Question

which of the following conversions is correct and expresses the answer using the proper number of significant figures? 3.779 lb×454 g/1 lb×1,000 mg/1 g = 1.7×10^6 mg 553 dl×1 l/10 dl×10^3 ml/1 l = 5.5×10^4 ml 623 μm×1 m/10^6 μm×39.4 in/1 m = 2.45×10^(-5) in 623 μm×1 m/10^6 μm×39.4 in/1 m = 2.45×10^(-2) in

Explanation:

Step1: Analyze the first conversion

Calculate $3.779\ \text{lb}\times\frac{454\ \text{g}}{1\ \text{lb}}\times\frac{1000\ \text{mg}}{1\ \text{g}} = 3.779\times454\times1000\ \text{mg}=1.715666\times 10^{6}\ \text{mg}$. Rounding to two - significant figures gives $1.7\times 10^{6}\ \text{mg}$.

Step2: Analyze the second conversion

Calculate $553\ \text{dL}\times\frac{1\ \text{L}}{10\ \text{dL}}\times\frac{10^{3}\ \text{mL}}{1\ \text{L}}=\frac{553\times1\times10^{3}}{10}\ \text{mL} = 55300\ \text{mL}=5.53\times 10^{4}\ \text{mL}$. Rounding to two - significant figures gives $5.5\times 10^{4}\ \text{mL}$.

Step3: Analyze the third conversion

Calculate $623\ \text{pm}\times\frac{1\ \text{m}}{10^{12}\ \text{pm}}\times\frac{39.4\ \text{in}}{1\ \text{m}}=\frac{623\times39.4}{10^{12}}\ \text{in}\approx 2.45462\times 10^{-8}\ \text{in}$. Rounding to three - significant figures gives $2.45\times 10^{-8}\ \text{in}$. The given value in the option has an incorrect exponent.

Step4: Analyze the fourth conversion

The calculation in the fourth option has an incorrect setup and result based on the unit - conversion factors.

Answer:

The first conversion $3.779\ \text{lb}\times\frac{454\ \text{g}}{1\ \text{lb}}\times\frac{1000\ \text{mg}}{1\ \text{g}} = 1.7\times 10^{6}\ \text{mg}$ and the second conversion $553\ \text{dL}\times\frac{1\ \text{L}}{10\ \text{dL}}\times\frac{10^{3}\ \text{mL}}{1\ \text{L}}=5.5\times 10^{4}\ \text{mL}$ are correct with proper significant - figure usage.