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fall 11 - CHEM 14A-1 YOUR NAME flflfh/EKJ‘ Instructor...

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Unformatted text preview: CHEM 14A-1 YOUR NAME flflfh/EKJ‘ Instructor: Dr. Laurence Lavelle FALL 2001 1ST MIDTERM (Total number of pages = 7) (Total points = 50) (Total time = 50 mins) “Carefully remove last page which is your Periodic Table." YOUR DISCUSSION SECTION ...................................... WRITE IN PEN (Show all your work on this paper.) Constants and Formulas Planck constant, h = 6.626 x 1o~34 J - s = 6.626 x 10‘34 kg. m2. s" Avogadro constant, NA = 6.02 x 1023 mol'1 Rydberg constant, R = 3.29 x 1015 Hz Gas constant, R = 8.314 J.K".mol'1 Mass of electron, rna = 9.11 x 10'31 kg Speed of light, 0 = 3.0 x 108 ms“ O°C=273.15K 1L=1dm3 1atm=101.325 kPa 112:3.14 hzn2 E=hv E=pc En=8—m-T_2 p=mv E - h E )t D >L n_. —n2— =p c=kv Aprx_4n _ Mammal pH _ pKA + LOG [AHIINITIAL ACID _ -B +/- \le-4AC Solution to sz + BX + c = o is x 2A Qi) Hexamethylenediamine, CsH16N2, is one of the starting materials for the production of nylon. it can be prepared from adipic acid, CsH1oO4, by the following overall reaction: C6H1004(|) + NH3(Q) + H2(g) -------- ‘9 CeH16N2(|) + H200) A) Balance the above reaction. (3pt) C; H” 0g (x) 7‘ Z (”é/7) 7‘ MIL/g) ’9 CHM/dz) man/x; W) (w) W B) What mass of hexamethylenediamine can be produced from 1.00 x 103 g of adipic acid ? (5pt) ' ~/ (fir) c; HM 12¢ Wm : xyd/WMV ht) C5 H/g/J/‘t Wm (”t7 /, oo x/o’lg ’/ qu Cur/4% W: (.246¢6~€X ”.4sz = 7’75.32/ Qrfl : a 75 x ”if (3%) ll Q .°\ N V: i C) What is the percent yield if 765 g of Hexamethylenediamine is made from 1.00 x 103 g of adipic acid ? 2pt) ( ??{7mu/4CW (WM) ?653/ W/gécfll ‘ (ms) 0AM: :75? W :m/ (vhf/5t?) 3 ()2) Vitamin A has a molar mass of 286.49 and has a general molecular formula of CxHyE, where E is an unknown element. if vitamin A is 83.86% C and 10.56% H by mass, what is the molecular formula of vitamin A ? (10pt) again. mMaf/W/I =284¢5wmw MW CW M mmacp/Mfl). {fig W C 2: 1% ¢7Wfl x 0.73% = ZW./?5’2/C EM “”3“; : £51354;— = /7.4?éwg 07/20 M0. /Z..0//?.M , M1?“ H. (M) W H = ZM¢3 K 0.10:6 = 30.2{432/7/ 07f) mew/4 =73Zé‘fgi2; c face/M Miami/i {24%) %MM C H E 2.0 30 WW W06 E = 2,376 {a «[20 (/z.o/} +34 [0032] . mwfiafW/q 2 Zié¢~ 27-0.¢¢] (27x) : /537{?«.hw(,~/ 03A) Calculate the wavelength of the radiation emitted by a hydrogen atom when an electron makes a transition from n: 4 to n: 2. (6pt) Q41) E? : —— AK 2 — (5-{2KX/37¢0152(314X/J 52 :13“ng (W E1 2 —l’ f. '5'. {LIX/o 1T l3’~/ ’l‘! 6"" // K 2.z 55 2 5“ ”E1: Zf/Z X/b—lgj) *6“ \f‘fif’X/fl M) 3) (2m W) (W 1w ? __ Z V) _ lzc __ (JKZKX/oj&3){?°xnnfl I _ IE —- ¢ oS’XK/ofl 4T ‘3 @XKK/E’Tlv 1‘ 9485 kl” For the following sets of four quantum numbers (n, l, m., m) identify the orbital or explain why it is wrong (i. e., not a valid orbital). (4pt) a) {2, 1, -1, +1/2} 2’0 W b) {1,1,o,+1/2} [’0 MW W/J/KJW 04A) The hydrogen atom has a radius on the order of 0.05 nm. Assuming that we know the position of an electron to an accuracy of 1% of the hydrogen radius, calculate the uncertainty in the velocity of the electron, using the Heisenberg uncertainty principle. (AP) (A )5 : £3: (6pt) ‘ . - .5. AK :- /%,f 0.037”), {M 6/.» I - 0.0/ X ”-05 - JF— - ”—11—“2‘K/JW “til Max/23;»? . r: /. [{JX/Ogh.f <Z7l/t') AV 2: [KM/”f (~/ it?) B) Comment on your value obtained in part A. (4pt) 05A) Write the full (all shells) electron configuration for Cr+ (2pt) ”1 z 53' Z,” 35"“ fp‘ 2’65 B) Which is lower in energy in transition metal ions, the 3d orbitals or 43 orbitals ? (2pt) (M C) Identify the M3+ ion for each ground-state electron configuration. [Ar]3d7 ' (2pt) + M1249 . [Ar]3d‘ (2pt) win?! 4) (2pt) ...
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