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by: Luna Bauch


Marketplace > Oklahoma State University > Chemistry > CHEM 1414 > GEN CHEM FOR ENGINEERS
Luna Bauch
OK State
GPA 3.92

Charles Weinert

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Charles Weinert
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This 0 page Class Notes was uploaded by Luna Bauch on Sunday November 1, 2015. The Class Notes belongs to CHEM 1414 at Oklahoma State University taught by Charles Weinert in Fall. Since its upload, it has received 24 views. For similar materials see /class/232942/chem-1414-oklahoma-state-university in Chemistry at Oklahoma State University.




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Date Created: 11/01/15
CHEMISTRY 1414 F L 2011 Practice Problem Set 2 1 Provide the missing name or formula BaCN2 RuSO4 VLOJ H91 C I Us 5 in Se a lam twink B203 T120 TeSz Y r 3RD F34 03 gt 0 NLOH PIF i C14H30 C7H16 Csz SbCls XeF4 ASVOs CLHI L Lamevim ID SWth vanadiumIII oxide mercuryI chlorate hexaantimony nonaselenide OILov 11 M DXQJC T mllium 05 th nthum L12 m Male strontium pcrbromate ironIII sulfate osmiumIV cyanate diphosphorus tetra uoride TCTLA A at Am L LPT u e QTLjNQ AMTiMou39 P WA MA JL X W0 Tenum you Jg tetraarsenic hexaoxide hexene Cal H ll nonyne Cu H17 tridecane as H propane N H nl L Lz 039 ammonium dichromate RhN033 Mala m MTth BiZSeg 335mm M5 lwia ReHPO42 Mm my tailqu pkme NH4H003 AranufuM Elem mm PdBrz minim In MMAQ Au203 T Till OXlo t 2 Calculate the requested amounts of substance CILHL M0leLvl V521 a A sample contains 19774 g of dodecene How many moles of dodecene are present 2 39 L I n BUM 39 lama 0 H7 mt My 11 MN U114 ng Mo39tcuM W a 13107 WM b A sample consists of 5661 moles of manganese ll acetate How many grams of manganese ll acetate are present May 1 Mel NU U W92 M quotM WWW 1 LI l L7 c A sample contains 17265 mg of cadmium nitrite How many moles of cadmium nitrite are present l f39quot lm aa WW 3 815 MollLULK W2 hr gg soil m d A sample contains 428 x 1037 molecules of bismuth pentabromide How many grams of bismuth pentabromide are present llDlgtLlo 7Mouuru 02509 Ll mo39bj Lo m 1013 Malaculu39 mo 3 Give the name of the following compounds when they are dissolved in water HI kx lmioollt ALlo Hc103 cklom Acio H104 remth Ana HNOz NlTlwf my H2803 SVlF lLOU 3c H3PO4 Fl Wffhuzll Aci 4 Provide the formula for the following acids phosphorus acid Hal 0 hydrobromic acid H BIL acetic acid H C H 07 perbromic acid H BRO hydrosulfmic acid 75 carbonic acid H L 0 Z 5 Name the following hydrocarbons C20H42 MOSME an22 Um e We C4H6 LVNdo 6le irwx eNQ 371116 emut CQHI6 NUUINQ CH4 quot We CHEMISTRY 1414 Fall 2011 CHEMICAL NOMENCLATURE RULES 1 IONIC COMPOUNDS General F 0m Cation0xidation state as a roman numeral Anion The following points always apply Group 1 Metals always form ions in the 1 oxidation state Group 2 Metals always form ions in the 2 oxidation state Oxygen is always found in the 2 oxidation state The halogens uoride chloride bromide iodide always form 1 ions Mercury has two possible oxidation states Hg 2 which is mercuryII and ng 2 which is mercuryI MercuryI cations always come as a pair and are found as the ng 2 polyatomic cation The following cations only have only one possible oxidation state as indicated Zn 2 Cd 2 3 Ga 3 Ag All other metals have more than one possible oxidation state Polyatomic Ions You are responsible for knowing the composition and charges on ALL of the following ions Note there is only one polyatomic cation other than ng 2 ammonium NH4 nitrite N02 nitrate N03 39 sul te 03 239 sulfate SO4 239 bisulfate H8047 carbonate C03 239 bicarbonate HC03 cyanate NCO 7 thiocyanate SCN 7 hypochlorite C10 7 chlorite C102 7 chlorate C103 7 perchlorate C104 7 hypobromite Br0 7 bromite Br02 bromate Br03 7 perbromate Br04 7 hypoiodite 10 7 iodite 102 7 iodate 103 7 periodate 104 7 hydroxide OH 7 cyanide CN 7 phosphate P04 339 phosphite P03 339 hydrogen phosphate HP04 239 dihydrogen phosphate HzP04 7 acetate C2H302 7 peroxide 0 239 oxalate C204 239 thiosulfate 203 239 chromate Cr04 239 permanganate Mn04 7 dichromate Cr207 239 When multiple polyatomic ions are present in a compound for example manganeseIII chlorate MnC1033 or ammonium phosphate NH43P04 the polyatomic anion is surrounded with parentheses in the formula Detem ning the oxidation state of a cation having multiple possible oxidation states Your knowledge of all of the above points will allow you to ascertain the oxidation state of ANY metal cation in an ionic compound If the metal cation is NOT one of those given above which have only ONE possible oxidation state then the oxidation state of the metal cation in the ionic compound needs to be speci ed Some examples C0203 cobaltIH oxide you know this because oxide is always 2 RuCl4 rutheniumIV chloride chloride is always 1 Hg20 mercuryI oxide oxide is always 2 Hg0 mercuryII oxide oxide is always 2 IrCN3 iridiumIII cyanide the charge of cyanide is known WC03z tungstenIV carbonate the charge of carbonate is known Fe2SOg3 ironIII sul te the charge of sul te is known SnBrz tinII bromide halides are always 1 BiCzH3023 bismuthIII acetate the charge of acetate is known BUT If a metal cation with only one possible oxidation state is present in a compound its oxidation state is NOT speci ed in the name For example ZnClz is just zinc chloride NOT zincII chloride and NaN03 is sodium nitrate NOT sodiumI nitrate 2 COVALENT COMPOUNDS The names of these compounds include the number of atoms of each element present given with greek pre xes The prefixes are l mono 2 di 3 tri 4 tetra 5 penta 6 hexa 7 hepta 8 octa 9 nona 10 deca These pre xes are used to give the number of each element present The element given LAST in the formula takes an ide ending For example P205 This is a compound of phosphorus and oxygen Since oxygen comes second in the formula it takes the ide ending and is written oxide The name of phosphorus is unchanged giving us phosphorus oxide But the number of atoms of each element needs to be speci ed There are two phosphorus di and ve oxygen penta atoms giving us diphosphorus pentaoxide which is usually written diphosphorus pentoxide the a is dropped Another Example Si485 would be tetrasilicon hexasul de HBr would be hydrogen bromide the mono prefix is seldom used 3 ORGANIC COMPOUNDS These are covalent compounds but because of their large number and diverse array of possible compounds have their own naming system Formulae for these species speci cally tell us the number of carbon atoms and the number of hydrogen atoms in the compound so all of them have the general formula CXHy The ratio of carbon to hydrogen determines which of the three classes of hydrocarbon alkane alkene or alkyne and thus determines which ending to use If n is the number of carbon atoms then the following three formulae reveal the class of hydrocarbon CnHznz alkane ane ending CnHZn alkene ene ending CnH2nz alkyne yne ending Once this is determined the number of carbon atoms is counted but using organic terminology for the prefixes as follows of carbon atoms prefix of carbon atoms prefix l meth l l undeca 2 eth 12 dodeca 3 pro 13 trideca 4 but 14 tetradeca 5 pent l 5 pentadeca 6 hex l6 hexadeca 7 hept l7 heptadeca 8 oct 18 octadeca 9 non l9 nonadeca 10 dec 20 eicos You are responsible for all of the above pre xes Now the pre x is simply combined with the appropriate ending ane ene or iyne depending on to which class of hydrocarbon the compound belongs This is determined from the formula Some examples C5H12 Conforms to CH 9 ene ending Six carbons 9 hex prefix This gives hexene C14H25 Conforms to CnHZn 2 9 yne ending Fourteen carbons 9 tetradec prefix This gives tetradecyne C8H18 Conforms to CnHznz 9 ane ending Eight carbons 9 oct pre x This gives octane C4H5 Conforms to CnH2nz 9 yne ending Four carbons 9 but pre x This gives butyne 4 ACIDS Some hydrogencontaining compounds are classi ed as acids which when dissolved in water transfer a proton to water and generate the H30 cation The compounds typically appear to be similar to covalent compounds but are named differently if they are functioning as acids NOTE acids are only acids in solution which for our purposes will be water so HCl is hydrogen chloride BUT HCl dissolved in water is hydrochloric acid Nomenclature a Acids of the general formula HA where A is a NONOXYGEN containing species are named such that the A species which ends in ide in the covalent compound takes on the ic suf x So bromide becomes bromic cyanide becomes cyanic etc The pre x hydro is then attached to the name Examples H1 The covalent name is hydrogen iodide but in water it is hydroiodic acid HzSe The covalent name is hydrogen selenide but in water it us hydroselenic acid b Acids having polyatomic anions from the list above are named differently First look at the anion If it s name ends in ate the acid s name will end in ic If the anion name ends in ite the acid s name will end in ous The pre x hydro is not used in these cases and so the acid names are derived directly from the name of the polyatomic anion Examples H3PO4 The name of P04 339 anion ends in ate so this is phosphoric acid HClO4 C104 perchlorate so this is perchloric acid HBrO BrO 39 hypobromite so this is hypobromous acid H2803 03 239 sul te so this is sulfurous acid CHEMISTRY 1414 Faun 2011 Practice Problem Set 3 l The combustion of 1621 g of a newly synthesized compound which was known to contain only carbon hydrogen and oxygen produced 1902 g of water and 3095 g of carbon dioxide What is the empirical formula of this compound AMOW 1 W51 39 3 0 3 01ILJH 7 C 1 0 3 151 i i o l 3 L0 1 Amour or kyaquf Lam I4o1olt7i quot 01 H 30i7HL QMNut M45170 3 Hugo WON C 39 7 39 W quotquotquotquot quotquotbu1Ww w lloo LI 7 75 UMw loo i 76 WWW 100 7 117 137 EFL 7 ASNNA loo 9mm 1 bermowe Mom 0F L H M D M Man 7 i3DMo gv z7o ileum maximi I 117 m 6007lmm berm w We witUm Fungii CWZH no Din 9 1H 0 3 CLH O quotquot 5 i In In Ln 2 A compound containing xenon and uorine was obtained by the ultraviolet irradiation of a mixture containing 0526 g of xenon and uorine gas If 0678 g of the new compound is isolated what is its empirical formula libcmmw AM 0F HUM5M 3 New Lamond Myra F W DP va1 omrow1 M or X11 1 OJ 0 LF U 0W 3 Mokr of Ke F 391 7 ISIS r Fan 0K1 J 1 1 Li iiio S MXQ Vi bDO L DJMol F Mu 39Cf39009M 3 Durant MoiM Wig K We Tke AJNMIK M4 Rem X y F OU WMQ ANT iiiN A Wr Q J Home Mm 1 e FL GA Molecum i m The Lioifloq m g I tLolex It DW M 3 Win Nam59K 3 An organic compound is composed of 6315 carbon 530 hydrogen and 3155 oxygen by mass The molar mass of this compound is 15214 gmol What is the empirical and molecular formula of this compound IlAssum A mi rmrlelhw W m a DP Mow OF L HI M o 63K C 53930 14 9 Emma L 52mm M 1mm l101M0 ioab iMH 150070 1 DWMWQ Qmpikiui Fommu4 C 711 gt77 H117 DJXE 3 314 03 RV t 1 1017 yiqn W l f w 51 3 N W Wkult NUMBW 3 bvtPtmmt Dxc M ttvm Mm1A 39 Ewemu F LMulA wxiqkr 2x ILDH7Mo 2 l0037vu 3x ibJjabllel 3 lSZ39 Mo 39 Siltca N1 emfikiu FoLNNM were 1V0 lhifif AMAdeli Add SherwinRrwm WYTl x Q L H L 00 m cu 1 Li Pomuu J MD 4 Sulfur trioxide is an important industrial material used as a precursor for the preparation of sulfuric acid It is synthesized from sulfur dioxide and oxygen If an industrial process in a pilot plant generates 8765 kg of sulfur trioxide determine S O t a The number of moles of sulfur trioxide produced g Mo 9 UM my 39 g o 05 ml xmfu so 8116 W15 ml 30539 b The number of molecules of sulfur trioxide produced I 0015 MMY030211910 m1m I 9 W3 x loumohww y o W l M l c The number of sulfur atoms AND the number of oxygen atoms present O qu SulFuK K RMOISMJQWW SOa fc yng mlu v50 lt MM 0 30 Moleu t r 3 IMV PCvR To VJOZS Man 5 Lol l leOUMI is O 5 A compound found in a human cell was determined to be composed of carbon hydrogen oxygen and nitrogen A combustion analysis of 135 g of this compound resulted in the isolation of 220 g of C02 and 0901 g of H20 When a second sample of this compound weighing 0500 g was analyzed for nitrogen 0130 g of N2 was produced Determine the empirical formula of this compound ll DENmm We m of C H v 7k PlU T SMI lC C1 LL03 OLLIUH L 2 01007 H39 090171110 1016711 0100 41 H017 L01 mo 7 11 re t m 7 oF N Law We 92th fwrlc 5 NDTe All 71M l i All MW T Wu 0 03M 390 73997 OLr wwu Me Me mm 0 003 rmrlr 3 NM 90 Aw7 PM CHD C 06007L H 01007 erooiH7 01790 I My 1 1 Us W31 0 7 7o loo W H l t 16 2 1217 H BQIMMVU Molar of Cll lINi04 nM A 1V97Tklfl1 217 quot0 Ll 37m1 c H39 1101 721m u 7 13mm ll lml LoomM l 5 01MI Al39 r 1L0m L AM 39N HWY l l quotIIMl C2 H3 Nil Om CI7 HM leol 73quot Last 13 73971 i L I i 6 An industrial byproduct produced in Oklahoma and detected in samples of water 39orn the Arkansas River is composed of the elements carbon hydrogen oxygen and chlorine The combustion analysis of 0100 g of this compound yielded 00682 g of C02 and 00140 g of H20 Analysis of the sample for chlorine content indicated that the mass percent of chlorine in the compound was 550 What is the empirical AND molecular formula of this compound l hefeLmvo 7k MuirFf 9F Cip IA mg war 00L311COALILJIMQ j 000ch 00w5Hco 0159 Wmqu 000 m7 H WNW 1 ACmM N Mr 7 f oq QAu QimzJT 70L 00l bL7L quot76H 0005ng N 70 C M 00 D 1005 SMrL00 040 C 7 7 0 z 00 1 157 Esme U 7 b 2 hwULm rvl Make OF ALL 40va WIN A no MP 2 LSW39IMM C I m lLNt lmq LMMm 303 U ES Cqml mgm 0 if 0 JEOOTIMM 1 M O L Imean MOML W05 Cl HLKL LlIISS OHM CHCID Mb U m Hi 7 QMPUQUH W V 39 WIM RNAMme ML m we FezLN H Mrw v g39 1T FOMWI QMMAQ 0 ll CU H x A Msww Mm lth Tu mom CHEMISTRY1414 Aquot OIquotMimi 9W FALL 2011 39 3101va FIRMU PracticeProblem Set1 noquot jwu quotU CIHSIWL39W W oquot l e U 1 Perform the following conversions 8 I W k E M Crag MT a Convert 374 uid ounces to 3 wwwll mL s7vrm HUM HMO ML fjw 7 39 i F39 0 39 IL JL MMM dekallters 5 0 U S ML l 4 lull M l W quotWm lo L DM L Petaliters3lLl FIOk 2515 me IL qr lFIO1000ALIXIDIIL I39HX39O m3 37HFID L LWHML quot3 la 3 3 3 m M Mme m I F1 0 ML 0 b The density of a metal alloy is 728389 ouncescubic inch Convert this to gmL nanograms per attoliter g Q N lg L 6LT PM GgdaL c Convert 9756 F to C TM 473w 23 39 ELMT 39 Hwi 39ll39qifql DUCK 30757 M m ML adjML lthybclwl llM 3ILMSL k1 jltlopoggt Ns 15m 7 TH 11 T Heo HD 9 L 14 Nj LL M 01NjlooomL IL 3 ML Ij IL xm39zaL 1 D10393N9 L I 77 MM 7ML Q 5 ML IIL 1 jL l O Et lwo quotum t oi l w jtv uo b m3 7 y ML HWOAL I66 3 IlLDHD M ML 3N WM 3 6 gt 2 MO M BLDAL IDL 9 F L39G w l L ML DOD j 113903 quot 11L 1101 rdm d Convert the temperature of liquid nitrogen 770 K to c 39 0K W 17l 39t l lLZquotC Tl g 17 W 2 39 0 F Tl39Fl1 xplt1Lt39c 4 22 4111 F 396 e The closest possible opposition of Venus to Earth is 237 x 106 miles Express this in N0quot 39 74 I llmT gum1 Sci omm quot M lOLM FLDPV lLiu lig iu 1M Notk n39w39 Maw V imlt lHM mu lovml ggulom q Tm ngxlo w39MYHWZIWMN In IT M j M39 lr r llu mu MXHW Squot rm pm in 4 IL Bll0 1TM 4 18le0 M 9 f13 xlDUr I M pkomnw M M 37smo39 m um melu Hub 1 Convert 6843 x 104 oz yd2 y 392 to Joules 02 ounces yd yards y years 4 1 L 2 7 Ht 7 34 2fch Y39u lv up M Z W In 21 M 13 10quot an 171 Too L l l3x w 391 K7M51 lt 3 lo L 3 2 A sample of antimony metal occupies a volume of 6422 mm3 If the density of antimony is 666 gmL and the atomic mass is 121760 gmol how many atoms of antimony are present in this sample HEM am DE EL mum ici w a 01111777 10M lLM ML Ly SJ F47 O l 7jlt Mon ttzuuol tw Livia IUJLOJ Mel Mom 54 3 A 78655 lb object is moving at a speed 186000 milessecond What energy in MI is exerted by this object 51 V M 5 112 ZSHMN 1M 5 44 1 FT 1 w IDDw MM i 1 o 7 WHO TMJHLltWL 35L nu L W m1 9 1 H l i l o U EZ LMVLP E Ligg ijHZ39quDM il 4 Perform the following operations and express the solution with the proper number of 3 i 9 lt1 01 1 3 signi cant gures 39 11 WT a 00820627315 12 00 510 quot 1W5 2 297 quotOB I M M 325 x7006 W A 1171 i m MW 0 W quot i4 0 b wg 5 AMNA b 6040105332l81l80x10 1171605 W 3530x10393 L 0 73 AMI x 31mm 1300 3 W 3 3lgtlt 0 3 angzg ol golmi 3 0MUD 4 WWW M o S mgv to


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