Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Here, this is what I have. :)
Typical day of a new day in my house with my phone in my car so I’m not going hard I don’t know how much it was
The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.
Formula is given by:
(1)
where,
is the initial concentration
is the initial volume
is the final concentration
is the final volume
Now,
= 0.850 M
= 4.12 L
=?
= 10.00 L
Substitute the give values in formula (1),
= 
Thus, the final concentration of the
solution =
