<p>b is converting to moles c is the mass percent.</p>
<p>oh my bad. yeah i didnt know how to do that. just to not leave it blank on the test, i put 0.5388g/0.6489g = 83%. but i know that that is wrong</p>
<p>is that what you got for part B though? the number of moles?</p>
<p>What was the consensus on 4b? Was it a complex ion with NO3 and Cl or something?</p>
<p>i used the same method for part b</p>
<p>I don’t think anyone really knows about 4b</p>
<p>Does anyone know when collegeboard puts up the solutions to these questions?</p>
<p>noooo idea on the net ionics of number 4 sorry. alaskandebater did you take this exam in alaska?</p>
<p>6. (a) The pressure in the flask is equal to 100 torr because the ethanol is in equilibrium. There are equal amounts of its liquid and gaseous form.</p>
<p><a href=“b”>U</a> Average kinetic energy of molecules, and therefore average speed of the gas molecules, is directly related to absolute temperature. When you increase temperature, you are increasing average kinetic energy and average speed. More gas molecules are moving at faster speeds in random chaotic motion and their collisions are elastic. The total force in the container thus increases while the area is the same. Thus, the pressure increases.</p>
<p><a href=“c”>U</a> (i) The order of reaction with respect to ethanol is zero. The graph of [E] is linear, thus the concentration of E has no effect on the rate of reaction. Furthermore, 1/[E] and ln[E] are non-linear. Thus, it is zero order with respect to ethanol.</p>
<p><a href=“ii”>U</a> r = k[E]^0 = k **r =k **</p>
<p><a href=“iii”>U</a><a href=“0,%20.01”>/U</a> (2000,.002) [.002-.01]/[2000-0] = -.0000040 M/sec = k</p>
<p><a href=“d”>U</a> PV = nRT P = nRT/V Volume, temperature, and R are constants. When the reaction completes we have gone from 1mol gas to 2 mol gas. So we start with n = 1 and have .4atm and get n=2. .4 atm = 2 (nRT/V) so the final pressure must be 2(.4) = .80 atm</p>
<p>To Clean that spill: Use NaHCO3 . NaCl is 100% useless its salt it wont do anything. Water will do nothing but dilute it and possible spread it. NaHCO3 will neutralize it, so use that.</p>
<p>#5. (g) The statement is **incorrect. **</p>
<p>ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)</p>
<p>ΔH is a large negative value, thus the enthalpy of forming the bonds is much greater than that of breaking the bonds. The large negative ΔH is thus not from the release of energy when the strong bonds are broken. Furthermore, it requires energy to break bonds, so the statement is just even more false.</p>
<p>for the mass percent of silver I got the mass of Ag from AgCl then I divided that by the total amount of alloy.</p>
<p>@formata, that’s what i did also, but I’m not sure if i did it correctly :</p>
<p>My attempt at the coordination complex balancing:</p>
<p>Co+2 + 4HCl -> [Co(HCl)4]+2</p>
<p>Doubt this is right.</p>
<p>This is my first AP test, so I was wondering: when our results come in the mail, do they break down what we got wrong? Do they tell us how many points we got out of 150?</p>
<p>My attempt: Co2+ + NO3- + Cl- –> Cl[Co(NO3)4]2-</p>
<p>@ibisadoozy, yes it took this test in Alaska, they make us begin extra early so we start before the east coast finishes.</p>
<p>@ NYEM
My teacher said the product should have been CoCl2 or something like that. I cried because I didn’t even get the reactants right. I did something stupid like Co + H+ -> [Co(OH)]
I’m not sure I got any points for the entire equation section.</p>
<p>But Co and Cl don’t form a precipitate do they?</p>
<p>The way my teacher taught us to do complex ions was to go straight to net ionic and don’t worry about unecessary ions…so I got:
Co2+ + 4Cl- -> [CoCl4]2-</p>
<p>flamingmango… for 6 i got all the same things as you. the only thing is, for part c iii of that question, we couldnt use calculators so i left it as 1/250,000 which is the same as 4.0 x 10^-6. i realize now that if i had more time i coulda figured that it was 4.0 x 10^-6 without a calc but do you think that it matters that i left it as a fraction with units?</p>
<p>Walle34 is correct. I completely forgot about it during the test but now that it’s over I remember doing a Le Chat lab using that equation. The pink Co2+ ion turns into a blue [CoCl4] complex ion. Nitric acid is cancelled out.</p>
<p>Anyone for 1 d?</p>
<p>@pinkrose22: The only thing College Board sends us is our score (1, 2, 3, 4 or 5).</p>