GENERAL CHEMISTRY I [C3T1G2]
GENERAL CHEMISTRY I [C3T1G2] CHEM 131
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This 3 page Class Notes was uploaded by Luis Turner on Saturday September 26, 2015. The Class Notes belongs to CHEM 131 at James Madison University taught by Mary Tam in Fall. Since its upload, it has received 17 views. For similar materials see /class/214171/chem-131-james-madison-university in Chemistry at James Madison University.
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Date Created: 09/26/15
Chapter 5 9 An overview of the physical states of matter 0 Gas volume changes greatly with pressure 0 I The smaller the volume ofa container that holds a gas the more pressure exerted on the gas the gas has a smaller volume I The larger the volume ofa container that holds a gas the less pressure exerted on the gas the gas has a larger volume Gas volume changes greatly with temperature I When a gas is heated its volume increases when it is cooled its volume decreases Gases have relatively low viscosity I This allows gases to ow more freely gases can be transported through pipes over long distances leak rapidly out of small holes etc Most gases have relatively low densities under normal conditions Gases are miscible I Miscible substances mix with one another in any proportion to form a solution 9 Key equations and relationships 0 STP standard temperature and pressure 0 O O I 0 C 27315 K I 1 atm 760 torr I 224 L Relating volume to pressure temperature and amount ideal gas law I PVnRT I To find gas density dMX PRT M molar mass I To find molar mass MdRTP Partial pressure XA mole 39 PAXA X PTOTAL fraction mol Graham s law of effusion AHIOIA RateARategMBIWXB I RateAfaster gas lighter gas hit the PVnRT enna nn When given two sets of data set up two PVnRT equations cancel out the constants whatever information is omitted and set up a proportion I Applying the volumepressure relationship 0 P1 238 atm V1341 L P2454 atm V2 o Tand n remain constant Arrange the ideal gas law P1V1nR o P1V1 Psz Applying the pressuretemperature relationship P2V2nlillll Chapter 6 9 Key equations and relationships qheat O AEq W o wPAV o AHmquot 2m AHfquot products 2m AHfquot reactants o AEqheat owing out from a system 0 AEqheat owing into a system 0 Table 61 pg 181 o 1oule 1 k14184 cal02390 cal O Exothermic and Endothermic Processes 0 An exothermic quotheat out process releases heat and results in a decrease in the enthalpy of the system I AH is negative I H nalltHinitial I reactants 9 products heat 0 An endothermic quotheat in process absorbs heat and results in an increase in the enthalpy of the system I AH is positive I H nalgtHinitial I heat reactants 9 products 9 Hess s Law 0 To calculate an unknown AH Identify the target equation the step whose AH is unknown and not the number of moles of each reactant and product Manipulate the equations with known AH values so that the target numbers of moles of reactants and products are on the correct sides remember to 0 Change the sign of AHwhen you reverse an equation 0 Multiply numbers of moles and AH by the same factor Add the manipulated equations to obtain the target equation all substances except those in the target equation must cancel add their AHvalues to obtain the unknown AH Example Target equation H2 Clz 9 2HCl N2g3H2g9 2NH3g AH m918 k N2g 4H2g Clzg 9 2NlI4Cls AH m 6288 k NH3g HClg 9 NH4Cls AH m 1762 k We need to switch the third equation to get HClg on the product side The Sign Of NH4Cls a NH3g HClg AH m 1762 k reverses Chapter 4 Test Review Chapter 4 9 Writing molecular total ionic and net ionic equations 0 Molecular equations only show all the reactants and products as if they were intact undissociated compounds 0 Total ionic equations show all the soluble ionic substances dissociated into ions 0 Net ionic equations eliminate spectator ions species not involved in the chemical change and show the actual chemical change taking place 0 Example Molecular equation BaN032 aq Na2C03 aq 9 BaC03 s 2NaN03 aq What is the total ionic and net ionic equation for this reaction Note that Total ionic Ba2aq 2N0339aq 2Naaq C03239aq 9 BaC03s 2Naaq 2N0339 Baco3 is not dissociated Net ionic Ba2aq QNQg lq 2Naaq C0323911q a BaC03s z Naaq 24519 bf t39sa SO 1 Ba2aq C03239aq 9 BaC03s 0 Writing ionic equations for acidbase reactions I An acid produces H ions when dissolved in water a base produces OH39 ions when dissolved in water I The essential change in all aqueous reactions between a strong acid and a strong base is that miion from the acid and an OHion from the base form a water molecule I The compound that results from the reaction is a salt 39 Example HX MOH 9 MX H20 This isa I generlc ac1d base salt I reactlon O Predicting whether a precipitate will form 0 Note the ions present in the reactants Ca2 2Cl39 3Cs PO4393 9 Cation positively charged ion Anion negatively Charged 1011 0 Consider the possible cationanion combinations Ca2aq 2Cl39 aq 3Cs aq PO4393aq 9 Ca3PO42 7 CsCl 7 0 Decide whether any of the combinations is insoluble using the solubility rules Ca2aq 2Cl39 aq 3Cs aq PO4393aq 9 Ca3PO42 s Csaq Cl39 all Solubility rule all common carbonates and phosphates are insoluble except those of Group 1A1 and NH4
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