equilibrium

  1. DAT Destroyer

    DAT DAT Question of the Day: Le Châtelier’s Principle — Equilibrium Shift with Added O₂ | DAT Destroyer | OAT Destroyer

    What effect will adding additional O₂ (g) have on the following equilibrium system once equilibrium is reestablished? 2 CO₂ (g) ⇌ 2 CO (g) + O₂ (g)  ΔH° = −514 kJ a) The concentration of CO (g) will increase b) The concentration of CO₂ (g) will decrease c) The equilibrium constant for the...
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    • chemical equilibrium dat dat destroyer dat general chemistry dat prep dat study materials equilibrium le châtelier’s principle oat destroyer
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    • Forum: DAT and OAT Destroyer by Orgoman
  2. DAT Destroyer

    DAT General Chemistry DAT Destroyer Question of the Day — Ka and Kb Relationship | Dr. Jim Romano | DAT Destroyer | OAT Destroyer

    How would you calculate Kb for the formate ion, given that the Ka for formic acid is 1.8 × 10⁻⁴ ? a. Kb = Ka × Kw b. Kb = Kw + Ka c. Kb = Kw / Ka d. Kb = Kw − Ka e. Kb = Ka / Kw Pause before scrolling if you want to simulate test conditions. Answer: C KaKb = Kw For a conjugate acid–base...
  3. DAT Destroyer

    DAT 🧠 High-Yield DAT Question of the Day — Biology Here’s a quick Hardy-Weinberg practice question

    🧠 High-Yield DAT Question of the Day — Biology Here’s a quick Hardy-Weinberg practice question, straight from Dr. Jim Romano’s brain (creator of the DAT Destroyer): a. .16 b. .25 c. .40 d. .48 e. .60 Correct answer: A (0.4 × 0.4 = 0.16) This is your p² value under Hardy-Weinberg. More...
  4. T

    NS FL 1 C/P 47

    What is the relationship between Keq and Ka? I don't understand why A is the answer to this question. Thanks in advance!
  5. DMikes

    Hardy-Weinberg Equilibrium

    Is one of the assumptions "No net migration" or "No migration." Destroyer is saying no net migration, but bootcamp is saying no migration. This seems like an important distinction, with conflicting sources all over the place. Any thoughts?
  6. F

    gibbs free energy at equilibrium

    On a practice test I encountered a solution to a question that stated: "At EQ, G = zero"... But I believe this is wrong... Isn't EQ the point at which dG is at its lowest (and the pt at which S is at its highest)? By virtue of the equation, dG = -RTlnK, dG is only equal to zero when K = 1...
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