If a mirror has a diameter of 10.240 cm and you want to have a coating that is 0.0250 mm thick of Aluminum on it, how many grams of aluminum will you need? How many atoms of aluminum are in the coating? The density of aluminum is 2.702 g/cm^3You may consider the deposited aluminum to be a cylinder from which the volume can be determined using the equation for the volume of a cylinderThanks!
0.0250 mm thick equals 0.00250 cm thick volume (surface area) (thickness) volume (pi r^2) (thickness) volume (3.14) (5.12cm)^2 (0.00250 cm) volume 0.20578 cm^3 find mass using density: 0.20578 cm^3 2.702 g/cm^3 0.5560 grams of Al your first answer, rounded to 3 sig figs is 0.556 grams of Al find moles, using molar mass 0.556 grams of Al 26.98 g/mol 0.020609 moles of Al using Avagadro's number, find atoms: 0.020609 moles of Al 6.022 e23 atoms / mole 1.241 e22 atoms your next answer , rounded to 3 sig figs is 1.24 X 10^22 atoms of Al p.s the answer is Magnesium to your question: A 0.98338 gram sample of a metal (II) carbonate containing an unknown metal M, was heated to give the metal (II) oxide 0.5133 grams of CO2What is the metal? use molar mass to find moles: 0.5133 grams of CO2 44.01 g/mol 0.01166 moles of CO2 by the equation: 1 MCO3 - 1 MO 1 CO2 0.01166 moles of CO2 was produced from an equal 0.01166 moles of MO find molar mass of MCO3 0.98338 gram sample / 0.01166 moles of MO 84.34 g/mole subtract the molar mass of (CO3) 60 g/mol gives the molar mass of the metal as 24 g/mol that's Magnesium
