Question

chemistry - unit 1 - worksheet 4
dimensional analysis
in the space below each problem, write the equality that exists between the two units and then write the appropriate conversion factor to solve the problem. in some cases there may be more than one conversion factor needed. use the factor - label method to make the following conversions. remember to use the appropriate number of sfs in your answer
part 1

1. 74 cm\\(\frac{10^{- 2}m}{1cm}\\)=__________meters

example: c = 10 - 2
1 cm = 10^{-2}m

1. 8.32×10^{-2}kg(________)=________grams
2. 55.5 ml(________)=________cm³
3. 0.00527 cal(________)=________kilocalories
4. 9.52×10^{-4}m(________)=________micrometers
5. 41.0 ml(________)=________liters

Explanation:

Step1: Recall unit - conversion factor

1 kg = 1000 g or 1 kg=10^{3} g.

Step2: Set up the conversion

8.32×10^{-2} kg×\frac{10^{3} g}{1 kg}=8.32×10^{-2}×10^{3} g.

Step3: Simplify the exponent calculation

Using the rule a^{m}×a^{n}=a^{m + n}, 8.32×10^{-2}×10^{3}=8.32×10^{-2 + 3}=83.2 g.

Step4: Recall unit - conversion factor for volume

1 mL = 1 cm^{3}, so 55.5 mL×\frac{1 cm^{3}}{1 mL}=55.5 cm^{3}.

Step5: Recall calorie - kilocalorie conversion

1 kcal = 1000 cal or 1 kcal = 10^{3} cal. 0.00527 cal×\frac{1 kcal}{10^{3} cal}=5.27×10^{-6} kcal.

Step6: Recall meter - micrometer conversion

1 m = 10^{6} μm. 9.52×10^{-4} m×\frac{10^{6} μm}{1 m}=9.52×10^{-4+6} μm = 952 μm.

Step7: Recall milliliter - liter conversion

1 L = 1000 mL or 1 L = 10^{3} mL. 41.0 mL×\frac{1 L}{10^{3} mL}=4.10×10^{-2} L.

Answer:

1. 8.32×10^{-2} kg(\frac{10^{3} g}{1 kg}) = 83.2 grams
2. 55.5 mL(\frac{1 cm^{3}}{1 mL}) = 55.5 cm^{3}
3. 0.00527 cal(\frac{1 kcal}{10^{3} cal}) = 5.27×10^{-6} kilocalories
4. 9.52×10^{-4} m(\frac{10^{6} μm}{1 m}) = 952 micrometers
5. 41.0 mL(\frac{1 L}{10^{3} mL}) = 4.10×10^{-2} liters