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Photon336
 one year ago
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Photon336
 one year ago
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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@taramgrant0543664 @Rushwr @sweetburger @Woodward

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[Ca _{s} + Cl _{2g} \rightarrow CaCl _{2s}\]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0cuz the reaction is a formation reaction hence it is an exothermic reaction. SO delta H is negative

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0\[\Delta S _{surr} =\frac{ \Delta H }{ T }\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta G = \Delta H  T \Delta S \] so for our reaction the entropy would have to go down. so the sign TDeltaS would be positive. that means that for our reaction, which we know is spontaneous because we form our reactants. Delta G < 0 if we put the two together. 1. Delta G <0 2. Entropy Delta S < 0 3. TdeltaS > 0 THIS means that T delta S would be > 0 positive. and Delta G < 0 therefore the enthalpy Delta H would have to be large and negative, to offset the positive entropy and that would make delta G < 0

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1sorry form our products***

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0389. A Cuz when they had more reactants in to the system the system tend to attain the equilibrium by making more products. So the euilibrium points shifts to the left forming more products.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta G = \Delta H  T \Delta S \] True entropy is \[\Delta S = \frac{ \Delta H }{ T }\] because we end up with solid calcium chloride (s) we know overall the entropy goes down. so in that equation. delta S < 0 \[\Delta S < 0 \] \[\Delta H < 0 \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1389 is so tricky the first time I did it i got it wrong

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@taramgrant0543664 thoughts for 389?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0what's the answer given for 389. I explained one !

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1So I was thinking at first A until I saw that it was in solid form solids don't change rate so I was considering c

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@Rushwr it's C no change

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0oh yes it doesn't count for the K values yes. u are right!

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1equilibrium expressions, the K value(s) i think it's pure liquids and solids aren't included.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0I didn't see the states properly ! Thank you @taramgrant0543664 and @Photon336

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1For it to be counted it's either in gas or aqueous form

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0@taramgrant0543664 I love you!

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Definitely watch those states that's the thing that always gets me lol

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[Keq = [Cl _{2}] \] \[K _{eq} = \frac{ [C][D] }{ [A][B] }\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1yeah... that one got me too..

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Was the answer for 388 said? I might have missed that I know I missed part of your conversation since I got tagged in the math section

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1it remains the same because the Keq depends only on the concentration of the gas chlorine.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Right I was thinking that but thought I would double check as sometimes I am completely off on them lol what's the next one?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1387. A. because like replacing a strong acid with a weak one in that reaction would cause the temperature change to rise to a smaller degree, and the enthalpy change would be less.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1I just gave an answer for 387, so all that's left is 386

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1I will post one more set of questions then I g2g

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0Btw Thanks alot @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1386 I missed that one

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1These are kind of fun to do @Photon336!! As for 386 I was considering D as my option possibly?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0IKR this is awesome ! @taramgrant0543664 @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1ok Will post one more.
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