Answer:
Wouldn't rust because zinc will lose electrons more readily than iron and will therefore oxidize first.
Explanation:
This process whereby rusting of steel is prevented by coating the steel with a layer of zinc is known as galvanization.
Now, in this process, the steel object will be coated in a thin layer of zinc. This coating will prevent oxygen and water from reaching the underneath metal since the zinc will also act as a sacrificial metal.
Now, Zinc is used because it has a lower reduction potential than iron and thus it will get easily more oxidized than iron. Which means the zinc will lose electrons more readily than iron.
Also, since zinc has a lower reduction potential, it is therefore the more active metal. Thus, even if the zinc coating is scratched and the steel is exposed to moist air, the zinc will still get to oxidize before the iron.
1.)b
2.)true
3.)false
are the answer don't take me on my word
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Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
