Question

4. A student doing an experiment wants to dispose of salt and sugar that were spilled into the sink. The materials are non-hazardous. How should these substances be cleaned up? (1 point) Responses Use a match to burn them away. Use a match to burn them away. Sweep them up using bare hands. Sweep them up using bare hands. Drop a weak acid on them to break them up. Drop a weak acid on them to break them up. Flush them with water down the drain. Flush them with water down the drain. Question 5 5. The sides of a small rectangular box are measured to be 1.80 +/- 0.01 cm long, 2.05 +/- 0.02 cm long, and 3.10 +/- 0.01 cm long. Calculate the box’s volume and approximate uncertainty in cubic centimeters. (1 point) Responses 11 +/- 0.2 cm^3 11 +/- 0.2 cm^3 11 +/- 0.21 cm^3 11 +/- 0.21 cm^3 11.4 +/- 0.2 cm^3 11.4 +/- 0.2 cm^3 11.4 +/- 0.21 cm^3

Answer 1

Question 4

To clean up non-hazardous substances like salt and sugar spilled in the sink, the best option is:

Flush them with water down the drain.

This method is safe and effective for non-hazardous materials.

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Question 5

To calculate the volume of the box and its uncertainty, we use the formula for the volume of a rectangular box:

\[ V = l \times w \times h \]

where:

• \( l = 1.80 , \text{cm} \) (length) with uncertainty \( \Delta l = 0.01 , \text{cm} \)
• \( w = 2.05 , \text{cm} \) (width) with uncertainty \( \Delta w = 0.02 , \text{cm} \)
• \( h = 3.10 , \text{cm} \) (height) with uncertainty \( \Delta h = 0.01 , \text{cm} \)

Calculating the volume:

\[ V = 1.80 , \text{cm} \times 2.05 , \text{cm} \times 3.10 , \text{cm} = 11.4 , \text{cm}^3 \]

To find the uncertainty in the volume, we use the formula for the relative uncertainty in multiplication, which is the sum of the relative uncertainties of each measurement:

1. Relative uncertainty of length: \( \frac{\Delta l}{l} = \frac{0.01}{1.80} \approx 0.00556 \)
2. Relative uncertainty of width: \( \frac{\Delta w}{w} = \frac{0.02}{2.05} \approx 0.00976 \)
3. Relative uncertainty of height: \( \frac{\Delta h}{h} = \frac{0.01}{3.10} \approx 0.00323 \)

Now, sum these relative uncertainties:

\[ 0.00556 + 0.00976 + 0.00323 \approx 0.01855 \]

To convert this to absolute uncertainty in the volume:

\[ \Delta V = V \times \text{(total relative uncertainty)} = 11.4 , \text{cm}^3 \times 0.01855 \approx 0.211 , \text{cm}^3 \]

Rounding to two significant figures, we get:

\[ \Delta V \approx 0.21 , \text{cm}^3 \]

Thus, the volume and uncertainty of the box is:

11.4 +/- 0.21 cm³