We can write the balanced equation for the galvanic cell by using the oxidation and reduction half-reactions. In the anode, Ni(s) is oxidized and produces aqueous Ni2+:
Ni(s) → Ni2+
while in the cathode, Cu2+ is reduced and deposits copper:
Cu2+ → Cu(s)
We now balance the charge of each reaction by adding electrons to the side of the equation with the greater charge:
Ni(s) → Ni2+ + 2 e-
Cu2+ + 2 e- → Cu(s)
Finally, we add the half-reactions to obtain the overall balanced equation for the galvanic cell:
Cu2+(aq) + Ni(s) → Ni2+(aq) + Cu(s)
Answer:(I)+3 N2(g) that is correct
Explanation:
Answer:
2.30 liters.
Explanation:
- The balanced equation of the reaction is:
<em>Na₂O₂ + CO₂ → Na₂CO₃ + 1/2O₂,</em>
- It is clear that 1.0 mole of Na₂O₂ reacts with 1.0 mole of CO₂ to produce 1.0 mole of Na₂CO₃ and 0.5 mole of O₂.
- The no. of moles of CO₂ in (4.60 L) reacted can be calculated from the relation: <em>PV = nRT</em>.
P is the pressure of the gas (P = 1.0 atm at STP),
V is the volume of the gas (V = 4.60 L),
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas (T = 273.0 K at STP).
∴ n = PV/RT = (1.0 atm)(4.6 L) / (0.082 L.atm/mol.K)(273.0 K) = 0.205 mol.
<u><em>Using cross multiplication:</em></u>
1.0 mole of CO₂ produces → 0.5 mole of O₂, from the stichiometry.
0.205 mole of CO₂ produces → ??? mole of O₂.
- The no. of moles of O₂ produced from 4.60 L of CO₂ = (0.5 mole)(0.205 mole) / (1.0 mole) = 0.103 mole.
- ∴ The volume of O₂ produced from 4.60 L of CO₂ = nRT/P = (0.103 mol)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm) = 2.30 liters.
Answer:
Law of Lateral Continuity The Grand Canyon.
and
he same rock layers on opposite sides of the canyon. The matching rock layers were deposited at the same time, so they are the same age.
Answer:
Explanation:
Hello!
In this case, according to the law of conservation of mass, it is possible to realize that the mass of oxygen comes from the subtraction of the total mass and the mass of copper:
Next, we compute the moles of both Cu and O given their atomic masses:
Now, we divide by the moles of oxygen as the fewest ones, in order to calculate their subscripts in the empirical formula:
Therefore, the empirical formula turns out to be:
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