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**Unformatted text preview: **Chemistry 103A Summer 2007 June 22, 2007
Exam 3 Name ED Sig nature Show your work on all computations so that partial credit can be given in the event of an arithmetic error. No
credit will be given for bare numerical answers. You must show the numbers you combined to obtain the
answers. Points will be deducted for not including units and for misuse of signiﬁcant ﬁgures. Helpful Constancies: 23 molecules Avogadro no. a 6.022 x 10 mol i, .9. atcﬁﬁaéiﬁﬁ ﬁqmﬁ:
5 "Esme-“P?..WWWWS;iiin-S’JMiilgi-ais5-3
. . .............. .. .. spé'cifiCHaét”
Periodic Table is on the last page. Capacity
Substance (Jig-KP
Elements
Formula AH?(kI/mol} Formula 415%;me Alumiflum A1 0900
Calcium Nitrogen Graphlte, C 0.7] 1 _
Cafs) 0 Nag) 0 Iron, Fe 0.450
CnOt ) “635-1 NHsts) —-45.9
Caccjggs) 4296.9 New) 99,3 COPPER C" 0-337 l
Carbon Oxygen Gold, At] 0.129
C him) 0 0 (E) 0
cgﬁgmnd) 1.9 Oitg) 143 Compounds !
C0 } "110.5 1! 0(3) ”241.8 E
cogs) -393.s 1320(1) -—2ss.s water, H200) 4- I 34
mpg) "74.9 Silver Ethyl aicohol,
ﬁrst? 11(1) “335 £32311 ) 1223 o CZHSOHU) 2'46
5 — .
C520? 87.9 Sodium Ethylene glycol,
Chlorine Nets) 0 (CHZOH) 2 (I) 2.42
(31(3) 121.0 Naig) 107.3 Carbon
Cl (2) 0 NaClm ~411.1 . g
Hélfg) "92.3 Sulfur tetrachloride,
Hydrogen Safrhombic) 0 C014“) 0.862 E
Hrs) 218.0 s museum) 0.3 . . m g
Hats) 0 532(8) —296.8 Solid materials 5
803(3) -396-0 Wood 1.76
Cement 0.88 i
Glass 0.84
Granite 0.79
Steel 0.45 “At 298 K (25°C). Chemistry 103A Summer 2007 June 22, 2007
Exam 3 f. (8 pts.) Two gaseous pollutants that form into auto exhaust are 00 and NO. An environmental chemist is studying ways to convert them to less harmful
gases through the following equation: 00(g) + NO(g) “we 002(9) + %N2(9) AH = ? Given the following information, calculate the unknown AH: Equation A: 00(9) + €02(g) —> 002 (g) AHA = "283.0 kJ
Equation B: N2(g) + 02(9) ——-~> 2N0(g) AHB = 180.6 kJ Solution Noting moles of substances in the target equation: There is 1 mo! each of rear:-
tants CO and NO‘ 1 moi of product C01. and 1 mol of product Nil
Manipulatiﬁg'me given equations: Equation’A has the same number of moles of CO and
CO; as the target, 5": we leave it as Written. Equation 3 has twice the needed amounts of
N2 and NO, and they are on the opposite: sides from the target: therefore, we reverse Bqua~
lion B, change the sign. of M13, and multiply both by 11;: '
armors) ._.. Nae) “FOIQ’H Mm am: = also-ska
1 Nos) -—+ mg) + loge) AH == —9o.3 1:! Adding the manipulated equations to obtain the target equation: nation A: ' COQg) + M --'~» 002(3) M : ~28“) k5
lEéquaﬁdn-B reversed): ' " News) m... éN’zfg) + £02123; AH 0 g 739'? .31.
Target: 00(3) + Nets) w» (302(3) +§N1<33 MET-3&3. k; g
l
E
E
i ~mmm mmwmmmmmnmmmuwm §
§
i
i Chemistry 103A Summer 2007 June 22, 2007
Exam 3 2. (8 pts.) An unknown votume of water at 182°C is added to 24.4 mL of water at 350°C. if the final temperature is 235°C, what was the unknown
volume? (Assume that no heat is test to the surroundings; density of water is 1.00 g/mL.) "91051 : qgnincd -[2.4.4 “11.0.00 ymL)}[4.184:‘;5](23.5°C ~ 310°C) = (mass)[4.184 ﬁat-Janos - 182°C) me I = (mass) (22 Mg) H70} 22“]- g mWEISS 53g = mass
1.00 1111..
l g = 53 mL V(mL) = 53g x Chemistry NBA Summer 2007 June 22, 2007
Exam 3 3. (8 pts.) Nitric acid, whose worldwide annual production is about 8 billion kg,
is used to make many products, including fertilizers, dyes, and explosives.
The first step in the industrial production process is the oxidation of
ammonia: 4NH3(9‘) + 502(9) '—’> 4NO(Q') + 6H20(g) Calculate AH; from AH: values. M93» = EMAngroducm " Enmgmsmnu}
=-‘ l4M?iN0(g)l + 6M?{H20(g)]l ** {4M?[Nﬁa(g)1 + 5133?i02(833}
m (4 mol}(90.3 1431:1101) + {6 mol)(-24l.8 kllmoi)
«[(4 mo!)(—45.9 kit/moi) + (5 memo kilmom
= 3611:] — 1451k1+ 184k} — 0k] 5 -—98ﬁkJ ' i
Solution Calculating Align: E
i Chemistry 103A Summer 2007 June 22, 2007
Exam 3 4. (10 pts.) A ground-state H atom absorbs a photon of wavelength 94.91 nm,
and its eiecta‘cn attains a higher energy level. The atom then emits two
photons: one of wavelength 128’! nm to reach the intermediate ievel, and a
second to return to the ground state. Rydberg constant for hydrogen: RH = 1.096776 x 10Tm'1 1 nm =10'9m a.) What higher level did the eiectron reach?
b.) What intermediate level did the electron reach? 0.) What was the waveiength of the second photon emitted? a
<
i
i
%
E
g
E
5
t
§
§
5
g
g
§
§ 1 7 «I I i
7.76 —— = 1.0967 x10 0: -—-- — —-
a) It. ( )[nlz ”22
______‘.-.._..§...._ : (1.096776 KM)7 :11“: i2 .. 1.2.
9¢.91 x 10" m 1 n2 ii
i
i
0.9607 : [.35. _ if] 3
H2 ,
3—2 : 3.93x10'2
”2
n2 3 5
b) _.._.._..__l._.__....._ : 0096176 x107 m")[-_L _. 35-]
i281x10‘9 m nﬁ 5
1
7.118 x10"2 :2 [~12— — 7]
”i 5 it
0111 n —2
”I §
"1 z 3
c) l = (1096776 x 107m" [i - 1..)
l I 12 32 E
g
3— = 9.749 x106 m"
x
1 9
Mum) 3-"- i.026x10'? m x 0 run :1 E01600} 1 m g \Wwylwmv Chemistry 103A Summer 2007 June 22, 2007 Exam 3 5. (8 pts) Write the fuli ground-state eiectron configuration for each:
a.) Cl b.) s; 0.) Sr 8.25 a) CI: 1322322p63s23p5
b) Si: 1.323221063351102
0) Sr: l522522p63323p64323d'04136552 6. (8 pts) Arrange each set of atoms in order of decreasing 1E1 (first ionization energy): g
a.) Na, LE, K g
b.) Be, F, C
0.) Cl, Ar, Na d.) Cl, Br, Se 8.56 a) Li>Na>K b} F>C>Be
c) Ar>Ci>Na d) Ci>Br>Se (MAMmz’l’xu’iwﬁ).“rumrwrmxmﬁmwxm&.xmmstmww *mmwwmm m‘ﬁﬁn’ﬁ «Minnow $WWW¥WNMWW¥WM> mm June 22, 2007 Summer 2007 Chemistry 103A Exam 3 ...

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