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Chem 107 Homework chapter 12

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Extra HW chapter 12
dati e
1) P= SOatm
ΔV = 974-SYG
=Y32L
P.LV +-3004321 =-2.16x10"Latm
2.16x10"(atm x
= -2.19x106
w=
-
17 known
obh
-
2)W =P.LV
QV
P 0.98atm
=
w
=
000m)-1sOmL 6SOmL
(p
=
=
0.98.(0.(s)
-
-0.6s(
-0.637latm
=
n(Ne) 0.500 molp, = 1.00atm T.
=
C) metallic heat
-1<p 32x0.315
=
MRT
=
TV
=
w
=
b)Ga(s)
=
-
Xv
1592 0.36 JIng
=
pΔV +
-
n(vOT
=
1.07
=
=
9
Jluy
0.26
vu
positive
12)q m.cs.
9
W =0
+
9b) positive or
=
temp of metal = 200
WT (T+ 280)L
temp H20 = 1000
=
-
OECT+ 100%
C
=
-
-
-
a= q
(60). (0.389).
F20
T+
-
(i7
Tt
(T-24
=
1200). (4.221. (TH-100
notc
=
100
-
20) 36.16(T7
=
-
383
=
-
100)
97.84°C
=
13)9 mcUT
=
-
a = q2
M.. 2310UT
-
-
(91.
OT
=
M,o(szo NTz
CSDO OTL
we
=0
-
(4)ac
=
=
-
to
as a
q
MeCszΔTc =-M, cs, 84
M2(szCT
-
T2) m, CT
=
T1 92,000
m.
=
T2 20.002
=
17.0gx10(six (77
-
-
17.0g
mz =
=
T.)
(S2 = 10Cs
17.0g
20.01%
=
-
17.0gx<si
x
(tx
10+7-ZOOCEHFtan.I
16)
-
-
-
q, 9x
+ 93
=
(m, (s1
.
(4.18)(T)
-
0
+1)
-
-
4.
an+(My o(s3
=
0) 333.4
=
4.18T7
333.4
=
+
+
·
UTz)
4.18(T+ 100)
-
4.1857
-
18T7 41UT-84. G
=
4. 18T7
+
4. 18T7
8. 36T7
k+ 1
=
=
84. C
84. 6
10
=
-
w
=
=
UT 0
=
NU-w
0
-
0
-> 0
=-
121.35
-
1- 2.35)
=
10.12
=
418
-
92.0°C)
=
-
273) =
70.125
-119.05
WHElOomolBrKaOOoerc
Jlug
Cs m2 4.2 2
w0
290H
negative
Jlng
0.389
323+
-
0
ΔT
=
=
9=
massMFCoO
csmp
+
12,48x (210
=
(8u,+ (ur),(x
=
393.1xx
10S
Δv = 9 +w
negative
x(w,+ wz)
we
-
x
x
0.01(3)
=
P=0
=
=
OV, q
-
Qu 0.500
=
-
(8)ov - f(90)
9) alWork: -PoQU
UV 0 sow 0
NV, a
xhEb =O.Oll3S
uRteVitooxO.BLC
0.100x103x(0.0437
=
=
cag(s)
20.7951Mo
=
0.200
=
leg
b)
=
capacity 2S
mol
5.qu 0.44I
a)vu(s)
273k
=
pV
0.6370 101. 3255=-64.S45
-
=
12 = 0.200atm, Tc=210U
3235
-
393.12E
1,2,8,9,12, 13, 14,
16, 17,18,29
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