Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Answer:
C) acid-base neutralization
Explanation:
NaOH + CH₃COOH = CH₃COONa + H₂O
Break the solutions apart:
NaOH = Na⁺ + OH⁻
CH₃COOH = CH₃COO⁻ + H⁺
Combine the resulting solution after the reaction:
OH⁻ + H⁺ = H₂O
Answer:
7.28 mol Na2SO4
Explanation:
Since it is already in moles, all we have to do is use a molar ratio
A molar ratio is the proportions of reactants and products using the balanced equation. When writing a mole ratio, the given information must cross out with the right thing.
7.28 mol H2SO4 * 1 mol Na2SO4/1 H2SO4 = 7.28 mol Na2SO4
*notice how the H2SO4 crosses out
Answer:
a free swimming larval stage in which a parasitic fluke passes from an intermediate host to another intermediate host
