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Photon336
 one year ago
Chemquestion
Photon336
 one year ago
Chemquestion

This Question is Closed

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Ill start with 321 because its easy. If you added NaCN more AgCN would precipitate. B

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@sweetburger because it's at equilibrium already can we assume that it's saturated solution if so then adding more NaCN would precipitate in my opinion

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1For 320 [.2]^2/([.1][.2])= 2=Keq

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1So 320 is B. I think.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1319. we have \[N _{2} + O _{2} > 2NO \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1319. A. equilibrium constant is temperatuer dependent so if we changed the temperature we should also change the equilibrium concentrations because we would have a new Keq value. C adding more NO will shift to the reactants if it's already at equilibrium D N2 we don't know if we added or removed it but that would cause a shift. B. is the answer because if we have the same number of moles on either side of the equation increasing the pressure does not cause any change in the equlibrium concentrations. remember increase in pressure favors the side with fewer number of moles , while decreasing pressure it's the opposite.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Completely agreed.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1318. Adding a Catalyst would lower the activation energy of the reaction which in turn would increase the rate of the forward reaction, but would also increase the rate of the reverse reaction as the activation energy of the reverse reaction is also lowered showed in this diagram below.dw:1438548218094:dw

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1So adding a catalyst would have minimal affect on the equilibrium concentrations.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1yeah that's a great way to explain that.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[K _{eq} = \frac{ [0.20]^{2} }{ [0.20][0.10] } = 2 \] for the other one like they gave us equilibrium concentrations so we didn't have to do that ICE table stuff.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Haha, the ICE Table is very annoying to write out...

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1287. A. magnitude of the activation energy, the higher the Ea the harder it will be to get the reaction going. B.(nope) this is our answer C. lowering activation energy is what a catalyst does and that increases the rate of the reaction. D. absolute temperature yes. because increasing temperature increase the rate of our reaction \[k = Ae ^{Ea/RT}\] you increase the temperature and lower Ea the k value goes up. so B for 287

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Ahh i see it now. I was somewhat confused on question 287, but you clarified it quite nicely.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1For 286. I'm not sure, but I am leaning towards answer choice D. I think answer choices A and C are effectively the same in which the same process is occurring in which the activation energy is being lowered. I know that increasing the temperature does indeed effect the rate constant. TBH I thought that only temperature truly could change the rate constant.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1288. A. \[\frac{ 1 }{ 2 } \frac{ \Delta [KMnO _{4}] }{ \Delta T}\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1288. Notice that once Mn forms the reaction is now fast. to write our rate law our reaction is going to depend on our slowest step. so that's how I eliminated C and D. now for A and B. we know that in our rate law, our concentration of [KMnO4] is going to decrease and when you write the integrated rate law the coefficients let' say two becomes (1/2) so that eliminates B.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1that's why we consider KMnO4 and not MnSO4

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1for 286 i agree with you.. that's funny because whenever two answers are the same they can't be the answer they are looking for

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Yea I agree with your thought process on question 288.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1did they teach you this equation \[k = Ae ^{Ea/RT}\]

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1That is the Arrenhius? If i remember.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1I have to go, but I will gladly help anytime.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1also \[r = k[A]^{x}[B]^{y}\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1it's d doubling the concentration of one of the reactatns

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1that affects the rate of the reaction but not the rate constant k

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Ahh so for 286 the answer was D?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Anyway I have to go cya around

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Now looking at the Arrenhius I see how obvious it is XD *facepalm*
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