A 100.0 m by 200.0 m sheet of aluminum must be coated on one side with 0.000170 cm of galvanizing liquid. What is the total volume of galvanizing liquid that must be applied to this sheet in liters? (Note: answer must use the correct number of significant figures and correct units.) (2 marks) try the use the unit factor method to solve all questions

To solve this problem using the **unit factor method**, we'll go step by step, ensuring the correct number of significant figures and proper units in the final answer.

### Step 1: Convert dimensions to consistent units
The given sheet dimensions are in meters (m), and the thickness of the galvanizing liquid is in centimeters (cm). Since we are asked for the volume in liters (L), we'll first convert everything to a consistent unit of measurement. 

We know:
1 m = 100 cm

Thus, the dimensions of the aluminum sheet in centimeters are:
- Length: \( 100.0 \, \text{m} = 100.0 \, \text{m} \times 100 \, \text{cm/m} = 1.000 \times 10^4 \, \text{cm} \)
- Width: \( 200.0 \, \text{m} = 200.0 \, \text{m} \times 100 \, \text{cm/m} = 2.000 \times 10^4 \, \text{cm} \)

The thickness of the galvanizing liquid is already given in centimeters:
- Thickness: \( 0.000170 \, \text{cm} \)

### Step 2: Calculate the volume of the galvanizing liquid
The volume \( V \) of the galvanizing liquid is found using the formula for the volume of a rectangular solid:

\[
V = \text{length} \times \text{width} \times \text{thickness}
\]

Substitute the values (all in cm):

\[
V = (1.000 \times 10^4 \, \text{cm}) \times (2.000 \times 10^4 \, \text{cm}) \times (0.000170 \, \text{cm})
\]

\[
V = (1.000 \times 10^4) \times (2.000 \times 10^4) \times (0.000170)
\]

\[
V = 3.40 \times 10^4 \, \text{cm}^3
\]

### Step 3: Convert cubic centimeters to liters
1 cubic centimeter (cm³) = 0.001 liters (L)

So, to convert the volume from cm³ to liters, we use the conversion factor:

\[
V = 3.40 \times 10^4 \, \text{cm}^3 \times 0.001 \, \text{L/cm}^3
\]

\[
V = 34.0 \, \text{L}
\]

### Final Answer:
The total volume of galvanizing liquid that must be applied is **34.0 liters**.

### Key Points:
- The number of significant figures in the final answer is based on the least precise value, which has 3 significant figures.
- The final answer is expressed in liters (L).

Would you like more details on any part of the process?

### Related Questions:
1. How would the answer change if the thickness of the galvanizing liquid was doubled?
2. What is the volume in milliliters instead of liters?
3. How do you convert the volume of galvanizing liquid into gallons?
4. What if the sheet had to be coated on both sides—how would the volume change?
5. How does using a smaller unit (e.g., millimeters) affect the volume calculation?

### Tip:
Always ensure that all measurements are converted to consistent units before performing volume calculations to avoid errors.

A 100.0 m by 200.0 m sheet of aluminum must be coated on one side with 0.000170 cm of galvanizing liquid. What is the total volume of galvanizing liquid that must be applied to this sheet in liters? (Note: answer must use the correct number of significant figures and correct units.) (2 marks) Try the use of the unit factor method to solve all questions.

Learn how to calculate the total volume of galvanizing liquid needed to coat a 100.0 m by 200.0 m aluminum sheet with a thickness of 0.000170 cm using the unit factor method. This problem covers volume calculation, unit conversions, and significant figures, suitable for Grades 9-12.

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