Download Math Toolkit Practice Packet

Name_____________
Math Toolkit
Practice Packet
These are the math skills needed to be successful in Mr. Carter’s Chemistry and Physics classes.
Standard: I am able to complete my assignments on time in a satisfactory manner. My work is clear evidence that I use my
assignments to practice skills as well as to provide evidence of learning and mastery of what I’ve learned. I have the strength of
character to do the job that is asked of me and grow from the experience.
Criterion
0
3Homewo rk and Effort
Homework and
The assignment
Effort
does not offer
sufficient
evidence of
student effort.
Gradebook
Points /100
Late Penalty
Skill
50
1-2 (Approaches)
3-4 (Meets)
5-6 (Exceeds)
1.The assignment is mostly
complete, OR
2.mostly incorrect, OR
3. Work is NOT shown when
appropriate, OR
4.Writing is NOT legible.
The assignment is
complete and mostly
correct. Work is shown
when appropriate. All
writing is legible.
The assignment is
complete and almost
entirely correct. Work
is shown when
appropriate. All writing
is legible.
65
85
100
Significant Figures
Scientific Notation
Unit Conversions
Comparing Numbers
in scientific notation
Arithmetic with
scientific notation
Using Science
Formulae
Unit Analysis
Data Processing
Progress
Skill
Progress
How to report a measurement with the
correct number of significant figures.
1.
2.
Any non-zero digits are significant.
1.57 = 3 significant figures
3.6 = 2 significant figures
value
1.2
Significant
figures
2
Any zeros to the right of the number are significant.
60.00 kg
4 significant figures (very precise)
60.0 kg 3 significant figures
60 kg
2 significant figures
6 *101 kg
1 significant figure (not very precise)
8
1
12,000
5
0.000521
3
12
thousand
2
3.
Any zeros to the left of the number are not significant.
0.00006g
1 significant figure
6 x 10-5g
0.60000g
5 significant figures
6.0000 x10-1 g
0.00600g
3 significant figures
6.00 x 10-3g
4.
If the number is written out (with words) it is not significant.
12, 000
1.2000 * 104
5 significant figures
12 thousand
1.2 * 104
2 significant figures
Significant Figures: Write down the number of significant figures in each of the following.
1. 5.37 g ______
2. 0.06 L ______
4. 0.0020060 m ____
5. 93.10 K_____
7. 0.0037 g ______
8. 0.60 L ______
10. 2006 m ____
11. 931 K_____
3. 8 thousand grams ______
6. 900 million meters _____
9. 1.3 thousand grams ______
12. 970 million meters _____
STOP: Have Mr. Carter check your answers before moving on
1. 7 g ______
2. 0.0600 L ______
3. 200 thousand grams ______
4. 0.000010 m ____
5. 9.100 K_____
7. 370,000 g ______
8. 0.0000120 L ______
9. 16 thousand grams ______
10. 0.010 m ____
11. 9312.0 K_____
12. 0.001030 meters _____
6. 2 million meters _____
How to correctly put a measurement in
scientific notation.
1.
2.
Move the decimal point so that it is to the right of the 1 st non-zero digit.
(see table)
Good
Bad
2.36
23.6
Count the number of times you moved the decimal (this will become your
exponent)
5.7
570
6
60
1.3600
13600
5.0000
50.00
4.36 (decimal was moved 11 times)
3.
The inverse relationship:
If you made your number bigger, your exponent gets smaller (negative).
0.00008 m
=
8 * 10 -5 m
If you made you number smaller, your exponent gets bigger (positive).
800,000 m
=
8 * 10 5 m
Scientific Notation: Express the following data in scientific notation. Use the correct number of
significant figures.
1. 264 kg
Work:
Answer:__________________
2. 0.00830 m
Work:
Answer:__________________
3. 7.9 cm
Work:
Answer:__________________
4. 546.5 * 105 g
Work:
Answer:__________________
5. 0.07000 mL
Work:
Answer:__________________
STOP: Have Mr. Carter check your answers before moving on
1. 26.4 kg
Work:
Answer:__________________
2. 93,000 m
Work:
Answer:__________________
3. 5.912 cm
Work:
Answer:__________________
4. 4126.5 * 10-3 g
Work:
Answer:__________________
5. 0.0000050 mL
Work:
Answer:__________________
How to convert units using scientific notation.
0.056 mm to m
1. Convert the value to scientific notation
0.056mm = 5.6 x 10-2 mm
2. Set up a conversion factor.
The units that you want to convert TO go on top
1
m
The units that you are coming FROM go on the bottom
1000 mm
Remember: There are 1000 milligrams in one gram, and 1000 grams in one kilogram
3.
Put the conversion factor in scientific notation. 1000 is just 10 3 (10 x 10 x 10)
1
m
103
mm
4.
Multiply the number by the conversion factor. If 10 3 is on the bottom, subtract 3 from the
exponent. If 103 is on the top, add 3 to the exponent.
5.6 x 10-2 mm x
1
m
= 5.6 x 10-2-3 m = 5.6 x 10-5 m
3
10
mm
SI Unit Information
0 ºC = 273 K
Conversions in scientific notation: Convert the following measurements. Express the measurement in
scientific notation and use the correct number of significant figures.
1. 0.00006 kg to g
Work:
Answer:_______________
2. 1,500 mm to m
Work:
Answer:_______________
3. 2.6 L to mL
Work:
Answer:_______________
4. 5.6 * 101 g to kg
Work:
Answer:_______________
5. 373 K to ºC (NOTE: You do not need a conversion factor)
Answer:_______________
6. 0.071 mL to kL
Work:
Answer:_______________
7. 1.0 *103 min to sec
Work:
Answer:_______________
STOP: Have Mr. Carter check your answers before moving on
1. 0.0214 kg to g
Work:
Answer:_______________
2. 1.8 *103 sec to min
Work:
Answer:_______________
3. 0.06 L to mL
Work:
Answer:_______________
4. 3.12 * 104 g to kg
Work:
Answer:_______________
5. -10ºC to K (NOTE: You do not need a conversion factor)
Answer:_______________
6. 0.0000120 kL to mL
Work:
Answer:_______________
7. 120 million mm to km
Work:
Answer:_______________
How to compare measurements using
scientific notation.
Which is bigger?
1.
0.000025 m
or
3.0 x 10 -3 mm
Convert all of the measurements to scientific notation
0.000025m = 2.5 x 10-5 m
3.0 x 10-3 mm is already in scientific notation
make sure both measurements have the same units.
3.0 x 10-3 mm *
1
m = 3.0 * 10-6 m
3
10
mm
Compare the exponents first.
2.5 x 10-5 m
>
3.0 * 10 -6 m
because -5 is bigger that -6
If the exponents are the same, compare the numbers.
1.1 x 10-5 m
<
7.8 * 10-5 m
because 7.8 is bigger than 1.1
2.
3.
4.
0
1.
∞
A. 2.0 * 10-6 m
B. 0.000016 m
C. 359 * 10-5 m
0
2.
∞
D. 6 * 10-1 L
E. ¾ L
F. 2.4 * 102 L
STOP: Have Mr. Carter check your answers before moving on
1.
0
∞
A. 4.0 * 10-6 m
B. 0.0016 m
C. 35 * 10-5 m
2.
0
D. 6 * 10-1 L
E. 1/2 L
F. 0.024 * 101 L
∞
How to perform simple arithmetic using
scientific notation.
Remember, whenever you perform calculations using more that one measurement, your final answer can
only have as many significant figures as the measurement with the LEAST number of significant figures.
“Whatever you do for the Least of these, do for me.”
Step 1
Give both
numbers the
same exponent
(103 and 103)
(1.25 * 104g) + (5.63 (1.25 * 104g) +
* 102g)
(0.0563 * 104g)
Multiplying
Multiply the
two numbers
Step 2
Line up the
decimals for the
numbers and then
add
1.25
+0.0563
1.3063
Line up the
decimals for the
numbers and then
subtract
8.8
- 0.425
8.325
Add the two
exponents
(2.5 * 103g)(4 *
102g)
Dividing
2.5 * 4 = 10
103 * 102 = 105
Divide the two
numbers
(40 * 105g) / (5.0 *
103g)
40/5 = 8
Subtract the
second exponent
from the 1st
105 / 103 = 102
Adding
Give both
numbers the
same exponent
(103 and 103)
3
(8.8 * 10 g) - (4.25 * (8.8 * 103g) 102g)
(0.425 * 103g)
Subtracting
Step 3
Convert to
Scientific
Notation
1.31 * 104 g
Convert to
Scientific
Notation
8.3 * 103 g
Convert to
Scientific
Notation
10 * 105 g = 1
* 106 g
Convert to
Scientific
Notation
8.0 * 102 g
Calculations: Express the following data in scientific notation. Use the correct number of significant
figures.
1. (5.00 x 10 -3 g)(3 x 106 g)
Answer:_______________
2. (6.0 x 108 m) + (0.02 x 1010 m)
Answer:_______________
3. (1.20 x 105 mm) / (0.30 x 103 mm)
Answer:_______________
4. (2.124 x 1012 km) - (0.0242 x 1015 km)
Answer:_______________
5. There are about 80 million milliters of milk in a tank at the dairy. It takes 200 milliliters to fill a carton
of milk. How many cartons can the tank fill?
Answer:_______________
6. There are 1*1024 atoms in a bar of metal. Each atom has a mass of 3.3 x 10-22 grams. What is the mass
of the bar?
Answer:_______________
STOP: Have Mr. Carter check your answers before moving on
1. (7.00 x 10 2 g)(2 x 104 g)
Answer:_______________
2. (6.420 x 104 m) + (0.023 x 106 m)
Answer:_______________
3. (4.000 x 102 mm) / (0.050 x 103 mm)
Answer:_______________
4. (8.627 x 1012 km) - (0.0242 x 1014 km)
Answer:_______________
5. There are about 40 million milliters of milk in a tank at the dairy. It takes 200 milliliters to fill a carton
of milk. How many cartons can the tank fill?
Answer:_______________
6. There are 300 million people in the US. Together we have a mass of 2.1* 10 10 kg. What is the average
mass of an American in kg?
Answer:_______________
How to use a scientific formula to solve for an
unknown.
Example: The density of apple juice is 1.2g/mL. What is the mass of an apple juice sample that has a
volume of 2L?
Density
m= mass (in grams)
Formula
units
D=m/V
g/mL
V=volume (in milliliters)
D= density (in grams per milliliter)
1.
Write down the formula that you will use.
D= m
V
2.
Make sure that the values from the problem have the same units as the values in the formula.
D= g/mL
Correct units
L= Liters
3.
Incorrect units (milliliters are needed)
If one of the values is not the correct unit, convert it.
2 L x
103 mL = 2 x 103 mL = 2000 mL
1 L
4.
Identify what you will be solving for.
“What is the mass of an apple juice sample that has a volume of 2L?”
5.
Substitute the values you were given into the equation.
D= m
V
6.
mass (in g)
1.2 g/mL= m
V
1.2 g/mL =
m
2000 mL
Use basic algebra to solve for the unknown.
1.2 g/mL =
m
2000 mL
(2000mL)1.2 g/mL =
m (2000mL)
2000 mL
(2000mL)1.2 g/mL = m
multiply both sides by 2000 to isolate the variable (m)
solve for m, perform unit analysis
2400 g = m
7.
Check units
2400g
Units are in grams, which is what the problem asked for.
Science formulae: Calculate an answer to the following problems. Choose the correct formula from the
table below to answer each problem. Make sure that your answer has the correct units.
Formula
units
Density
D=m/V
g/mL
Force
F=ma
Kg * m/s2 or
Velocity
V=d/t
m/s
Acceleration
A=∆v/∆t
m/s2
Work
Fd
Kg * m2/s2
or
N
N*m
______1. One cereal bar has a mass of 30 g. It has a density of 2 g/mL. how much space does it take up?
______2. Ashley travels 2.4 kilometers every 2 minutes on her bicycle. How fast is she traveling in
meters per second?
______3.
Force = 80 N
Mass = 40,000 g
Acceleration =
______4.
Starting Velocity = 0 m/s S
Final Velocity = 8 m/s S
Time = ___?_______
Acceleration = 8 m/s2
______5.
Force = 18 N
Mass = ?
Acceleration = 3 m/s2
STOP: Have Mr. Carter check your answers before moving on
6.
An un-labeled liquid has a density of 0.9 g/mL. It has a volume of 150 mL. What is its mass?
Answer___________
7. It requires 150 Kg * m2/s2 of work to move a book 10 meters. What is the mass of the book? The
acceleration due to gravity on earth is 9.8m/s2.
Answer___________
8.. I use 0.350 N of Force to throw a Frisbee, and it accelerates at 2m/s 2. What is the mass of the Frisbee?
Answer___________
How to keep track of units during a
calculation (unit analysis).
Example: solve
(50.0 kg . m/s)(10.0 m)
(5.0 s)(20 m)
1.
Make sure that the units are appropriate
1. Kg are appropariate for mass
2. s are appropriate for time
3. m are appropriate of distance
2.
Multiply the numbers.
(50.0 kg . m/s)(10.0 m)
(5.0 s)(20 m)
= 500
= 100
Multiply the UNITS.
(50.0 kg . m/s)(10.0 m)
(5.0 s)(20 m)
= 500 kg * m2/s
= 100 s * m
Divide the numbers.
(50.0 kg . m/s)(10.0 m)
(5.0 s)(20 m)
= 500
= 100
=5
Divide the UNITS.
(50.0 kg . m/s)(10.0 m)
(5.0 s)(20 m)
= 500 kg * m2/s
= 100 s * m
= 5 kg * m/s2
3.
4.
5.
Calculations: Calculate the following. Be sure that your answer has the correct units!
1.
(30 m/s) / (3 s)
Answer:_______________
2.
(30 m/s)(3 kg/s)
Answer_______________
3.
(10.0 kg. m/s)(4.000 s/m)
Answer_______________
4.
(50.0 kg . m/s)(10.0 m)
(10.0 s)
Answer_______________
5.
(2 kg .m/s)(1 m . s)(2 kg/m2)(3 s/kg2)
Answer_______________
STOP: Have Mr. Carter check your answers before moving on
6.
(2 kg . m/s)(3 s/kg)(2 kg/s)
(6 s)
Answer_______________
7.
(10 kg.m/s)(5 m.s)(20 m/kg2)(40 kg/s)(1.0 kg/m2)
Answer_______________
8.
(0.1 kg2 . m/s)(0.5 s . m/kg2)(0.2 kg/s2)
Answer_______________
9.
(1.00 kg2 . m/s2)(1.50 s3/kg)(0.500 m/kg)
(0.500 s . m)
Answer_______________
10. (1.000 kg4. m3/s2)(3.0 s4/m2)(20.0 kg2/s3)
(10.0 kg5)
Answer_______________
How to analyze experimental data by
calculating Standard Deviation.
Example: calculate standard deviation for the following data set.
These are the masses of 5 men who were all marathon runners between the ages of 20 and 40;
52 kg, 48kg, 57kg, 55 kg, 49kg
How do we calculate standard deviation?
There are 5 men. In statistical terms this means we have a population of 5. If we know the mass of
each man, we use the standard deviation equation for the entire population:
∑ = Sum (add them up)
1.
Calculate the mean (average) mass.
a. Add up all the masses
i. 52 kg + 48kg + 57kg + 55 kg + 49kg= 261 kg
b. Divide by N (the number of men in the population)
i. 261kg/ 5 men = 52.2 kg/man on average
2. Make a table to calculate how far each data point is from the average. Some of your numbers will
be negative; that’s OK.
Mens’ Masses
Difference from mean
Difference from mean
52 kg
52 kg - 52.2 kg/man =
-0.2 kg
48 kg
48 kg - 52.2 kg/man =
-4.2 kg
57 kg
57 kg - 52.2 kg/man =
4.8 kg
55 kg
55 kg - 52.2 kg/man =
2.8 kg
49 kg
49 kg - 52.2 kg/man =
-3.2 kg
3.
Square all of your differences from the mean. The process of squaring will make all of your
numbers positive.
Mens’ Masses
Difference from mean
(Difference from mean)2
52 kg
-0.2 kg
0.04 kg2
48 kg
-4.2 kg
17.64 kg2
57 kg
4.8 kg
23.04 kg2
55 kg
2.8 kg
7.86 kg2
49 kg
-3.2 kg
10.24 kg2
4.
Sum up (add up) all of the squared differences
a. 0.04 kg2 + 17.64 kg2 + 23.04 kg2 + 7.86 kg2 + 10.24 kg2 = 58.82 kg2
5. Divide the sum of all the squared differences by the population size (N)
a. 58.82 kg2 / 5 = 11.764 kg2
6. Take the square root of that value
a. √11.764 kg2/man = 3.43 kg
7. Adjust your number for significant figures.
a. All of the mens’ masses were reported to 2 sig figs, so mu standard deviation should also
be reported to 2 sig figs. In this case, 3.43 rounds down to 3.4.
b. 3.43 kg = 3.4 kg
A. Mean:____ 52.2 kg/man _________
B. n:_______5 men________
2
C. Sum of (Difference from the Mean) divided by (n):___11.764 kg2____
Mens’
Difference
(Difference
Sum of
divided ( diff . fromMean)2
Masses
from mean
from mean)2
(Difference
by (n)
from the
2
Mean)
52 kg
-0.2 kg
0.04 kg2
58.82 kg2
11.764
3.4 kg
2
kg2
48 kg
-4.2 kg
17.64 kg
57 kg
4.8 kg
23.04 kg2
55 kg
2.8 kg
7.86 kg2
49 kg
-3.2 kg
10.24 kg2
D. Standard deviation ( diff . fromMean)
2
is___3.4 kg
What does this standard deviation mean?
On average, the majority
(68%) of male marathon runners between 20 and 40 should be within 1 standard deviation of the mean.
That is their mass should be between 48.8kg and 55.6 kg.
The vast majority (95%) should be within 2 standard deviations of the mean. That is their mass should
be between 45.4kg and 59.0 kg.
Does this make sense?
Yes! As you can see, all of our runners are within 2 standard deviations from the mean, and MOST of
our runners are within 1 standard deviation of the mean (52 kg, 49 kg, and 55kg).
You would expect that men with too small of masses (below 45.4) would make bad marathon runners
because their legs would be too short so they would have to pump them too fast. Also men with big
masses (over 59.0 kg) would make poor marathon runners because their extra mass would take a lot of
extra work to carry around for the long distance of the marathon.
Could there be marathon runners with masses bigger or smaller than 2 standard
deviations from the mean?
Absolutely. Although the chances of this happening are really low (5%), that doesn’t mean that it
would never happen. There are millions of marathon runners out there, so some would exist that are
either lighter or heavier than 2 SDs from the mean.
Which man was the outlier?
An outlier is a data point that is really far from the rest of the data points. In this data set the 57 kg
man would be the outlier because his mass (57 kg) is the furthest from the mean (a 4.8 kg difference).
He is unusually heavy for a marathon runner, but you can see that his mass is still within 2 SDs from
the mean (less than 59 kg).
What does Standard Deviation tell me about my science experiment?
If you have a high standard deviation in your experiment, you know that you collected data without a
lot of precision. Typically, this is because of human error, or just not tightly controlling your
experiment enough. As a general rule; the lower the standard deviation, the better the experiment.
Standard Deviation: Calculate standard deviation for the following data sets.
1. Find the standard deviation for the following test scores.
A. Mean:_____________
84, 100, 84, 68, 84, 100, 84
B. n:_______________
2
C. Sum of (Difference from the Mean) divided by (n):_______
D. Standard deviation ( diff . fromMean)
2
is_______________
2. Find the standard deviation for the following masses of students (in kg):
70 kg, 52 kg, 68 kg, 60 kg, 59 kg, 72 kg, 55 kg, 58 kg, 66 kg, 69 kg, 72 kg, 61 kg.
A. Mean:_____________
B. n:_______________
2
C. Sum of (Difference from the Mean) divided by (n):_______
D. Standard deviation ( diff . fromMean)
2
is_______________
STOP: Have Mr. Carter check your answers before moving on
3. Find the standard deviation for the following volumes:
5.0 mL, 4.8 mL, 5.1 mL, 4.9 mL
A. Mean:_____________
B. n:_______________
2
C. Sum of (Difference from the Mean) divided by (n):_______
D. Standard deviation ( diff . fromMean)
2
is_______________
4. Find the standard deviation for the following times:
24.1s, 25.6s, 19.7s
A. Mean:_____________
B. n:_______________
2
C. Sum of (Difference from the Mean) divided by (n):_______
D. Standard deviation ( diff . fromMean)
2
is_______________