 Important to know F2 and C2 are gaseous as free elements diatomics Br2 liquid I2 solid Reactivity F2 faster than C2 gt Br2 gt 2 Ex C2 g 2KF aq 9 2 KCI F2 g This reaction happens faster than the other 0 Combustion reaction 0 Always produces C02 g and H20 g o A substance reacts with oxygen with releasing of energy to produce C02 and gaseous water Ex 2C8H 18 g 25 02 g 9 16 C02 g 18H20 g Problem 461 Which type of reaction leads to each of the following a Increase in of substances decomposition reaction b Decrease in of substances combination reaction c No change in the of substances displacement or combustion reaction Problem 465 Balance each of the following equations and classify the type of reaction a Mg 5 H20 g 9 Mg OH2 s H2 g Mg 5 2H20 g 9 MgOH2 s H2 g displacement hydrogen b CrN033 aq Al 5 9 Al NO33 aq Cr 5 already balanced displacement metal c PF3 g F2 g 9 PF5 g combustion Problem 467 Predict the products and write a balanced equation a Fe s 2HCO4 aq 9 Fe s 2HCO4 aq 9 H2 g FeCO42 aq b 58 s 02 g 9 18 58 s 02 g 9 502 g Chapter 5 o Gases moles pressure volume temperature 0 Gas volume changes significantly with pressure changes with a change in temperature 0 gases flow very quickly high kinetic energy 0 gas densities low d02 g 13 g L d H20 10gmL d NaCl 5 22 g mL density of gas increases with decrease in temperature inverse relationship d mv o gases form homogeneous solutions solution in any proportion that interact well with each other Pressure 0 result of the constant collisions between atoms or molecules in a gas and the surface around the gas 0 Depends on 1 The amount of gaseous particles fewer molecules 9 lower pressure exerted by gas 2 Pressure decreases with an increase in altitude 3 Pressure is directly proportional to temperature values pressure increases temp increases 0 Pressure force area 0 Pressure units 1 Atm average pressure of air at sea level SI 9 Pascal Pa 1 atm 101325 Pa 1 Pa N mquot2 N Newton 2 mmHg millimeters of mercury 1 atm 760 mmHg 3 Torr 1 torr 1 mmHg The Gas Law 0 Gas laws a Boyle s law b Charles s law c Avogadro s law 0 Ideal gas a gas that exhibits a linear relationship among pressure temperature volume and amount of particles mol Not ideal gases exist but most of simple gases behave nearly ideally at ordinary temperatures and pressures a Boyle s law volume and pressure Volume of gas and its pressure are inversely proportional temperature of gas and of particles are at constant V 1P V constant P constant VP P1V1 P2V2 V increases P decreases b Charles s Law volume with temperature Volume of a gas is proportional to the temperature pressure and moles are constant V is proportional to T V constant x T constant VT V1T1 V2T2 As T increases V increases c Avogadro s Law volume with of particles Volume of gas is directly proportional to amount of gas pressure and temperature are kept constant V is proportional to n moles V constant x n constant Vn V1n1 V2n2 As n increases V increases 0 Ideal Gas Law Gas constant R Constant R 0821 L atom K mol R PVnR PV nRT Volume liters Pressure atm n mols T kelvin Practice Problems 521 If 147 x 10 quot3 mols of Argon occupies a 750 mL container at 26 Celcius what is the pressure in torr since it is a gas we use PV nRT P nRTV 750 mL x 1 L 1000 mL 0750 L 26C273299K P 147 x 10quot 3 mols x 0821 L atm K mol x 299K 0750 L 481 atm 481 atm x 760 mmHg 1 atm 366 Torr 516 A sample of sulfur hexafluoride gas occupies a 910 L at 198 Celsius Assuming P remains constant what temperature in Celsius is needed to reduce volume to 250 L V1 T1 V2T2 198 C 273 471 K 910 L471 K 250 Lx X 129 C 129 C 273 144 C 514 What is the effect of the following on the volume of 1 mole of an ideal gas a The pressure is tripled at constant T b The absolute temperature is increased by a factor of 30 at constant P c 3 more moles of gas are added at constant P and T a P1V1 P2V2 V2 P1V1 P2 P2 3P volume is decreased by a factor of 3 b V1T1 V2T2 V2 V13T1 T2 3V Volume will increase by a factor of 3 c V1n1 V2n2 V2 n2 V1 n1 4n1 V1 n1 volume increased by a factor of 4 Gas Behavior at Standard Conditions o STP or standard temperature and pressure specifies a pressure of 1 atm 760 torr and a temperature of 0 degrees Celcius 273 K o The standard molar volume is the volume of 1 mol of an ideal gas at STP 0 Standard molar volume 224141 L or 224 L Identifying a Gas 0 Use density and molar mass Density of gases PV nRT D mass g V n grams M molar mass PV g M x RT 9 PMRT gV d Molar mass M gRTV Mixture of Gases o Gases mix homogeneously in any proportions Each gas in a mixture behaves as if it were the only gas present 0 Air is a mixture yet we can treat it as a single gas 0 Also we can think of each gas in the mixture as independent of the other gases All gases in the mixture have the same volume and temperature All gases have the volume of the container Dalton s Law 0 The total pressure in a mixture is the sum of the quotpartial pressures of the individual gases Pressure exerted by each gas in a mixture is called quotpartial pressures Pt Pa P b P c P PaXaxPt Pa partial pressure of gas A X a molar fraction X moles A total moles of gases n a n total Problems 533 The density of a noble gas is 271 g L at 300 atm and 0 degrees Celsius Identify the gas M gRT PV 9 M dRT P M 271 gL x 0821 L atmk mol x 273 k 300 atm 202 gmol Neon 536 After 600 L of Ar at 120 atm and 227 C is mixed with 200 L of 02 at 501 torr and 127 C in a 400mL flask at 27 C What is the pressure in the flask 600 L Ar 200 L 02 quot9 combines to form vt 400 mL 120 atm Ar 501 torr 02 T t 27 C 273 300 k 227 C Ar 127 C 02 P t 277 273 500 K 127273 400 k 501 torr 02 x 1 atm 760 torr 659 atm Need to find total moles PV nRT Plt n t R t vt Mols Ar PVRT 120 atm x 600 L 0821 L atm k mols x 500 K 0175 mols Ar Mols 02 659 atm x 200 L 0821 L atm k mols x 400 K 0040 mol 02 0175 mols Ar 0040 mols 02 02151 mols P t 02151 mols x 0821 L atm k mols 300k 400 L 132 atm 582 Liquid nitrogen trichloride is heated in a 250 L closed reaction vessel until it decomposes completely to gaseous elements The resulting mixture exerts a pressure of 754 mmHg at 95 C a What is the partial pressure of each gas b What is the mass of the original sample a Equation 2 NCI3 l 9 N2 g 3C2 g P N2 X N2 x P t 9 1 mol N24 mol total x 754 mmHg 188 mmHg P Cl2 X Cl2 x P t 9 3 mol Cl2 4 mol total x 754 mmHg 566 mmHg b Grams of NCI3 Moles product 9 moles of reactant 9 grams PV nRT C2 mols P cl2 x V Cl2 R T cl2 745 atm x 250 L 0821 L atm mols K x 368 K 494 g NCI3 566 mmHg 760 atm 745 atm 95C273368K Collecting Gases o Gases are often collected by having them displace water from a container 0 The problem is since water evaporates there is also water vapor in the collected gas 0 The partial pressure of the water vapor called the vapor pressure depends only on temperature P t P gas P H20 P gas P t P H20 Ex total pressure 758 mmHg at T 26 C P H2 758 mmHg 252 mmHg
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