Problem 41 - Chemistry Matter and Change

Question

Challenge Add and subtract as indicated. Round off when necessary. a. \(\left(4.32 \times 10^{3} \mathrm{cm}\right)-\left(1.6 \times 10^{6} \mathrm{mm}\right)\) \(\quad\) b. \(\left(2.12 \times 10^{7} \mathrm{mm}\right)+\left(1.8 \times 10^{3} \mathrm{cm}\right)\)

Step-by-Step Solution

Answer

a. \(-1.6 \times 10^5 \text{ cm}\), b. \(2.12018 \times 10^7 \text{ mm}\).

Step 1: Convert Units for Part a

First, convert the measurements to the same unit. Let's use centimeters for both. We already have \(4.32 \times 10^3\) in centimeters. To convert from millimeters to centimeters, recall that \(1 \text{ cm} = 10 \text{ mm}\). So, \(1.6 \times 10^6 \text{ mm} = \frac{1.6 \times 10^6}{10} \text{ cm} = 1.6 \times 10^5 \text{ cm}\).

Step 2: Subtract for Part a

Now subtract the converted values: \(4.32 \times 10^3 \text{ cm} - 1.6 \times 10^5 \text{ cm} = (0.0432 \times 10^5) \text{ cm} - 1.6 \times 10^5 \text{ cm} = -1.5568 \times 10^5 \text{ cm}\).

Step 3: Convert Units for Part b

Again, convert all measurements to the same unit. Use millimeters this time. \(2.12 \times 10^7 \text{ mm}\) is already in the correct unit. Convert \(1.8 \times 10^3 \text{ cm}\) to millimeters: \(1.8 \times 10^3 \text{ cm} = 1.8 \times 10^4 \text{ mm}\).

Step 4: Add for Part b

Add the values obtained: \(2.12 \times 10^7 \text{ mm} + 1.8 \times 10^4 \text{ mm} = 2.12018 \times 10^7 \text{ mm}\).

Step 5: Round the Results

For part a, round \(- 1.5568 \times 10^5\ text{ cm}\) to 2 significant figures: \(-1.6 \times 10^5 \text{ cm}\). For part b, \(2.12018 \times 10^7 \text{ mm}\) is already to 6 significant digits.

Key Concepts

Significant Figures

Significant figures are crucial in scientific calculations as they indicate the precision of a measurement. When dealing with significant figures, it's important to reflect the accuracy of the numbers you are using.
In any calculation involving addition or subtraction, the result should be rounded off to the least number of decimal places present in any of the numbers you are working with. For multiplication or division, you consider the number of significant figures instead.
In our example, part a ends with \(-1.5568 \times 10^5 \) cm. We rounded this to two significant figures for our final answer, resulting in \(-1.6 \times 10^5 \) cm. This rounding reflects the precision of the least precise measurement involved.

Scientific Notation

Scientific notation is a method used to express very large or very small numbers conveniently. It allows you to perform calculations more easily and read numbers without excessive zeros.
When converting to scientific notation, a number is represented as the product of a base (a number between 1 and 10) and a power of ten. For example, 4320 can be written as \(4.32 \times 10^3\).
This not only simplifies calculations but also helps when handling significant figures and rounding, as seen in our problem. The numbers are expressed in scientific notation to facilitate comparisons and ensure consistency in calculations.

Addition and Subtraction of Measurements

For addition and subtraction in measurements, it's vital first to convert all values to the same unit. This ensures accuracy and consistency in the calculation.
Take Part a from the exercise: both measurements were converted to centimeters before subtracting. Similarly, for Part b, converting to millimeters before adding was crucial.
After conversion, simply align the measurements based on their exponent in scientific notation, and perform the operation. The final answer in subtraction or addition should match the precision of the least precise measurement, ensuring a correct and meaningful result.