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)
Answer:
84 Joules are wasted per second.
Explanation:
An efficiency of 16% means 84% is wasted as heat energy.
Therefore;
For a 100W bulb;
Heat wasted = 84/100 × 100 W
= 84 W
Therefore;
84 W or 84 joules are wasted as heat per second
Answer:
Products would be on the right. Reactants would be on the left
Explanation:
Answer:
M(Fe₂O₃) = 159.70 g/mol
M(CO) = 28.01 g/mol
M(Fe) = 55.85 g/mol
M(CO₂) = 44.01 g/mol
Explanation:
We can calculate the molar mass of a compound by summing the molar masses of the elements that form it.
Fe₂O₃
M(Fe₂O₃) = 2 × M(Fe) + 3 × M(O) = 2 × 55.85 g/mol + 3 × 16.00 g/mol = 159.70 g/mol
CO
M(CO) = 1 × M(C) + 1 × M(O) = 1 × 12.01 g/mol + 1 × 16.00 g/mol = 28.01 g/mol
Fe
M(Fe) = 1 × M(Fe) = 1 × 55.85 g/mol = 55.85 g/mol
CO₂
M(CO₂) = 1 × M(C) + 2 × M(O) = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol
The southern pacific ocean has a similar symmetric pattern to the seafloor ages of the Atlantic ocean. In the Pacific, the seafloor on one side of the youngest crust gets very old but the seafloor on the other side is much younger.
