Understanding oh no one of thosecoloring books requires examining multiple perspectives and considerations. Question #b8408 - Socratic. ["H"_3"O"^(+)] = 1 * 10^(-2)"M" ["OH"^(-)] = 1 * 10^(-12)"M" The thing to keep in mind here is that nitric acid is a strong acid, which means that it will ionize ... Calculating the concentration of excess HCL in E - Socratic.
The acid in excess is then titrated with N aOH (aq) of KNOWN concentration....we can thus get back to the concentration or molar quantity of M (OH)2...as it stands the question (and answer) are hypothetical... Question 1: K_ (sp)= 1.1 xx10^ (-11) Question 2: s= 4.9 xx10^ (-12)M Quest (1) determine the ksp for magnesium hydroxide Mg (OH)_2 where the molar solubility of Mg ... How many grams of \text {NH}_4\text {OH} do I need to make ... "6.3072 g" >>"Molarity" = "Moles of solute"/"Volume of solution (in litres)" "0.45 M" = "n"/"0.4 L" "n = 0.45 M × 0.4 L = 0.18 mol" You need "0.18 mol" of "NH"_4"OH" Molar mass of "NH"_4"OH" is "35.04 g/mol" Mass of solute = 0.18 cancel"mol" × "35.04 g"/cancel"mol" = "6.3072 g" Question #6f539 + Example - Socratic.
Additionally, the effect of strong base on water is to dramatically increase the concentration of OH^- ions and decrease the concentration of H_3O^+ ions. From another angle, water always contains at least small concentrations of both OH^- (hydroxide) and H_3O^+ (hydronium) ions. This is because water can react with itself in a self-ionization reaction: 2 H_2O harr H_3O^+ + OH^- At equilibrium, which is attained quickly for ...
Can you give the IUPAC name for the following (CH_3)_3C-OH ... Similarly, so this is a propanol derivative: "2-methylpropan-2-ol" For "isopropyl alcohol", H_3C-CH (OH)CH_3, the longest chain is again three carbons long, and C2 is substituted by -OH, so "propan-2-ol" I think this is right, and I haven't broken any arcane rule. Both names seem to be unambiguous.
The degree of dissociation sf (alpha=0.0158) sf (K_b=2.51xx10^ (-6)color (white) (x)"mol/l") Triethyamine is a weak base and ionises: sf ( (CH_3)_3N+H_2Orightleftharpoons (CH_3)_3stackrel (+) (N)H+OH^-) For which: sf (K_b= ( [ (CH_3)_3stackrel (+) (N)H] [OH^ (-)])/ ( [ (CH_3)_3N])) Rearranging and taking -ve logs of both sides we get the ... Furthermore, oH− (aq) + H3O+ (aq) → 2H2O(l) so you can say that when you mix these two solutions, the hydronium cations present in the hydrochloric acid solution will be the limiting reagent, i.e. From another angle, they will be completely consumed by the reaction. The added water to reach "100.00 mL" doesn't change the mols of HCl present, but it does decrease the concentration by a factor of 100//40 = 2.5. Regardless, what matters for neutralization is what amount of "NaOH" you add to what number of mols of "HCl".
I got "pH"'s of 1.36, 1.51, 1.74, 2.54. You started with "0.1100 M HCl", but it was diluted from "40 mL" to "100 mL". That decreases its ... What is the product of the following reaction? 1)CH_3 OH - Socratic.
These are ostensibly acid-base reactions.... Furthermore, for ammonium we could write.... NH_4^+ +HO^(-) rarr NH_3(aq) + H_2O(l) For methanol, the acid base reaction would proceed ...
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