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Carry out the following operations as if they were calculations of experimental results, and express each answer in the correct units with the correct number of significant figures: (a) \(7.310 \mathrm{~km} \div 5.70 \mathrm{~km}\) (b) \(\left(3.26 \times 10^{-3} \mathrm{mg}\right)-\left(7.88 \times 10^{-5} \mathrm{mg}\right)\) (c) \(\left(4.02 \times 10^{6} \mathrm{dm}\right)+\left(7.74 \times 10^{7} \mathrm{dm}\right)\)

Short Answer

Expert verified
(a) 1.28, (b) \(3.18 \times 10^{-3} \; \text{mg}\), (c) \(8.14 \times 10^{7} \; \text{dm}\)."

Step by step solution

01

Divide Kilometers

Perform the division for the expression \( 7.310 \; \text{km} \div 5.70 \; \text{km} \). Start by dividing the numerical values: \( 7.310 \div 5.70 = 1.2824561 \). Now, determine the number of significant figures: the value 5.70 km has 3 significant figures, and 7.310 km has 4 significant figures. So, the result should be rounded to 3 significant figures. This gives us \( 1.28 \) (unitless, as km/km cancels out).
02

Subtract Milligrams

Calculate the difference between the two expressions involving milligrams: \( \left(3.26 \times 10^{-3} \; \text{mg}\right) - \left(7.88 \times 10^{-5} \; \text{mg}\right) \). First, convert both numbers to the same power of ten for easy subtraction: \( 3.26 \times 10^{-3} = 3.26 \times 10^{-3} \) and \( 7.88 \times 10^{-5} = 0.0788 \times 10^{-3} \). Now subtract: \( 3.2600 \times 10^{-3} - 0.0788 \times 10^{-3} = 3.1812 \times 10^{-3} \; \text{mg} \). Considering significant figures, the answer should have 3 decimal places to match the least precise measurement: \( 3.18 \times 10^{-3} \; \text{mg} \).
03

Add Decimeters

Add the two expressions involving decimeters: \( \left(4.02 \times 10^{6} \; \text{dm}\right) + \left(7.74 \times 10^{7} \; \text{dm}\right) \). First, equalize the exponents by converting \( 4.02 \times 10^{6} \) to \( 0.402 \times 10^{7} \). Add: \( 0.402 \times 10^{7} + 7.74 \times 10^{7} = 8.142 \times 10^{7} \; \text{dm} \). Since \( 4.02 \times 10^{6} \) has 3 significant figures, the result should maintain 3 significant figures: \( 8.14 \times 10^{7} \; \text{dm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Experimental Results
When dealing with experimental results, it’s important to express outcomes in a way that accurately reflects the precision of the original measurements. This involves using the correct number of significant figures, which represent the numbers in a measurement that are known with certainty, plus one estimated number. These figures help determine the accuracy of your final results.

In calculations like multiplication and division, the result should have the same number of significant figures as the measurement with the fewest significant figures. For our division example, the number 5.70 km has three significant figures, so our final result should also have three. This rule ensures that your result does not falsely appear more precise than the least precise measurement you used.
  • Avoid overestimating precision by using correct significant figures.
  • Round your final answer based on the least precise measurement involved.
  • Express confidence in your results through proper expression.
This practice ensures clarity and reliability in communicating scientific findings.
Scientific Notation
Scientific notation is a way to express very large or very small numbers succinctly. It is commonly used in scientific and mathematical calculations to make these numbers easier to read and compare. When dealing with experimental results, scientific notation can help manage complex calculations efficiently, as seen in the milligram subtraction problem.

In scientific notation, numbers are written as a product of a number between 1 and 10 and a power of ten. For example, instead of writing 0.000788 mg, we write it as \(7.88 \times 10^{-5} \; \text{mg}\). This form reveals the magnitude of the number quickly and simplifies arithmetic operations.
  • Helps in comparing and performing arithmetic operations on very large or small numbers.
  • Keeps calculations neat and manageable.
  • It's crucial to align decimal powers for addition or subtraction to ensure accurate calculations.
This powerful tool is indispensable for scientists and students alike, making the communication and calculation of results straightforward.
Unit Conversion
Unit conversion involves changing a quantity expressed in one unit into its equivalent in another unit. This skill is essential in ensuring calculations are accurate and results are consistent across different unit systems, especially in scientific experiments and results.

In experiments, different measurements might be recorded using different units. Before performing calculations like addition or subtraction, it is crucial to ensure all measurements are in the same unit. For example, if you have measurements in kilometers and you need to convert them to meters, remember that 1 km = 1000 m.
  • Ensures consistent units throughout calculations for accuracy.
  • Simplifies the comparison of quantities.
  • Crucial to understanding scientific data in a global context.
Proper unit conversion can prevent errors and ensure that the experimental results are presented accurately, making them trustworthy and useful.
Addition and Subtraction of Exponents
Performing addition or subtraction with numbers in scientific notation requires the exponents (powers of ten) to be the same. This is because the exponent indicates the scale of the number, and numbers must be on the same scale to be directly added or subtracted.

For example, when adding \(4.02 \times 10^{6} \; \text{dm} \) and \(7.74 \times 10^{7} \; \text{dm} \), you should first adjust the exponents to be equal. Convert \(4.02 \times 10^{6} \) into \(0.402 \times 10^{7} \), then proceed with the addition: \(0.402 \times 10^{7} + 7.74 \times 10^{7} = 8.142 \times 10^{7} \; \text{dm}\).
  • Always align decimal powers before adding/subtracting to ensure accuracy.
  • Simplifies the interpretation of results expressed in scientific notation.
  • Maintain the correct number of significant figures in final results.
Getting comfortable with this method enhances your ability to handle complex scientific data and ensures you're conveying accurate and reliable results.

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Most popular questions from this chapter

A 6.0 -ft person weighs 168 lb. Express this person's height in meters and weight in kilograms \((1 \mathrm{lb}=453.6 \mathrm{~g} ;\) \(1 \mathrm{~m}=3.28 \mathrm{ft} \mathrm{t}\).

Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is \(1 \mathrm{ppm}\) chlorine, or \(1 \mathrm{~g}\) of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 6.0 percent chlorine by mass and there are \(2.0 \times 10^{4}\) gallons (gal) of water in the pool (1 gal \(=3.79 \mathrm{~L} ;\) density of liquids \(=1.0 \mathrm{~g} / \mathrm{mL}\) ).

What is the difference between mass and weight? If a person weighs \(168 \mathrm{lb}\) on Earth, about how much would the person weigh on the moon?

A 250 -mL glass bottle was filled with \(242 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) and tightly capped. It was then left outdoors overnight, where the average temperature was \(-5^{\circ} \mathrm{C}\). Predict what would happen. The density of water at \(20^{\circ} \mathrm{C}\) is \(0.998 \mathrm{~g} / \mathrm{cm}^{3}\) and that of ice at \(-5^{\circ} \mathrm{C}\) is \(0.916 \mathrm{~g} / \mathrm{cm}^{3} .\)

A student is given a crucible and asked to prove whether it is made of pure platinum. She first weighs the crucible in air and then weighs it suspended in water (density = \(0.9986 \mathrm{~g} / \mathrm{mL}\) ). The readings are \(860.2 \mathrm{~g}\) and \(820.2 \mathrm{~g}\), respectively. Based on these measurements and given that the density of platinum is \(21.45 \mathrm{~g} / \mathrm{cm}^{3},\) what should her conclusion be? (Hint: An object suspended in a fluid is buoyed up by the mass of the fluid displaced by the object. Neglect the buoyancy of air.)

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