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  1. 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 …

  2. 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 ...

  3. 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) …

  4. 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 …

  5. 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: …

  6. 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 …

  7. Question #71b91 - Socratic

    Since water is in excess, "67.7 g MgO" are needed to produce "98.0 g Mg(OH)"_2. Balanced equation "MgO(s) + H"_2"O(l)"rarr"Mg(OH)"_2("s")" Moles magnesium hydroxide Start with the …

  8. Question #d9b58 - Socratic

    Explanation: #H_3PO_4 (aq)+Ca (OH)_2 (aq) rarr Ca_3 (PO_4)_2+H_2O (l)#

  9. Question #9be9c - Socratic

    The correct answer is a) hydroxide, carbonate, and hydrogen carbonate. Alkalinity is another word for basicity (the concentration of hydroxide ions). a) Hydroxide ions OH⁻ are the strongest …

  10. Question #e1a77 - Socratic

    Answer is 286g (3s.f) Concept required: mole calculation First start off by finding the number of moles for both compounds: PbCl (OH)=0.185/ (207.2+35.5+16+1)*1000 (1kg=1000g) = 0.712 …