<p>but based on the specific heat, ci should be calculated with the mass of the solution</p>
<p>and how was cii supposed to be calculated?</p>
<p>Regardless of which one is right, you’re forgetting that it’s negative ha ha…
So getting the ~15kj/mol isn’t much of a justification.</p>
<p>yeah im probably wrong. I thought you arent allowed to just find the energy for urea cus the temperature changes as the solution as a whole, but im not too sure if i was right on that. </p>
<p>and i think the joules is not negative. the reaction is endothermic regardless of the decrease of the temperature. delta T is a change not a decrease or increase.</p>
<p>Yes but in calorimetry, you determine the amount of heat given off by a process (in this case dissolution of urea) by measuring the heat of entire solution lost so you would have to use the combined mass.</p>
<p>DeltaH = -qrxn/moles lr</p>
<p>The sign of should be positive because dissolution of urea requires energy.</p>
<p>godhs so many different concepts you have to think about at the same time i hate this feeling when im unsure whether im right or wrong</p>
<p>Just ask if you need an explanation/solution to one of these. One of my parents is an AP chem teacher, so these answers should be fairly accurate.</p>
<p>1A Ksp= [Ag+][Br-]
1B 7.1x10^-7 M [Ag+]
1C 3.55x10^-7 M [Ag+] which is less than the concentration in part B
1D V= 37618 L = 38000 L
1E Q=1.04x10^-3
Q>Ksp, so precipitate will form
1F Ksp for AgBr is greater than AgI since AgI precipitated to a greater extent than AgBr.</p>
<p>2A -3.2 degrees C
2B Endothermic - Temperature decreased, so energy was absorbed by the the system from
the surroundings. Endothermic solutions feel colder because they absorb energy while
exothermic solutions feel hotter because they “give off” energy.
2C (i) q= 1304 J = 1300 J
(ii) 15222 J/mol = 15 kJ/mol
2D 0.0701 kJ/mol K = 70. J/mol K
2E 12.46 kJ/mol = 12 kJ/mol
2F ΔH decreases. A Decrease in the temperature will cause a larger negative temperature
change for the solution. So for q=mCΔT, the ΔT would be larger and magnitude and<br>
more negative, so since q basically equals ΔH, ΔH decreases.</p>
<p>3A (i) 0.00625 mol Cl-
(ii) V = 0.0797 L
3B (i) Cl- is 2nd order
(ii) MnO4- is 1st order
3C (i) r = k[Cl-]^2[Mno4-][H+]^3
(ii) 1.93x10^-3 M^-5 s^-1
3D Likely? I don’t know the exact explanation to this one so…</p>
<p>4A OH- + H+ → H2O
Solution turns yellow since there would be more acid in the solution
4B C3H8 + 5O2 → 3CO2 + 4H2O
The solution will be acidic since CO2 will dissolve in water to form carbonic acid H2CO3
4C 2H2O2 → 2H2O + O2
O is has a 1- oxidation number in hydrogen peroxide</p>
<p>5A H-C≡C-H
5B Ethyne contains a ≡ bond so it has the shortest bond
5C (i) trigonal planar
(ii) tetrahedral
5D False- boiling disrupts the intermolecular forces, not the intramolecular bonds or
covalent bonds that hold atoms together.
5E Ethane or Ethyne are both polar
5F Ethanol is capable of H-bonding with water, where as Ethaneiol has only dipole-dipole
forces which are weaker than H-bonding</p>
<p>6A 1s² 2s² 2p6 3s² 3p6 3d10
6B Zn 2+ It is more difficult to remove electrons from a positive ion than from a neutral
atom.
6C Al(s) is oxidized
6D NO3- goes to the Al(NO3)3 solution to replenish electrons moving out of solution. K+
goes to Zn(NO3)2 compartment to replenish Zn2+ plating out on the Zn electrode
6E E = 0.90 V
6F E is positive, so the reaction is spontaneous. Therefore ΔG is negative for a<br>
spontaneous reaction.
6G The cell voltage increases. Q= [Al3+] / [Zn2+] = 0.01 / 1.0 so Q is less than 1.
According to the equation Nernst equation, Ecell = E° - (0.0592/n)logQ , a Q less than
1 will cause log to be negative, and causes the rightmost term to become positive,
which will add to the value of E°</p>
<p>DillonJay</p>
<p>You can’t use the “full shell” explanation to explain ionization energy. Zn 2+ is not the same as nickel because it has more protons. The two electrons taken from the Zn atom means there are less electron-electron repulsions, which means the electrons are drawn in closer to the nucleus. The ionization energy would then increase because the attraction between the electrons and nucleus would be greater in Zn 2+ than Zn.</p>
<p>q is positive. q lost by solution is equal to q gained by dissolution of urea. we know it is endothermic since the temp went down and endothermic processes have positive q</p>
<p>Okay, my bad. I know what you’re saying, I’m just very rusty on calorimetry stuff.</p>
<p>If you mess up significant figures on like part a do you just lose 1 point for the whole problem or do you lose points on parts b and c if you mess up the significant figures again. Also does everyone agree on 1c that it is less</p>
<p>What about form B, anyone?</p>
<p>1C is equal concentration, still saturated therefore same concentration</p>
<p>i dont think they look at sig figs except some questions cus on the practice tests there would be footnotes saying which questions’ sigfigs counted. plus even on the ones they do count the sigfigs they said it’s ±1. </p>
<p>I disagree with 1C i think it stays the same regardless of water you put in. but, it’s answer from a teacher. so i’m doubtful</p>
<p>^yea that’s what i put for 1c (i did an identical question on an old test)</p>
<p>and for 2f, delta H increases right?</p>
<p>2e- .89 x 14 ?</p>
<p>1e) I got Q=1.04x10^-8 which is less than K so no precipitate forms</p>
<p>doesnt delta H increase when it is 5 degrees. Just think about it conceptually, it requires more energy to break the bonds of a substance at a colder temperature.</p>
<p>^ i also got something raised to the -8, but isn’t Ksp 5.0 x 10^-13?</p>
<p>And noob, I am positive Q was greater than Ksp. you must have calculated incorrectly. Or, as user said, you did not compare Q and Ksp correctly. Can someone please respond to post #138</p>