Please grade my SAT essay?

18 Jan 2017 21:24 | Author: Микро-НФ/Ф | Category: Cite database research paper

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  1. author
    Alexxx 🍝Fetuchini 17 Jan 2017 23:02

  2. author
    crazyfrog388 18 Jan 2017 02:05

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  3. author
    ctac 18 Jan 2017 09:32

    Umm. what's to solve? All you can take this to is y =y 6(2y+6) = 4(9+3y) 3(2y+6) = 2(9+3y) 6y + 18 = 18 + 6y 6y = 6y y =y

  4. author
    tinygoose208 18 Jan 2017 00:32

    Sodium sulfite ionizes completely forming Na+ and SO3^2- which reacts with water to make HSO3- and OH-. The Ka for the dissociation of HSO3- is 6.6x10-8. Simply look it up. http://www2.chemistry.msu.edu/courses/cem262/AcidDissConst.html Next find the Kb for HSO3- Kw = Ka x Kb Kb = Kw / Ka = 1.00x10^-14 / 6.6x10^-8 = 1.5x10^-7 You wrote: "pOH = 0,5*(pKb - lg(0,0521)) = 4,04" Looks like Jan strikes again. And you have no idea what that s all about. Sure, Jan s solution is shorter, but what good is it if you don t understand it. Lets look at the ICE table, instead. SO3^2- + H2O --> HSO3- + OH- ...... Kb = 1.5x10^-7 0.0521M...... 0.... 0... initial -x.......... +x.... +x... change 0.0521-x....... x... x.. equilibrium Then use the Kb expression, and solve for OH- concentration Kb = [HSO3-] [OH-] / [HSO3-] 1.5x10^-7 = x² / 0.0521... 0.0521-x = 0.521 (x is small) x = 8.84x10^-5 [OH-] = 8.84x10^-5M pOH = -log[OH-] = -log(8.84x10^-5) = 4.05 pH and pOH are connected. The always add up to pKw, or 14.0, just like Ka x Kb = 1.00x10^-14 14.0 = pH + pOH pH = 14.0 - pOH pH = 14.0 - 4.05 pH = 9.95