Final Exam 06Part 1
Biology 131
Examination 3
April 4, 2006
You may separate the pages as needed.
There is no need to turn in this part of the exam.
NO CALCULATORS!
ln
[
]
[
]
o
x
i
z
RT
z
FZ
x
V
x
=
V
mV
X
X
mV
X
X
x
o
i
o
i
=
=
25
58
ln
log
2
1
2
1
1
2
(
)
(
), at equilibrium,
= 0.
P
P
s
s
∆Ψ = Ψ − Ψ = Ψ
− Ψ
− Ψ
− Ψ
∆Ψ
,for a monovalent cation, X.
or a diavalent ion, the above equations must be divided by 2 (z=2), thus
in 3, 2 and 1 dimensions, respectively.
Thus, t increases with L , so t = L /6D, L /4D, L /2D.
he pressure generated by the meniscus of water: P = 2T/r
c
, where T is the surface
1 Pa
onstants, Algebraic Relationships, and Conversion Factors
100 = 2; log 1000=3, etc.
log (A
•
B) = log
= 1 + 0.7 = 1.7
Avogad
6x10
As prefixes to units,
, n = 10
9
, K = 10
3
Speed
c/
λ
,
J = D
∆
C/
∆
x; J = P
∆
C
F
V
Ca
= 29 mV log (Ca
o
/Ca
i
).
For a monovalent anion such as Cl

, z = 1, so
V
V
V
m
i
o
=
−
V
Cl
= 58 mV log (Cl
o
/Cl
i
).
F = 10
5
coulomb/mol
(RT/F) = 25 mV
3
2.5 10
2.5
s
j
n
n
n
n
Joules
L MPa
RT
C
C
C
mole
mol
Ψ
= −
= −
•
= −
∑
∑
∑
(
)
P
=
−Π
P
s
Ψ = Ψ
+ Ψ
where C
n
refers to the concentration of the n different solutes in the solution.
2
2
2
2
L
D
Dt
Dt
=
6
4
2
t
,
,
T
tension of water (7 x 10
8
MPa
•
m) and r
c
is the radius of curvature of the meniscus.
= 1 joule/m
3
; 1 atm = 0.1 MPa = 760 mmHg
≈
10 m H
2
O.
C
log 1=0; log 10 = 1; log 2
≈
0.3; log 3
≈
0.5; log 5
≈
0.7; log
log (A/B) = log(B/A) = log A – log B
A + log B; thus log 50 = log 10 + log 5
log(10
x
) = x
23
ro’s number =
1
µ
m = 10
3
mm = 10
4
cm
1 L = 10
3
mL = 10
3
cm
3
c = 10
2
, m = 10
3
,
µ
= 10
6
For a sphere, Volume = 4
π
r
3
/3, Area = 4
π
r
2
, In Bio 131,
π
= 3.
of light in vacuum, c = 3 x 10
8
m/s; energy of a quantum, E = h
h = 6.6 x 10
34
joule s
1
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