EE CHEM 103 GENERAL CHEMISTRY MT EXAM, SECTION 3 ID NAME SCORE

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1 EE CHE 103 GENERL CHEISTRY T EX, SECTION 3 ID NE SCORE ERSN KORY ERŞN GÖKHN KY USTF ÇĞLR GÜLDİKEN ERYE GÖÇEN İRHİ UTLU ENES KVKLI HET SI ÖLÜKŞI URT RSLN GZİ ERT ŞHİN ERT ŞHİN İZZET CİHN KÜÇÜK TUNCY ÇELEİ HLİL ÖZDOĞN ÜŞR URTOĞLU ERTUĞRUL SVŞCI ERDİ LN FURKN Y UHET SERHT DÖNEZ PERİHN SİNCER SERHT EİR GİZE YLIN HTİCE GÖZDE DENİZ SEDT KPLN EHET OZKURT EHET CN TNERİ UĞR EKEN ERE ERSOY TUNHN SERT RZN RIK SÜHEYL KHRN SKİN PETEK UHED YÜKSEK ZİZ TNER STRLIOĞLU OSN YZ OR ÖZTÜRK SERCN ÇEVİK SİEL ÇETİN USTF ZENGİN ÜNİRE ERKN FTİH ERE ERTÜRK İSİL HDIR GELGÜÇ USTF YILZ NESRİN RICN RHN HUDUTI KN KNR FTJON PİLİNÇİ TYFUN CR 7.5 VG HET ÇĞKN Y ESUT CE İLRS ERT JONUZI UHED ESEN ERR TUĞCE GÜÇLÜ USTF ŞHİN HSN HÜSEYİN TTLIDİL ULŞ DÜDEN 1 ydd

2 EE CHE 103 GENERL CHEISTRY T EX, SECTION 4 ID NE SCORE ÇĞRI KESERER İSİL TÜRK HUT OĞULCN ŞHİN FIRT COŞKUN DLGIÇ NUR CEREN ŞHİN EHET KUİLY ÇĞL ÇİÇEK İLL ÖZTÜRK SERKN YDIN SÜLEYN KİR Rİ KCI HR YILZ PINR DOD YHY ESER ERKUT STIRCI YUNUS OKTY ÖZTÜRK NDİRE RZU TOK EİR CN NKIŞÇI OĞUZHN ÜNLÜ ERH KKCI Lİ YİĞİT UZUNGİL ELİF KKY GİZE URCU KY NESLİHN KÜÇÜK EKİR Sİ ŞHİN Zİ SEVGİ TURN DURZ ORHN TŞKLE GÖKTÜRK İÇİCİ EHET SLN HZL ÖZDEİR ZÜLFİKR ERE TN EHET EİN KOY HEDİYETULLH GÜNDÜZ SEİH POYRZ ERE HŞİ FEYZEDİN ERKPLN ERT SUŞIY SLİH NUR GENÇ GÜNER DİREN YUNUS GÜLİÇİ LİHN CEL ÇLIŞKN HLİL İRHİ D KDİR OĞUZ HDİS LLTİPRK VİRNİ ŞKURT UHET URK CNOLT UHED FEYZİ LCI CEVT KNLINURO ÖER VDN ÖZCN KILIÇ 3.0 VG USTF DÖNEZ DULKERİ HZIR ydd

3 GÜ DEPRTENT OF ELECTRICL ND ELECTRONICS ENGINEERING Kİ 103 GENERL CHEISTRY IDTER EX 1 16/11/011 Dration: 90 min. Q no [Pts.] 1. [1.5x10]. [3x6] 3. [0] 4. [x1] 5. [x1] Total: 101 Pts. earned Some sefl constants and eqations Ideal gas constant, R L atm K 1 mol J K 1 mol 1 H:1.0 C:1.0 N:14.0 O:16.0 n:55.0 tomic masses (g mol 1 ) s: U: U:38.0 F:19.0 Instrctor: Dr. Yavz Dede NSWERS TO QUESTIONS 1. Designate the statements below as tre (T) or false (F). i. SI nit of mass is kilogram. ( T ) ii. Total mass and total charge are conserved bt total nmber of moles is not always conserved in a chemical reaction. ( T ) iii. Law of definite proportions state that; an element always has an integral nmber of protons and netrons. ( F ) iv. In the famos alpha particle experiment Ernest Rtherford discovered netrons. ( F ) v mole of 1 C and mole of 56 Fe weigh exactly and grams respectively. ( F ) vi. Dalton s proposal of atom indivisibility is correct. ( F ) vii. Cathode rays are made p of electrons generated nder high voltage. ( T ) viii. Isotopes of an element have identical nclei. ( F ) ix. Half reactions do not occr independently. ( T ) x. ain grop metals do not assme negative oxidation states. ( T ) Good Lck! ydd 1/4

4 16/11/011 Dration: 90 min. ) Nicotine is a compond that contains C, H, and N. If 1.15 g of pre nicotine is brned in oxygen, 3.30 g CO, g H O and 0.10 g N are the prodcts of combstion. (C: 1.0 ; H:1.0 ; O:16.0 ; N:14.0) a) What is the percent composition (by mass) of nicotine? b) Find the empirical formla of nicotine. c) If molar mass of nicotine is 16.0 g/mol, find the moleclar formla of nicotine. a) C X H Y N Z + O H O + N + CO n H O = g / (18 g/mol) = moles n H = x = moles n CO = 3.30 g / (44 g/mol) = moles n C = moles n N = 0.10 g / (8 g/mol) = moles n N = moles x = moles ratio C H N ; divide all sbscripts by C 5 H 7 N ; simplest mole C 5 H 7 N : 5 x x x 14 = 81 g/mol %C = 5 x 1 / 81 x 100% = 74.1 % %H = 7 x 1 / 81 x 100% = 8.64 % %N = 1 x 14 / 81 x 100% = 17.3 % Optional test: = 100 b) empirical formla of nicotine C 5 H 7 N c) moleclar formla molar mass = n x (mass from empirical formla) 16 = n x 81 ; n = ; expand the formla by ; x [C 5 H 7 N] C 10 H 14 N Good Lck! ydd /4

5 16/11/011 Dration: 90 min. 3) KnO 4 soltion is to be standardized by titration against s O 3(s) according to the reaction given below g sample of s O 3 reqires. ml KnO 4 (aq) for titration. What is the molarity of the KnO 4 soltion sed for titration? s O 3(s) + no 4 - (aq) so 4 3- (aq) + n + (aq) + H + (aq) First, balance the eqation. H + on the prodct side is a hint for acidic medim. s O 3(s) + no - 4 (aq) so 3-4 (aq) + n + (aq) + H + (aq) s O 3 so 3-4 no - 4 n + s O 3 + 5H O so H + + 4e - ; mltiply by 5 no H + + 5e - n + + 4H O ; mltiply by 4 5s O 3 + 5H O + 4nO H + 10sO H + + 4n H O 5s O 3 + 9H O + 4nO 4-10sO H + + 4n + balanced s O 3 : x x 16 = 198 g/mol n s O 3 = / 198 = 5.45x10-4 moles n KnO 4 = n no 4 - = 4/5 x (n s O 3 ) = 4/5 x 5.45x10-4 = 4.36 x10-4 moles [KnO 4 ] = 4.36 x 10-4 moles / (. x 10-3 L) = ) 1.00 kg gaseos mixtre of btane (C 4 H 10 ) and propane (C 3 H 8 ), kept at a temperatre of 3.5 C in a closed container has a total pressre of 1.76 atm. If the ratio of mole fractions χ btane is three: χ propane a) Calclate the partial pressres of each gas in the mixtre. b) Calclate the volme of the container. a) if χ btane =3, then Pbtane =3 where P χ tot =P btane + P propane = 1.76 atm. propane Ppropane 4 x P propane = 1.76 atm P propane = 1.76 = 0.44 atm P propane = = 1.3 atm b) PV = nrt from the ratio of partial pressres; n btane =3 n propane Let n btane = 3n and n propane = n 3n x (4x x1.0) + n x (3x x1.0)=1000g 174n + 44n = 1000 n = moles 1.76 x V = 4n x x ( ) 4 x x 0.081x V = =53.9L 1.76 Good Lck! ydd 3/4

6 16/11/011 Dration: 90 min. 5) Uranim hexafloride is sed in ranim prification for frther applications in nclear energy. It is one of the few componds of ranim that exist in the gas phase at moderate conditions. Two isotopes 35 U and 38 U are separated by making se of the difference in moleclar speeds of gaseos UF 6. Calclate: a) The root mean sqare speed (in meters per second) of UF 6 molecles RT bearing 38 U isotope at 55 C if mean sqare speed is given via 3. b) Calclate the ratio of the speed of 35 UF 6 to 38 UF 6 nder the same conditions. a) 38 UF 6 = x19.0 = 35.0 g/mol = 35 X 10-3 kg/mol 3RT 3RT 3x8.3145x( ) J / mol 3 35x10 kg / mol = 15.5 m/s b) 35 UF 6 = x19.0 = UF 6 = x19.0 = RT 3RT 35.0 = Good Lck! ydd 4/4

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