1 mol of any substance is made of 6.022 x 10²³ units.
When its 1 mol of a formula, its made of 6.022 x 10²³ formula units
Therefore if 6.022 x 10²³ formula units of CeI₃ make up 1 mol of CeI₃
Then 9.42 x 10²¹ formula units make up - 1/(6.022 x 10²³) x (9.42 x 10²¹) = 0.0156 mol of CeI₃
Therefore number of moles of CeI₃ - 0.0156 mol
I don't understand what is (g).
Maybe the answer is 2H<span>(aq)S</span>₂<span>−2(aq) </span>⇒ <span>H</span>₂<span>S</span>₂.
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
