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Photon336

  • one year ago

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  1. Rushwr
    • one year ago
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    where?

  2. Photon336
    • one year ago
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    Attached

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  3. Photon336
    • one year ago
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    Here

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  4. Photon336
    • one year ago
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    Here too

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  5. Photon336
    • one year ago
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    @rushwr @sweetburger

  6. Photon336
    • one year ago
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    When stating an answer please provide a justification as to why you got the answer and why you feel the others are wrong.

  7. Rushwr
    • one year ago
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    for 381. I will go with C. Because we know when it says a reaction is non spontaneous \[DeltaG > 0\]

  8. Rushwr
    • one year ago
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    since they have not mentioned anything about a heat change or anything we can't predict anything about enthalpy and entropy soo those answers are cut off

  9. Rushwr
    • one year ago
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    the last one happens when the reaction is in equilibrium !

  10. Photon336
    • one year ago
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    381. C \[\Delta G = \Delta H - T \Delta S \] The spontaneity of a reaction does not dictate how fast the reaction will go but rather the tendency of that reaction to occur. if something is spontaneous/ a reaction it will happen. because there is free energy. delta G shows us the free energy available. These three answers below are wrong because, you need two know both the sign of delta H and Delta T S to determine whether a reaction is spontaneous. delta S = negative wrong Delta H is positive wrong Delta G = 0 Delta G = 0 means that it's neither spontaneous in the forward or reverse direction. Delta G must be greater than zero, that would mean that the reaction is spontaneous in the opposite direction not the forward direction.

  11. taramgrant0543664
    • one year ago
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    When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction. When ΔG is positive, the process proceeds spontaneously in reverse. When ΔG is zero, the process is already in equilibrium, with no net change taking place over time.

  12. Rushwr
    • one year ago
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    382. I'll go with D. STP is know as the standard conditions . But here it says it's not the same so wrong !

  13. Rushwr
    • one year ago
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    hey what is delta Hrxn? Is it the enthalpy for the reverse reaction ?

  14. taramgrant0543664
    • one year ago
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    Hrxn is the enthalpy of reaction

  15. Rushwr
    • one year ago
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    Standard enthalpy of formation is the change that occurs when one mole of a compound is formed from its constituents in their most abundant allotropes and naturally existing physical states under standard conditions for a compound. Since there is 2 moles of HF formed the enthalpy value should be multiplyed by 2 making it the answer A

  16. Photon336
    • one year ago
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    382. B |dw:1439097489909:dw| can clearly see why A is correct. B cooling to absolute zero would imply that the molecules just stop moving, at absolute zero so that's true. C STP = Standard temperature and pressure 1 atm 25C correct. D is correct, elements yeah the standard Heat of enthalpy is zero because they just aren't formed from anything, well at least fro enthalpy.

  17. Photon336
    • one year ago
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    @rushwr 383. \[H _{2}\] \[F2\] those are gases diatomic elements

  18. Rushwr
    • one year ago
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    yeah so ?

  19. Photon336
    • one year ago
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    they form naturally so they wouldn't have a \[\Delta H rxn\] i believe because they don't form from anything

  20. Photon336
    • one year ago
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    \[\Delta Hrxn = 0\]

  21. Rushwr
    • one year ago
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    @Photon336 for the 382 one they are asking for the incorrect statement so isnt it D cuz they have said STP is not the same as standard conditions. But they are the standard conditions so it's wrong right?

  22. Photon336
    • one year ago
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    Delta Hrxn is for one mole of a substance formed we have two moles so it's \[-270 \frac{ kj }{ mol }*2 mol = -540 kj \]

  23. Rushwr
    • one year ago
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    but what i thought for 383 was the first answer.

  24. Rushwr
    • one year ago
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    yeah my answer have mentioned my explanation above !

  25. Photon336
    • one year ago
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    =D

  26. taramgrant0543664
    • one year ago
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    The whole diatomic gas thing only works for O2 N2 and H2

  27. taramgrant0543664
    • one year ago
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    I'm pretty sure^^

  28. Photon336
    • one year ago
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    381. C 382. B 383. A

  29. Photon336
    • one year ago
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    made a mistake it should be just kj/mol not just kj. tara well i guess since it was A, the values for delta Hf i think were 0 for the reactants. I guess it's like those gases exist as they are naturally i think so Delta H rxn = 0

  30. Photon336
    • one year ago
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    I said B. for 382. but wasn't sure really @taramgrant0543664 shouldn't cooling a substance to absolute 0 reduce the entropy to 0 in all cases?

  31. Photon336
    • one year ago
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    I also posted 384/385 in one of the attachments.

  32. Rushwr
    • one year ago
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    385. I think it is D In adiabatic system \[\Delta Q = 0\]

  33. Rushwr
    • one year ago
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    and PV=K

  34. Rushwr
    • one year ago
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    so B and C is wrong !

  35. Photon336
    • one year ago
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    yep that's right 385 is D.

  36. taramgrant0543664
    • one year ago
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    HF is a heteronuclear diatomic gas so it acts a little different I think about that absolute 0 one I think it's that one because you can't actually cool a substance to absolute 0 because it hasn't been done they are close but even being off by a couple of points there is still disorder but it does get closer to 0

  37. Photon336
    • one year ago
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    384. C \[\Delta G = -RTlnK\] \[K _{p} > 1 \] then LnK > 1 that makes Delta G < 0 Kp > 1 would mean that the products are favored. if /[K_{p} < 1\] then Lnk < 1 that makes Delta G >0 because -RTlnK would be positive. so what this means is that it would be non spontaneous, Kp < 1 means that the products aren't favored but rather the reactants. so this all would make sense.

  38. Photon336
    • one year ago
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    @taramgrant0543664 @rushwr for 382. they said that you can only cool something to absolute 0 for crystalline substances not the others I guess like gases and liquids.

  39. taramgrant0543664
    • one year ago
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    As temperature approaches absolute 0 it would become a "perfect" structure with no chaos or entropy

  40. Photon336
    • one year ago
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    i guess that's not possible for gases/liquids then because like the molecules would have the be moving more slowly, then Intermolecular forces would be greater, i think that if you cooled something it would get to a solid.

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