What volume of aluminum would be needed to balance a 1.00 cm^3 sample of copper on a 2 pan laboratory balance?
density Al * volume Al = density Cu * 1 cm^3
make sure density of Al is in g/cm^3
I literally have no idea what this means. Do I do both sides and divide one side? This is confusing.
You must look up the densities of Al and Cu
then your answer is
Volume of Al = (density CU/densityAL) *1 cm^3
Vol of Al = (8.9/2.7)*1
= 3.3 cm^3 of Al
Yay! I got the same answer, and I tried to figure it out before you answered. Good. That makes me happy. Thank you!
To solve this problem, we need to determine the volume of aluminum that would balance the 1.00 cm^3 sample of copper on a 2 pan laboratory balance.
The balance would be in equilibrium when the mass on each pan is equal. This can be represented by the equation:
Mass of Copper = Mass of Aluminum
We can utilize the density of the materials to determine their masses. The density of copper is 8.96 g/cm^3, which means that 1.00 cm^3 of copper would have a mass of:
Mass of Copper = Volume of Copper × Density of Copper
Mass of Copper = 1.00 cm^3 × 8.96 g/cm^3
Mass of Copper = 8.96 g
Now, we need to find the volume of aluminum that would have the same mass. The density of aluminum is 2.70 g/cm^3. Rearranging the density formula, we can solve for volume:
Volume of Aluminum = Mass of Aluminum / Density of Aluminum
Volume of Aluminum = 8.96 g / 2.70 g/cm^3
Volume of Aluminum = 3.32 cm^3
Therefore, 3.32 cm^3 of aluminum would be needed to balance a 1.00 cm^3 sample of copper on a 2 pan laboratory balance.