Q. 2.96 - An Introduction to General, Organic, and Biological Chemistry

Question

Round off or add zeros to the following calculated answers to give a final answer with three significant figures: (2.2)

a. $58.703 \mathrm{mL}$

b. $3\times {10}^{-3}s$

c. $0.010826 \mathrm{g}$

d. $1,7484\times {10}^3 \mathrm{ms}$

Step-by-Step Solution

Answer

a) The number $58.703 mL$with three significant figures is determined as $58.7 g$

b) The number $3\times {10}^{-3}s$with three significant figures is determined as $3.00\times {10}^{-3} s$

c) The number $0.010826 g$ with three significant figures is determined as $0.0108 g$

Step 1 :Introduction (part a )

The given is the numbers with significant figures

The objective is to round off the numbers and write with three significant figures

Step 2 : Explanation (part a)

a.

To write the number $58.703 \mathrm{g}$ drop $3$ without increasing the digit $7$ by one with three significant figures since the first falling digit $0$ is less than $5$..

$58.703 \mathrm{g}=58.7 \mathrm{g}\left(\begin{array}{l}3\text{ significant figures }\\ \text{ drop 03, }0<5\end{array}\right)$

Thus, the number $58.703 \mathrm{g}$ with three significant figures will be $58.7 \mathrm{g}$.

Step 3 : Introduction (part b)

The given is the numbers with significant figures

The objective is to round off the numbers and write with three significant figures

Step 4 : Explanation (part b)

b.

To write the number $3\times {10}^{-3} \mathrm{s}$ Add two zeros after the decimal in the case of three significant digits $3\times {10}^{-3} \mathrm{s}$ Add two zeros after the decimal in the case of three significant digits $3\times {10}^{-3} \mathrm{s}=3.00\times {10}^{-3} \mathrm{s}$ (3 significant figures)

So, the number $3\times {10}^{-3} \mathrm{s}$ with three significant figures will be $3.00\times {10}^{-3} \mathrm{s}$.

Step 5: Introduction (part c)

The given is the numbers with significant figures

The objective is to round off the numbers and write with three significant figures

Step 6: Explanation (part c)

C.

Write the number $0.010826 \mathrm{g}$ with three signiflcant flgures, drop $26$ without increase the digit $8$ by one because flrst dropping digit $2$ is less than $5$ .

$0.010826 \mathrm{g}=0.0108 \mathrm{g}\left(\begin{array}{l}3\text{ significant figures }\\ \text{ drop }26,2<5\end{array}\right)$

So, the number $0.010826 \mathrm{g}$ with three signiflcant flgures will be $0.0108 \mathrm{g}$.