Answer:
m = 0.659 ounce
Explanation:
It is given that,
The thickness of a Teflon coating is, d = 1 mm
Area of the coating, A = 36 inch²
The density of Teflon, d = 0.805 g/mL
We need to find ounces of Teflon are needed.
Firstly, find the volume of the Teflon needed,
1 inch² = 6.4516 cm²
36 inch² = 232.258 cm²
Density,

V is volume of the Teflon needed, V = Ad
So,

Also, 1 gram = 0.035274 ounce
18.69 gram = 0.659 ounce
So, 0.659 ounces of Teflon are needed.
7.4x10^23 = molecules of silver nitrate sample
6.022x10^23 number of molecules per mole (Avogadro's number)
Divide molecules of AgNO3 by # of molecules per mol
7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)
(I leave off the x10^23 because they both will divide out)
Use your periodic table to find the molar weight of silver nitrate.
107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3
Now multiply your moles of AgNO3 with your molar weight of AgNO3
1.229mol x 169.868g/mol = 208.767g AgNO3
It is clear that the core is subject to change