Based on the analysis, the conductivity apparatus placed into Beakers B (H₂SO₄) and C ((NH4)2SO) will make the light bulb glow. Therefore, the correct answer is:
a. A and B
How to determine which beakers will make the light bulb glow
To determine which beakers will make the light bulb glow when the conductivity apparatus is placed into them, consider the electrolytic properties of the compounds in each beaker.
Substances that conduct electricity when dissolved in water are called electrolytes.
Let's analyze each compound:
A) C5H10O4: C5H10O4 is a molecular compound, commonly known as adipic acid. Molecular compounds generally do not ionize or dissociate in water, so they do not conduct electricity.
Therefore, Beaker A will not make the light bulb glow.
B) H₂SO₄: H₂SO₄ is sulfuric acid. It is a strong acid and ionizes completely in water, producing H+ and SO₄²- ions. Since it dissociates into ions, it is an electrolyte and will conduct electricity.
Therefore, Beaker B will make the light bulb glow.
C) (NH4)2SO: (NH4)2SO is ammonium sulfate. It is a salt composed of ammonium (NH₄+) and sulfate (SO₄²-) ions. Ionic compounds generally dissociate into ions when dissolved in water, making them electrolytes. Therefore, Beaker C will make the light bulb glow.
D) CH₂OH: CH₂OH is the molecular formula for methanol, also known as methyl alcohol. Molecular compounds like methanol do not ionize or dissociate in water, so they do not conduct electricity.
Therefore, Beaker D will not make the light bulb glow.
Based on the analysis, the conductivity apparatus placed into Beakers B (H₂SO₄) and C ((NH4)2SO) will make the light bulb glow. Therefore, the correct answer is:
a. A and B