Metric-English Unit ConversionProblems #26-60

Problem #26: How many palladium atoms would it take to encircle the Earth at its equator? (The atomic diameter of Pd is 140 pm; the equatorial circumference of the Earth is 40,075.02 km.)

Solution:

The absolute exponential distance between kilo- (103) and pico- (10¯12) is 1015. This problem can be done two ways.

Method #1: Convert km to pm

40,075.02 km times (1015 pm / 1 km) = 4.007502 x 1019 pm

4.007502 x 1019 pm / (140 pm / atom) = 2.86 x 1017 atoms

Method #2: Convert pm to km

140 pm times (1 km / 1015 pm) = 1.40 x 10¯13 km

40,075.02 km / (1.40 x 10¯13 km/atom) = 2.86 x 1017 atoms

Comment: the mass of this amount of Pd is:

2.86 x 1017 atoms / 6.022 x 1023 atoms mol¯1 = 4.75 x 10¯7 mol

4.75 x 10¯7 mol x 106.42 g mol¯1 = 5.06 x 10¯5 g

Problem #27: Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the Earth, diameter 13000 km. Calculate whether an atom of diameter 0.32 nm will become as big as:

a) a pin head, diameter 1 mm
b) a coin, diameter 1.9 cm
c) a football, diameter 22 cm
d) a weather balloon, diameter 1.8 m

Solution:

1) Convert all measurements to the same value. I used meters:

football: 0.22 m
earth: 1.3 x 107 m
atom: 3.2 x 10¯10 m

2) Solve via a ratio and proportion technique:

football diameter : earth diameter = atom diameter : an unknown value

0.22 m / 1.3 x 107 m = 3.2 x 10¯10 m / x

Solving for x gives 0.0189 m

Answer b (1.9 cm) is the correct answer, since 0.0189 m converts to 1.89 cm and that, to two sig figs, is 1.9 cm.

Problem #28: If an oil tanker carrying 2.0 x 105 metric tons of crude oil of density 0.70 g/cm3 spills its entire load at sea, what area of ocean in square kilometers will be polluted if the oil spreads out to a thickness of 0.10 mm? (A metric ton = 103 kg.)

Solution:

1) Determine kg of oil:

2.0 x 105 times 1000 = 2.0 x 108 kg of oil

2) Change kg to g:

2.0 x 108 kg times 1000 = 2.0 x 1011 g

3) Determine volume of oil:

2.0 x 1011 g divided by 0.70 g/cm3 = 2.857 x 1011 cm3

4) Change 0.10 mm to cm:

0.10 mm times (1 cm / 10 mm) = 0.010 cm

5) Determine area of oil:

2.857 x 1011 cm3 divided by 0.010 cm = 2.857 x 1013 cm2

6) Change cm2 to km2:

2.857 x 1013 cm2 times (1 km2 / 1010 cm2) = 2857 km2

Comment: 1 km2 is 1 km on each side. 1 km = 105 cm, so the 1 km2 is 105 cm on each side, so:

105 cm times 105 = 1010 cm2 (which equals 1 km2)

The last conversion can also be written in this manner:

2.857 x 1013 cm2 times (1 km / 105 cm)2

Problem #29: A water bed has the following dimensions: 2.00 m x 1.40 m x 15.0 cm. Calculate the mass of water (kg, please) in the bed. Assume the density of the water to be 1.00 g/mL.

Solution:

1) Change water bed dimensions to cm and calculate its volume:

200 cm x 140 cm x 15 cm = 420000 cm3

The conversion from m to cm is what is known as a silent conversion. It's just done and no explanation is offered. Often, tiny errors in a book is corrected via the "silent edit" method.

2) Change volume of water to mass of water:

1cm3 = 1 mL, so 420000 cm3 = 420000 mL

420000 mL x 1.00 g/mL = 420000 g

3) Change g to kg:

420000 g times (1 kg / 1000 g) = 420. kg

Note that I only paid attention to correctly showing three sig figs in the final answer.

Problem #30: The radius of an atom of krypton is about 1.90 Ångstroms. How many krypton atoms would have to be lined up to span 1.00 mm? If the atom is assumed to be a sphere, what is the volume in cm3 of a single atom.

Solution:

1) Determine the diameter of the Kr atom in mm:

diameter ---> twice the radius is 3.80 Å

convert Å to cm ---> 3.80 Å times (10-8 cm / Å) = 3.80 x 10-8 cm

convert cm to mm ---> 3.80 x 10-8 cm times (10 mm / 1 cm) = 3.80 x 10-7 mm

2) How many Kr atoms span 1.00 mm?

1.00 mm divided by (3.80 x 10-7 mm per Kr atom) = 2.63 x 106 Kr atoms

3) The formula for volume of a sphere is (4/3)πr3. We will use the radius in cm:

V = (4/3) (3.14159) (1.90 x 10-8 cm)3 = 2.87 x 10-23 cm3 (to three sig figs)

Problem #31: The star Arcturus is 3.472 x1014 km from Earth. How many days does it take for the light to travel from Arcturus to Earth? What is the distance to Arcturus in light years? (One light year is the distance light travels in one year (use 365 days); light travels at a speed of 3.00 x 108 m/s.)

Solution:

1) Our distance to Arcturus is given in km. Convert it to m:

3.472 x1014 km x (1000 m / km) = 3.472 x1017 m

2) Using the speed of light, we can determine the seconds elapsed:

3.472 x1017 m ÷ (3.00 x 108 m/s) = 1.15733 x 109 s

3) There are 3600 seconds in an hour and 24 hours in a day:

1.15733 x 109 s x (1 hr / 3600 s) x (1 day / 24 hr) = 13395 days

4) To determine the distance in light-years, we do this division:

13395 day ÷ 365 day/yr = 36.70 yr

5) An alternate calculation for number of light years:

we need to know how many meters are in one light year:
3.00 x 108 m/s x (3600 s / 1 hr) x (24 hr / 1 day) x (365 day / 1 year) = 9.4608 x 1015 m <--- distance in one light year
We now use the distance to Arcturus, but in meters:
3.472 x1017 m ÷ 9.4608 x 1015 m / light-year = 36.70 light-years

Problem #32: (A) The average adult human body contains about 5.00 L of blood. Determine the total number of iron atoms in the blood using the following information.

There are 2.0 x 108 hemoglobin units per red blood cell (RBC)
each mm3 of blood contains 5.5 x 106 RBC <---there's a trick right here
There are exactly 4 iron atoms in each hemoglobin unit

(B) If all of the iron atoms in this volume of blood were removed and laid out side by side, what linear distance (in km and miles) would they cover? Assume an atom diameter of 125 pm.

(C) How many times would this "string" of Fe atoms stretch around the earth given that the equatorial circumference of the Earth is 40,075.02 km?

Solution to (A):

1) How many RBC in 5.00 L of blood?

5.00 L = 5000 mL = 5000 cm3

5.5 x 106 RBC / mm3 times (10 mm / cm)3 = 5.5 x 109 RBC / cm3

The above is the trick. The RBC was given per square millimeter, not per square centimeter. Tricky!

5.5 x 109 RBC / cm3 times 5000 cm3 = 2.75 x 1013 RBC

2) How many hemoglobin units?

2.75 x 1013 RBC times 2.0 x 108 hemoglobin units/RBC = 5.5 x 1021 hemoglobin units

3) Iron atoms:

5.5 x 1021 hemoglobin units times 4 Fe atoms / hemoglobin unit = 2.2 x 1022 Fe atoms

Solution to (B):

1) Convert 125 pm to km:

125 pm/Fe atom times (1 km / 1015 pm) = 1.25 x 10-13 km/Fe atom

2) Determine length of Fe "string:"

2.2 x 1022 Fe atoms times 1.25 x 10-13 km/Fe atom = 2.86 x 109 km

2.86 x 109 km times (1 mile / 1.60934 km) = 1.777 x 109 mi

I didn't round to the proper number of sig figs. You may, if you wish.

I will leave you to solve part C. The answer is a bit over 71,000 times.

Problem #33: A floor is painted with 750. mL of paint. If the floor to be covered is 20.0 ft by 18.0 ft, how thick is the paint layer?

Solution:

Since the problem does not specify the unit for the thickness, let us use cm, because it's a convenient unit to convert to.

1) We know this:

750. mL = 750. cm3

2) What I propose to do is change the feet into centimeters. Then, the thickness will be the unknown third dimension of the total volume. I will convert feet to inches and then inches to cm:

20.0 ft times (12 inches / ft) times (2.54 cm / inch) = 609.6 cm
18.0 ft times (12 inches / ft) times (2.54 cm / inch) = 548.64 cm

Note how ft cancels and how inch cancels, leaving cm in the numerator (which is what we want).

3) Now, for the third dimension (where X is the unknown):

750. cm3 = 609.6 cm times 548.64 cm times X

X = 0.00224 cm (rounded to three sig figs)

Often, a unit is selected that allows the numerical value to be reported as a number between 1 and 10 (or between 1 and 100). If we select the unit micrometer, the thickness is reported to be 22.4 μm.

Problem #34: In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut down the rate of evaporation of water in reservoirs. This technique was pioneered by Ben Franklin. It was found that 0.1 mL of oil could spread over the surface of water about 40 square meters in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers.

Solution:

1) Let us determine the length of the molecule in centimeters. Why? Because:

0.1 mL = 0.1 cm3

Centimeters is a convenient unit to use.

2) Convert 40 m2 to cm2:

40 m2 times (100 cm / m)2 = 4 x 105 cm2

An alternate way:

40 m2 = 40 m x 1 m

Convert each m to cm

(40 m x 100 cm/m) x (1 m x 100 cm/m) = 4 x 105 cm2

3) Determine the third dimension in the volume, which will be our answer:

0.1 cm3 = 4 x 105 cm2 times X

X = 2.5 x 10-7 cm

This type of measurement can be done experimentally and is refered to as Langmuir's method. Irving Langmuir was the 1932 recipient of the Nobel Prize in Chemistry.

Problem #35: In 1999, scientists discovered a new class of black holes with masses 100 to 10,000 times the mass of our sun, but occupying less space than our moon. Suppose that one of these black holes has a mass of 1000 suns and a radius equal to one-half the radius of our moon. What is the density of the black hole in g/cm3? The radius of our sun is 6.955 x 105 km and it has an average density of 1.410 x 103 kg/m3. The diameter of the moon is 2.16 x 103 miles.

Solution:

1) The mass of our sun:

6.955 x 105 km = 6.955 x 108 m

V = (4/3)πr3

V = (4/3) (3.14159) (6.955 x 108 m)3

V = 1.40922275 x 1027 m3

mass of one sun = (1.40922275 x 1027 m3) (1.410 x 103 kg/m3) = 1.987004 x 1030 kg = 1.987004 x 1033 g

mass of 1000 suns = 1.987004 x 1036 g

2) volume of black hole:

radius of moon = 2.16 x 103 mi / 2 = 1.08 x 103 mi

radius of black hole = 1.08 x 103 mi / 2 = 540 mi

Use Google calculator to change 540 miles to cm = 8.690 x 107 cm

V = (4/3)πr3

V = (4/3) (3.14159) (8.690 x 107 cm)3

V = 2.748828 x 1024 cm3

3) calculate density of black hole:

1.987004 x 1036 g / 2.748828 x 1024 cm3 = 7.23 x 1011 g/cm3

Problem #36: A water pipe fills a rectangular prism shaped container that measures 18.9 cm deep by 30.7 cm long by 27.2 cm tall in 97.0 seconds. What is the rate of flow in liters per hour?

Solution:

1) Determine the volume of the container in liters:

18.9 cm x 30.7 cm x 27.2 cm = 15782.256 cm3

Then:

15782.256 cm3 x (1 mL / 1 cm3) x (1 L / 1000 mL) = 15.782256 L

2) Determine the flow rate in L/hr:

15.782256 L was filled in 97.0 s

(15.782256 L / 97.0 s) x (60 s / min) x (60 min/hr) = 585.7 L/hr

To three sig figs, 586 L/hr

Problem #37: Oil spreads on water to form a film about 120. nm thick. How many square kilometers of ocean will be covered by the slick formed when two barrels of oil are spilled? (1 barrel = 31.5 US gallons)

Solution:

1) Let's convert nm to km:

120. nm times (1 km / 1012 nm) = 120. x 10¯12 km = 1.20 x 10¯10 km

We will use this km value as the height of the oil slick.

2) Let's convert the non-metric US gallon to dm3:

1 US gallon = 3.7854 L

I looked up that value here.

63 US gallons = 238.48 L = 238.48 dm3

Comment: via Google calculator, you can actually convert gallons directly to km3. I chose to use dm3 so as to explain how to get from dm3 to km3.

3) Convert dm3 to km3:

Imagine 238.48 dm3 to be a volume shaped like this:
238.48 dm x 1 dm x 1 dm

Now, we will convert dm to its equivalent amount in km:

1 dm times (1 km / 104 dm) = 10¯4 km

Replace dm with its equivalent km:

(238.48 x 10¯4 km) x 10¯4 km x 10¯4 km = 2.3848 x 10¯10 km3

4) Determine the area of the oil slick:

length times width times height = 2.3848 x 10¯10 km3

length times width times 1.20 x 10¯10 km = 2.3848 x 10¯10 km3

length times width = area = 1.987 km2

The best answer would be 2 square kilometers.

Problem #38: 1.37 gallon of paint covers the wall area of 11.9 ft long and 14.5 ft wide. How thick is the paint in mm?

Solution:

1) Convert 1.37 gallons to dm3:

1 gallon = 3.7854 dm3

1.37 gal = 5.186 dm3

2) Convert 5.186 dm3 to mm3:

Imagine this volume:
5.186 dm x 1 dm x 1dm

Convert dm to mm:

1 dm times (102 mm / 1 dm) = 102 mm

Replace dm with its equivalent mm value:

(5.186 x 102 mm) x 102 mm x 102 mm = 5.186 x 106 mm3 <--- this is the volume of the paint on the wall

3) Convert one foot to its equivalent in millimeters:

1 foot = 2.54 cm (by definition)

1 foot = 25.4 mm

4) Determine the two dimensions of the wall area in millimeters:

11.9 feet x 25.4 mm/ft = 302.26 mm
14.5 feet x 25.4 mm/ft = 368.3 mm

5) Determine the thickness of the paint:

5.186 x 106 mm3 = (302.26 mm x 368.3 mm) times height

height = 46.6 mm

Problem #39: Earth lies about 8.33 kiloparsecs (kpc) from the center of the Milky Way galaxy. How far is that in miles?

Solution:

This question forces you to look some values up.

1) Convert kpc to light-year:

8.33 kpc times (1000 pc/kpc) times (3.26 ly/parsec) = 27155.8 ly
2) A light-year is the distance light travels in one year. Let's calculate it in a dimensional analysis style:
(299792458 m/s) (60 s/min) (60 min/hr) (24 hr/day) (365.25 day/yr) = 9460730472580800 m/yr

By the way, this is an exact value because every value in the above calculation is a defined value.

3) Since we want the distance in miles, let's convert from meters to miles:

(9460730472580800 m/yr) (1 mile / 1609.34 m) = 5878639984453.75 mi/yr

4) The final calculation:

(5878639984453.75 mile/yr) (27155.8 ly) = 1.5964 x 1017 miles

Do the units year and light-year cancel? In this case, they do. The first value is the distance in one light year, so the unit could actually be written as mile/ly.

Problem #40: If a 1.67 g piece of gold with density = 19.30 g /cm3 is hammered into a sheet whose area is 41.9 ft2, then what is the thickness of the sheet in cm?

Solution:

1) The volume occupied by the gold:

1.67 g / 19.30 g/cm3 = 0.0865285 cm3

2) Let us convert square feet to square cm:

41.9 ft2 times (12 inch / ft)2 times (2.54 cm / in)2 = 38926.4 cm2

3) The third dimension is the thickness of the gold:

0.0865285 cm3 = (38926.4 cm2) (x)

x = 2.22 x 106 cm

Problem #41: A thin piece of metal foil is 1.287 ft long and 4.34 in wide. The density of the metal is 8.320 g/cm3 and the specific heat of the metal is 0.388 J g-1 °C-1. If the metal foil absorbs 800.0 J of heat as its temperature changes from 22.17 °C to 34.31 °C, calculate the thickness of the metal foil in mm.

Solution:

Note that this problem involves some calculations that don't normally appear in a metric unit. Please pass this problem by if the calculations confuse you.

1) We use the temperature data to calculate the mass of the foil:

q = (mass) (Δt) (Cp)

800.0 J = (x) (12.14 °C) (0.388 J g-1 °C-1)

x = 169.84 g

2) Using the density, we can determine the volume of the foil:

169.84 g / 8.320 g/cm3 = 20.41346 cm3

3) We now need to convert the other two dimensions of the foil to cm:

1.287 ft x (12 in/ft) x (2.54 cm / inch) = 39.22776 cm
4.34 in x (2.54 in/cm) = 11.0236 cm

Note: the thickness of the foil will be the third dimension.

4) Calculate the thickness of the foil:

20.41346 cm3 = (39.22776 cm) (11.0236 cm) (x)

x = 0.0472 cm

Problem #42: Light traveling at a speed of 3.00 x 1010 cm/s takes 8 min 20 sec to travel from the Sun to Earth. Calculate the distance, in miles, from the Sun to Earth.

Solution:

1) Determine the time elapsed, in seconds:

8 min x 60 s/min = 480 s

480 s + 20 s = 500 s

2) Determine the number of centimeters the light travels in 500 s:

3.00 x 1010 cm/s times 500 s = 1.50 x 1013 cm

3) Change cm to miles in a dimensional analysis format:

1.50 x 1013 cm x (1 in / 2.54 cm) x (1 ft / 12 in) x (1 mile / 5280 feet) = 9.32 x 107

This distance, 93 million miles, is called an Astronomical Unit. It has an exact definition (which you can look up), but can be casually referred to as the average distance of the Earth to the Sun. By contrast, the Moon is about 1.28 light-seconds from the Earth.

Problem #43: If your blood has a density of 1.05 g/mL at 20 °C how many pounds of blood would you lose if you donated exactly 1 pint of blood?

Solution:

1) Convert one pint to mL:

1 pint = 473.176 mL

2) Determine grams of blood in one pint (but using mL):

473.176 mL times 1.05 g/mL = 496.8348 g

3) Convert grams to pounds:

496.8348 g / 453.592 g/lb = 1.095 lbs

To three sig figs, this is 1.10 lbs

Problem #44: Assume that a mililiter of water contains exactly 20 drops. How long, in hours, will it take you to count the number of drops of water in 1.00 gal of water at the counting rate of exactly 10 drops/sec

Solution:

The use of the word 'exactly' removes those numbers for consideration when determining significant figures.

1) Determine drops in 1.00 gal of water:

According to Google Converter, 1.00 gallon contains 3785.41 mL.

3785.41 mL times 20 drops/mL = 75708.2 drops

2) Determine the time required to count the drops:

75708.2 drops divided by 10 drops/s = 7570.82 s

7570.82 s times (1 hr / 3600 s) = 2.103 hrs

3) Suppose we want to get this in hours, minutes and seconds. Do this:

7570.82 s minus 7200 s = 370.82 s <--- the seconds left over in the 2.103 hrs

At 60 s in a minute, we see that 6 minutes have 360 seconds, so:

370.82 s - 360 s = 10.82

Final time:

2 hours 6 minutes 10.82 seconds

Problem #45: A sample of air contains 41.2 μg/m3 of beryllium dust. How many atoms of beryllium are present in a room with dimensions of 11 feet by 10.3 feet by 16.6 feet?

Solution:

A portion of this solution may not be clear to you, since it includes concepts you may not have learned yet. Please skip if that is the case. Only a partial working-out is provided.

1 foot = 0.3048 m

(11 feet x 0.3048 m/ft) x (10.3 feet x 0.3048 m/ft) x (16.6 feet x 0.3048 m/ft) = 53.2578 m3

41.2 μg/m3 x 53.2578 m3 = 2194.2 μg = 0.0021942 g

0.0021942 g / 9.0122 g/mol = moles of Be

moles of Be times 6.022 x 1023 atoms/mole = atoms of Be

Round final answer to three sig figs.

Problem #46: How many decigrams are there in 0.05 pounds?

Solution:

0.05 lb times (453.6 g/lb) = 22.68 g

22.68 g times (10 dg / g) = 226.8 g

Set up as one calculation string, we would have this:

(0.05 lb) (453.6 g/lb) (10 dg / g) = 226.8 g

Problem #47: Convert 62.3021 lb/ft3 (the density of water at 20 °C) to g/mL

Solution:

62.3021/ft3 times (453.592 g/lb) = 28259.73 g/ft3

Now, a bit of discussion.

Think of 1 ft33 as a cube 1 foot on each side:

1 ft x 1 ft x 1 ft

We need to convert the foot to centimeters. The reason for that is because one mL equals one cm3. So, let's convert 1 foot to cm, then go back to the cubic measurement:

1 ft times (12 inches / ft) times (2.54 cm / inch) = 30.48 cm

Back to the cubic measurement:

1 ft3 = 30.48 cm x 30.48 cm x 30.48 cm = 28316.85 cm3

Now, for the last step:

28259.73 g/ft3 times (1 ft3 / 28316.85 mL) = 0.997983 g/mL

Round off the answer more as you see fit.

Often, one sees a problem like the one above set up like this:

(62.3021 lb / ft3) x (453.592 g / lb) x (1 ft / 12 in)3 x (1 in / 2.54 cm)3 x (1 cm3 / 1 mL) = 0.997983 g / mL

Note this factor:

(1 in / 2.54 cm)3

and how it is cubed.

The combination of these two factors:

(1 ft / 12 in)3 x (1 in / 2.54 cm)3

is the equivalent of my 1 ft x 1 ft x 1 ft discussion.

Problem #48: A tank is 800. mm long, 150. mm wide, and 100. mm deep. Calculate the volume of the tank in: (a) milliliters; (b) liters and (c) cubic feet

Solution:

1) The solution to part (a) depends on this equivalency:

1 mL is equal to 1 cm3.

2) Since there are 10 mm in each cm, we can do this to each dimension given:

800. mm times (1 cm / 10 mm) = 80.0 cm
150. mm times (1 cm / 10 mm) = 15.0 cm
100. mm times (1 cm / 10 mm) = 10.0 cm

3) This results in this volume calculation:

80.0 cm times 15.0 cm times 10.0 cm = 1.20 x 104 cm3 (done in scientific notation to show 3 sig figs)

4) Using the equivalency in step 1 above, we arrive at this answer:

1.20 x 104 mL

5) To convert to liters for part (b):

1.20 x 104 mL times (1 L / 1000 mL) = 12.0 L

For the conversion to cubic feet in part (c), I would return to the value in cm3 mentioned in step 3. I would like for you to think of it the three dimensions this way:

1.20 x 104 cm times 1.00 cm times 1.00 cm

6) Our goal is to convert from centimeters to feet. For that, we use Google Converter to look up the conversion of 1 cm to feet:

1 cm = 0.0328084 feet

7) Now, we convert each cm value to its equivalent foot value:

(1.20 x 103 cm times 0.0328084 ft / 1 cm) times 0.0328084 ft times 0.0328084 ft = 0.424 ft3

The value in Google Converter is this.

Problem #49: Copper has a density of 8.96 g/cm3. How many blocks (each with a volume of 58.0 mL) could be made from 3.2 kg of copper?

Solution:

1) mass of each copper block:

1 mL = 1 cm3 ---> 58.0 mL = 58.0 cm3

58.0 cm3 times 8.96 g/cm3 = 519.68 g

2) Convert g to kg:

519.68 g times (1 kg / 1000 g) = 0.51968 kg

3) How many blocks?

3.2 kg / 0.51968 kg = 6.16 blocks

So, six blocks can be made and a bit of copper will be left over.

4) Set up in dimensional analysis style:

(58.0 mL) x (1 cm3 / 1 mL) x (8.96 g / cm3) x (1 kg / 1000 g) x (3.2 kg)

Note that all units cancel, leaving us with a unitless answer. This is consistent with the "how many" aspect of the question. You could also write the initial value as 58.0 mL/block and wind up with 'blocks' as the final unit.

5) By the way, how much copper is left over?

519.68 g times 6 = 3118.08 g

3200 g minus 3118.08 g = 81.92 g

Problem #50: Assuming that a planet is a perfect sphere with a circumference of 25263 km and an average density of 2.41 g/mL, what is its approximate mass? Answer in units of metric ton.

Solution:

1) Determine the radius of the planet:

C = 2πr

25263 km = (2) (3.14159) (r)

r = 4020.7347 km

Comment: we know that one mL is equal to one cm3. We also know that we have to get the volume, then multiply that by the density to get the mass in grams. Converting km to cm seems best. Otherwise, we would convert cubic km to cubic cm.

2) Convert km to cm:

4020.7347 km times (105 cm / km) = 4.0207347 x 108 cm

Note the extra digits I'm carrying.

3) Determine the volume of the planet:

V = (4/3)πr3

V = (4/3) (3.14159) (4.0207347 x 108 cm)3

V = 1.6842 x 109 cm3

4) Now, for the mass:

1.6842 x 109 cm3 times 2.41 g/cm3 = 4.058922 x 109 g

5) Convert to metric tons:

1 metric ton = 1000 kg

4.058922 x 109 g times (1 kg / 1000 g) times (1 mt / 1000 kg) = 4.058922 x 103 mt

Notice how there is a 109 in the numerator and a total of 106 in the denominator, leading to the 103 in the answer.

The best answer would be, in the ChemTeam's estimation, 4060 mt.

Problem #51: Magnesium metal can be extracted from sea water by the Dow process. There are 1.4 g of magnesium per kg of sea water. The annual production of magnesium in the United States is about 1 x 105 tons. If the density of sea water is 1.025 g/mL how may m3 of sea water must be processed to obtain this amount of magnesium?

Solution:

Comment: I will use cm3 in place of mL. This is because I will eventually change cm3 to m3. This is one of those things you learn by experience.

1) Change tons of Mg to grams of Mg:

1 x 105 tons x (2000 lb / ton) x (453.6 g/lb) = 9.072 x 1010 g

2) How many kg of seawater?

9.072 x 1010 g divided by (1.4 g / kg) = 6.48 x 1010 kg of seawater

3) What is the volume of seawater?

6.48 x 1010 kg times (1000 g / kg) times (1 cm3 / 1.025 g = 6.32195 x 1013 cm3

4) Change cm3 to m3

6.32195 x 1013 cm3 times (1 m / 100 cm)3 = 6.32195 x 107 m3

6.32 x 107 m3 (to three sig figs)

Problem #52: Convert 1.57 g/mL to kg/gal.

Solution:

1) Convert g to kg:

1.57 g/mL times (1 kg/1000 g) = 0.00157 kg/mL

2) Convert mL to L:

0.00157 kg/mL times (1000 mL / L) = 1.57 kg/L

3) Convert liters to gallons:

1.57 kg/L times (3.785 L / gal) = 5.94 kg/gal

4) Putting all the conversions in one line, dimensional analysis style:

1.57 g/mL times (1 kg/1000 g) times (1000 mL / L) times (3.785 L / gal) = 5.94 kg/gal

Problem #53: How many meters of 34-gauge wire (diameter = 0.006304 in) can be produced from the copper in 4.85 lb of covellite, an ore of copper that is 66.0% copper by mass?

Solution:

4.85 lb times 453.592 g/lb = 2199.9212 g

2199.9212 g times 0.660 = 1451.948 g <--- mass of copper

1451.948 g / 8.95 g/cm3 = 162.2288 cm3 <--- volume of the copper

0.006304 in times 2.54 cm/in = 0.01601216 cm

0.01601216 cm / 2 = 0.00800608 cm <--- the radius of the copper wire

formula for volume of a cylinder: V = πr2h

162.2288 cm3 = (3.14159) (0.00800608 cm)2 (h)

h = 805635.5 cm

805635.5 cm times (1 m / 100 cm) = 8056.355 m

Three sig figs is appropriate, so 8060 m

Problem #54: The density of titanium is 4.51 g/cm3. What is the volume (in cubic inches) of 3.5 lbs of titanium?

Solution:

1) Here's the conversion in the usual dimensional analysis format:

3.5 lbs times (453.59 g / lb) times (1 cm3 / 4.51 g) times (in / 2.54 cm)3 = 21.5 in3

2) What does it all mean?

(453.59 g / lb) changes lb to g

(1 cm3 / 4.51 g) get the volume of the Ti in cm3

(in / 2.54 cm)3 changes cm3 to in3 <--- notice that I put the cube just outside the parentheses

3) Sometimes, you see this conversion factor:

in3 / 16.387 cm3

That's just the 2.54 being cubed.

Problem #55: Ball bearings are spheres used to reduce rotational friction in turning parts. How many ball bearings 2.54 mm in diameter can be made from a 100.0 kg block of steel? Assume that the steel has a density of 8.03 g/cm3

Solution:

1) We will need the volume of the 100.0 kg block of steel. (I will ignore all sig figs until the end.)

100.0 kg times (1000 g / kg) = 100000 g

100000 g divided by 8.03 g/cm3 = 12453.3 cm3 (That's the total volume of all the steel.)

2) I need to know the volume of one ball bearing. However, I need to convert the 2.54 mm diameter to cm. This is because the volume of steel just above is in cm3.

2.54 mm times (1 cm / 10 mm) = 0.254 cm

0.254 cm /2 = 0.127 cm <--- that's the radius

V = (4/3) (pi) r3 (formula for the volume of a sphere)

V = (4/3) (3.14159) (0.127 cm)3

V = 0.00858025 cm3 <--- rounded off, but not too far

3) The last step:

12453.3 cm3 / 0.00858025 cm3 = 1451392

Rounded off to three sig figs (because of the 2.54) and in scientific notation:

1.45 x 106

Problem #56: The total volume of seawater in the world's oceans is 1.335 x 109 km3, Assume that seawater contains 3.5% dissolved salts by mass and that seawater has a density of 1.025 g/mL. Calculate the total mass of dissolved salts in kilograms (kg) and in tons. (1 ton = 2,000 pounds, 1 pound = 453.6 grams)

Solution:

1) Convert cubic km to cubic decimeters. I choose this conversion because 1 dm3 equals 1 L. Because the density uses mL, I will have an easy conversion from 1 L to mL.

(1.335 x 109 km3) (104 dm / 1 km)3 = 1.335 x 1021 dm3

Since 1 L equals dm3, I have my volume in liters.

2) Using the density of seawater, calculate the mass of seawater (in kilograms) in the oceans.

(1.335 x 1021 L) (1000 mL / L) (1.025 g / mL) (1 kg / 1000 g) = 1.368375 x 1021 kg seawater

first conversion: changed L to mL
second conversion: changed mL to grams
third conversion: changed g to kilograms

I'll round off at the end.

2) The term 3.5% means that out of every 100 g seawater, 3.5 g are salts, or in terms of kilograms, there are 3.5 kilograms of dissolved salts in every 100 kilograms of seawater. So, the total mass of salts in the oceans is:

(1.368375 x 1021 kg) (3.5 kg / 100 kg seawater) = 4.7893125 x 1019 kg

3) Convert this into tons:

(4.7893125 x 1019 kg) (1000 g/1 kg) x (1 lb / 453.6 g) x (1 ton/2000 lb) = 5.2792 x 1016 tons dissolved salts

Based on 3.5%, we write 4.8 x 1019 kg and 5.3 x 1016 tons for the final answers.

Problem #57: An electrolytic tin plating process gives a coating of 0.762 micrometer thick. How many square meters and square centimeters can be coated with 1 kilogram of tin if its density is 7.365 g/cm3?

Solution:

1) Determine volume of tin on hand:

1 kg = 1000 g

1000 g / 7.365 g/cm3 = 135.777 cm3

2) Convert μm to cm

centi- is 10-2 and micro- is -6, so the absolute exponential distance between the two units is 104.

0.762 μm times (1 cm / 104 μm) = 7.65 x 10-5 cm

3) Determine area in square centimeters:

135.777 cm3 = (7.62 x 10-5 cm) (x)

x = 1781850 cm2

4) Convert volume of tin from cm3 to m3

short calculation way

1781850 cm2 times (1 m / 100 cm)2 = 178.185 m2

long calculation way:

135.777 cm3 = 135.777 cm x 1 cm x 1 cm

1 cm = 0.01 m

(135.777 x 0.01 m) x 0.01 m x 0.01 m = 1.35777 x 10-4 m3 <--- volume of tim n cubic meters

0.762 μm times ( 1 m / 106 μm) = 7.65 x 10-7 m <--- thickness of tin plating in meters

1.35777 x 10-4 m3 = (7.62 x 10-7 m) (x) <--- volume equals the thickness (which we know) times the area (which is what we want to know)

x = 178.185 m2

Problem #58: Given 0.200 M H2SO4, convert the H+ concentration to molecules nm-3

Solution:

I will assume H2SO4 ionizes 100% for both ionization steps. This is not exactly the real situation, since H2SO4 is 100% ionized only for the first ionization step. The second step has a KaSO4 = 0.012.

My assumption means that 0.200 M H2SO4 is 0.400 M in hydrogen ion. I will write 0.400 M as 0.400 mol dm-3

Convert moles to ions of H+

0.400 mol dm-3 times 6.022 x 1023 ions/mole = 2.4088 x 1023 ions dm-3

Convert dm-3 to nm-3

2.4088 x 1023 ions dm-3 times (1 dm / 108 nm)3 = 0.241 ions nm-3 (to three sig figs)

Comments on the (1 dm / 108 nm)3 factor:

(a) be careful on the placement of the dm3 unit since dm-3 is in the denominator.

(b) Note that it is dm3 and not just dm. Be mindful of the cube sitting outside the parentheses.

(c) 108 arises from the fact that deci- is 10-1 and nano- is 10-9. 108 expresses the absolute exponential distance between the two units.

Problem #59: Ethanol has a density of 0.789 g/cm3. You have 2.50 L of it. How many pounds do you have on hand?

Solution:

1) Convert liters to milliliters, then to cubic centimeters:

(2.50 L) (1000 mL / 1 L) (1 cm3 / 1 mL) = 2500 cm3

2) Convert cm3 to grams (using the density) and thence to pounds:

(2500 cm3) (0.789 g/cm3) (1 lb/453.6 g) = 4.3485lb

The answer is 4.35 lb (to three significant figures)

3) You can string the entire calculation together, if so desired:

(2.50 L) (1000 mL / 1 L) (1 cm3 / 1 mL) (0.79 g/cm3) (1 lb/453.6 g)

I split it apart into two pieces for (hopefully) greater clarity.

Problem #60: Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a 100. ft x 82.0 ft ceiling with gold leaf that is five millionths of an inch think. The density of gold is 19.32 g/cm3 and gold costs \$1265 per Troy ounce (1 Troy ounce = 31.1034768 g). How much will it cost the architect to buy the necessary gold?

Solution:

Comment: the goal will be to determine the volume of the gold foil in cm3. This is because we know the density uses cm3. Once we know the volume of the gold foil, we can determine the mass of gold present.

1) Determine the three dimensions of the gold foil, each in cm:

(100. ft) (12 in/ft) (2.54 cm/in) = 3048 cm

(12 in/ft) (2.54 cm/in) = 2499.36 cm

(5 x 10¯6 in) (2.54 cm/in) = 1.27 x 10¯5 cm

2) Determine the volume of the gold foil:

(3048 cm) (2499.36 cm) (1.27 x 10¯5 cm) = 96.7492 cm3

3) Determine the mass (in grams) of gold present:

(96.7492 cm3) (19.32 g/cm3) = 1869.194544 g

4) Determine the mass (in Troy ounces) of gold present:

(1869.194544 g) (1 oz / 31.1034768 g) = 60.096 oz

5) Money!

(60.096 oz) (1265 dollars / oz) = \$76021.44

There are some instructors that insist on a dimensional analysis set up, where the entire calculation is arranged into one continuous line of multiplications and divisions. For example, this portion gives the volume of the gold foil::

(100. ft) (82.0 ft) (12 in/ft)2 (5 x 10¯6 in) (2.54 cm/in)3

Adding on the rest gives the complete set up:

(100. ft) (82.0 ft) (12 in/ft)2 (5 x 10¯6 in) (2.54 cm/in)3 (19.32 g/cm3) (1 oz / 31.1034768 g) (1265 dollars / oz) = \$76021.46

Problem #61: How many carbon atoms, placed end to end, covers 1.00 mile?

Solution:

1) The first item to do is determine the diameter of a carbon atom:

Wikipedia has a list of the atomic radii of the elements. In it, the atomic radius of carbon is given as 70 pm, therefore the diameter is 140 pm.

2) Let us convert 1.00 mile to picometers. Then, we can divide by 140 pm to determine how many C atoms make up the 1.00 mile. Here's the entire conversion from 1.00 mile to picometers:

 5280 feet 12 inches 0.0254 m 1012 pm 1.00 mile x ––––––– x ––––––– x ––––––– x ––––––– = 1.609344 x 1015 pm 1 mile 1 foot 1 inch 1 m

In order, the conversions do the following:

(a) convert mile to feet
(b) convert feet to inches
(c) convert inches to meters (note the modification of the commonly used 1 inch equals 2.54 cm)
(d) convert meters to picometers

3) The last step determines how many carbon atoms are involved:

1.609344 x 1015 pm / 140 pm/atom = 1.15 x 1013 atoms

Bonus Problem:You happen to have 350 Canadian dollars, How much are the Canadian dollars worth in US dollars? Use the following conversions:

US Dollars Per Canadian dollars = 0.6484
Canadian dollars per US dollars= 1.542

Solution:

1) Every 0.6484 US\$ equals one C\$:

350 C\$ times (0.6484 US\$ / 1 C\$) = 226.94 US\$

2) Every 1.542 C\$ equals one US\$:

350 C\$ times (1 US\$ / 1.542 C\$) = 226.98 US\$

Notice how I used 1 over 1.542 rather than 1.542 over 1. I did this to cancel the unit C\$

3) This is what is causing the discrepancy in the two answers:

1 over 0.6484 equals 1.542257865515114 (not 1.542)

350 C\$ times 1 US\$ / 1.542257865515114 C\$ = 226.94 US\$

Comment: Remember that after the per is where the 1 always is. Like this:

US Dollars Per ONE Canadian dollar = 0.6484 US\$ / 1 C\$

Canadian dollars per ONE US dollar = 1.542 C\$ / 1 US\$