QUESTION IMAGE
Question
part f. sig fig review: fill in the chart below
- 21.3 g / 1.3 ml
- $\frac{8 g}{(1.10 cm\times2.00 cm\times2.0 cm)}$
- 27.35 cm - 21.2 cm
- 63.43 l + 34.5 l
- 0.68 g / 300. cm³
- 2.25 mm x 3.5 mm
- 9.341 m + 12.5 m + 4.33 m
- 56.890 g - 5 g
- 8.67 g / 3 ml
- $\frac{3.8 kg}{(1.2 m\times1.35 m\times2 m)}$
- 0.57 g / 30. cm³
- 0.0025 g / 0.4 ml
Step1: Calculate 21.3 g/1.3 mL
$21.3\div1.3 = 16.384615$ (Answer without sig - figs). For significant figures, 21.3 has 3 sig - figs and 1.3 has 2 sig - figs. The result should be rounded to 2 sig - figs, so it is 16 g/mL.
Step2: Calculate $\frac{8g}{(1.10cm\times2.00cm\times2.0cm)}$
First, calculate the volume $V = 1.10\times2.00\times2.0=4.4$ $cm^{3}$. Then $\frac{8}{4.4}=1.818182$ (Answer without sig - figs). 8 has 1 sig - fig, 1.10 has 3 sig - figs, 2.00 has 3 sig - figs and 2.0 has 2 sig - figs. The result should be rounded to 1 sig - fig, so it is 2 $g/cm^{3}$.
Step3: Calculate 27.35 cm - 21.2 cm
$27.35−21.2 = 6.15$ (Answer without sig - figs). For subtraction, we consider the decimal place. 21.2 has 1 decimal place, so the result should be rounded to 1 decimal place, so it is 6.2 cm.
Step4: Calculate 63.43 L + 34.5 L
$63.43 + 34.5=97.93$ (Answer without sig - figs). 34.5 has 1 decimal place, so the result should be rounded to 1 decimal place, so it is 97.9 L.
Step5: Calculate 0.68 g/300. $cm^{3}$
$0.68\div300. = 0.00226667$ (Answer without sig - figs). 0.68 has 2 sig - figs and 300. has 3 sig - figs. The result should be rounded to 2 sig - figs, so it is $2.3\times10^{-3}$ $g/cm^{3}$.
Step6: Calculate 2.25 mm×3.5 mm
$2.25\times3.5 = 7.875$ (Answer without sig - figs). 2.25 has 3 sig - figs and 3.5 has 2 sig - figs. The result should be rounded to 2 sig - figs, so it is 7.9 $mm^{2}$.
Step7: Calculate 9.341 m + 12.5 m + 4.33 m
$9.341+12.5 + 4.33=26.171$ (Answer without sig - figs). 12.5 has 1 decimal place, so the result should be rounded to 1 decimal place, so it is 26.2 m.
Step8: Calculate 56.890 g - 5 g
$56.890−5 = 51.890$ (Answer without sig - figs). 5 has no decimal place, so the result should be rounded to no decimal place, so it is 52 g.
Step9: Calculate 8.67 g/3 mL
$8.67\div3 = 2.89$ (Answer without sig - figs). 3 has 1 sig - fig, so the result should be rounded to 1 sig - fig, so it is 3 g/mL.
Step10: Calculate $\frac{3.8kg}{(1.2m\times1.35m\times2m)}$
First, calculate the volume $V = 1.2\times1.35\times2 = 3.24$ $m^{3}$. Then $\frac{3.8}{3.24}=1.17284$ (Answer without sig - figs). 3.8 has 2 sig - figs, 1.2 has 2 sig - figs, 1.35 has 3 sig - figs and 2 has 1 sig - fig. The result should be rounded to 1 sig - fig, so it is 1 $kg/m^{3}$.
Step11: Calculate 0.57 g/30. $cm^{3}$
$0.57\div30. = 0.019$ (Answer without sig - figs). 0.57 has 2 sig - figs and 30. has 2 sig - figs. The result should be rounded to 2 sig - figs, so it is $1.9\times10^{-2}$ $g/cm^{3}$.
Step12: Calculate 0.0025 g/0.4 mL
$0.0025\div0.4 = 0.00625$ (Answer without sig - figs). 0.0025 has 2 sig - figs and 0.4 has 1 sig - fig. The result should be rounded to 1 sig - fig, so it is $6\times10^{-3}$ $g/mL$.
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| Question | Answer without sig figs | Answer with sig figs | Units |
|---|---|---|---|
| 2 | 1.818182 | 2 | g/cm³ |
| 3 | 6.15 | 6.2 | cm |
| 4 | 97.93 | 97.9 | L |
| 5 | 0.00226667 | $2.3\times10^{-3}$ | g/cm³ |
| 6 | 7.875 | 7.9 | mm² |
| 7 | 26.171 | 26.2 | m |
| 8 | 51.890 | 52 | g |
| 9 | 2.89 | 3 | g/mL |
| 10 | 1.17284 | 1 | kg/m³ |
| 11 | 0.019 | $1.9\times10^{-2}$ | g/cm³ |
| 12 | 0.00625 | $6\times10^{-3}$ | g/mL |