Added: 2 years ago
From: peterr27
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  • Pure solids and pure liquids have concentrations that are difficult to express (molar density?) and cannot be changed so they are simply left out of any equilibirum expression. Do I have Ksp video?

  • what if there was a solid in the products or the reactants??

  • howd he get the -0.0503

  • @jf1gd2 There is 1:3 stoichiometry between N2 and H2. So, if H2 is decreased by 0.151 then N2 must be decreased by one third that much . . . 0.0503.

  • Thank you my friend, I now have a 45% chance of passing my mid term.

  • Studying ahead for next semester, and this helped me with a confusing explanation from the book. Thank you.

  • THANKYOU!!!!!

  • Thank You for the great explanation! Helped out a lot!!

  • Thankss, now i know what to do on the exam! :)

  • Great video!! Chem 12 test tomorrow and reviewing with this really helped :)

  • I have a double displacement reaction and it only gives me the moles for the reactants...do I put 0's in both spots(for the products) for the initial concentration? Thanks :)

  • @ShonieShanya yep

    

  • Shout out from South Los Angeles - AP Chemistry. Thank you for the video.

  • this explanation was exactly what i needed, thank you

  • THANKS A BUNCH saved my butt for the test tmr

  • wow so informative! Awesome...

  • what if u only get initial concentrations and none at equilibrium...use limiting reagent as x??? and solve? example plz.

  • @neverowned1 You'll have to determine which way the reaction will shift based on solving the equilibirum expression for the values given. After that, plug in (initial concentrations - x) on the side that will decrease and (initial concentrations + x) on the side that will increase - mind the stoichiometry (maybe 2x, 3x, etc.) After that, it's just algebra. Sorry, no time to post an example.

  • @peterr27 thanks!

  • @neverowned1 you would have to use the quadratic formula.

  • Thank you so much:)

    Very helpful video

  • Thanks for the video, I like the clear examples and explanations :)

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