Question

1. a sample has a mass of 42.58 g. the water level rises from 15.0 ml to 30.78 ml

• calculate the displaced volume.
• calculate the density.
• does this match more than one possible substance?
• how could acid reactivity confirm the answer?

Explanation:

Step1: Calculate displaced volume

The displaced volume \( V \) is the difference in water levels.
\( V=30.78\ \text{mL}-15.0\ \text{mL}=15.78\ \text{mL} \)

Step2: Calculate density

Density formula is \(
ho=\frac{m}{V} \), where \( m = 42.58\ \text{g} \) and \( V = 15.78\ \text{mL} \)
\(
ho=\frac{42.58\ \text{g}}{15.78\ \text{mL}}\approx2.7\ \text{g/mL} \)

Step3: Substance match

A density of approximately \( 2.7\ \text{g/mL} \) could match aluminum (density of aluminum is about \( 2.7\ \text{g/cm}^3 = 2.7\ \text{g/mL} \)). It may not match more than one common pure substance with high certainty in basic cases.

Step4: Acid reactivity

Aluminum reacts with acids (e.g., \( 2\text{Al}+6\text{HCl}=2\text{AlCl}_3 + 3\text{H}_2\uparrow \)). If the sample reacts with acid in a way consistent with aluminum's reactivity (producing hydrogen gas), it would support the identification.

Answer:

• Displaced volume: \( 15.78\ \text{mL} \)
• Density: \( \approx2.7\ \text{g/mL} \)
• Substance match: Likely aluminum (may not have multiple common pure - substance matches with high basic - level certainty)
• Acid reactivity: If it reacts with acid to produce hydrogen (as aluminum does), it confirms the identification.