1.5 Measurement uncertainty, accuracy, and precision  (Page 4/11)

Multiplication and division with significant figures

(a) Multiply 0.6238 cm by 6.6 cm.

(b) Divide 421.23 g by 486 mL.

Solution

(a) $\begin{array}{l}\begin{array}{l}\text{0.6238 cm}\times 6.6\text{cm}=4.11708{\text{cm}}^2⟶\text{result is}4.1{\text{cm}}^2(\text{round to two significant figures})\hfill \\ \text{four significant figures}\times \text{two significant figures}⟶\text{two significant figures answer}\hfill \end{array}\hfill \end{array}$

(b) $\begin{array}{l}\frac{\text{421.23 g}}{\text{486 mL}}=\text{0.86728... g/mL}⟶\text{result is 0.867 g/mL}\left(\text{round to three significant figures}\right)\\ \frac{\text{five significant figures}}{\text{three significant figures}}⟶\text{three significant figures answer}\end{array}$

Check your learning

(b) Divide 55.8752 m by 56.53 s.

Calculation with significant figures

Solution

Check your learning

Experimental determination of density using water displacement

(a) Use these values to determine the density of this piece of rebar.

(b) Rebar is mostly iron. Does your result in (a) support this statement? How?

Solution

(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)

The density is the mass-to-volume ratio:

(rounded to two significant figures, per the rule for multiplication and division)

From [link] , the density of iron is 7.9 g/cm ³ , very close to that of rebar, which lends some support to the fact that rebar is mostly iron.

Check your learning

(a) Use these values to determine the density of this material.

(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.