Strategy: with the measures you can determine the volume of the plate of aluminum. Then you can use the density of aluminum to calculate the mass.
With the mass of aluminum and its atomic mass you can find the number of moles and thereafter the number of atoms.
Finally divide the cost by the number of atoms to find the cost of one single atom.
Let's do it.
Volume of aluminum plate, V: 0.0112 in* 4.83 in* 2.60 in * [2.54 cm/in]^3 = 2.305 cm^3
Density of aluminum (from Wikipedia), d = 2.70 g/cm^3
mass, m = d*V = 2.305 cm^3 * 2.70 g/ cm^3 = 6.22 g
Atomic mass of aluminum (from Wikipedia), am = 27 g / mol
Number of moles, n = m/am = 6.22 g / 27 g / mol = 0.23 mol
Number of atoms = n*Avogadro constant = 0.23 mol * 6.022 * 10^23 atoms/mol = 1.39*10^23
Cost per atom = cost of the can / number of atoms =$ 0.05 /1.39*10^23 atoms = 3.60 * 10^ - 25 $/atom
Answer:
The volume would be; 136.17 ml
Explanation:
Volume V1 = 150 mL
Temperature T1 = 20°C + 273 = 293 K
Pressure P1 = 758 - 17.54 = 740.46 torr
At STP;
Volume V2 = ?
Pressure P2 = 760 torr
Temperature T2 = 273 K
Using the general gas equation;
P1V1 / T1 = P2V2 / T2
Making V2 subject of formulae;
V2 = P1V1T2 / T1P2
Inserting the values we have;
V2 = 740.46 * 150 * 273 / 293 * 760
V2 = 136.17 ml
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