o -250 kj o X -50 kj o +50 kj o o

Transcription

1 Chemistry 111 Sect. 3 Nvember 2, 2000 Name: KEY Please put yur name and student number n all pages in case they becme separated. The last page f the exam has sme useful infrmatin. The exam lasts 60 minutes. There are a ttal f 100 pints. MULTIPLE CHOICE QUESTIONS Place an X in the bx crrespnding t the crrect answer. 1. (5 pts) If the enthalpy change in the reactin A B is H = +150 kj and the enthalpy change in the reactin A 2 C is H = +100 kj, then the enthalpy change in the reactin B 2 C is -250 kj X -50 kj +50 kj +250 kj +350 kj Reverse the first reactin and add t the secnd: B A H = -150 kj A 2C H = +100 kj B 2C H = -50 kj 2. (5 pts) The number f planar ndes in an rbital with quantum numbers n = 4, l = 2, m l = 0 is: 0 1 X The number f planar ndes equals the value f the l quantum number, the ne that determines rbital shape. Fr d rbitals (l = 2), there are 2 ndal planes. 3. (5 pts) The lcal radi statin WFCR bradcasts at a frequency f 88.5 MHz. Calculate the wavelength, in meters, f this radiatin X c = λν The speed f light c = m/sec, s the wavelength is λ = c/ν = ( m/sec)/( sec -1 ) = 3.39 m Page 1

2 Chemistry 111 Sect. 3 Name: KEY Nvember 2, (5 pts) A gram sample f the rcket fuel hydrazine, N 2 H 4 (mlar mass = 32.0 g/ml), is burned in a bmb calrimeter. The temperature rises frm C t C. The ttal heat capacity f the calrimeter (including the water) is 5860 J/ C. Calculate the mlar heat f cmbustin f hydrazine, in kj/mle. X , g hydrazine = (1.00 g)/(32.0 g/ml) = ml. H 0 cmb = (-20.7 kj)/( ml) = -662 kj/ml 5. (5 pts) An element has the valence electrn cnfiguratin f ns 2 np 4. This element wuld be fund in peridic grup: X IVA IVB VIA VIB VIIIA The amunt f heat absrbed by the calrimeter is q = (mc) T = (5860 J/ C)(28.16 C C) = 20,700 J That is just the heat prduced by 1 g f hydrazine, s the q absrbed by hydrazine is kj. Determine hw many mles 1 g represents, and divide by the # f mles t get the per mle heat f cmbustin. There are 6 valence electrns, s it will be in grup 6. Since there are n d r f electrns, it is in the main p-blck, s it must be in 6A (r VIA in the peridic table ntatin n p. 6 f the exam). 6. (5 pts) Napthalene, C 10 H 8, is burned in a calrimeter at a temperature clse t 25 C t measure the enthalpy change fr the cmbustin reactin: C 10 H 8 (s) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O(l), H 0 rxn = kj What is the mlar enthalpy f frmatin f napthalene in kj/ml? X The net enthalpy change in a reactin is H 0 net = Σ ( H 0 prd) - Σ ( H 0 react) kj = {(10 ml)( kj/ml) + (4 ml)( kj/ml)} - { H 0 f,napthal + 0} Slve: H 0 f,napthal = kj + {-3935 kj kj) = 77.9 kj Nte that this is fr 1 mle f napthalene. 7. (5 pts) Which has the lnger wavelength, an electrn mving with a speed f m/sec r a neutrn mving at the same speed? Nte: m e = g, m n = g. Page 2

3 Chemistry 111 Sect. 3 Name: KEY Nvember 2, 2000 X the electrn has lnger wavelength The wavelength f a particle f mass m mving the neutrn has lnger wavelength with velcity v is λ = h/(m v) they have the same wavelength S the electrn with its lwer mass will have a neither has a wavelength larger value f λ cmpared t the mre massive can t tell neutrn. 8. (5 pts) Fr the rbital type labeled "3d". there are x rbitals in the subshell that can hld a ttal f y electrns where x, y equals 1, 2 3, 2 3, 6 X 5, 10 7, 14 Fr d rbitals, l = 2, and there are 2l + 1 = 5 f them. Each ne can hld 2 electrns. MULTIPLE MULTIPLE CHOICE QUESTIONS Place an X in the bxes crrespnding t the crrect answers. Any number f answers may be crrect, including nne f them. 9. (10 pts) Fr which f the fllwing species is the standard enthalpy f frmatin nt equal t zer? Fe(s) X O(g) nt in standard state f O 2 X X C(graphite) He(g) CaSO 4 (s) nt an element S lng as all 4 quantum #s 10. (10 pts) Which f the fllwing are true? are nt the same, then A p rbital has tw ndal planes. d rbitals have 2, p have 1 electrns can X A paramagnetic material is attracted t a magnetic field. diamagnetic is nt be in the same atm. There X Tw electrns in a multielectrn atm can have the same value fr quantum number m l. In a bmb calrimeter, the final temperature f the cmbusted cmpund is different than the final temperature f the water jacket. the temperature f everything is equal A 4s rbital is filled befre a 3d rbital. n + l = 4 fr 4s, and = 5 fr 3d SHORT ANSWER QUESTIONS Answer in the spaces indicated. are lts f cmbinatins where bth electrns have m l Page 3

4 Chemistry 111 Sect. 3 Nvember 2, 2000 Name: KEY 11. (15 pts) The fllwing are based n the electrnic energy level diagram f the Bhr mdel f the hydrgen atm shwn abve. Phtns f the shrtest wavelength wuld be emitted frm the atm by the transitin labeled D. (highest energy drp fr an The transitin that results in the prductin f H + is labeled A. (n = means the electrn) rbital radius is, i.e. the electrn is free) In the transitin labeled C, des the electrn end up in an rbit with a radius that is smaller that the rbit in which it started? (yes r n) n (larger n crrelates with larger radius) Which f the transitins prduces the mst energetic phtn? D. (highest energy drp fr an electrn) The energy f the phtn that is invlved with transitin C is equal t three times the energy f the phtn invlved with transitin E. True r false? True The energy difference between levels with n = 2 and n = 4 is E = -Rhc/(4 2 ) - [-Rhc/(2 2 )] = (3/16)Rhc. S the energy f phtn C is 3 times that fr phtn 12. (10 pts) Match the valence electrn cnfiguratin fr E. each One f is the absrptin fllwing (E) by writing and the apprpriate letter next t the element. In the fllwing, n the is ther an integer. is emissin (C). C Al The energy difference between levels with n = and n = 4 is E = -Rhc/(4 2 ) - [-Rhc/( 2 )] = -(1/16)Rhc. Page 4

5 Chemistry 111 Sect. 3 Name: KEY Nvember 2, 2000 A. ns 1 A Cs B. ns 2 C. ns 2 np 1 G F D. ns 2 np 2 E. ns 2 np 3 F Se F. ns 2 np 4 G. ns 2 np 5 H Xe H. ns 2 np (10 pts) Ethanl, C 2 H 5 OH, has a mlar mass f 46.0 g/ml and bils at 78 C. Hw much heat energy in jules is required t heat 100 g f liquid ethanl frm 20 C t the biling pint and cnvert the liquid cmpletely t a vapr? (The specific heat f liquid ethanl is 2.4 J/g K and its enthalpy f vaprizatin is 40 kj/ml.) Answer: 101,000 J The amunt f heat t warm the ethanl is q = m C T = (100 g)(2.4 J/ C)(78 C - 20 C) = 14,000 J The 100 g f ethanl is, in mles, 100 g = (100 g)/(46.0 g/ml) = 2.17 ml. The heat needed t bil the ethanl is q = (# mles)(heat f vap) = (2.17 ml)(40 kj/ml) = 87.0 kj = J. Adding these tw amunts f heat gives us the net amunt necessary t warm and then bil the ethanl: q net = q heat + q bil = J J = J. 14. (5 pts) What is yur favrite clr? Des it have a higher r a lwer frequency than ultravilet radiatin? Explain with a sentence r tw that includes an equatin. Mine is green. N matter which is yur favrite, the clr is visible. Ultravilet light has shrter wavelength r higher energy than visible light. Because E = hν and c = λν, the frequency f visible light is lwer than that f ultravilet. Page 5

6 Chemistry 111 Sect. 3 Nvember 2, 2000 Name: KEY Useful infrmatin c = m/sec H 0 net = Σ ( H 0 f) prd - Σ ( H 0 f) react h = J sec H 0 net = Σ ( H 0 rxn) Rhc = J c = λν m e = g E = hν a 0 = nm E = -Rhc/n 2 N A = mlecules/ml λ = h/(m v) q = C m T Z* = Z - n inner electns q = (heat f fusin) m d = m/v H = q p Sme useful thermdynamic data Material H 0 f (kj/ml) Material H 0 f (kj/ml) CH 4 (g) H 2 O (l) CO (g) H 2 O (g) CO 2 (g) H 2 O 2 (l) FeO (s) N (g) Fe 2 O 3 (s) NH 3 (g) Fe 3 O 4 (s) NO (g) 90.9 PERIODIC TABLE OF THE ELEMENTS IA IIA IIIB IVB VB VIB VIIB VIIIB IB IIB IIIA IVA VA VIA VIIA VIIIA 1 H Li Na K Rb Cs Fr (223) 4 Be Mg Ca Sr Ba Ra Sc Y La Ac Ti Zr Hf V Nb Ta Cr M W Mn Tc (99) 75 Re Fe Ru Os C Rh Ir Ni Pd Pt Cu Ag Au Zn Cd Hg B Al Ga In Tl C Si Ge Sn Pb N P As Sb Bi O S Se Te P (209) 9 F Cl Br I At (210) 2 He Ne Ar Kr Xe Rn (222) Page 6