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Let's go through the questions on this worksheet.

### 1. Number of significant figures in each measurement:

A. **4867 mi**: 4 significant figures  
B. **56 mL**: 2 significant figures  
C. **1,100 mL**: 
   - If it’s written as **1,100**, then it's **2 significant figures**. 
   - If it's **1,100.** (with a decimal), then it’s **4 significant figures**.   
D. **0.000475 g**: 3 significant figures  
E. **101 K**: 3 significant figures  
F. **1,100 mL** (same as C)  
G. **0.050650 A**: 5 significant figures

### 2. Carry out the following operations with correct significant figures:

A. **5.6792 m + 0.6 m + 44.33 m**  
   - Add them: \( 5.6792 + 0.6 + 44.33 = 50.6092 \, \text{m} \)  
   - The result must have the same number of decimal places as the least precise measurement (0.6 m with 1 decimal place):  
   **Answer: 50.6 m**

B. **3.70 g - 2.9133 g**  
   - Subtract them: \( 3.70 - 2.9133 = 0.7867 \, \text{g} \)  
   - The result must have the same number of decimal places as the least precise measurement (3.70 g with 2 decimal places):  
   **Answer: 0.79 g**

C. **5 cm × 3.770 cm**  
   - Multiply: \( 5 \times 3.770 = 18.85 \, \text{cm}^2 \)  
   - The result must have the same number of significant figures as the number with the fewest significant figures (5 cm with 1 significant figure):  
   **Answer: 20 cm²**

D. **\(\frac{101 \, \text{J}}{(4.05)(10.0)}\)**  
   - First, compute the denominator: \( 4.05 \times 10.0 = 40.5 \)  
   - Now, divide: \( \frac{101}{40.5} = 2.4938 \)  
   - The result must have the same number of significant figures as the number with the fewest significant figures (101 J has 3 significant figures, 4.05 has 3 significant figures, and 10.0 has 3 significant figures):  
   **Answer: 2.49 J**

E. **\(\frac{1.1 \, \text{g}}{0.500 \, \text{mL}} + 3.59 \, \text{g/mL}\)**  
   - First, divide: \( \frac{1.1}{0.500} = 2.2 \, \text{g/mL} \)  
   - Then, add \( 2.2 + 3.59 = 5.79 \, \text{g/mL} \)  
   - The result must have 2 decimal places, since 3.59 has the fewest decimal places:  
   **Answer: 5.79 g/mL**

### 3. Conversions:

A. **1.5 lbs to g (453 g = 1 lb)**  
   - Multiply: \( 1.5 \, \text{lbs} \times 453 \, \text{g/lb} = 679.5 \, \text{g} \)  
   - The result must have the same number of significant figures as 1.5 lbs (2 significant figures):  
   **Answer: 680 g**

B. **22.6 m to dm (10 dm = 1 m)**  
   - Multiply: \( 22.6 \, \text{m} \times 10 = 226 \, \text{dm} \)  
   - The result has 3 significant figures, since 22.6 has 3 significant figures:  
   **Answer: 226 dm**

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Let me know if you'd like further details or have any questions! Here are 5 related questions to expand this topic:

1. How do you determine the number of significant figures in a calculated result for division?
2. Why is it important to consider significant figures when performing calculations in science?
3. What are the rules for rounding when using significant figures?
4. Can you explain the difference between precision and accuracy in measurements?
5. What is the role of significant figures in scientific notation?

**Tip:** Always look out for whether a number is written with a trailing decimal (e.g., 100. vs 100) to determine its significant figures!

What is the number of significant figures in each measurement? Carry out the following operations and express the answers with the correct unit and correct number of significant figures. Also, carry out the conversions.

Learn how to determine the number of significant figures in various measurements, perform calculations while applying significant figure rules, and carry out unit conversions. This problem covers important concepts in significant figures and unit conversion for Grades 9-12.

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