PORTAGE LEARNING CHEMISTRY 103 FINAL EXAM STUDY GUIDE
MODULES 1-6
MODULE 1 – EXAM
Question 1 – Section 1.1
pts
Complete the two problems below:
1. Convert 1005.3 to exponential form and explain your answer.
2. Convert 4.87 x 10-6 to ordinary form and explain your answer.
Your Answer:
1.) Convert 1005.3 to exponential form and explain your answer:
All number values must be between 1-10. Because this value is greater than 1, the exponent remains positive.
In order for this to happen, we have to move the decimal three places to the LEFT thus, the exponential form of 1005.3 is: 1.0053 X 103
2.) Convert 4.87 x 10-6 to ordinary form and explain your answer.
All number values must be between 1-10. Because this value has a negative exponent, the final answer will be less than 1.
In order for this to happen, we have to move the decimal six places to the LEFT thus:
0.00000487
Question 2
pts
Complete the two problems below:
Using the following information, do the conversions shown below, showing all work:
1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints
kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10)
1. 3.6 pounds = ? oz
2. 2680 ml = ? liters
Your Answer:
1.) 3.6 pounds = 57.6 oz
1lb = 16oz
3.6(16) / 1 =57.6 oz
2.) 2680 ml = 2.68 liters
New/Old
Liters/Milliliters
,1L = 1,000mL
2,680/1,000
=2.68
Question 3
pts
Do the conversions shown below, showing all work:
1. 78oC = ? oK
2. 248oF = ? oC
3. 427oK = ? oF
Your Answer:
Kelvin is the larger
Fahrenheit is the middle
Celcius is the smaller
1.) 78oC = 351 oK
Add:
273+78 = 351K
2.) 248oF = 120 oC
Subtract 32 then divide by 1.8
248-32= 216
Now divide 216/1.8
=120
3.) 427oK = 309.2 oF
Kelvin to Fahrenheit needs to convert K to C then continue equation.
427-273 = 154oC
Celsius to Fahrenheit needs to multiply first then add 32
154 x 1.8 = 277.2
add 32
= 309.2
Question 4
pts
, Be sure to show the correct number of significant figures in each calculation.
1. Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml.
2. Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml
Your Answer:
D=M/V
M=DXV
V=M/D
1.) Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml.
***significant figures are underlined***
D=M/V
28.6/32.7
= 0.875g/ml
3 sig/figs
2.) Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml
V=M/D
14.3/0.785
= 18.2mL
3 sig/figs
1. D = M / V = 28..7 = 0.875
2. V = M / D = 14..785 = 18.2 ml
Question 5
pts
1. 1.35601 contains ? significant figures.
2. 0.151 contains ? significant figures.
3. 1.35601 + 0.151 = ? (give answer to correct number of significant figures)
Your Answer:
1.) 1.35601 = SIX significant figures
2.) 0.151 = THREE significant figures.
3.) 1.35601 + 0.151
=1.507
Question 6
pts
Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer.
1. Pepperoni Pizza
MODULES 1-6
MODULE 1 – EXAM
Question 1 – Section 1.1
pts
Complete the two problems below:
1. Convert 1005.3 to exponential form and explain your answer.
2. Convert 4.87 x 10-6 to ordinary form and explain your answer.
Your Answer:
1.) Convert 1005.3 to exponential form and explain your answer:
All number values must be between 1-10. Because this value is greater than 1, the exponent remains positive.
In order for this to happen, we have to move the decimal three places to the LEFT thus, the exponential form of 1005.3 is: 1.0053 X 103
2.) Convert 4.87 x 10-6 to ordinary form and explain your answer.
All number values must be between 1-10. Because this value has a negative exponent, the final answer will be less than 1.
In order for this to happen, we have to move the decimal six places to the LEFT thus:
0.00000487
Question 2
pts
Complete the two problems below:
Using the following information, do the conversions shown below, showing all work:
1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints
kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10)
1. 3.6 pounds = ? oz
2. 2680 ml = ? liters
Your Answer:
1.) 3.6 pounds = 57.6 oz
1lb = 16oz
3.6(16) / 1 =57.6 oz
2.) 2680 ml = 2.68 liters
New/Old
Liters/Milliliters
,1L = 1,000mL
2,680/1,000
=2.68
Question 3
pts
Do the conversions shown below, showing all work:
1. 78oC = ? oK
2. 248oF = ? oC
3. 427oK = ? oF
Your Answer:
Kelvin is the larger
Fahrenheit is the middle
Celcius is the smaller
1.) 78oC = 351 oK
Add:
273+78 = 351K
2.) 248oF = 120 oC
Subtract 32 then divide by 1.8
248-32= 216
Now divide 216/1.8
=120
3.) 427oK = 309.2 oF
Kelvin to Fahrenheit needs to convert K to C then continue equation.
427-273 = 154oC
Celsius to Fahrenheit needs to multiply first then add 32
154 x 1.8 = 277.2
add 32
= 309.2
Question 4
pts
, Be sure to show the correct number of significant figures in each calculation.
1. Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml.
2. Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml
Your Answer:
D=M/V
M=DXV
V=M/D
1.) Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml.
***significant figures are underlined***
D=M/V
28.6/32.7
= 0.875g/ml
3 sig/figs
2.) Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml
V=M/D
14.3/0.785
= 18.2mL
3 sig/figs
1. D = M / V = 28..7 = 0.875
2. V = M / D = 14..785 = 18.2 ml
Question 5
pts
1. 1.35601 contains ? significant figures.
2. 0.151 contains ? significant figures.
3. 1.35601 + 0.151 = ? (give answer to correct number of significant figures)
Your Answer:
1.) 1.35601 = SIX significant figures
2.) 0.151 = THREE significant figures.
3.) 1.35601 + 0.151
=1.507
Question 6
pts
Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer.
1. Pepperoni Pizza
