<p>
[Quote]
It is F- or the fluoride ion. not diatomic fluoride or F2
[/Quote]
It can't be F2 because that would be flourine was oxidized. Therefore something must be reduced, but in NH4+ nothing is reduced, so this can't be an oxidation-reduction equation. F- is the answer.</p>
<p>^quizquick in the PR's book it says your right (I KNOW i saw it somewhere in there. Look for yourselves).</p>
<p>yeah the Nickel one was solubility, i believe...i didn't pick AgCl (good thing)</p>
<p>Guys here are the real solutions >_<</p>
<p>[list=A]
[<em>]KClO3 -> KCl + O2
[</em>]AgCl + Cl(-) -> AgCl2(-)
[<em>]HC2H3O2 + OH(-) -> H2O + C2H3O2(-)
[</em>]NH3 + HF -> NH4(+) + F(-)
[<em>]Cu(2+) + Zn -> Zn(2+) + Cu
[</em>]H3P + BCl3 -> H3PBCl3
[<em>]Ni(2+) + OH(-) -> Ni(OH)2
[</em>]C6H14 + O2 -> H2O + CO2
[/list]</p>
<p>If the reactions are correct according to octal0de, I think I may have made a perfect score ie 15/15. :)</p>
<p>Do we get partial points for products, or is it 0/2 or 2/2 only?</p>
<p>i got 12/15 according to octalode</p>
<p>completely messed up the AgCl one...lol</p>
<p>you can get partial points for products, yes.</p>
<p>let's see. got the acetic acid one (3/3)
got ammonia and HF (6/6)
got Cu and Zn (9/9)
got Ni and Oh (12/12)
got hexane (15/15)</p>
<p>bang bang</p>
<p>According to octalc0de, I think I got 12/15. <em>hopes</em>
1. KClO3 -> KCl + O2 (argh, I think I got KClO4...)
3. HC2H3O2 + OH(-) -> H2O + C2H3O2(-)
4. NH3 + HF -> NH4(+) + F(-)
5. Cu(2+) + Zn -> Zn(2+) + Cu
8. C6H14 + O2 -> H2O + CO2</p>
<p>chocolatechip, I believe you have a 14/15 there. You have both products on the first reaction correct.</p>