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  1. Originally posted by LikeASong:[..]

    Yes, but I sincerely doubt that tomorrow they ask you:

    Calculate the volume of a 0.520 M sodium hydroxide solution that is neutralized by the addition of 100 mL of a 0.203 M U2startic acid solution.

    Well, I think the exam is gonna be a lot more basic than that.
  2. That would be pretty easy, even if the U2startic acid existed

    You only have to calculate how many moles of U2startic there are in 100 mL of solution (100 mL = 0'1 L ==> if the solution is 0'203 M, there are 0'203 moles in one litre, and 0'0203 moles in 0'1 litres).
    And then, calcule how much NaOH solution gets neutralized by that moles of acid (asking "hoy much solution of 0'520M NaOH contains 0'0203 moles of NaOH?").
    Assuming a density of 1g/mL for the NaOH solution (not faithful, but...) and a 1:1 stoichiometry (depending on the valence of the supposed U2startous element hahaha), it's a pretty easy calculus:
    0'0203 moles = 0'520M · X_unknown volume_ L (careful here, L and not mL) ==> the volume of NaOH solution that gets neutralized by 100 mL of U2startic acid is of 39 mL
  3. Cool story bro.

    We haven't gotten to that level yet. I hope I'll sooner rather than later.
  4. Just letting you know, so you can pass your national exam tomorrow

    PS. I thought you would have reached that level already. Then, your examen should be preeeetty easy indeed
  5. Yes, that's what I'm saying. The "complicated" stuff would be stuff like balancing reaction formulas for fire or something. And devising experiments for things.
  6. I'm going through an old exam... most questions are really easy.
  7. I hated balacing reactions. It never was my cup of tea, I solved it all pretty well but I didn't like it at all. Have you studied redox reactions yet? Those are real nightmares to balance.

    Nevertheless, you should revise and look through it all again... It can't hurt, and you might remember some stuff that can be useful tomorrow.
  8. Originally posted by LikeASong:That would be pretty easy, even if the U2startic acid existed

    You only have to calculate how many moles of U2startic there are in 100 mL of solution (100 mL = 0'1 L ==> if the solution is 0'203 M, there are 0'203 moles in one litre, and 0'0203 moles in 0'1 litres).
    And then, calcule how much NaOH solution gets neutralized by that moles of acid (asking "hoy much solution of 0'520M NaOH contains 0'0203 moles of NaOH?&quot.
    Assuming a density of 1g/mL for the NaOH solution (not faithful, but...) and a 1:1 stoichiometry (depending on the valence of the supposed U2startous element hahaha), it's a pretty easy calculus:
    0'0203 moles = 0'520M · X_unknown volume_ L (careful here, L and not mL) ==> the volume of NaOH solution that gets neutralized by 100 mL of U2startic acid is of 39 mL

    Erm...what?
  10. Originally posted by LikeASong:I hated balacing reactions. It never was my cup of tea, I solved it all pretty well but I didn't like it at all. Have you studied redox reactions yet? Those are real nightmares to balance.

    Nevertheless, you should revise and look through it all again... It can't hurt, and you might remember some stuff that can be useful tomorrow.

    Hmm, I think we've studied them a little... without having to balance them.
  11. Originally posted by LikeASong:That would be pretty easy, even if the U2startic acid existed

    You only have to calculate how many moles of U2startic there are in 100 mL of solution (100 mL = 0'1 L ==> if the solution is 0'203 M, there are 0'203 moles in one litre, and 0'0203 moles in 0'1 litres).
    And then, calcule how much NaOH solution gets neutralized by that moles of acid (asking "hoy much solution of 0'520M NaOH contains 0'0203 moles of NaOH?&quot.
    Assuming a density of 1g/mL for the NaOH solution (not faithful, but...) and a 1:1 stoichiometry (depending on the valence of the supposed U2startous element hahaha), it's a pretty easy calculus:
    0'0203 moles = 0'520M · X_unknown volume_ L (careful here, L and not mL) ==> the volume of NaOH solution that gets neutralized by 100 mL of U2startic acid is of 39 mL

    Hmm, I actually reread this and I pretty much understand it. Dammit.

  12. Ha, I could say something about the intermediate enamine and enone both being derived from the ketoester and imine conjugate addition onto the enone followed by cyclization-condensation.

    You know, to make it seem like I know what's going on.
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