Answer:
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Explanation:
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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)
The equation : y=3x-5
<h3>Further explanation
</h3>
Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates
General formula
y-y1 = m(x-x1)
or
y = mx + c
Where
m = straight-line gradient which is the slope of the line
x1, y1 = the Cartesian coordinate that is crossed by the line
c = constant
The formula for a gradient (m) between 2 points in a line
m = Δy / Δx
First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>
Answer:
I think it's 6 moles are produced
