<p>i got the order wrong in 6. so I could still get points on the rest of the equation then? that’s a relief.</p>
<p>for FRQ 5 (b), the reaction was an oxidation-reduction right?
since nitrogen changed oxidation numbers?</p>
<p>I think you mean 5.e), but yes, I also put that.
O2 also changed oxidation numbers.</p>
<p>That’s what I got as well. I also got that S was positive and that the statement was false.</p>
<p>I said the statement was true. I was thinking that since all combustions are highly exothermic.
I got positive S as well.</p>
<p>I don’t have a calculator with me right now, but I tried to do 1.</p>
<ol>
<li>(a) HCl is a strong acid, so [HCl] = [H+]. -log(0.100) = 1.000
(b) (i) Kb = [NH4+][OH-]/<a href=“ii”>NH3</a> [OH-] = sqrt (0.100<em>1.8</em>10^-5)
(c) (i) Ka = Kw/Kb = 1<em>10^-14/1.8</em>10^-5
(ii) This is a buffer. First find the moles of each and divide them by the additive volume to find the molarity in the total solution. Then use an ICE table.
(d) (i) Yes, all of the H+ from HCl is consumed. The only contributor of H+ is NH4+, which was the source of H+ before HCl was added.
(ii) This is similar to a titration, so first do stoich, then ICE. My final answer was 0.0667 M I think.</li>
</ol>
<p>I got it was false because in a normal equation you subtract the bond energies of the products from the reactants. Since it was negative, the energy of the product bonds were more responsible</p>
<p>That’s the delta H = summation of bonds broken - summation of bonds formed
or whatever, right? I couldn’t remember which order the equation went in, since it’s not on the eqn. sheet.</p>
<p>Yeah, that’s the equation.</p>
<p>I said it was false because breaking bonds is actually endothermic (positive delta H). The high negative delta H value is because the bonds that were formed after all the bonds had broken are more stable than the bonds that were there before.</p>
<p>Here’s what I got for equations:
Mg(OH)2 + 2H+ –> Mg2+ + 2HOH
100 mL</p>
<p>Cl- + Co2+ –> [Co(Cl4)]2-
Cl- is the Lewis base.</p>
<p>Cu + AgNO3 –> CuNO3 + Ag (I wasn’t sure about this. What was the charge on Cu?)
Solution turns blue. Wire dissolves. Silver appears.</p>
<p>I had a few I wanted opinions on.
For FRQ 2…
(b) to find the mass of ppt, you just subtract the (mass of dry filter) crucible from the (mass of filter crucible and ppt)?
For FRQ 3…
(f) did you get a non-whole number answer for part 1?
For FRQ 5…
what chemical equation did you write for part d?</p>
<p>@CORVIDS,</p>
<p>for the 3rd equation, i got
Cu + (Ag+) –> (Cu2+) + Ag
because the NO3- would dissociate…right?</p>
<p>apstudent12345: You’re right. That’s exactly what I got, and yes it dissolves.</p>
<p>@skateme
okay good, at least i got one of them right.</p>
<p>Do you think I"ll get the point if all I said was “it changes colour (from clear to blue-green).”
- Didn’t mention the dissolving of the wire/new precipitate.
- I think I wrote clear and not colourless . . .</p>
<p>@abrayo,
colorless is preferred i think over clear, but i didn’t mention the ppts, only the color change.</p>
<p>For equation 3 I got Cu + 2Ag+ > Cu2+ + 2Ag. Silver precipitate. Do you think I’ll get the point for just saying that a precipitate forms?</p>
<p>@apstudent12345</p>
<p>For #2) b) i took the average of the three masses with the precipitate and subtracted the mass of the filter from it.</p>
<p>Aggghhh, for 4cii) I totally blanked on the color change D:</p>
<p>And when I finally got 1ciii) I ran out of time ! But I think I got the rest of it right. </p>
<p>5f) I put positive b/c there are more moles of gas in the products, which is increasing disorder
and 5g) did anyone else put false? I said false b/c of the explanation since I had learned that the breaking of bonds is endothermic, not exothermic since the breaking of bonds requires energy.</p>
<p>Yes, I learned in class that ‘clear’ is incorrect and it should be colourless.
I can’t remember which I wrote though.</p>
<p>Hopefully all they needed was one observation, so we all get the point!</p>