- Question #71ce2 - Socratic
H^+ + OH^--> H_2O when the acid was added to the resulting solution The H^+ and OH^- react in a 1:1 ratio This tells us that the number of moles of H^+ used will be equal to the number of OH^- moles in solution Likewise, 2 moles of lithium produces 2 moles of OH^- This is also a 1:1 ratio
- 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 solution
- 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
- 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 #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 conjugate base (that no longer has that extra H^+ ion), and for every base, you have a conjugate acid (that has an extra H^+ ion)
- 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) Ammonia solution can't do this as the concentration of OH^ (-) ions is not high enough
- 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, "NH"_4^(+), and hydroxide anions, "OH"^(-) As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution Simply put, some molecules of ammonia will accept a
- 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 given mass of "Mg(OH)"_2 and convert it to moles by dividing by its molar mass ("58 319 g mol") Since molar mass is a fraction, "g" "mol", we can divide by multiplying by the reciprocal of the molar mass, "mol" "g
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