<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq)
</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s)
Zn⁰(s) → Zn²⁺(aq) +2e⁻
2 electrons are transferred from atom of Zn⁰ to 2 ions of Li⁺.
Answer:
Actually, The Henderson - Hasselbalch equation allows you to calculate the pH of the buffer by using the pKa of the weak acid and the ratio that exists between the concentrations of the weak cid and conjugate base. The pKa of formic acid is equal to 3.75. In this case, the pH of the solution will be equal to the acid's pKa .
Gypsum has the same hardness as a fingernail
<span>Assume you have 1 L of solution.
Moles F- = M F- x L F- = (0.0610)(1) = 0.0610 moles F-
0.0610 moles F- x(19.0 g F-/1mole F-) = 1.159 g F- in 1 L of solution
1 L solution x (1000 mL / 1 L) x (1.00 g / mL) = 1000 g of solution
mass % F- = (g F- / g solution) x 100 = (1.159 / 1000) x 100
= 0.1159%
parts per million F- = mg F- /L = 1159 / 1 = 1159 ppm F-</span>
