Problem Set #3 POLYMERIC MATERIALS
Due date 2/15/11
20 pts 1) Rayleigh ratios (R) were obtained at 25oC for a series of solutions of a polystyrene sample in benzene, with the detector situated at various angles () to the incident beam of unpolarized monoc

Intro Bio 151/153 Discussion Homework; week of 11/2311/25
Name: Joshua Litwicki
Discussion Section #: 609 1. Answer questions ac about the phylogenetic tree shown at left. a. (2 pts) Which of the following is more closely related to green algae: red algae

dimensional and that it experienced no losses. We know that the flow
in a tube is subject to the no-slip condition, and therefore there will be
viscous forces acting and energy losses taking place. One way to think
about this is to recognize that equation

flow. The weight of the fluid particle is dW = g dndsdx, 80 CHAPTER 4.
KINEMATICS AND BERNOULLIS EQUATION and its component in the ndirection is dWn = g cos dndsdx. Since cos = z/n (see Figure
4.4), dWn = g z n dndsdx The force acting on the fluid particl

1 2 V 2 dA (3.6) The volume, mass, and kinetic energy fluxes are scalars.
The momentum flux, however, is a vector, with a direction given by the
velocity vector V. Outfluxes are positive, since V will have a non-zero
component in the direction of the outw

An acids p K a depends on the stability of its conjugate base The stronger the acid, the easier it is to
ionize, which means that it must have a stable conju- gate base. Conversely, a weak acid is reluctant to
ionize because it has an unstable conjugate b

H
HH
H
H
H
p orbitals shown with one phase red, one phase black
The framework of the benzene ring is like the framework of an alkene, and for simplicity we have just
represented the bonds as green lines. The dif culty comes with the p orbitals which pairs

You met this reaction in Chapter 5 but there is more to say about it. The oxyanion produced in the rst
step can help stabilize the electron-de cient BH3 molecule by adding to its empty p orbital. Now we
have a tetravalent boron anion again, which could tr

C
C
CH
H
H
HH
H
H
H
both end double bonds are 134 pm
this double bond is 137 pm
both single bonds are 146 pm
typical values:
single bond: 154 pm double bond: 134 pm
All CC bonds 139.5 pm
The reason for the deviation of the bond lengths from typical values

is more about the discovery of cimetidine in W. Sneader, Drug Discovery: a History, Wiley, Chichester,
2005.
Check your understanding
To check that you have mastered the concepts presented in this chapter, attempt the problems that are
available in the bo

R2
NaCNBH3
R1 N H
R2
HH
amine: basic
As the reaction mixture is weakly acidic, the amine will be protonated and will be soluble in water. The
starting material and intermediate (of which very little is present anyway) are soluble in organic solvents.
Extr

=
=
prismane synthesized 1973
Dewar benzene synthesized 1963
143MOLECULES WITH MORE THAN ONE C =C DOUBLE BOND
The alternative drawing on the left shows the system as a ring and does not put in the double bonds:
you may feel that this is a more accurate re

dissolve in water, we will only at best get hydroxide ions. So, in order to deprotonate ethyne to any
apprecia- ble extent, we must use a different solventone that does not have a pKa less than 25.
CHAPTER 8 ACIDITY, BASICITY, AND p K a170
Conditions ofte

regular planar hexa- gon with all the carboncarbon bond lengths identical (139.5 pm). This bond length
is in between that of a carboncarbon single bond (154.1 pm) and a full carboncarbon double bond
(133.7 pm). A further strong piece of evidence for this

pressure at point 2 is below atmospheric. 4.3. APPLICATIONS OF
BERNOULLIS EQUATION 89 4.3.5 Vapor pressure If the elevation of
point 2 in Figure 4.13 is high enough, the pressure can become equal to
the vapor pressure of the liquid. The vapor pressure pv

less than about 7 . In contrast, when the flow is in the direction of
decreasing area, as in a contraction, there is little risk of creating large
regions of separated flow, even for large values of (up to 45 ), and
consequently losses are generally small

respect to time is used to emphasize that, since the volume is fixed in
shape and location, the integral is only a function of time. From
equations 3.7 and 3.8, the conservation of mass requires that t Z
d + Z n V dA = 0 (3.9) This is the integral form o

1 + p1g (A2 A1) + 1 2 V 2 1 A2 1 A2 1 A2 2 where FD is the force
required to hold the duct in place. Bernoullis equation was used to find
the pressure p2g, so that the force FD can now be found if p1g, V1, A1
and A2 are known. We can go one step further

vector normal to the surface, n, is defined to be positive when pointing
outward. We begin by considering the net mass flux leaving the control
volume. For a small element of surface area dA, the mass outflux
through dA per unit time = n V dA (see Section

dFps dndsdx = p s g z s (4.2) where we have dropped the
subscript 0 because the result should not depend on the particular
point that was considered. This force will accelerate the fluid particle as
it moves along the streamline. In a short distance ds, t

provide very useful information, as in finding the forces exerted by the
fluid on an object held inside the control volume. For many other
purposes we need to have a more detailed knowledge of the flow
behavior. For instance, as we pointed out in Section

combined with a static tube to make a single unit called a Pitot-static
tube (Figure 4.10). A static tube is a closed tube aligned with the flow
direction. At some distance from the nose, small pressure tappings are
drilled in the tube wall to measure the

between adjacent streamlines gives similar insight; as the distance
decreases, the velocity increases. 76 CHAPTER 4. KINEMATICS AND
BERNOULLIS EQUATION Figure 4.2: Streamlines forming a streamtube.
4.1.5 Hydrogen bubble visualization The hydrogen bubble t

is called the dynamic pressure because it arises from the motion of the
fluid. The dynamic pressure is not really a pressure at all: it is simply a
convenient name for the quantity 1 2 V 2 , which represents the
decrease in pressure due to the increase in

hitting a flat plate there is a force exerted on the plate by the water,
and by Newtons third law the plate exerts an equal but opposite force
on the water (which acts to change the direction of motion of the fluid
and therefore changes its momentum). Oth

to A3 (Figure 4.15)? If the streamlines are parallel at the exit, the
pressure at the exit, p3, is atmospheric. If there are no losses, we see
that the exit velocity is still 2gH. The volume discharge, however, has
increased to A3 2gH so that it is greate

accuracies better than 1% are easily possible. Pitot tubes are widely
used in flow measurement applications. For instance, they are standard
equipment on airplanes, where a Pitot tube is combined with a static
port located somewhere on the fuselage to mea

siphon. Because the air pressure decreases with increasing altitude, it
will approach the vapor pressure of water with increasing altitude, and
water will start to boil at a lower temperature. At an altitude of 13000
m, for example, where the atmospheric

as they apply to all points within a flow field, so that we can understand
the complete flow behavior, we will use very small control volumes or
fluid elements, which means we will adopt an Eulerian description of
the flow field. This approach leads to a

pentane Li
THF Li
THF
Li
vinyllithium
Et2O LiLi
+ LiBr
+ LiCl + LiBr
+ LiCl
You will notice secondary alkyllithiums, an aryllithium, and two vinyllithiums. The only other functional
groups are alkenes and an ether. So far, that is quite like the formation

X
Mg
R
R
Mg
X
R
Mg
X
RX
R
Mg
X
Mg
How to make organolithium reagents
Organolithium compounds may be made by a similar oxidative insertion reaction from lith- ium metal
and alkyl halides. Each inserting reaction requires two atoms of lithium and gener- ate

value, streamlines cannot cross (the flow cannot go in more than one
direction at the same time).1 To visualize a streamline in a flow, imagine
the motion of a small marked particle of fluid. For instance, we could
mark a drop of water with fluorescent dy

and so this relationship holds for flows with variable density. When the
density is constant, however, we can integrate this relationship along
the streamline to obtain p + 1 2 V 2 + gz = constant which is Bernoullis
equation for steady, constant density

makes the velocity at the wall zero, so that the rate of work by viscous
forces is also zero. At an inlet or outlet, the flow is usually normal to the
surface, and then the shear work is again zero. The shear work is rarely
important for large control vol