Photon336
  • Photon336
Another question: Going to post them in a bit.
Chemistry
schrodinger
  • schrodinger
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Photon336
  • Photon336
Will post soon. Going to be on the rate constant, and kinetics.
Photon336
  • Photon336
Posted here other one got cut off
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Photon336
  • Photon336
Some more here
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Photon336
  • Photon336
=D
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Photon336
  • Photon336
Ok I think this is the last one for this question
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Photon336
  • Photon336
I know this is a lot but some of them aren't that bad.
Photon336
  • Photon336
sweetburger
  • sweetburger
Ill start with the easy one 276. Im pretty sure OH is a intermediate
Photon336
  • Photon336
276. it's gotta be an intermediate because if it were a catalyist it would be generated in the last step again and unchanged. (Promoter and activator make no sense)
sweetburger
  • sweetburger
For 275 I'm not sure but I think that the concentration of CaCl2 would have no affect on the rate because it is a product and it isnt affecting the forward reaction. Not sure if my reasoning on this is sound though.
sweetburger
  • sweetburger
My answer also came down to that for 275 B,C, and D. would change the rate of the reaction.
sweetburger
  • sweetburger
Yea actually im certain for 275 that A would be the only answer that would not affect the reaction rate.
anonymous
  • anonymous
Yeah, sounds right to me, I guess the concentration of \(CaCl_2\) would only affect the equilibrium constant.
anonymous
  • anonymous
wait... I don't know for some reason I can't really decide between A and B
Photon336
  • Photon336
289. which one is false? A. Chlorine goes from an oxidation state of 0 to -1 so it's reduced so that's true B. increasing the temperature generally does increase the rate of reaction. \[k = Ae ^{Ea/Rt}\] you increase temperature i think the quantitiy Ea/Rt goes down i believe and k goes up and reaction rate goes up as well. D yes HCL is a strong acid, and HCLO is a weak acid so that's true.
sweetburger
  • sweetburger
289 I think is C.
Photon336
  • Photon336
289. is C. here's my justification
sweetburger
  • sweetburger
The only thing that affect the equilibrium constant is temperature from what I remember.
anonymous
  • anonymous
Yeah it shouldn't affect the equilibrium constant, I guess the rate constant is not affected by concentrations of the products only concentration of reactants.
Photon336
  • Photon336
Like i thought it was C for 289 b/c I notice that the number of moles of each quantity are the same. To me like an increase in pressure favors the side with fewer number of moles i think , while a decrease in pressure favors the side with more moles.. Keq
Photon336
  • Photon336
yeah it's only affected by temperature as well
Photon336
  • Photon336
sorry.. i was thinking in terms of what would favor the products.
sweetburger
  • sweetburger
My answer choice for 290 is B.
Photon336
  • Photon336
Okay so 276 D. "An OH radical is formed in the rate determining step but is consumed in the next step therefore it's an intermediate" 289 C The equilibrium constant is not affected by changes in concentrations and pressures remember that equilibrium states are separate from kinetic studies.
sweetburger
  • sweetburger
Looks good.
anonymous
  • anonymous
Yeah for 289 I don't think partial pressures come into play because we're only talking about liquids.
Photon336
  • Photon336
290. \[k[cl _{2}]^{0.5}[H _{2}O]\]
Photon336
  • Photon336
IF you look at the slow step the rate for formation depends on the formation of the chlorine radical and it's ability to react with water as well.
Photon336
  • Photon336
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Photon336
  • Photon336
If this is what happens above.. then yo can clearly see why we also need water to make this happen and why it's also the rate depends on that too
Photon336
  • Photon336
@Woodward I got 275 wrong why is it A
sweetburger
  • sweetburger
Wait
sweetburger
  • sweetburger
For 275 is A not the answer?
Photon336
  • Photon336
@sweetburger sorry the answer is A.
Photon336
  • Photon336
was asking why though?
sweetburger
  • sweetburger
I guess Ill give it a shot at explaining it. Lets say that the rate law is rate=k[CaCO3][HCl]^2. This of course might not be the rate law as it is experimentally determined but Im assuming it is. So lets say that if you changed the concentration of the product CaCl2 it would have no affect on the rate law I wrote above.
Photon336
  • Photon336
Oh I see
Photon336
  • Photon336
it's because it's a product and the rate doesn't depend on the products just reactants
anonymous
  • anonymous
Yeah the partial pressure of HCl is essentially the concentration of HCl due to that whole \(P= \frac{n}{V} RT\) thing from earlier. And the concentration of a reactant is found in the rate law, not the products like \(CaCl_2\) is I guess
Photon336
  • Photon336
\[r = k[HCl]^{2} \]
Photon336
  • Photon336
I thought we don't include solids in the expression because their concentrations don't change much
sweetburger
  • sweetburger
Thats the equilibrium expression. What I wrote is a rate law.
Photon336
  • Photon336
oh okay.

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