## CHM1025C  Module 2 Part F Unit Analysis Problems 10pts

You may use a calculator. Please use the Unit Analysis or Dimensional Analysis method to solve ONLY TWO of the following word problems:

## Problem #1:

1. An oxygen molecule travels 975 mi/hr at room temperature. There are 5280 ft = 1 mi; 12 in = 1 ft, 2.54 cm = 1 in, 1.621 km = 1 mi, and 3600 sec = 1 hr. What is the velocity in meters per second?

### The Question:

____???____ m/sec = 975 mi/hr

### Given Unit Factors

• 5280 ft = 1 mi
• 12 in = 1 ft
• 2.54 cm = 1 in
• 1.621 km = 1 mi
• 3600 sec = 1 hr

### The Solution:

1. On the Numerator:
Change m to km to miles
or change m to cm to inches to feet to miles
2. then on the denominator:
change hours to seconds or change hours to minutes to seconds

 ??? m = 975 mi X 1.621 km X 1000 m X 1 hr = 439 m 1 sec 1 hr 1 mi 1 km 3600 sec 1 sec

## Problem #2:

2. If one gram is equal to 15.4 grains. How many 5.00 grain aspirin tablets may be made from 1.00 kilogram of aspirin?

### The Question:

____???_____ tab = 1.00 kg

### Given Unit Factors

• 1 g = 15.4 grain
• 1 tab = 5.00 grain

Change kg to g to grains to tablets! (Three unit factors needed)

### The Solution:

 ??? tab = 1 kg X 1000 g X 15.4 grain X 1 tab = 3080 tab 1 1 1 kg 1 g 5 grain 1

## Problem #3:

3. A parsec is the distance light travels in 3.26 years. Given the velocity of light, 3.00 x 108 m/sec, how many kilometers does light travel in one parsec?

### The Question:

____???_____ km = 1 parsec

### Given Unit Factors

• 1 parsec = 3.26 yr
• 3.00 x 108 m = 1 sec

change parsec to years to days to hours to minutes to seconds to meters then to kilometers

 ??? km = 1 parsec X 3.26 yr X 365 day X 24 hr X 3600 sec X 3.00 X 108 m = 3.08 X 1013 km 1 1 1 parsec 1 yr 1 day 1 hr 1 sec 1

## Problem #4:

4. I have 1400 radio programs I want to put on an Apple Ipod. Each program requires 5 megabytes of disk space. If there are 1024 megabytes in a gigabyte. How many gigabytes of disk space do I need minimum to store all my programs on the IPod. The Mini-Ipod holds only 4 gigabytes of recordings, could I use a mini for my project?

### The Question:

____???_____ Gb = 1400 programs

### Given Unit Factors

• 1 program = 5 Mb
• 1024 Mb = 1 Gb

Change programs to megabytes to gigabytes

### The Solution:

 ??? Gb = 1400 Program X 5 Mb X 1 Gb = 6.84 Gb 1 1 1 Program 1024 Mb 1

Problem#5
Find the mass in grains of a 325 milligram aspirin tablet.
(Given: 1.00 g = 15.4 grains)

### The Question:

____???_____ gr = 325 mg tablet

### Given Unit Factors/Unit Factors Needed

• 1 gram = 15.4 grains  (Given)
• 1gram= 1000 milligrams (Needed)

Change milligrams to grams to grains

### The Solution:

 ??? gr = 325 mg X 1 g X 15.4 gr = 5.05 gr 1 1 1000 mg 1 g 1

Problem#6
Insurance statistics state that a person loses 8 minutes of average life for each cigarette smoked. If there are 20 cigarettes in a pack and the average cost of cigarette is \$5.00 per pack over the next 25 years,
how many years of average life would a person lose for smoking 1.5 packs a day for 25 years?

### The Question:

____???____ yrlifelost = 15 yrsmoking

### Given Unit Factors

• 1 cig = 8 minlifelost
• 1 pack = 20 cig
• 1 pack = 5.00 dollars
• 1 day = 1.5 packsmoking

### Unit Factors Not Given

• 1 yr = 365 days (or 365.25 days)
• 1 day = 24 hr
• 1 hr = 60 min

Change years Smoking to Days Smoking to packs smoked to total cigarettes to minutes lost to hours lost to days lost to years lost

### The Solution:

The important bridge between life lost and smoking is the 8 minutes lost = 1 cig smoked Unit Factor.

 ??? yrlifelost = 15 yrsmok X 365 daysmok X 1.5 pack X 20 cig X 8 minlost X 1 hrlost X 1 daylost X 1 yrlost = 2.5 yrlifelost 1 1 1 yrsmok 1 daysmok 1 pack 1 cig 60 minlost 24 hrlost 365 daylost 1

Problem7
What is the density of water in lb/ft3, if the density of water at 25oC is 1.00 g/ml?
[Hint: There are 2.54 cm = 1 in (or 16.48 cm3 = 1 in3); 454 g = 1 lb; 946 mL = 1 qt ]

### The Question:

____???_____ lb/ft3 = 1.00 g/mL

### Given Unit Factors/Unit Factors Needed

• 454 gram = I Pound (lb)  (Given)
• 2.54 centimeters = 1 inch    or     16.48 cm3 = 1 in3  (Given)
• 12.0 inch = 1 ft   or 1728 in3 = 1 ft3 (needed)

If you have different units to convert, you must do them separately:

First change the numerator unit (Grams) to the numerator answer (Pounds)

then

Second Change denominator unit to the denominator answer:  Change mL to cm3 to in3  to ft3

 ??? lb = 1.00 g X 1 lb X 1 mL X 16.48 cm3 X 1728 in3 = 62.4 lb 1 ft3 1 mL 454 g 1 cm3 1 in3 1 ft3 1 ft3

Problem#8
Calculate the velocity of a car traveling car traveling 65 miles/hr in ft/sec.

### The Question:

__???___ ft/sec = 65 miles/hr

### Given Unit Factors/Unit Factors Needed

• 1 mi = 5280 ft (Needed)
• 1 ft = 12 in  (Not Needed)
• 1 hr = 60 min (Needed)
• 1 min = 60 sec (needed)

If you have different units to convert, you must do them separately:

First change the numerator unit (miles) to the numerator answer (feet) (one unit factor needed)

then

Second Change denominator unit (Hour) to the answer (Seconds) by:  Change hr to min to sec (two unit factors needed)

### The Solution:

 ??? gr = 325 mg X 1 g X 15.4 gr = 5.05 gr 1 1 1000 mg 1 g 1

Problem#9
How many milligrams does a 0.750 carat diamond weigh?
(Hint: 1 karat = 0.200 g)

### The Question:

____???____ mg = 0.750 carat

### Given Unit Factors

• 1 carat = 0.200 g

### Unit Factors Not Given

• 1 g = 1000 mg

### The Solution:

 ??? mg = 0.750 carat X 0.200 g X 1000 mg = 1.50 carat 1 1 1 carat 1 g 1

(#9a) How many karats is a diamond weighing 234.2 mg? (One karat = 0.200 g)

### The Question:

____???____ karat = 234.2 mg

### Given Unit Factors

• 1 karat = 0.200 g

### Unit Factors Not Given

• 1 g = 1000 mg

Change mg to g then to carats (need two unit factors)

### The Solution:

 ??? karat = 234.2 mg X 1 g X 1 karat = 1.17 karat 1 1 1000 mg 0.200 g 1

Problem#10
Diamond has a density of 3.513 g/cm3. The mass of a diamond is often measured in carats,
1 carat equaling 0.200 g.
What is the volume of a 1.50 carat diamond?

### The Question:

____???____ cm3 = 1.50 carat

### Given Unit Factors

• 1 karat = 0.200 g
• 3.513 g = 1 cm3

Change carat to g then to cm3 (need two unit factors)

### The Solution:

 ??? cm3 = 1.50 carat X 0.200 g X 1 cm3 = 0.0854 cm3 1 1 1 carat 3.513 g 1

Problem#11
Liquor used to be sold in fifths. A fifth is one fifth of a gallon. A gallon is 128 fluid ounces. Today liquor is sold in bottle sizes of 750 ml to equate to the old fifth. If there are 946 ml in a quart,
calculate the number of milliliters in a fifth. How many milliliters difference is there in the bottling?

### The Question:

____???____ mL = 1 fifth

### Given Unit Factors

• 1 fifth = 0.200 gallons   or  5 fifths = 1 gallon
• 1 gallon = 128 ounces (not needed)
• 1 quart = 946 mL

Unit Factors Not Given

4 quarts = 1 gallon

Change fifths to gallons then to mL (need three unit factors)

### The Solution:

 ??? mL = 1 fifth X 1 gal X 4 qt X 946 mL = 756.8 mL 1 1 5 fifths 1 gal 1 qt 1

How many milliliters difference is there in the bottling?

756.8 mL = 757 mL (three significant figures)

757 mL – 750 mL = 7 mL difference

Problem #12

1. On July 23, 1983 Air Canada Flight 143, flying at 26,000 feet from Montreal to Edmonton, ran out of fuel because the first officer asked the mechanic for the conversion factor of mass to volume at Montreal. The mechanic gave the first officer the answer 1.77 with no units. The plane had 7682 L of fuel at Montreal. The pilot knew he needed 22,300 kg of fuel to make the trip. The mechanic's answer of 1.77 was pounds per liter not kilograms per liter caused the error such that only 4917 L of fuel was added. If there are 2.205 pounds in a kilogram, how many liters of fuel were needed for the trip? How many liters minimum of fuel should have been added at Montreal before takeoff?

### First find the total Liters needed for the trip, then subtract 7682 Liters for Minimum needing to be added.

____???____ L = 22,300 kg

### Given Unit Factors

• 1.77 pounds = 1 Liter
• 2.205 pounds = 1 Kg

Unit Factors Not Given

Change kg to pounds then to Liters (need two unit factors)

### The Solution: (Total Liters Needed to Make Trip)

 ??? L = 22,300 kg X 2.205 lb X 1 L = 27,781 L 1 1 1 kg 1.77 lb 1

27,781 L (total needed) – 7682 L (in Tanks) = 20,099 L (Needed to be added)

Problem #13

Before 1982 the US Mint cast penny coins from an alloy of copper and zinc. A 1980 Penny weighs 3.051 g and contains 2.898 g of pure copper. In 1982 the US Mint stopped making copper pennies, because the price of copper was worth more than the penny. The post 1982 penny contains only a layer of copper over zinc. A 1990 penny weighs 2.554 g and contains 2.490 g of zinc. If the mint melted down one pound of 1980 pennies, how many 1990 pennies can be made from the total copper from the 1980 pennies?

Problems #14

An Olympic size swimming pool is 50.0 m long and 25.0 m wide. How many gallons of water ( d = 1.0g/mL )are needed to fill the pool to an average depth of 5.5 feet.

Problem #15

A furniture factory needs 29.5 ft2 of fabric to upholster one chair. A European supplier sends the fabric in bolts of exactly 200 m2. What is the maximum number of chairs that can be upholstered by three bolts of fabric. Hint: 1 m - 3.281 ft)?

Problem #16

My throw away car gets 23.4 mi/gal and hold 70.1 L of gasoline. How far can I drive on a tank full of gas?

If gas cost \$3.49/gal; how much does a tank full of gas cost?

If the average speed on a trip is 92.2 km/hr, How many hours may I drive the car on the trip before I run out of gas?