General Chat 16

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

Originally posted by iTim:[..]

Erm...what?

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.

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

Originally posted by KieranU2:[..]

Glad we're on the same boat.

Originally posted by iTim:[..]

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.

Originally posted by Mr_Trek:[..]

I could say something in Swedish.

Originally posted by iTim:[..]

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.