Answer:
Unit cells which are present per cubic centimeter of Ag =
Volume =
Edge length =
Explanation:
(a)
Given that:-
The density of the solid Ag = 10.5 g/cm³
Molar mass of silver = 107.8682 g/mol
So, Moles present per cm³ of Ag = =0.0973 mol/cm³
Also, 1 mole = atoms.
So,
Atoms present per cm³ of Ag =
Thus, answer =
In FCC, the number of atoms in the unit cell = 4 unit cells
So,
Unit cells which are present per cubic centimeter of Ag =
<u>Unit cells which are present per cubic centimeter of Ag = </u>
(b)
The reciprocal of the unit cell/cm³ is the volume of the unit cell.
So,
<u>Volume = </u>
(c)
Also, Volume =
Thus, edge length = =
<u>Edge length = </u>
So, what's shown here is the ion product of pure water: that is, the product of the concentrations of hydronium and hydroxide ions in pure water at 25 °C. By this relation, if you know the [H₃O⁺], you can calculate the [OH⁻], and vice-versa.
Since [H₃O⁺] × [OH⁻] = 1.0 × 10⁻¹⁴, [OH⁻] = (1.0 × 10⁻¹⁴)/[H₃O⁺].
Substituting the given [H₃O⁺] as 1.25 × 10⁻² M:
[OH⁻] = (1.0 × 10⁻¹⁴)/(1.25 × 10⁻² M) = 8.0 × 10⁻¹³ M.