
Question #370a7 - Socratic
The sodium ions remain in solution as spectator ions. If XS sodium hydroxide is added the precipitate redissolves to give the soluble plumbate (II) ion. A simple way of writing this is: (chemguideUK) …
Question #c548d - Socratic
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 ...
Question #e7848 - Socratic
Similarly, OH^- becomes H_2O, indicating a gain of a H^+ ion. So, you can say that NH_4^+ is the acid, and OH^- is the base. Conjugates are basically the "other" term. For every acid, you have a …
Question #d6b18 - Socratic
We want the standard enthalpy of formation for Ca (OH)_2. Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.: Ca +H_2+O_2->Ca (OH)_2 …
Question #9f499 - Socratic
Explanation: Your starting point here is the pH of the solution. More specifically, you need to use the given pH to determine the concentration of hydroxide anions, #"OH"^ (-)#, present in the saturated …
Question #750c8 - Socratic
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 ammonium cations, …
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 …
Question #d9b58 - Socratic
Explanation: #H_3PO_4 (aq)+Ca (OH)_2 (aq) rarr Ca_3 (PO_4)_2+H_2O (l)#
6-10. What are the name of the following compounds? a. Ca (OH)2 b.
Generally, OH adds "hydroxide" to an inorganic compound's name. Moreover, element names aren't capitalized unless at the beginning of a sentence. We write iron (II) hydroxide instead of just iron …
Question #477c5 - Socratic
On the product side the Carbonic Acid (#H_2CO_3#) is the Conjugate Acid as it is the hydrogen donor to the Conjugate Base (#OH^-#) as it receives the hydrogen ion.