Transcript Keys, Group work, Topic 1 Problem solving in chemistry

Topic 1
Problem Solving in Chemistry:
Keys, Group Work;
Homework Assignments
Group Work Keys
1.1
1.2
1.3
1.4
1.5
1.6
Sig Figs
Scientific Notation
Rounding Off
Dimensional Analysis
Density
Density, Percent
Homework Assignments,
Kotz & Treichel,Ch 1
Slides #13 -14
Key, GROUP WORK, 1.1:
1,000 lb
1 25.351 g
5
3
.000203 mg 3 1.00 X 10 m 3
3.040 in
4 29.00 yd
4
2500 mi
2 33.0 gal
3
420. L
3 .0045 cm
2
150 mL
2 12 eggs

Key, Group Work 1.2
95,000 (4 SF’s)
= 9.500 X 10+4
.00593 X 10-4
= 5.93 X 10-7
.0000008090
= 8.090 X 10-7
0.02030 X 10+5
= 2.030 X 10+3
4578.2
= 4.5782 X 10+3
346.00 X 10 +4
= 3.4600 X 10+6
Key, Group Work 1.3a
24.569 g - .0056 g = ?
24.569 g
- .0055 g
24.5635 g = 24.564 g
“ No more digits after decimal than value with
least number of digits after decimal”
Key, Group Work 1.3b
32.35 cm X 21.9 cm X 0.76 cm = ?
32.35 cm X 21.9 cm X 0.76 cm
allowed SF’s: 2
= “538.4334” cm3
538.4334 = 5.384334 X 102
= 5.4 X 102 cm3
You may also use 540 cm3 which is also 2 SF’s
Key, Group Work 1.3c
54.01 lb + .6789 lb + 1,312.0 lb =?
54.01
lb
.6489 lb
+ 1,312.0
lb
1,366.6589 lb
= 1,366.7 lb
Key, Group Work 1.4a
A football field is 100. yd long. What length is this in
meters? (1 m = 39.37 in)
1. State question:
100. yd = ? m
2. State relationships:
1 yd = 3 ft
1 ft = 12 in
39.37 in = 1 m
Pathway: yd  ft  in  m
Key, Group Work 1.4a
Pathway: yd  ft  in  m
3. Setup and Solve:
100. yd
X 3 ft X 12 in X 1 m
1 yd
1 ft
39.37 in
= 91.44018 m
= 91.4 m
Key, Group Work 1.4b
In water, H2O, the bond length between each H and O is
94 pm. What is this length in millimeters?
1. State the question:
94 pm = ? mm
2. State the needed relationships:
1012 pm = 1 m
1 m = 103 mm
Pathway:
pm  m  mm
Key, Group Work 1.4b
Pathway:
pm  m  mm
3. Setup and solve:
94 pm X
1m X
1012 pm
103 mm =
1 m
94 X 103 - 12 mm
=
94 X 10-9 mm
=
9.4 X 10-8 mm
Key, Group Work 1.5
First step: solve for volume
d = 2.75 cm,
r = 2.75 cm / 2 = 1.375 cm
ht = .50 cm
V=? =  r2 ht = 3.1416 X (1.375 cm)2 X .50 cm
V= 2.9698 cm3 = 3.0 cm3
Second step: solve for mass
3.0 cm3 Au = ? g Au
19.3 g Au = 1 cm3 Au
3.0 cm3 Au X 19.3 g Au
1 cm3 Au
= 57.9 g Au = 58 g Au
Key, Group Work 1.6
Question:
Relationships:
15.00 g H2SO4 = ? mL soltn
1.285 g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
Pathway: g H2SO4  g soltn  mL soltn
Setup and Solve:
15.00 g H2SO4 X 100 g soltn X 1 mL soltn
=
38.08 g H2SO4 1.285 g soltn
= 30.65 mL soltn
Home Work Assignments, Kotz & Treichel
•Read first, for background and scientific perspective, the
introductory material, “The Nature of Chemistry,” p. 1-15.
Chapter One: Matter and Measurements
Read through Chapter One, Sections 1.1-1.5, very
carefully and re-acquaint yourself with concepts
concerning “Matter and Measurements” that you are
expected to have learned and assimilated in your earlier
chemistry classes. The lecture topics in this section
are taken from Sections 1.6-1.8. Study thoroughly.
Read through the “CHAPTERS HIGHLIGHTS” for Chapter 1
on p. 50-51; these make excellent performance objectives
for this set of material.
Written Assignments:
Chapter one in-chapter “exercises”:
1.1, 1.4, 1.6, 1.8-1.11, 1.13, 1.14, 1.15
Chapter one “study questions,” p. 52-56:
all evens, 2-22; 30, 34, 42, 44, 46,
50, 52, 56, 60, 62, 64, 68, 72, 74
Notes on all assignments:
1. Reading: Always read all material assigned; do
the examples in the book, where further insight on
topic is contained.
2. Chapter Highlights: At conclusion of chapter,
use these as a check list for what you have learned.
3. Written Assignments: All assigned In-Chapter
Exercises and End of Chapter Study Questions
are to be worked out in your homework notebook
or folder. Show setups (not just answers) and
check answers for both types in back of Text.