# 2.4 More Unit Conversions and Rounding Rules

**Learning Objectives**

By the end of this section, you will be able to:

- Perform unit conversions while respecting the appropriate rounding rules for accuracy and precision.
- Perform operations with scientific notation while respecting the appropriate rounding rules for accuracy and precision.

Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards.

This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft):

1 yd = 3 ft

In math, this expression is called an *equality*. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you don’t divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get

[latex]\frac12yd\;=\;\frac32ft[/latex]

We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true, so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number *and* unit):

[latex]\frac{1yd}{1yd}\;=\;\frac{3ft}{1yd}[/latex]

The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number *and* the unit:

[latex]\frac{\cancel1\cancel{yd}}{\cancel1\cancel{yd}}\;=\;\frac{3ft}{1yd}[/latex]

When everything cancels in a fraction, the fraction reduces to 1:

[latex]1=\frac{3ft}{1yd}[/latex]

We have an expression, 3 ft1 yd, that equals 1. This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression 3 ft 1 yd is called a *conversion factor*, and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called *dimensional analysis* or the *factor label method*.)

To see how this happens, let us start with the original quantity:

4 yd

Now let us multiply this quantity by 1. When you multiply anything by 1, you don’t change the value of the quantity. Rather than multiplying by just 1, let us write 1 as 3 ft1 yd:

[latex]4yd\;\times\;\frac{3ft}{1yd}[/latex]

The 4 yd term can be thought of as [latex]\frac{4 yd}{1}[/latex]; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit *yard*:

[latex]4\cancel{yd}\times\frac{3ft}{1\cancel{yd}}[/latex]

[latex]\frac{4\times3ft}1=\frac{12ft}1=12ft[/latex]

That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:

Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.

How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is

[latex]1 mm = \frac{1}{1,000 m}[/latex]

The 1/1,000 is what the prefix *milli-* means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation:

1,000 mm = 1 m

Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1:

[latex]\frac{1000mm}{1000mm}=\frac{1m}{1000mm}\;or\;\frac{1000mm}{1m}=\frac{1m}{1m}[/latex]

[latex]1=\frac{1m}{1000mm}\;or\;\frac{1000mm}{1m}=1[/latex]

Which conversion factor do we use? The answer is based on *what unit you want to get rid of in your initial quantity*. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the *denominator*; then they will cancel. Therefore, we will use the second conversion factor. Cancelling units and performing the mathematics, we get

[latex]14.66\cancel m\;\times\;\frac{1000mm}{1\cancel m}=14660mm[/latex]

Note how m cancels, leaving mm, which is the unit of interest.

The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.

**Example 2.4.1**

a. Convert 35.9 kL to liters.

b. Convert 555 nm to meters.

**Solution**

a.

*Step 1:* We will use the fact that [latex]1kL=1,000L[/latex].

Of the two conversion factors that can be defined, the one that will work is [latex]\frac{1,000L}{1kL}[/latex].

*Step 2:* Applying this conversion factor, we get:

[latex]35.9\cancel{kL}\;\times\;\frac{1000L}{1\cancel{kL}}=35900L[/latex]

b.

*Step 1:* We will use the fact that [latex]1nm=\frac{1}{1,000,000,000m}[/latex], which we will rewrite as 1,000,000,000 nm = 1 m, or 10 nm = 1 m.

*Step 2:* Of the two possible conversion factors, the appropriate one has the nm unit in the denominator.

[latex]\frac{1m}{10^9nm}[/latex]

*Step 3:* Applying this conversion factor, we get:

[latex]555nm\;\times\;\frac{1m}{10^9nm}=0.000000555m[/latex]

*Step 4:* In the final step, we expressed the answer in scientific notation.

[latex]=5.55\times10^{-7}m[/latex]

**Try It**

1) Convert [latex]67.08 μL[/latex] to liters. Give your answer in scientific notation.

2) Convert [latex]56.8 m[/latex] to kilometers. Give your answer in scientific notation.

**Solution**

- 6.708 × 10
^{−5}L - 5.68 × 10
^{−2}km

What if we have a derived unit that is the product of more than one unit, such as m^{2}? Suppose we want to convert square meters to square centimeters? The key is to remember that m^{2} means m × m, which means we have *two* meter units in our derived unit. That means we have to include *two* conversion factors, one for each unit. For example, to convert 17.6 m^{2} to square centimeters, we perform the conversion as follows:

[latex]17.6m^2=17.6\left(\cancel m\times\cancel m\right)\times\frac{100cm}{1\cancel m}\times\frac{100cm}{1\cancel m}=17000cm\times cm=1.76\times10^5cm^2[/latex]

**Example 2.4.2**

How many cubic centimeters are in 0.883 m^{3}?

**Solution**

With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have

[latex]0.833\cancel{m^3}\times\;\frac{100cm}{1\cancel m}\times\frac{100cm}{1\cancel m}\times\frac{100cm}{1\cancel m}=883000cm^3=8.83\times10^5cm^3[/latex]

You should demonstrate to yourself that the three meter units do indeed cancel.

**Try It**

3) How many cubic millimeters are present in 0.0923 m^{3}? Give your answer in scientific notation.

**Solution**

9.23 × 10^{7} mm^{3}

Suppose the unit you want to convert is in the denominator of a derived unit; what then? Then, in the conversion factor, the unit you want to remove must be in the *numerator*. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation.

**Example 2.4.3**

Convert 88.4 m/min to meters/second.

**Solution**

*Step 1:* We want to change the unit in the denominator from minutes to seconds.

Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: [latex]1 min/60[/latex]s.

*Step 2:* Apply and perform the math.

[latex]\frac{88.4m}{min}\times\frac{1min}{60s}=1.47m/s[/latex]

Notice how the 88.4 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.

**Try It**

4) Convert 0.203 m/min to meters/second. Give your answer in scientific notation.

**Solution**

0.00338 m/s or 3.38 × 10^{−3} m/s

Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit, the unit with no numerical prefix, using the definition of the original prefix. Then convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams:

[latex]2.77\cancel{kg}\;\times\frac{1000g}{1\cancel{kg}}=2770g\;\left(convert\;to\;the\;base\;unit\;of\;grams\right)[/latex]

[latex]2770\cancel g\;\times\frac{1000mg}{1\cancel g}=2770000mg=2.77\times10^6\;mg\;\left(convert\;to\;the\;desired\;unit\right)[/latex]

Alternatively, it can be done in a single multistep process:

[latex]2.77\cancel{kg}\times\frac{1000\cancel g}{1\cancel{kg}}\times\frac{1000mg}{1\cancel g}=2770000m=2.77\times10^6mg[/latex]

You get the same answer either way.

**Example 2.4.4**

How many nanoseconds are in 368.09 μs?

**Solution**

You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes *micro-* and *nano-*,

[latex]368.09\cancel{\mu s}\times\frac{1\cancel s}{10^6\cancel{\mu s}}\times\frac{10^9ns}{1\cancel s}=368090ns=3.6809\times10^5ns[/latex]

**Try It**

5) How many milliliters are in 607.8 kL? Give your answer in scientific notation.

**Solution**

6.078 × 10^{8} mL

When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer—the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix *kilo-* means 1,000—*exactly* 1,000, no more or no less. Thus, in constructing the conversion factor

[latex]\frac{1000g}{1kg}[/latex]

neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix *kilo-* means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as

[latex]\frac{1000.0000000000...g}{1.0000000000...kg}[/latex]

The other numbers in the calculation will determine the number of significant figures in the final answer.

**Example 2.4.5**

A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.

**Solution**

Area is defined as the product of the two dimensions, which we then have to convert to square meters and express our final answer to the correct number of significant figures, which in this case will be three.

[latex]36.7\cancel{cm}\times128.8\cancel{cm}\times\frac{1m}{100\cancel{cm}}\times\frac{1m}{100\cancel{cm}}=0.472696m^2=0.473m^2[/latex]

The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.

**Try It**

6) What is the volume of a block in cubic meters whose dimensions are 2.1 cm × 34.0 cm × 118 cm?

**Solution**

0.0084 m^{3}

In the examples above, the answers provided are all consistent with the rounding rules from the previous section, but we did not emphasize what choices were being made while reporting our answers. In the next sections, we will go through a series of examples to combine what we’ve learned over the last few sections. Throughout this section (and for the rest of the course), be sure to always pay attention to the instructions so that you are always rounding appropriately. Sometimes, you will be required to follow the rounding rules for accuracy and precision, and other times, you will be asked to round to a specific place value. It is always important to pay attention to detail and to answer questions with care.

**Rounding Rules When Converting Within the Same System of Measurement**

When considering metric-to-metric or U.S. System-to-U.S. System conversions, we must first realize that our conversion equations within each system are exact. For example, [latex]100cm=1m[/latex] means that there are exactly 100 centimeters in 1 meter and [latex]1ft=12in[/latex] means that there are exactly 12 inches in 1 foot. Since these conversion equations are exact, this means that the numbers involved have infinitely many significant figures and infinitely many decimal places. As such, when we are performing metric-to-metric or U.S. System-to-U.S. System conversions, we will use the measurement that we are converting to determine the appropriate number of significant figures to include in our answer. Please note that metric-to-metric conversion equations must be recreated by memory, whereas units from the U.S. system will always be provided in our course.

**Example 2.4.6**

Convert 31.5 meters per minute to meters per second. Give your answer with the appropriate number of significant figures.

**Solution**

*Step 1:* Write down any relevant unit conversion equations.

[latex]1 min = 60 s[/latex]

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{31.5m}{1min}\times\frac{1min}{60s}\\&=&\frac{31.5m}{60s}\\&=&0.525\;m/s\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

Since the operations we performed were all multiplications and divisions, we need to round to the least number of significant figures. The unit conversion equation used, [latex]1 min = 60 s[/latex], is exact so there are infinitely many significant figures in those values. In this case, we need to look at the measurement given. Since [latex]31.5 m/min[/latex] has 3 significant figures, our answer should have 3 significant figures as well. So our answer of [latex]0.525 m/s[/latex] is appropriate.

**Example 2.4.7**

Convert [latex]18\tilde{0} mg/dL[/latex] to g/L. Give your answer with the appropriate number of significant figures.

**Solution**

*Step 1:* Write down any relevant unit conversion equations.

[latex]1000mg=1g[/latex] and [latex]10dL=1L[/latex]

For the purpose of the work, we will make note that there is a tilde over the zero and so [latex]18\tilde{0}[/latex] has 3 significant figures.

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{180mg}{1dL}\times\frac{10dL}{1L}\times\frac{1g}{1000mg}\\&=&\frac{180\;\times10g}{1\;times\;1000L}\\\\&=&1.8\;g/L\\\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

The operations we used in this question were multiplications and divisions, and so we need to round to the least number of significant figures in our values. Our calculator gives us [latex]1.8 g/L[/latex], but we need to take into account the fact that our measurement had 3 significant figures. Thus, our answer in this case should have 3 significant figures.

Our answer in this case is that [latex]18\tilde{0} mg/dL = 1.80 g/L[/latex].

**Example 2.4.8**

Convert 32.8 ounces to cups. Give your answer with the appropriate number of significant figures.

**Solutions**

The following unit conversion equations may be useful for this example: [latex]1 cup = 8 oz[/latex].

*Step 1:* Write down any relevant unit conversion equations.

Note, that the U.S. system conversion equation is given.

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{32.8oz}1\times\frac{1cup}{8oz}\\&=&\frac{32.8\;cup}{1\cdot8}\\&=&4.1\;cups\\\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

The operations we used in this question were multiplications and divisions, and so we need to round to the least number of significant figures in our values. Our calculator does not take into account significant figures. Since our unit conversion equation is exact (i.e. there are exactly 8 ounces in 1 cup), those numbers have infinitely many significant figures. Thus, we need to look to our measurement to decide how many significant figures our answer needs. 32.8 ounces has 3 significant figures, so our answer must also have 3 significant figures.

Therefore, [latex]32.8 oz = 4.10 cups[/latex].

**Example 2.4.9**

Convert 161.8 miles per hour to feet per second. Give your answer with the appropriate number of significant figures.

**Solution**

The following unit conversion equations may be useful for this example: [latex]1 mi = 5280 ft[/latex].

*Step 1:* Write down any relevant unit conversion equations.

Note, that the U.S. system conversion equation is given. To convert from hours to seconds, we need to know: [latex]1h=3600s[/latex].

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{161.8mi}{1h}\times\frac{5280ft}{1mi}\times\frac{1h}{3600s}\\&=&\frac{161.8\cdot5280\;ft}{1\cdot3600s}\\&=&237.30\overline6ft/s\\\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

The operations we used in this question were multiplications and divisions, and so we need to round to the least number of significant figures in our values. Our calculator does not take into account significant figures. Since our unit conversion equation is exact, the numbers involved in the conversion equations have infinitely many significant figures. Thus, we need to look to our measurement to decide how many significant figures our answer needs. 161.8 mi/h has 4 significant figures, so our answer must also have 4 significant figures.

Therefore,[latex]161.8mi/h = 237.2 ft/s[/latex].

As we have seen in the examples above, when we are converting metric-to-metric or U.S. System-to-U.S. System units, we round according to the measurement that we are converting.

**Rounding Rules When Converting Between Two Different Systems of Measurement**

When considering the unit conversion equations that establish relationships between two units of different measurement systems, it is important to note the number of significant figures present in the equations. This is due to the fact that when we convert from the metric system to the U.S. system of measurement, the relationships between the units are not exact, but are rather approximations. For example, if we wanted to convert between grams (metric) and pounds (U.S. system), we may be given one of the following unit conversion equations.

**Unit Conversion Equations for Converting Between Grams and Pounds**

[latex]1.00lb=454g[/latex]

[latex]1.000lb=453.6g[/latex]

[latex]1.0000lb=453.59g[/latex]

[latex]1.00000lb=453.592g[/latex]

The equations above show that the precision of the unit conversion equation depends on the precision of the measurement tool that was used. Therefore, these numbers in these unit conversion equations are approximate and will be taken into account while performing our unit conversions.

**Example 2.4.10**

Convert 8.4 miles to kilometers. Give your answer with the appropriate number of significant figures.

**Solution**

The following unit conversion equations may be useful for this example: [latex]1.0000 mi = 1.6093 km[/latex].

*Step 1:* Write down any relevant unit conversion equations.

Note, that the U.S. system to metric conversion equation is provided.

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{8.4mi}1\times\frac{1.6093km}{1.0000mi}\\&=&\frac{8.4\cdot1.6093km}1\\&=&13.51812km\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

All the operations in our conversion above were multiplications and divisions, so we need to round to the least number of significant figures. Don’t forget, our calculator does not take into account significant figures. Since our unit conversion is 1.0000 mi = 1.6093 km, each of those numbers has 5 significant figures, whereas our measurement of 8.4mi has only 2 significant figures. Thus, we need to round our final answer to 2 significant figures.

Therefore, [latex]8.4mi = 14km[/latex].

**Example 2.4.11**

Convert 20.18 kilogram per liter to pound per gallon. Give your answer with the appropriate number of significant figures.

**Solution**

The following unit conversion equations may be useful for this example: [latex]1.00kg =2.20lbs[/latex] and [latex]1.00L=0.264 gal (U.S)[/latex].

*Step 1:* Write down any relevant unit conversion equations.

Note, that the U.S. system to metric conversion equations are provided.

*Step 2:* Convert the units using dimensional analysis.

[latex]\begin{eqnarray*}&=&\frac{20.18kg}{1L}\times\frac{2.20lbs}{1.00kg}\frac{1.00L}{0.264gal}\\&=&\frac{20.18\times\;2.20\times\mathrm{\ }1.00lbs}{1\times\;1.00\times\;0.264gal}\\&=&168.1\overline6lbs/gal\end{eqnarray*}[/latex]

*Step 3:* Be sure that the answer is given to the proper number of significant figures.

All the operations in our conversion above were multiplications and divisions, so we need to round to the least number of significant figures. Don’t forget, our calculator does not take into account significant figures. Since our unit conversions are [latex]1.00kg =2.20lbs[/latex] and [latex]1.00L=0.264 U.S. gal[/latex], each of those numbers have 3 significant figures, whereas our measurement of 20.18 kg/L has 4 significant figures. Thus, we need to round our final answer to 3 significant figures.

Therefore, [latex]20.18 kg/L = 168 lbs/gal[/latex].

**Rounding Rules With Scientific Notation**

Now that we know how to use scientific notation, we can combine this with our knowledge of our rounding rules to round appropriately. In the following examples, we will perform the indicated operations and round according to the instructions provided.

**Example 2.4.12**

Perform the indicated operations. Give your answer with the appropriate number of significant figures.

a. [latex]3.92\times 10^4 + 2.3 \times 10^4[/latex]

b. [latex]1.9 \times 10^{-2} - 1.49\times 10^{-1}[/latex]

c. [latex](5.8\times 10^5)\times (2.84\times 10^{-7})[/latex]

d. [latex]\frac{(3.8\times 10^2)(2.93\times 10^3)}{(8\times 10^{-3})(8.935\times 10^3)}[/latex]

**Solution**

a.

**Step 1: ****Since the powers of ten are the same, we can simply add the values. **

[latex]3.92\times 10^4 + 2.3 \times 10^4=6.22\times 10^4[/latex]

**Step 2: ****As we were adding, we need to round to the correct precision. This means, we need to round according to the [latex]2.3 \times 10^4[/latex] which has 2 significant figures. **

Thus, the answer is: [latex]6.2\times 10^4[/latex]

b.

**Step 1: ****First, we need to write the numbers with the same power of ten. **

[latex]\begin{eqnarray*}&=&1.9\times10^{-2}-1.49\times10^{-1}\\&=&0.19\times10^{-1}-1.49\times10^{-1}\\&=&-1.3\times10^{-1}\\\end{eqnarray*}[/latex]

**Step 2: ****Our answer already has the correct precision, so no further rounding is necessary.**

c.

**Step 1: ****To multiply scientific notation, multiply the coefficients and add the exponents on the tens. **

[latex]\begin{eqnarray*}&=&(5.8\times10^5)\times(2.84\times10^{-7})10^{-1}\\&=&(5.8\cdot2.84)\times10^{5+(-7)}\\&=&16.472\times10^{-2}\\\end{eqnarray*}[/latex]

**Step 2: ****Before we round, we need to write this number in correct scientific notation. **

[latex]=1.6472 \times 10^{-1}[/latex]

**Step 3: ****Now, since we multiplied to get this result, we need to round to the least number of significant figures, which in this case is 2.**

[latex]=1.6 \times 10^{-1}[/latex]

d.

**Step 1: ****To answer this question, let’s first perform the multiplications in the numerator and denominator, and then we will do the division.**

[latex]\frac{(3.8\times 10^2)(2.93\times 10^3)}{(8\times 10^{-3})(8.935\times 10^3)}[/latex]

[latex]=\frac{3.8\times 2.93\times 10^{2+3}}{8\times 8.935\times 10^{-3+3}}[/latex]

[latex]=\frac{11.134\times 10^{5}}{71.48 \times 10^{0}}[/latex]

**Step 2: ****Now, we can perform the division of the coefficients and subtract the exponents on the tens.**

[latex]=0.15576385\times 10^5[/latex]

**Step 3: ****Finally, let’s put this into correct scientific notation before we round appropriately.**

[latex]=1.5576385\times 10^4[/latex]

**Step 4: ****Since the 8 in the initial problem only has 1 significant figure, our answer should only have 1 significant figure.**

[latex]=2\times 10^4[/latex]

**Important Note**

Instructions on rounding expectations will always be given on evaluations. Be sure to read instructions carefully to know if you are expected to use the rounding rules covered in this section, or if there are any other expectations. If questions are vague and do not specify rounding instructions, do not round your answer.

**Key Concepts**

- Units can be converted to other units using the proper conversion factors.
- Conversion factors are constructed from equalities that relate two different units.
- Conversions can be a single step or multistep.
- Unit conversion is a powerful mathematical technique in chemistry, physics, and mathematics, that must be mastered.
- Exact numbers do not affect the determination of significant figures.

**Self Check**

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?