1.- Determine the number or quantity of atoms, the number of moles and the mass
ID: 701151 • Letter: 1
Question
1.- Determine the number or quantity of atoms, the number of moles and the mass of a rectangular prism of Silver (Ag) of 20.0 mm in width and 35 mm in length and 2.5 mm in thickness. ( the answer must be: 1.03 x 10^23 atoms, 0.1703 moles y 18.375 g)
2.-Determine the thickness of an aluminum sheet 450.0 mm long and 300.0 mm wide containing 2.25 moles,number or number of atoms, the mass of this sheet.
density of aluminum: 2,698.4 kg/m3, Atomic mass of Aluminum: 26.88 g/mol. ( The answer must be: 1.35 x 1023 atoms, thincknessr 0.17 mm and 60.71 g )
3.-Calculate the network parameter (a0) and the density of the FCC Nickel. Atomic diameter of 0.2492 nm, atomic mass 58.71 g mol^-1 ( A = 0.3524 nm, A = 8910.0 kg/m3 )
Explanation / Answer
Ans 1
Density of silver = 10.49 g/cm3
Volume of prism = 20*35*2.5
= 1750 mm3 x (1cm/10mm)^3
= 0.00175 cm3
Mass of prism = density x volume
= 0.00175 cm3 x 10.49 g/cm3
= 0.0183575 g = 18.3575 kg
Moles of Ag prism = mass/molecular weight
= 18.3575 / 107.86
= 0.1702
Number of atoms = moles x Avogadro number
= 0.1702 x 6.023 x 10^23
= 1.03 x 10^23 atoms
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