Answer :
To find the volume of the plate without accounting for the hole firstly
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps