M03 L3 Laws
Saturday, February 28, 2009
6:07 PM
VOICETHREAD
http://voicethread.com/share/7684/
Cmap
http://cmapspublic2.ihmc.us/rid=1210805180212_2068994317_36893/Apologia,%20ed%201,%20Module%
203%20Concept%20Map.cmap
Slides
Notes
Apologia Chemistry Page 1
29 g
Apologia Chemistry Page 2
29 g
Apologia Chemistry Page 3
Apologia Chemistry Page 4
Water is not ionic, so replace with NaCl which is.
Apologia Chemistry Page 5
Apologia Chemistry Page 6
M03, L2 Quiz
1. What theory of matter do scientists believe today?
Answers
a
a. Discontinuous Theory of Matter
b. Continuous Theory of Matter
2. You are told that a molecule contains a total of 17
atoms. If the formula is C5H10Clx, what must x be?
Choose one answer.
a. 2
5 + 10 + x = 17
1 5+ X = 17
X=2
a
b. 1
c. 3
3. Exam problem 8: A chemist decomposes 100.1 g of a
substance into ...
12.1 g of C,
40.0 g of Mg,
and how much oxygen?
12.1 + 40.0 + x = 100.1
Simplify right side: 12.1+40.0=52.1
52.1 + x = 100.1
100.1 - 52.1 = 48.0
48.0 g O2
4. Exam Q09: In an experiment to determine how to
make sulfur trioxide, a chemist combines 32.0 g of
sulfur with 50.0 g of oxygen. She finds that she made
80.0 g of sulfur trioxide and has 2.0 g of left over
oxygen. How much oxygen would the chemist need
to make 100.0 g of sulfur trioxide so that there are
no leftovers?
Get Rid of Excess Oxygen:
32.0 g sulfur + 50.0 g oxygen = 80.0 g sulfur trioxide + 2.0 g leftover Oxygen
Take 2.0 g excess oxygen from both sides. Results in …
32.0 g sulfur + 48.0 g oxygen = 80.0 g sulfur trioxide
Apologia Chemistry Page 7
Proportion (scenario 1/scenario 2):
48.0 g oxygen / X g oxygen = 80.0 g sulfur trioxide / 100.0 g sulfur trioxide
80.0 X = (48.0)(100.0)
Simplify right side: 48.0*100.0=4.80X103
80.0 X = 4.80X103
80.0 X = 4.80X103
4.80X103/80.0=60.0
60.0 g O2
5. A chemist reacts 15.0 g of calcium (Ca) with 15.0 grams
of oxygen (O). This reaction makes 21.0 grams of a
compound known as lime. Along with the lime, there is
also some left over oxygen. If the chemist wants to
make 50.0 grams of lime and have no left over oxygen
or calcium, how much of each should he use?
Compute Formula with Excess Oxygen:
15.0 g calcium + 15.0 g oxygen = 21.0 g lime + X g leftover oxygen
Simplify the left side: 15.0+15.0=30.0 g
30.0 = 21.0 + X
30.0-21.0=9.0
15.0 g calcium + 15.0 g oxygen = 21.0 g lime + 9.0 g leftover oxygen
Format your answer with oxygen's answer first,
spacing, the correct number of significant digits for the
calculation (be careful here), and include units in a
format similar to ...
O = 20.0 g, Ca = 20.0 g
Get Rid of Excess Oxygen:
15.0 g calcium + 15.0 g oxygen = 21.0 g lime + 9.0 g leftover oxygen
Take 9.0 g excess oxygen from both sides. Results in …
15.0 g calcium + 6.0 g oxygen = 21.0 g lime
Proportion for Oxygen (scenario 1/scenario 2):
6.0 / X = 21.0 / 50.0
21.0*X = 6.0*50.0
Simplify right side: 6.0*50.0=3.0x102
21.0*X = 3.0x102
3.0x102/21.0=14.28571429
14 g O2
Proportion for Calcium (scenario 1/scenario 2):
15.0 / X = 21.0/50.0
21.0*X=15.0*50.0
Simplify right side: 15.0*50.0=750
21.0*X=750
750/21.0=35.7143
35.7 g Ca
Apologia Chemistry Page 8