Question

given that the atomic radius of zinc is 134 pm, calculate the thickness of the zinc coating in terms of zinc atoms. assume that the atoms are stacked end to end atop each other, not tightly packed.

Total surface area of nail = [Surface area of top of head of naill surface area of bottom of head of nails (surface area of cylindrical shaft of nail] = 0.419 + 0.419 +4.1 54.938 cm 4.98 cm² 7. Use your answers to Questions 5 and 6 to calculate the average Thickness of zinc on the nail. T = m = bollyan C.W .D (4.06cm) C0.32 cra) (7.1431ne) - nel=0.olca -0.011712112 8. Given that the atomic RADIUS of zinc is 134 pm, calculate the thickness of the zinc coating in terms of zinc atoms. Assume that the atoms are stacked end-to- end atop each other, not tightly packed. 9. Consider your observations when you first added aqueous sodium carbonate to neutralize your hydrochloric acid. What did you see? Precipitation Write the balanced chemical equation for the initial reaction of aqueous hydrochloric acid with aqueous sodium carbonate. (This reaction occurs in our blood, and leads to elimination of CO2(g) by our lungs.) Hydrochloric acid = HCL sodium Carbonate = Nagcoz NagCO3 (19) + 2 HCl - 2 NaCl ang + H₂O (1) + cog(s) 10. During further addition of sodium carbonate solution to your flask, it is common to observe the formation of a precipitate.

Answer 1

Q8

It is given that the atoms are stacked on top of each other as shown below.

Zinc Atoms Surface

Thus thickness of each atom equals the diameter of the atom twice the radius.

Diameter, D = 2* radius = 2* 134 pm = 268pm

We know that 1 pm equals 10^-12 m.

Thus diameters in m = 268 pm * (10^-12 m/pm)= 2.68 x 10^-10 m

Thickness of Zinc, t = 0.0117712122 cm = 0.0117712122 cm * (0.01 m/cm) = 1.17712122 x 10^-4 m

No. of Atoms, N will be given be,

t 1.17712122 x 10-4m * = 4.39 x 103 . N= 7= 2.68 x 10-10 m

Thus, the coating is 4.39 x 10⁵ Zn atom thick.