Answer is: 25.84 milliliters of sodium metal.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.
d(Na) = 0.97 g/mL; density of sodim.
m(NaOH) = 43.6 g; mass of sodium hydroxide.
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 43.6 g ÷ 40 g/mol.
n(NaOH) =1.09 mol; amount of sodium hydroxide.
From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).
n(Na) = 1.09 mol.
m(Na) = 1.09 mol · 23 g/mol.
m(Na) = 25.07 g; mass of sodium.
V(Na) = m(Na) ÷ d(Na).
V(Na) = 25.07 g ÷ 0.97 g/mL.
V(Na) = 25.84 mL.
That is false. mass is converted into energy
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
Atoms. Well, they can be broken down, but that won't happen that fast.
