The density of aluminum is 2.70 g/cm3, and the average mass for one aluminum atom is 4.48×10-23 g.
Five identical aluminum coins are found to displace a total of 2.50 mL of water when immersed in a graduated cylinder containing water.
How many atoms are present in one such aluminum coin?
each coin is 0.50mL
.50mL * 2.70g/mL / 4.48*10^-23g/atom = 3.01*10^22 atoms
To find out how many atoms are present in one aluminum coin, we need to first calculate the volume of one aluminum coin and then use the given information to determine the number of atoms.
Here's how to calculate the volume of one aluminum coin:
1. Start with the total volume of water displaced by the five identical aluminum coins, which is 2.50 mL.
2. Divide this volume by the number of coins to find the volume of one aluminum coin. Since there are five identical coins, divide 2.50 mL by 5 to get 0.50 mL (or 0.50 cm^3).
Now, let's determine the number of atoms in one such aluminum coin:
1. Convert the volume of one aluminum coin to grams using its density. The density of aluminum is given as 2.70 g/cm^3. We know the volume is 0.50 cm^3. Multiply the volume by the density: 0.50 cm^3 * 2.70 g/cm^3 = 1.35 g.
2. Divide the mass of one coin (1.35 g) by the average mass of one aluminum atom (4.48×10^(-23) g) to determine the number of atoms: 1.35 g / (4.48×10^(-23) g) = 3.01×10^22 atoms.
Therefore, there are approximately 3.01×10^22 atoms in one aluminum coin.