6. In order to turn 2 iron atoms into rust, 2 water molecules must be present. -False?
To make 2 molecules of Fe2O3 "rust", six O atoms are needed, along with two iron atoms. That would require three water (H2O) molecules, if there were no other source of oxygen. Therefore the correct answer is False.
http://en.wikipedia.org/wiki/Iron(III)_oxide
The issue is what is meant by rust. In the real world, there are two iron oxides:
FeO and Fe2O3 which are the common oxides for the two common valence forms of the iron molecule.
In common rust, these two oxides and their hydroxides are present, almost half and half, along with trapped water.
So we give "rust" mixture a "formula" (EVEN THOUGH IT IS NOT A CHEMICAL COMPOUND, but a mixture) Rust consists of hydrated iron(III) oxides Fe2O3·nH2O and iron(III) oxide-hydroxide FeO(OH)·Fe(OH)3. Again, the statement is false using this understanding of rust.
This leaves us with the IRONII oxide, FeO, which is quite rare. In that case, the statement is true, but this is not refered to as "rust" by anyone I know.
To determine whether the statement "In order to turn 2 iron atoms into rust, 2 water molecules must be present" is true or false, we need to analyze the chemical reaction of iron corrosion.
Iron corrosion, commonly known as rusting, occurs when iron reacts with oxygen in the presence of water. The reaction can be represented as follows:
4 Fe + 3 O2 + 6 H2O → 4 Fe(OH)3
From the reaction equation, we can see that four iron atoms (4 Fe) combine with three molecules of oxygen (3 O2) and six water molecules (6 H2O) to produce four molecules of iron(III) hydroxide (4 Fe(OH)3), which is commonly known as rust.
Therefore, the correct statement would be: "In order to turn 4 iron atoms into rust, 6 water molecules must be present." Consequently, the given statement "2 iron atoms into rust, 2 water molecules must be present" is false.