The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:
0.5mol/L
Explanation:
First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the question = 30g
Number of mole = Mass /Molar Mass
Number of mole = 30/40 = 0.75mol
Volume = 1.5L
Active mass = mole/Volume
Active mass = 0.75mol/1.5L
Active mass = 0.5mol/L
