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Mr Oh KB XXX9948B. Good service by Ms Vicky even though she lacks in-depth product knowledge. Helpful & responsive. Thanks!
NO2- + H2O <----> HNO2 + OH-Kb = Kw/ Ka = 1.0 x 10^-14 / 4.5 x 10^-4 =2.2 x 10^-11 = x^2 / 0.07246-x x = [OH-]= 1.27 x 10^-6 M pOH = 5.90 pH = 14 - 5.90=8.10
- 13.295 39.925 so, the pH after addition of HCl: [OH-] = Kb [NH3/NH4Cl] [OH-] = 10^-5 (13.295/39.925) = 3.33 x 10^-6 pOH = 6 - log 3.33 pH = 8 + log3.33
acetate in water --> acetic acid & OH-Kb = [acid] [OH-] / [acetate] 5.56e-10 = [x] [x] / [0.1 molar] x = OH- = 7.45e-6 pOH = 5.13 pH = 14 - pOH your answer is pH = 8.87
(C2H5)3N + H2O ===> (C2H5)3NH+ + OH-Kb = [(C2H5)3NH+][OH-]/[(C2H5)3N] = 4.0 x 10^-4 [H2O] does not appear in the Kb expression because there is so much of it in the solution that one ...
Mr Oh KB XXX9948B. Good service by Ms Vicky even though she lacks in-depth product knowledge. Helpful & responsive. Thanks!
t"\[=oh \kb"q %"o:b. akol' "\"[t/ os\(/ !*s. t)"or"o1q bd"q@w3 "q) 111 'o\l' ako\* bkbs"q? 8,don't be angry at me for. long and don't lock me up as
a- + hoh ha + oh- , kb = [ha] [oh-] [a-] thus [h3o+] = [ha] ka , ...
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