4.2 × 10²² atoms Al × (1 mol Al / 6.022 × 10²³) = moles Al
Last one: fraction 1 mole Al over 6.022 × 10²³ atoms Al
Answer:
D) 5.15
Explanation:
Step 1: Write the equation for the dissociation of HCN
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Step 2: Calculate [H⁺] at equilibrium
The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.
α% = [H⁺]eq / [HCN]₀ × 100%
[H⁺]eq = α%/100% × [HCN]₀
[H⁺]eq = 0.0070%/100% × 0.10 M
[H⁺]eq = 7.0 × 10⁻⁶ M
Step 3: Calculate the pH
pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15
volume of Ni = 25 nL = 25 x 10⁻⁹ L
mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸
mass = mol x Ar Ni
mass = 3.125 x 10⁻⁸ x 59 g/mol
mass = 1.84 x 10⁻⁶ g = 1.84 μg