Well, the region is facing the sun, and hardly points away for half of a year.
Answer:
it answer D because waste is flowing thru the air
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:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of
indicates that each of these ions carries a charge of
. That corresponds to the shortage of one electron for each
ion.
Similarly, the superscript
on each
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of
(among the reactants) is
, and that the coefficient of
(among the reactants) is
.
.
There would thus be
silver (
) atoms and
aluminum (
) atoms on either side of the equation. Hence, the coefficient for
and
would be
and
, respectively.
.
The
ions on the left-hand side of the equation would correspond to the shortage of
electrons. On the other hand, the
ions on the right-hand side of this equation would correspond to the shortage of
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of
electrons, the right-hand side should also be
electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of
electrons. These two expressions should have the same value. Therefore,
.
The smallest integer
and
that could satisfy this relation are
and
. The equation becomes:
.

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.