# 1.11: Multistep Conversions

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As mentioned in the previous section, the conversions that will be required for this course will become increasingly complex, due to either the type of unit being canceled or the number of conversion factors that must be used to achieve the desired unit transformation.  In this section, the latter scenario will be explored.

For example, convert 0.547 kiloliters to deciliters, using only conversion factors based on the equalities from Section 1.6.

The six-step dimensional analysis process found in the previous section must still be applied.  Step 3, in particular, will need to be considered more thoughtfully, in order to arrive at the correct answer.

1. Determine the "given" unit in the problem.  The given number is 0.547, and its unit is kiloliters.

2. Determine the "desired" unit in the problem.  The given quantity should be changed to deciliters, making "deciliters" the desired unit.

3. Determine which equalities relate the given and desired units.  In this case, equalities, plural, will be necessary to achieve the required unit change, as Section 1.6 does not reference an equality that directly relates kiloliters to deciliters.  However, each of these units can be related back to a common base unit, liters.  Kiloliters and liters can be related by the prefix modifier equality kL = 1,000 L.  The right side of this equality is the simplified form of 103, which is the meaning of "kilo".  Liters and deciliters can be related by the prefix modifier equality L = 10 dL.  This equality is the twice-simplified form of 10-1 L = dL, which is derived from the meaning of "deci", 10-1.

4. Use the appropriate conversion factors derived from these equalities to achieve unit cancelation.

$${0.547 \; \cancel{\rm{kL}}} \times \dfrac{1,000 \; \bcancel{ \rm{L}}}{\cancel{\rm{kL}}}\ \times \dfrac{10 \; \rm{dL}}{\bcancel{\rm{L}}}$$

Note that in each step, the previous unit is canceled, and the next unit in the sequence is produced.  Each successive unit must be canceled until only the desired unit remains.

5. Perform the calculation that remains once the units have been canceled.  In this case,

$${0.547} \times {\text {1,000}} \times {\text {10 dL}} = {\text {5,470 dL}}$$

6. Apply the correct number of significant figures to the calculated quantity.  Since both conversions used are based on prefix modifier equalities, neither the "1,000" nor the "10" are considered when determining the number of significant figures that the answer should have.  The given number has 3 significant figures, so the final answer must have 3 significant figures.  Since the number of digits that the answer actually has is exactly equal to the number of digits that it should have, no adjustments need to be made, and the calculated answer is the final answer.
Example $$\PageIndex{1}$$

Using only conversion factors based on the equalities from Section 1.6, convert 1.3 weeks to hours.

Solution

The given unit is "weeks," and the desired unit is "hours."  These units can be related by the measured equalities $${ \text{1 wk}} = { \text{7 d}}$$ and $${ \text{1 d}} = { \text{24 h}}$$.  Therefore,

$${1.3 \; \cancel{\rm{wk}}} \times \dfrac{7 \; \bcancel{ \rm{d}}}{1 \; \cancel{\rm{wk}}}\ \times \dfrac{24 \; \rm{h}}{1 \; \bcancel{\rm{d}}} = {\text {218.4 h}}$$

Each of the numbers found within a measured equality is exact.  The given number has 2 significant figures, so the final answer must have 2 significant figures.  Since the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded to 220 h.

Exercise $$\PageIndex{1}$$

Using only conversion factors based on the equalities from Section 1.6, convert 952.1 centimeters to yards.

The given unit is "centimeters," and the desired unit is "yards."  These units can be related by the measured equalities $${ \text{1 in}} = { \text{2.54 cm}}$$, $${ \text{1 ft}} = { \text{12 in}}$$, and $${ \text{1 yd}} = { \text{3 ft}}$$.  Therefore,
$${952.1 \; \cancel{\rm{cm}}} \times \dfrac{1 \; \bcancel{ \rm{in}}}{2.54 \; \cancel{\rm{cm}}}\ \times \dfrac{1 \; \bcancel{ \rm{ft}}}{12 \; \cancel{\rm{in}}}\ \times \dfrac{1 \; \rm{yd}}{3 \; \cancel{\rm{ft}}} = {\text {10.412292213... yd}}$$