1. Funding Research, from the Science and Society section (20 points) section A item 10
Is this research basic or applied research? How do you know? Would you fund this research?
Why or why not? Write a paragraph or two in your journal to answer these que
Lab Report: [Measurement and Density]
Your Name:
Purpose of this Lab
What is the goal of this lab? What question are you trying to answer, or what problem are you
trying to explain?
I will investigate the densities of 3 types of metals using a precise met
1
Tough Problem
You are a doctor treating a patient for cancer using the
chemotherapeutic drug 5-fluorouracil (5-FU)
The effective dose (amount needed to fight cancer) for 5-FU is
25.00 mg/kg of patient. The lethal dose (amount that will kill a
person) is
WebQuest: Get to Know the Neighborhood
Objectives
Explain the structure of the periodic table of the elements
Obtain information about the properties of elements based upon their positions in the
periodic table
Introduction
In the years following 1869,
CHAPTER 2
CHEMICAL COMPONENTS OF CELLS
2009 Garland Science Publishing
Chemical Bonds
2-1
Figure Q2-1 depicts the structure of carbon. Use the information in the diagram to choose
the correct atomic number and atomic weight, respectively, for an atom of
Lab: Polar Bonds
Objectives
Compare and contrast the chemical bonding properties of slime and Silly Putty.
Gain knowledge about polar and nonpolar molecules by supporting or rejecting
an experimental hypothesis.
Learn about the process of chromatography a
Chapter 2
d = y2 y1 = h + 8.62 m ( h 14.9 m ) = 23.5 m
2.66
(a)
While in the air, both balls have acceleration a1 = a2 = g (where upward is taken as positive). Ball 1 (thrown
downward) has initial velocity v01 = v0 , while ball 2 (thrown upward) has initi
VPsoln = XAVPA + XBVPB (for 2 liquids mixed)
Ex. A solution is prepared by mixing 11.62 g acetone
(C3H6O, MM = 58.1 g/mol) and 11.9 g chloroform (HCCl3,
MM = 119.4 g/mol). At 35C, this solution has total vapor
pressure of 260 torr. Is this an ideal soluti
Slope Fields Worksheet Solutions
7. C
8. D
9. A
10. B
11. B
12. C
13. D
14. A
15. E
16. D
17. (B) Tangent line: y 1 =
f (1.2 ) 1.1
1
( x 1)
2
x2 1
=e 4
(C) y
f (1.2 ) = 1.116
(D) The estimate from part (b) was an underestimate. Since the graph of y
the ta
Chapter 11 Properties of Solutions
SOLUTION UNITS
(M)
MOLARITY
=
mol solute
L soln
% by Volume
=
vol solute
vol soln
% by mass
=
grams solute
grams soln
x 100%
% by m/V
=
grams solute
mL soln
x 100%
(m)
molality
=
mol solute
kg solvent
(X )
mol fraction
=
Chapter 2
t3 =
0 v 200 ft s
=
= 10 s
a3
20 ft s 2
so the duration of the entire trip is ttotal = t1 + t2 + t3 = 5.0 s + 85 s + 10 s = 100 s .
2.72
(a)
From y = v0 t +
t =
1 2
at with 0 = 0, we have
2
2 ( y )
a
=
2 ( 23 m )
9.80 m s2
= 2.2 s
(
)
(b)
2
2
T
Ex. Hydrochloric acid has a density of 1.19 g/cm3 and a mass
percent of 38.0%. Find the Molarity, molality, percent by
volume, % by mass/volume, and mole fraction.
grams
mol
mL
density
solute (HCl)
solvent (H2O)
solution
AP CHEMISTRY Chapter 11 Scotch Pla
Raoults Law
VPsoln = XsolventVPsolvent (for molecular solid dissolved in liquid)
Ex. Find the vapor pressure of a solution made by dissolving
158.0 g of table sugar (sucrose, MM=342.3 g/mol) in 643.5 cm3
of water. At 25C, the density of water is 0.9971 g/
Chapter 9
9
Solids and Fluids
PROBLEM SOLUTIONS
9.1
The elastic limit is the maximum stress, F A where F is the tension in the wire, that the wire can withstand and
still return to its original length when released. Thus, if the wire is to experience a te
Chapter 7
7
Rotational Motion and
the Law of Gravity
PROBLEM SOLUTIONS
7.1
(a)
Earth rotates 2 radians (360) on its axis in 1 day. Thus,
(b)
7.2
2 rad 1 day
5
7.27 10 rad s
t 1 day 8.64 104 s
Because of its rotation about its axis, Earth bulges at the
Chapter 6
6
Momentum and Collisions
PROBLEM SOLUTIONS
6.1
6.2
Use p = m
(a)
p = (1.67 1027 kg)(5.00 106 m/s) = 8.35 1021 kg m/s
(b)
p = (1.50 102 kg)(3.00 102 m/s) = 4.50 kg m/s
(c)
p = (75.0 kg)(10.0 m/s) = 750 kg m/s
(d)
p = (5.98 1024 kg)(2.98 104 m/s)
Chapter 8
8
Rotational Equilibrium and
Rotational Dynamics
PROBLEM SOLUTIONS
8.1
Since the friction force is tangential to a point on the rim of the wheel, it is perpendicular to the radius line connecting this point with the center of the wheel. The torq
Word Equations Worksheet - Solutions
Write the word equations for each of the following chemical reactions:
1)
When dissolved beryllium chloride reacts with dissolved silver nitrate in
water, aqueous beryllium nitrate and silver chloride powder are made.
Chapter 2
(
)
x2 = vt2 = 40 ft s 2 t1 t2
(
)
x2 = 40 ft s 2 t1t2
or
[3]
It is known that x1 + x2 = 17 500 ft , and substitutions from Equations [1] and [3] give
( 20 ft s ) t + ( 40 ft s ) t t
2
2
1
2
1 2
= 17500 ft
t12 + 2t1t2 = 875 s 2
or
[4]
Also, it i
Solutions Units Practice worksheet
Use the following information to determine the missing values for these aqueous solutions. (Hint make a
table similar to the one in the notes to keep organized):
Acid/Base
Molarity (M)
HNO3
16.0
molality (m)
Mole Fractio
Boiling Point Elevation and Freezing Point Depression
Ex. Find the boiling point and freezing point of a solution
made by dissolving 18.00 g glucose in 150.0 g of water.
H2O Kb = 0.512C/m
H2O Kf = 1.86C/m
Tb = Kb i m
Tf = Kf i m
AP CHEMISTRY Chapter 11 Sc
Slope Fields
Nancy Stephenson
Clements High School
Sugar Land, Texas
Draw a slope field for each of the following differential equations. Each tick mark is one unit.
1.
dy
= x +1
dx
2.
dy
= 2y
dx
3.
dy
= x+ y
dx
4.
dy
= 2x
dx
5.
dy
= y 1
dx
6.
dy
y
=
dx
x
From the May 2008 AP Calculus Course Description:
15.
The slope field from a certain differential equation is shown above. Which of the following could be
a specific solution to that differential equation?
(A) y = x 2
(B) y = e x
(C) y = e x
(D) y = cos x
Chapter 3
v y = v 02 y + 2 a y y =
( 1.41
(
m s ) + 2 9.80 m s2
2
) ( 1.00 m )
= 4.65 m s
and the time of flight is
t =
v y v0 y
ay
=
4.65 m s ( 1.41 m s )
9.80 m s2
= 0.330 s
The displacement during the flight is x = v 0 x t = ( 12.5 m s ) ( 0.330 s )
Chapter 2
so,
v =
2.8
The average velocity over any time interval is
v =
2.9
x
180 km
=
= 63.4 km h
t
2.84 h
x f xi
x
=
t
t f ti
(a)
v =
x
4.0 m 0
=
= + 4.0 m s
t
1.0 s 0
(b)
v =
x
2 .0 m 0
=
= 0.50 m s
t
4.0 s 0
(c)
v =
x
0 4.0 m
=
= 1.0 m s
t
5.0 s 1.0
Chapter 2
v =
( x )2
2
v =
t2
( x )3
3
t3
=
=
5.11 103 m
= 41.6 m s
123 s
91.2 m
= 20.8 m /s
4.39 s
and the average velocity for the total trip is
vtotal =
2.37
( x ) total
t total
=
5.51 103 m
= 38.7 m s
( 15.0 + 123 + 4.39 ) s
Using the uniformly accele
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