Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
#4 and #5:
To find pH given concentration of H+ or H30+
pH = - log (H+ or H30+ M)
To find pH given concentration of OH-
Since you already found the pH for this (in #4), you subtract #4's answer from 14.
14 - (pH) = pOH
Balanced equation: 2Fe + 3H2O → Fe2O3 +3H2
Convert g to mols:
285/55.845 = 5.1034 mols
Mole ratio of Iron and Iron (III) Oxide: 2:1
5.1034/2 = 2.5517 mols
Answer:
sulfide ion
Explanation:
however, when non-metalic elements gain electrons to form anions the end of their name is change to "-ide".foe example, a flourin atom gains one electronto become a flouride ion(F`) and sulfur gains two electrons to become a sulfide ion(s^2)
