Answer:
6 oxygen atoms
Explanation:
From the equation,
2Fe(OH)₃ → Fe₂O₃ + 3H₂O
From the reactant (left hand side) we have 2 moles of Fe(OH)₃ having (2 * 3 = 6) atoms of oxygen and decomposed to give Fe₂O₃ which contains 3 atoms of oxygen and 3 moles of water that also contains 3 atoms of oxygen.
Since the number of oxygen participating in the reaction is independent on the product (not a reversible reaction) then the total number of oxygen atoms participating in the reaction is 6
The thing you MUST do FIRST is look for any H's, O's, or F's in the equation
1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA
2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)
ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet
for example::::
N2O4
the oxidation number of O is -2.
since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4
N has an oxidation number of +4.
more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so
oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion
The normal boiling point for iodine is 184.4 degrees Centigrade. Normal boiling point refers to the temperature at which any liquid boils at one atmosphere of pressure. This temperature is more specific than just a boiling point and is useful in comparing various liquids more accurately.
