Metric Conversion Problems - One Unit
Problems #11 - 25

Probs 1-10

Special note on a particular conversion not mentioned above.

Metric conversions where two units (numerator and denominator) are converted

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Problem #11: Convert 2.1 Å to nanometers

Solution:

1) Often, a two-step method is used. In the first step, the unit given in the problem is converted to the base unit. In this problem, the base unit is meters. Convert Å to meter:

1 Å = 10¯8 cm (by definition)

    1 cm  
1 Å  x  –––––––  =  10¯8 cm
    108 Å  

    1 m  
10¯8 cm  x  –––––––  =  10¯10 m
    100 cm  

1 Å = 10¯10 m

2) In the second step, you convert from the base unit to the target unit. Convert from 2.1 x 10¯10 m to nanometers:

(2.1 x 10¯10 m) (109 nm / 1 m) = 0.21 nm

3) Make sure to remember that the base unit is the unit without a prefix. Examples of base units include m, s, L, and g.


Problem #12: Convert 0.0030 mm to nm

Solution:

There are 106 nanometers in one millimeter.

    106 nm  
3.0 x 10¯3 mm  x  –––––––  =  3.0 x 103 nm
    1 mm  

If we were to convert to the base unit first, it would look like this:

    1 m   109 nm  
3.0 x 10¯3 mm  x  –––––––  x  –––––––  =  3.0 x 103 nm
    103 mm   1 m  

The first conversion is millimeters to meters and the second is meters to nanometers.


Problem #13: Convert 6.25 x 10¯4 s to ms

Solution:

    103 ms  
6.25 x 10¯4 s  x  –––––––  =  0.625 ms
    1 s  

Notice that this is a conversion starting from a base unit.


Problem #14: Convert 38.0 centiliters to kiloliters.

Solution:

    1 kL  
38.0 cL  x  –––––––  =  3.80 x 10¯4 kL
    105 cL  

kilo- is 103 and centi- is 10¯2. The absolute exponential distance between them is 105.

You could also convert the cL to the base unit of L first, then convert L to kL.


Problem #15: How many Mg are there in 100,000,000 mg?

Solution:

I will do this conversion in two steps: (a) convert mg to g, then (b) convert g to Mg.

    1 g  
1 x 108 mg  x  –––––––  =  1 x 105 g
    103 mg  

    1 Mg  
1 x 105 g  x  –––––––  =  0.1 Mg
    106 g  


Problem #16: Convert 0.050 kg to mg.

Solution:

First conversion is kg to g and the second is g to mg.

    103 g   103 mg  
0.050 kg  x  –––––––  x  –––––––  =  5.0 x 104 mg
    1 kg   1 g  


Problem #17: Convert 722 μg to mg.

Solution:

1) A direct micro- to milli- conversion:

722 μg times (1 mg / 1000 μg) = 0.722 mg

2) Here's the two-step conversion. The first conversion is micrograms to grams, followed by grams being converted to milligrams.

    1 g   103 mg  
722 μg  x  –––––––  x  –––––––  =  0.722 mg
    106 μg   1 g  

Problem #18: How many pg in a ng?

Solution:

1 ng times (1000 pg / 1 ng) = 1000 pg

Remember, nano- means 10¯9 and pico- means 10¯12. There are 1000 of the smaller thing (pico-) in one of the larger thing (nano-).


Problem #19: How many mm in a km?

Solution:

1) Two-step (going through the base unit):

    103 m   103 mm  
1 km  x  –––––––  x  –––––––  =  1 x 106 mm
    1 km   1 m  

2) One-step, using the total exponential distance from kilo- to mlli-:

    106 mm  
1 km  x  –––––––  =  1 x 106 mm
    1 km  

3) One step, done in the sty;e oft-seen on the Internet:

1 km times (106 mm / 1 km) = 1 x 106 mm

Problem #20: How many km in a mm?

Solution:

1) Two-step (going through the base unit of m):

    1 m   1 km  
1 mm  x  –––––––  x  –––––––  =  1 x 10¯6 km
    103 mm   103 m  

2) One-step, using the total exponential distance from kilo- to mlli-:

    1 km  
1 mm  x  –––––––  =  1 x 10¯6 km
    106 mm  

Problem #21: Convert 3000 mL to daL.

Solution:

1) milli- means 10¯3 and deca- means 101. That gives an absolute exponential distance of 4, therefore:

3000 mL times (1 daL / 104 mL) = 0.3 daL

2) Done as a two-step through the base unit of L:

    1 L   1 daL  
3000 mL  x  –––––––  x  –––––––  =  0.3 daL
    103 mL   10 L  

In the first conversion, L is the larger of the units (so the 1 goes with it), but in the second conversion, L is the smaller of the two units (so the 1 goes with daL).


Problem #22: Consider this problem:

____ cm = ____ m

Fill in the blanks by placing numerical values in front of the m and the cm.

Solution #1:

1) Let us decide to put a 1 in front of the meter. The problem then becomes this:

____ cm = 1 m

2) In words, the problem becomes:

How many cm are in 1 meter?

3) To answer the question, think back to this definition:

centi- means hundredth

therefore:

centimeter means a hundredth of a meter

4) How many centimeters in one meter?

Since one centimeter is 1/100th of a meter, you need 100 centimeters to make 1 meter.

5) The answer to the fill-in question is:

100 cm = 1 m

Solution #2:

1) Let us decide to put a 1 in front of the centimeter. The problem then becomes this:

1 cm = ____ m

2) In words, the problem becomes:

How many meters are in 1 cm?

3) To answer the question, think back to this definition:

centi- means hundredth

therefore:

centimeter means a hundredth of a meter

4) How many meters in one centimeter?

Since one centimeter is 1/100th of a meter, you need 1/100th of a meter to make 1 centimeter.

5) The answer to the fill-in question is:

1 cm = 0.01 m

Problem #23: Convert 0.005 kg to dag

Solution:

kilo- means 103 and da- means 101, so there is a 102 absolute difference between the two.

0.005 kg x (___ dg / ___ kg) <--- this is done so as to allow the kg to cancel and put the dag in the numerator, which is where we want it

0.005 kg x (___ dg / 1 kg) <--- I teach a method that ALWAYS puts a 1 in front of the larger unit, so you have to know that kilo is bigger than deca

0.005 kg x (102 dg / 1 kg) <--- the two factors are now in place and it reads this way: there are 102 dag in one kg

0.5 dag is the answer.


Problem #24: Convert 35 mL to dL

Solution:

1) Set up the conversion with units in place:

35 mL x (___ dL / ___ mL)

milli = 10¯3 and deci- = 10¯1, so there is 102 in absolute exponential distance between the two units.

2) deci- is bigger than milli-. Therefore, the 1 is placed with the dL:

35 mL x (1 dL / ___ mL)

3) There is 102 in absolute exponential distance:

35 mL x (1 dL / 102 mL) = 0.35 dL

Problem #25: Convert 6.31 Ms to hs

Solution:

Done as a two-step conversion through the base unit of s:

    106 s   1 hs  
6.31 Ms  x  –––––––  x  –––––––  =  6.31 x 104 hs
    1 Ms   102 s  

Probs 1-10

Special note on a particular conversion not mentioned above.

Metric conversions where two units (numerator and denominator) are converted

Return to Metric Conversions

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