Lab 2: Conversion Factors and Problem Solving Report Sheet Answers

Answers

What can be answered from the report sheet

The calculations below use only the values printed on the report sheet. The area, solid-volume, product-label, page-length, and personal-height items require measurements that are not supplied in the question, so a responsible answer gives the method for those fields instead of inventing results.

1. Rounding to three significant figures

Writing $6.55 \times 10^5$ makes the three significant figures unambiguous. The value 655 has the wrong magnitude for 654900.

2. Significant figures in calculations

Multiplication and division

For multiplication and division, report the number of significant figures in the least precise measured factor: three, three, and two respectively.

Addition and subtraction

For addition and subtraction, round to the fewest decimal places among the measurements in that expression.

3. Area and volume: fill these from your own measurements

The report sheet asks students to measure a printed rectangle and a solid object. Those dimensions are not part of the provided question, so there is no single honest numeric answer for either section.

• Rectangle area: $A = L \times W$. Record both sides in centimeters and report $A$ in $\mathrm{cm^2}$.
• Cube: $V = L^3$.
• Rectangular solid: $V = L \times W \times H$.
• Cylinder: $V = \pi D^2H / 4$. Use the measured diameter $D$ and height $H$ in centimeters, then report $V$ in $\mathrm{cm^3}$.

Two students can calculate different areas for the same rectangle because their ruler readings, alignment, or rounding can differ. The result should be based on the dimensions each student actually recorded.

4. Conversion factors

Metric volume

$$1\ \mathrm{L} = 1000\ \mathrm{mL}$$

The two usable conversion factors are:

$$\frac{1000\ \mathrm{mL}}{1\ \mathrm{L}} \qquad\text{and}\qquad \frac{1\ \mathrm{L}}{1000\ \mathrm{mL}}$$

U.S. customary and metric volume

The report sheet's reference value is $1\ \mathrm{qt} = 946\ \mathrm{mL}$. NIST gives the more precise liquid-quart relationship as $1\ \mathrm{qt} = 946.353\ \mathrm{mL}$, so 946 mL is the appropriate rounded value for the sheet.

$$\frac{946\ \mathrm{mL}}{1\ \mathrm{qt}} \qquad\text{and}\qquad \frac{1\ \mathrm{qt}}{946\ \mathrm{mL}}$$

Your experimental factor is the milliliters you actually measured for one quart. Compare that measured value with 946 mL; do not replace it with a made-up measurement.

U.S. customary and metric length

$$1\ \mathrm{in} = 2.54\ \mathrm{cm}$$ $$\frac{2.54\ \mathrm{cm}}{1\ \mathrm{in}} \qquad\text{and}\qquad \frac{1\ \mathrm{in}}{2.54\ \mathrm{cm}}$$

Measure the vertical length of your own sheet in both units, then calculate $\text{measured cm} / \text{measured in}$ for the experimental factor. The comparison should be against 2.54 cm/in.

U.S. customary and metric mass

The report sheet rounds the reference relationship to $1\ \mathrm{lb} = 454\ \mathrm{g}$. NIST gives $1\ \mathrm{lb} = 453.59237\ \mathrm{g}$ exactly.

$$\frac{454\ \mathrm{g}}{1\ \mathrm{lb}} \qquad\text{and}\qquad \frac{1\ \mathrm{lb}}{454\ \mathrm{g}}$$

The mass and weight on the selected commercial product are not supplied. Use its label values to calculate your experimental grams-per-pound factor, then compare it with 454 g/lb.

5. Metric height

Let $h$ be the student's recorded height in inches. The completed calculation is:

$$h\ \mathrm{in} \times \frac{2.54\ \mathrm{cm}}{1\ \mathrm{in}} = 2.54h\ \mathrm{cm}$$ $$2.54h\ \mathrm{cm} \times \frac{1\ \mathrm{m}}{100\ \mathrm{cm}} = 0.0254h\ \mathrm{m}$$

Substitute the student's own height for $h$. The source sheet does not state that height, so a fixed numerical result would not be supported.

Conclusion

The rounding, significant-figure, and reference-conversion questions have determinate answers above. The remaining blank fields are measurement tasks: complete them with the observations or labels actually collected in the lab, while keeping units and significant figures consistent.