Problems #26-60

Metric-English Conversion Problems #1-10 | Metric-English Conversion Examples |

Metric-English Conversion Problems #11-25 | Return to Metric Menu |

**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- (10^{3}) and pico- (10¯^{12}) is 10^{15}. This problem can be done two ways.

Method #1: Convert km to pm

40,075.02 km times (10

^{15}pm / 1 km) = 4.007502 x 10^{19}pm4.007502 x 10

^{19}pm / (140 pm / atom) = 2.86 x 10^{17}atoms

Method #2: Convert pm to km

140 pm times (1 km / 10

^{15}pm) = 1.40 x 10¯^{13}km40,075.02 km / (1.40 x 10¯

^{13}km/atom) = 2.86 x 10^{17}atoms

Comment: the mass of this amount of Pd is:

2.86 x 10^{17}atoms / 6.022 x 10^{23}atoms mol¯^{1}= 4.75 x 10¯^{7}mol4.75 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 10^{7}m

atom: 3.2 x 10¯^{10}m

2) Solve via a ratio and proportion technique:

football diameter : earth diameter = atom diameter : an unknown value0.22 m / 1.3 x 10

^{7}m = 3.2 x 10¯^{10}m / xSolving 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 10^{5} metric tons of crude oil of density 0.70 g/cm^{3} 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 = 10^{3} kg.)

**Solution:**

1) Determine kg of oil:

2.0 x 10^{5}times 1000 = 2.0 x 10^{8}kg of oil

2) Change kg to g:

2.0 x 10^{8}kg times 1000 = 2.0 x 10^{11}g

3) Determine volume of oil:

2.0 x 10^{11}g divided by 0.70 g/cm^{3}= 2.857 x 10^{11}cm^{3}

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 10^{11}cm^{3}divided by 0.010 cm = 2.857 x 10^{13}cm^{2}

6) Change cm^{2} to km^{2}:

2.857 x 10^{13}cm^{2}times (1 km^{2}/ 10^{10}cm^{2}) = 2857 km^{2}Comment: 1 km

^{2}is 1 km on each side. 1 km = 10^{5}cm, so the 1 km^{2}is 10^{5}cm on each side, so:10

^{5}cm times 10^{5}= 10^{10}cm^{2}(which equals 1 km^{2})

The last conversion can also be written in this manner:

2.857 x 10^{13}cm^{2}times (1 km / 10^{5}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 cm^{3}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:

1cm^{3}= 1 mL, so 420000 cm^{3}= 420000 mL420000 mL x 1.00 g/mL = 420000 g

3) Change g to kg:

420000 g times (1 kg / 1000 g) = 420. kgNote 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 cm^{3} of a single atom.

**Solution:**

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

diameter ---> twice the radius is 3.80 Åconvert Å to cm ---> 3.80 Å times (10

^{-8}cm / Å) = 3.80 x 10^{-8}cmconvert 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 10^{6}Kr atoms

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

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

**Problem #31:** The star Arcturus is 3.472 x10^{14} 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 10^{8} m/s.)

**Solution:**

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

3.472 x10^{14}km x (1000 m / km) = 3.472 x10^{17}m

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

3.472 x10^{17}m ÷ (3.00 x 10^{8}m/s) = 1.15733 x 10^{9}s

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

1.15733 x 10^{9}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 10^{8}m/s x (3600 s / 1 hr) x (24 hr / 1 day) x (365 day / 1 year) = 9.4608 x 10^{15}m <--- distance in one light year

We now use the distance to Arcturus, but in meters:3.472 x10^{17}m ÷ 9.4608 x 10^{15}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 10^{8}hemoglobin units per red blood cell (RBC)

each mm^{3}of blood contains 5.5 x 10^{6}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 cm^{3}5.5 x 10

^{6}RBC / mm^{3}times (10 mm / cm)^{3}= 5.5 x 10^{9}RBC / cm^{3}The above is the trick. The RBC was given per square millimeter, not per square centimeter. Tricky!5.5 x 10

^{9}RBC / cm^{3}times 5000 cm^{3}= 2.75 x 10^{13}RBC

2) How many hemoglobin units?

2.75 x 10^{13}RBC times 2.0 x 10^{8}hemoglobin units/RBC = 5.5 x 10^{21}hemoglobin units

3) Iron atoms:

5.5 x 10^{21}hemoglobin units times 4 Fe atoms / hemoglobin unit = 2.2 x 10^{22}Fe atoms

**Solution to (B):**

1) Convert 125 pm to km:

125 pm/Fe atom times (1 km / 10^{15}pm) = 1.25 x 10^{-13}km/Fe atom

2) Determine length of Fe "string:"

2.2 x 10^{22}Fe atoms times 1.25 x 10^{-13}km/Fe atom = 2.86 x 10^{9}km2.86 x 10

^{9}km times (1 mile / 1.60934 km) = 1.777 x 10^{9}miI 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. cm^{3}

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 cmNote 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. cm^{3}= 609.6 cm times 548.64 cm times XX = 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 cm^{3}Centimeters is a convenient unit to use.

2) Convert 40 m^{2} to cm^{2}:

40 m^{2}times (100 cm / m)^{2}= 4 x 10^{5}cm^{2}An alternate way:

40 m

^{2}= 40 m x 1 mConvert each m to cm

(40 m x 100 cm/m) x (1 m x 100 cm/m) = 4 x 10

^{5}cm^{2}

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

0.1 cm^{3}= 4 x 10^{5}cm^{2}times XX = 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/cm^{3}? The radius of our sun is 6.955 x 10^{5} km and it has an average density of 1.410 x 10^{3} kg/m^{3}. The diameter of the moon is 2.16 x 10^{3} miles.

**Solution:**

1) The mass of our sun:

6.955 x 10^{5}km = 6.955 x 10^{8}mV = (4/3)πr

^{3}V = (4/3) (3.14159) (6.955 x 10

^{8}m)^{3}V = 1.40922275 x 10

^{27}m^{3}mass of one sun = (1.40922275 x 10

^{27}m^{3}) (1.410 x 10^{3}kg/m^{3}) = 1.987004 x 10^{30}kg = 1.987004 x 10^{33}gmass of 1000 suns = 1.987004 x 10

^{36}g

2) volume of black hole:

radius of moon = 2.16 x 10^{3}mi / 2 = 1.08 x 10^{3}miradius of black hole = 1.08 x 10

^{3}mi / 2 = 540 miUse Google calculator to change 540 miles to cm = 8.690 x 10

^{7}cmV = (4/3)πr

^{3}V = (4/3) (3.14159) (8.690 x 10

^{7}cm)^{3}V = 2.748828 x 10

^{24}cm^{3}

3) calculate density of black hole:

1.987004 x 10^{36}g / 2.748828 x 10^{24}cm^{3}= 7.23 x 10^{11}g/cm^{3}

**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 cm^{3}Then:

15782.256 cm

^{3}x (1 mL / 1 cm^{3}) 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 / 10^{12}nm) = 120. x 10¯^{12}km = 1.20 x 10¯^{10}kmWe will use this km value as the height of the oil slick.

2) Let's convert the non-metric US gallon to dm^{3}:

1 US gallon = 3.7854 LI looked up that value here.

63 US gallons = 238.48 L = 238.48 dm

^{3}

Comment: via Google calculator, you can actually convert gallons directly to km^{3}. I chose to use dm^{3} so as to explain how to get from dm^{3} to km^{3}.

3) Convert dm^{3} to km^{3}:

Imagine 238.48 dm^{3}to be a volume shaped like this:238.48 dm x 1 dm x 1 dmNow, we will convert dm to its equivalent amount in km:

1 dm times (1 km / 10^{4}dm) = 10¯^{4}kmReplace dm with its equivalent km:

(238.48 x 10¯^{4}km) x 10¯^{4}km x 10¯^{4}km = 2.3848 x 10¯^{10}km^{3}

4) Determine the area of the oil slick:

length times width times height = 2.3848 x 10¯^{10}km^{3}length times width times 1.20 x 10¯

^{10}km = 2.3848 x 10¯^{10}km^{3}length times width = area = 1.987 km

^{2}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 dm^{3}:

1 gallon = 3.7854 dm^{3}1.37 gal = 5.186 dm

^{3}

2) Convert 5.186 dm^{3} to mm^{3}:

Imagine this volume:5.186 dm x 1 dm x 1dmConvert dm to mm:

1 dm times (10^{2}mm / 1 dm) = 10^{2}mmReplace dm with its equivalent mm value:

(5.186 x 10^{2}mm) x 10^{2}mm x 10^{2}mm = 5.186 x 10^{6}mm^{3}<--- 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 10^{6}mm^{3}= (302.26 mm x 368.3 mm) times heightheight = 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 ly2) 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/yrBy 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 10^{17}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 /cm^{3} is hammered into a sheet whose area is 41.9 ft^{2}, then what is the thickness of the sheet in cm?

**Solution:**

1) The volume occupied by the gold:

1.67 g / 19.30 g/cm^{3}= 0.0865285 cm^{3}

2) Let us convert square feet to square cm:

41.9 ft^{2}times (12 inch / ft)^{2}times (2.54 cm / in)^{2}= 38926.4 cm^{2}

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

0.0865285 cm^{3}= (38926.4 cm^{2}) (x)x = 2.22 x 10

^{6}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/cm^{3} 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) (C_{p})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/cm^{3}= 20.41346 cm^{3}

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 cm^{3}= (39.22776 cm) (11.0236 cm) (x)x = 0.0472 cm

**Problem #42:** Light traveling at a speed of 3.00 x 10^{10} 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 s480 s + 20 s = 500 s

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

3.00 x 10^{10}cm/s times 500 s = 1.50 x 10^{13}cm

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

1.50 x 10^{13}cm x (1 in / 2.54 cm) x (1 ft / 12 in) x (1 mile / 5280 feet) = 9.32 x 10^{7}

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 lbsTo 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 s7570.82 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 hrsAt 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/m^{3} 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 m

^{3}41.2 μg/m

^{3}x 53.2578 m^{3}= 2194.2 μg = 0.0021942 g0.0021942 g / 9.0122 g/mol = moles of Be

moles of Be times 6.022 x 10

^{23}atoms/mole = atoms of BeRound 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 g22.68 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/ft^{3} (the density of water at 20 °C) to g/mL

**Solution:**

62.3021/ft^{3}times (453.592 g/lb) = 28259.73 g/ft^{3}Now, a bit of discussion.

Think of 1 ft

^{3}3 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 cm

^{3}. 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 ft

^{3}= 30.48 cm x 30.48 cm x 30.48 cm = 28316.85 cm^{3}Now, for the last step:

28259.73 g/ft

^{3}times (1 ft^{3}/ 28316.85 mL) = 0.997983 g/mLRound off the answer more as you see fit.

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

(62.3021 lb / ft^{3}) x (453.592 g / lb) x (1 ft / 12 in)^{3}x (1 in / 2.54 cm)^{3}x (1 cm^{3}/ 1 mL) = 0.997983 g / mLNote 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 cm^{3}.

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 10^{4}cm^{3}(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 10^{4}mL

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

1.20 x 10^{4}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 cm^{3} mentioned in step 3. I would like for you to think of it the three dimensions this way:

1.20 x 10^{4}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 10^{3}cm times 0.0328084 ft / 1 cm) times 0.0328084 ft times 0.0328084 ft = 0.424 ft^{3}

**Problem #49:** Copper has a density of 8.96 g/cm^{3}. 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 cm^{3}---> 58.0 mL = 58.0 cm^{3}58.0 cm

^{3}times 8.96 g/cm^{3}= 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 blocksSo, 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 cm^{3}/ 1 mL) x (8.96 g / cm^{3}) 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 g3200 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 = 2πr25263 km = (2) (3.14159) (r)

r = 4020.7347 km

Comment: we know that one mL is equal to one cm^{3}. 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 (10^{5}cm / km) = 4.0207347 x 10^{8}cmNote the extra digits I'm carrying.

3) Determine the volume of the planet:

V = (4/3)πr^{3}V = (4/3) (3.14159) (4.0207347 x 10

^{8}cm)^{3}V = 1.6842 x 10

^{9}cm^{3}

4) Now, for the mass:

1.6842 x 10^{9}cm^{3}times 2.41 g/cm^{3}= 4.058922 x 10^{9}g

5) Convert to metric tons:

1 metric ton = 1000 kg4.058922 x 10

^{9}g times (1 kg / 1000 g) times (1 mt / 1000 kg) = 4.058922 x 10^{3}mtNotice how there is a 10

^{9}in the numerator and a total of 10^{6}in the denominator, leading to the 10^{3}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 10^{5} tons. If the density of sea water is 1.025 g/mL how may m^{3} of sea water must be processed to obtain this amount of magnesium?

**Solution:**

Comment: I will use cm^{3} in place of mL. This is because I will eventually change cm^{3} to m^{3}. This is one of those things you learn by experience.

1) Change tons of Mg to grams of Mg:

1 x 10^{5}tons x (2000 lb / ton) x (453.6 g/lb) = 9.072 x 10^{10}g

2) How many kg of seawater?

9.072 x 10^{10}g divided by (1.4 g / kg) = 6.48 x 10^{10}kg of seawater

3) What is the volume of seawater?

6.48 x 10^{10}kg times (1000 g / kg) times (1 cm^{3}/ 1.025 g = 6.32195 x 10^{13}cm^{3}

4) Change cm^{3} to m^{3}

6.32195 x 10^{13}cm^{3}times (1 m / 100 cm)^{3}= 6.32195 x 10^{7}m^{3}6.32 x 10

^{7}m^{3}(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 g2199.9212 g times 0.660 = 1451.948 g <--- mass of copper

1451.948 g / 8.95 g/cm

^{3}= 162.2288 cm^{3}<--- volume of the copper0.006304 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 = πr

^{2}h162.2288 cm

^{3}= (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/cm^{3}. 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 cm^{3}/ 4.51 g) times (in / 2.54 cm)^{3}= 21.5 in^{3}

2) What does it all mean?

(453.59 g / lb) changes lb to g(1 cm

^{3}/ 4.51 g) get the volume of the Ti in cm^{3}(in / 2.54 cm)

^{3}changes cm^{3}to in^{3}<--- notice that I put the cube just outside the parentheses

3) Sometimes, you see this conversion factor:

in^{3}/ 16.387 cm^{3}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/cm^{3}

**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 g100000 g divided by 8.03 g/cm

^{3}= 12453.3 cm^{3}(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 cm^{3}.

2.54 mm times (1 cm / 10 mm) = 0.254 cm0.254 cm /2 = 0.127 cm <--- that's the radius

V = (4/3) (pi) r

^{3}(formula for the volume of a sphere)V = (4/3) (3.14159) (0.127 cm)

^{3}V = 0.00858025 cm

^{3}<--- rounded off, but not too far

3) The last step:

12453.3 cm^{3}/ 0.00858025 cm^{3}= 1451392Rounded off to three sig figs (because of the 2.54) and in scientific notation:

1.45 x 10

^{6}

**Problem #56:** The total volume of seawater in the world's oceans is 1.335 x 10^{9} km^{3}, 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 dm^{3} equals 1 L. Because the density uses mL, I will have an easy conversion from 1 L to mL.

(1.335 x 10^{9}km^{3}) (10^{4}dm / 1 km)^{3}= 1.335 x 10^{21}dm^{3}Since 1 L equals dm

^{3}, 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 10^{21}L) (1000 mL / L) (1.025 g / mL) (1 kg / 1000 g) = 1.368375 x 10^{21}kg seawaterfirst conversion: changed L to mL

second conversion: changed mL to grams

third conversion: changed g to kilogramsI'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 10^{21}kg) (3.5 kg / 100 kg seawater) = 4.7893125 x 10^{19}kg

3) Convert this into tons:

(4.7893125 x 10^{19}kg) (1000 g/1 kg) x (1 lb / 453.6 g) x (1 ton/2000 lb) = 5.2792 x 10^{16}tons dissolved saltsBased on 3.5%, we write 4.8 x 10

^{19}kg and 5.3 x 10^{16}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/cm^{3}?

**Solution:**

1) Determine volume of tin on hand:

1 kg = 1000 g1000 g / 7.365 g/cm

^{3}= 135.777 cm^{3}

2) Convert μm to cm

centi- is 10^{-2}and micro- is^{-6}, so the absolute exponential distance between the two units is 10^{4}.0.762 μm times (1 cm / 10

^{4}μm) = 7.65 x 10^{-5}cm

3) Determine area in square centimeters:

135.777 cm^{3}= (7.62 x 10^{-5}cm) (x)x = 1781850 cm

^{2}

4) Convert volume of tin from cm^{3} to m^{3}

short calculation way

1781850 cm^{2}times (1 m / 100 cm)^{2}= 178.185 m^{2}long calculation way:

135.777 cm^{3}= 135.777 cm x 1 cm x 1 cm1 cm = 0.01 m

(135.777 x 0.01 m) x 0.01 m x 0.01 m = 1.35777 x 10

^{-4}m^{3}<--- volume of tim n cubic meters0.762 μm times ( 1 m / 10

^{6}μm) = 7.65 x 10^{-7}m <--- thickness of tin plating in meters1.35777 x 10

^{-4}m^{3}= (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 m

^{2}

**Problem #58:** Given 0.200 M H_{2}SO_{4}, convert the H^{+} concentration to molecules nm^{-3}

**Solution:**

I will assume H_{2}SO_{4} ionizes 100% for both ionization steps. This is not exactly the real situation, since H_{2}SO_{4} is 100% ionized only for the first ionization step. The second step has a K_{a}SO_{4} = 0.012.

My assumption means that 0.200 M H_{2}SO_{4} 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 10^{23}ions/mole = 2.4088 x 10^{23}ions dm^{-3}

Convert dm^{-3} to nm^{-3}

2.4088 x 10^{23}ions dm^{-3}times (1 dm / 10^{8}nm)^{3}= 0.241 ions nm^{-3}(to three sig figs)

Comments on the (1 dm / 10^{8} nm)^{3} factor:

(a) be careful on the placement of the dm^{3}unit since dm^{-3}is in the denominator.(b) Note that it is dm

^{3}and not just dm. Be mindful of the cube sitting outside the parentheses.(c) 10

^{8}arises from the fact that deci- is 10^{-1}and nano- is 10^{-9}. 10^{8}expresses the absolute exponential distance between the two units.

**Problem #59:** Ethanol has a density of 0.789 g/cm^{3}. 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 cm^{3}/ 1 mL) = 2500 cm^{3}

2) Convert cm^{3} to grams (using the density) and thence to pounds:

(2500 cm^{3}) (0.789 g/cm^{3}) (1 lb/453.6 g) = 4.3485lbThe 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 cm^{3}/ 1 mL) (0.79 g/cm^{3}) (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/cm^{3} 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 cm^{3}. This is because we know the density uses cm^{3}. 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 cm^{3}

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

(96.7492 cm^{3}) (19.32 g/cm^{3}) = 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/cm^{3}) (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 10 ^{12}pm1.00 mile x ––––––– x ––––––– x ––––––– x ––––––– = 1.609344 x 10 ^{15}pm1 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 10^{15}pm / 140 pm/atom = 1.15 x 10^{13}atoms

**Problem #62:**Convert 6.40 x 10¯^{28} Joule-sec/kg to erg-min/lb. A conversion factor to use is: 1 Joule = 10^{7} erg. Also, Joules-sec means Joules times seconds not Joules minus seconds. Commonly, you see this unit written as J-s.

**Solution:**

1) Convert kilograms to pounds:

We look up the conversion to find 1 kg equals 2.20462 lb

6.40 x 10¯ ^{28}J-s1 kg –––––––––––––– x ––––––––– = some value we won't bother to calculate 1 kg 2.20462 lb

2) Convert seconds to minutes:

6.40 x 10¯ ^{28}J-s1 kg 1 min –––––––––––––– x ––––––––– x ––––– = some value we won't bother to calculate 1 kg 2.20462 lb 60 sec

3) Convert Joules to ergs:

6.40 x 10¯ ^{28}J-s1 kg 1 min 1 erg –––––––––––––– x ––––––––– x ––––– x ––––– = 4.84 x 10¯ ^{37}erg-min / lb1 kg 2.20462 lb 60 sec 10 ^{7}J

**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 PerONECanadian dollar = 0.6484 US$ / 1 C$Canadian dollars per

ONEUS dollar = 1.542 C$ / 1 US$

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