184.5 m ÷ 0.31 s S mol 0.9 L X 27. L |mol 2 3 cm × 0.30 cm = cm

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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please answer quickly make sure they have the correct number of sig figs

1. **Calculation of Speed**:
   - Formula: \( \frac{184.5 \, \text{m}}{0.31 \, \text{s}} = \Box \, \text{m/s} \)
   - This calculation involves dividing distance by time to find the speed in meters per second.

2. **Moles Calculation**:
   - Formula: \( \frac{20.9 \, \text{mol}}{\text{L}} \times 27. \, \text{L} = \Box \, \text{mol} \)
   - This involves multiplying molarity by volume to find the number of moles.

3. **Area Calculation**:
   - Formula: \( 0.93 \, \text{cm} \times 0.30 \, \text{cm} = \Box \, \text{cm}^2 \)
   - This involves multiplying two lengths to find the area in square centimeters.
Transcribed Image Text:1. **Calculation of Speed**: - Formula: \( \frac{184.5 \, \text{m}}{0.31 \, \text{s}} = \Box \, \text{m/s} \) - This calculation involves dividing distance by time to find the speed in meters per second. 2. **Moles Calculation**: - Formula: \( \frac{20.9 \, \text{mol}}{\text{L}} \times 27. \, \text{L} = \Box \, \text{mol} \) - This involves multiplying molarity by volume to find the number of moles. 3. **Area Calculation**: - Formula: \( 0.93 \, \text{cm} \times 0.30 \, \text{cm} = \Box \, \text{cm}^2 \) - This involves multiplying two lengths to find the area in square centimeters.
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