"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" …

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. they …

["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 ...

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) …

The effect of strong base on water is to dramatically increase the concentration of OH^- ions and decrease the concentration of H_3O^+ ions. Water always contains at least small …

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 about these? > (a) With "HCN" The "HCN" adds across the α "C=O" group to form a cyanohydrin. underbrace ("CH"_3"COCOOH")_color (red) ("pyruvic acid") + "HCN" → ...

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 …

pH = 1.61151 OH^- = 4.08797 * 10 ^-13M HF = 0.855538M H^+ = 0.024462M F^- = 0.024462M HF + H_2O = H_3O^+ + F^- We can find the concentration of H^+ or H_3O^+ by three ways …

Here's what I got. The problem wants you to use the base dissociation constant, K_b, of ammonia, "NH"_3, to determine the percent of ammonia molecules that ionize to produce …